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Bernard2
2004-Jan-03, 03:07 PM
first off i don't have a clue about physics. ok with that established, here's my question

how come radiowaves can go through walls and lightwaves cannot?

ty in advance for answering

Amadeus
2004-Jan-03, 03:16 PM
First off i'am like you I not an expert so dont take this as concrete fact but I would guess its done to the frequencey / wavelength. Its the same for radiation some types can pass through pretty much anything except special shielding and some types can be stoppped by a sheet of paper.

(yes I know that light and radio waves are also types of radiation but i'am using laymans terms!) :lol:

Some light can pass through walls though for instance infra red can.Or has tv lied to me again? I remember the scene in robo cop where he sees the guy holding people hostage through a wall.

P.S Welcome to the board!

Bernard2
2004-Jan-03, 03:59 PM
thank you for your answer and kind welcome. i know radiowaves have a a bigger wavelength than lightwaves but could someone please be more specific as to the exact workings. why does larger wavelength mean it can go through a wall?

Sam5
2004-Jan-03, 04:59 PM
why does larger wavelength mean it can go through a wall?

I can not answer your question, but I can observe that since I have aluminum siding on my house, radio and TV waves have a difficult time coming through that metal. I have to run a radio antenna outside the house so I can get a better short-wave signal, and I have to do that with my TV antenna too.

I’ve heard that a satellite dish won’t work if you put it in an attic that has a metal roof on the house. Also, when it rains on my outside satellite dish, my TV picture breaks up and sometimes goes out completely.

I have a radio-transmitter outdoor thermometer, and I have to keep it near a window so the radio signals can come into my house. It has trouble transmitting through the aluminum siding.

swansont
2004-Jan-03, 05:15 PM
Because the material in the walls doesn't readily absorb the radio waves. (This isn't meant to be smart-alecky, even if it sounds that way.)

All materials absorb radiation at frequencies determined by the electronic structure of their constituent molecules. We use glass as windows because it doesn't redily absorb photons in the visible range, but we choose the material for wall based on other criteria (e.g. looks, availability and structural properties. And we don't actually want light to go through!)

Metals tend to be really good at absorbing radiation in that range, as are (generally) any materials that have a large number of states for electron excitation.

Sam5
2004-Jan-03, 05:21 PM
Metals tend to be really good at absorbing radiation in that range, as are (generally) any materials that have a large number of states for electron excitation.

Do atomic clocks have to be insulated and shielded to keep out unwanted EM waves? If so, if they were not shielded, would some EM waves alter the “tick rates” of the clocks?

Sandor
2004-Jan-03, 05:56 PM
OK, so radio waves (fotons) just go through the empty space of most bodies, sometimes bouncing against electrons, but not having enough energy to really move these electrons up to another level? In other words, they do not get absorbed because the energy cannot be USED in any way, something like that?

But I don't understand, because the fotons will at least move the electrons to which they bounce and loose their energy that way? While light fotons, which have more energy, will likely have a bigger chance to survive all the bouncings and have enough energy left to cross the entire body?

Well, obviously I don't understand this at all.

Sandor
2004-Jan-03, 05:59 PM
The fact is that I don't understand at all what a WAVE is. Yes, it's some sort of electronic or magnetic FIELD, in which a particle (foton) moves, but this foton moves in a straight line.

Now I know that we must regard light as sometimes being a wave and sometimes a particle, but what is JUST a wave? What is waving then?

BTW, I know this other guy, Bernard, we were chatting about this earlier.

Glom
2004-Jan-03, 06:12 PM
OK, so radio waves (fotons) just go through the empty space of most bodies, sometimes bouncing against electrons, but not having enough energy to really move these electrons up to another level? In other words, they do not get absorbed because the energy cannot be USED in any way, something like that?

Basically that. A photon is only useful if it has the exact energy needed to move the atom to an excited state (possibly the infinitieth excited state which involves ionisation). If no transition is requires the energy of incident photons, then it will pass through with little incident.


But I don't understand, because the fotons will at least move the electrons to which they bounce and loose their energy that way? While light fotons, which have more energy, will likely have a bigger chance to survive all the bouncings and have enough energy left to cross the entire body?

That's the classical idea. But the quantum idea is all about packets of energy. If a transition between energy states requires 100eV, the classical idea is that if a wave containing 150eV comes along, it will do the excitation thang and carry on with 50eV left. But according to quantum theory, which is more accurate at this level, a 150eV photon will not have any effect because it doesn't have the exact energy level required.

Glom
2004-Jan-03, 06:13 PM
The fact is that I don't understand at all what a WAVE is. Yes, it's some sort of electronic or magnetic FIELD, in which a particle (foton) moves, but this foton moves in a straight line.

Now I know that we must regard light as sometimes being a wave and sometimes a particle, but what is JUST a wave? What is waving then?

BTW, I know this other guy, Bernard, we were chatting about this earlier.

I believe the way to interpret it is that a large number of photons travelling together will approximate to a wave like thing.

Sam5
2004-Jan-03, 06:21 PM
The fact is that I don't understand at all what a WAVE is. Yes, it's some sort of electronic or magnetic FIELD, in which a particle (foton) moves, but this foton moves in a straight line.

Now I know that we must regard light as sometimes being a wave and sometimes a particle, but what is JUST a wave? What is waving then?


If I understand the situation correctly, there are two different theories about “what is waving”. I think the most common theory is that it is the electric and magnetic fields that are “waving”, and they zip through space at 186,000 mps as they “wave”. In other words, a light bulb and a radio transmitter generate these tiny little alternating electric and magnetic fields, and they "wave" or "wiggle" as they travel through space.

The other theory is the minority opinion, and it says that what “waves” are the electric and magnetic fields that are already everywhere in space, and that all a light bulb and a radio transmitter does is start the local electric and magnetic fields to “waving” at the light bulb and the transmitter, and it is the “wave”, not the fields, that go zipping through space. This is a type of “ether” theory, so it is not very popular among physicists, although they will often talk about the “waves”, rather than the “fields”, traveling through space.

Sam5
2004-Jan-03, 06:36 PM
I believe the way to interpret it is that a large number of photons travelling together will approximate to a wave like thing.


The way I understand it is that a bunch of little waves traveling together in a “wave packet” group, constitutes a single “photon”. And so, several different groups of little “wave packets” constitute several “photons”, with one group out front and the others following that one. I don’t know if this is correct, but this is the way I imagine what the physicists are saying.

With this idea, a large number of photons traveling one after another, constitute a light “ray” or “beam”. While a lot of them traveling side by side, constitute a wide light “wavefront”. If a large number travel side by side, and more are following behind those, then we have a wider “beam” or “ray” of light.

It is the individual “wave trains” of several photons traveling in a straight line, one behind the other, that constitutes a very narrow “beam”. And when a light bulb generates a lot of photons in all directions, then a lot of individual “beams” diverge and go out in an expanding “spherical” manner, and this is what the “inverse square law” describes, that is, the gradual divergence of all these individual narrow photon “beams”.

In other words, the individual “beams” of single photon “wave trains” get further and further apart, the further they move away from a light bulb. Because of this phenomenon, a flashlight has a curved mirror behind the bulb, which changes the direction of some of these photon “beams” and focuses them to all go out together in a fairly straight line in the direction in front of the flashlight bulb.

A single photon:

~

A single photon “beam”:

~~~~~~~~~~~~

Several photon beams coming out of a flashlight as a “light ray”:

~~~~~~~~~~~~~
~~~~~~~~~~~~~
~~~~~~~~~~~~~

Sandor
2004-Jan-03, 06:39 PM
Glom, that makes sense! It's the 'quantum amount' that will determine if there is no interaction at all, in an energy-kind of way.

Sam, I still don't understand, because an electric field is only measurable (or maybe even existing) when MATTER is involved, so a particle must be IN this field. or doesn't it (in light theory)? I mean, when they say that light can EITHER be regarded as wave OR as particle, how can it be ONLY a wave? Or is that not what they mean then?

Sandor
2004-Jan-03, 06:47 PM
Sam, that also was what I was thinking: I guess it cannot be that 1 photon is really 'something', or a particle, because they say that a radio photon has less energy (longer wave) than a light photon, but all photons have no mass (or not detectable) and all photons travel at the same speed (speed of light), so one photon cannot have more or less energy than another.

But the shorter the wave, the more photons will be fired in 1 time frame, so we can say: the number of photons in 1 time frame is the energy of the 'photon'.

Tensor
2004-Jan-03, 07:22 PM
I guess it cannot be that 1 photon is really 'something', or a particle, because they say that a radio photon has less energy (longer wave) than a light photon, but all photons have no mass (or not detectable) and all photons travel at the same speed (speed of light), so one photon cannot have more or less energy than another.

What you are missing here Sandor is that photons have momentum which, because a photon has no rest mass, is the photons energy. A photon of visible light wave has more momentum than a photon of a radio wave.

Sandor
2004-Jan-03, 07:30 PM
Tensor, but momentum is mass x velocity. Velocity is always the same and there is no real mass, or what? Mass is increasing when velocity increases, but velocity of all photons is the same and there is no mass? I don't understand.

Tensor
2004-Jan-03, 07:53 PM
Tensor, but momentum is mass x velocity. Velocity is always the same and there is no real mass, or what?


You are using the Classical (Newtonian) formula. In Relativity, momentum for massless particles is defined as p = E/c Where p is the momentum, E is the total energy, and c is the speed of light.


Mass is increasing when velocity increases, but velocity of all photons is the same and there is no mass? I don't understand.

Under relativity, the total energy of a system increases (not mass) as velocity increases, but the total energy also includes momentum as well as mass. In the case of photons, because they have no rest-mass, the total energy of a photon is equal to its momentum.

Sandor
2004-Jan-03, 08:10 PM
You are using the Classical (Newtonian) formula. In Relativity, momentum for massless particles is defined as p = E/c Where p is the momentum, E is the total energy, and c is the speed of light.

OK, but can you explain why a light photon has more momentum than a radio photon? The electrical/magnetic wave 'around it' is shorter, so there are more waves in 1 space frame, but is this wave actually 'pulling it' forwards? Photons have no electrical charge I believe... (I'm just guessing around)


Under relativity, the total energy of a system increases (not mass) as velocity increases

I read that it is impossible for an object with any mass to travel at the speed of light, because when it comes closer to c, mass will increase, and it will increase so much that you will need all the energy in universe and more to get it at c. I understand what you're saying about ENERGY increasing and not mass, but the explanation above doesn't make sense then.

Russell
2004-Jan-03, 09:35 PM
Hello:
An analogy for an electromagnetic wave is an audio wave-the peaks and the troughs represent amplitude, the number of frequencies represents pitch. The difference though is that lower frequency wave can past through walls much easier than higher freqs, it's just the opposite with electromagnetic waves.

swansont
2004-Jan-04, 12:20 AM
Metals tend to be really good at absorbing radiation in that range, as are (generally) any materials that have a large number of states for electron excitation.

Do atomic clocks have to be insulated and shielded to keep out unwanted EM waves? If so, if they were not shielded, would some EM waves alter the “tick rates” of the clocks?

Yes, they have to be shielded for a few reasons. EM radiation will cause an AC Stark shift in the clock states, so you either have to ensure they aren't there or measure them if they are. It's not easy, since room temperature blackbody radiation is a significant effect for the best clocks.

Static magnetic field cause Zeeman shifts as well. Even though you use the transition that has zero magnetic moment (no linear shift) there is a second order effect. You put a known bias field in place to split the levels, since you don't want all of the states to mix - they are degenrate in zero field. The key is that it's a known bias, so you know the frequency shift. You shield out the earth's field and the associated fluctuations as best you can.

Tensor
2004-Jan-04, 02:56 AM
You are using the Classical (Newtonian) formula. In Relativity, momentum for massless particles is defined as p = E/c Where p is the momentum, E is the total energy, and c is the speed of light.

OK, but can you explain why a light photon has more momentum than a radio photon?

Photons (like all massless particles) have all their energy in momentum. As a result, higher energy photons (eg light) have more momentum than low energy photons (eg radio).


The electrical/magnetic wave 'around it' is shorter, so there are more waves in 1 space frame, but is this wave actually 'pulling it' forwards?

Do you know about wave-particle duality. Waves or particles (a photon is the particle aspect of an EM wave) are different aspects of the same thing. An analogy would be vapor and dry ice are different aspect of carbon dioxide


Photons have no electrical charge I believe... (I'm just guessing around)

Photons carry the EM force from one particle to another, but (and you are correct here) have no charge themselves.



Under relativity, the total energy of a system increases (not mass) as velocity increases

I read that it is impossible for an object with any mass to travel at the speed of light, because when it comes closer to c, mass will increase, and it will increase so much that you will need all the energy in universe and more to get it at c. I understand what you're saying about ENERGY increasing and not mass, but the explanation above doesn't make sense then.

The mass increasing expanation you've read is a simplified explanation (as a matter of fact, mine is also simplified, though less so). The two explanations can be somewhat reconciled if you remember energy and mass are related by c (E=mc^2) I can, if you want, post how relativistic equations modify an objects acceleration thus preventing it from reaching c.

Gsquare
2004-Jan-04, 03:01 AM
[
OK, but can you explain why a light photon has more momentum than a radio photon? The electrical/magnetic wave 'around it' is shorter, ...

Well, Sandor, have you ever heard of E = hf ? (where h is planck's constant & f = frequency)

It's that simple; the greater the frequency, the higher the energy. 8)

So:
momentum p = E/c = hf/c ...........the greater the frequency (shorter wavelenth) means greater the momentum.

G^2 :D

Sam5
2004-Jan-04, 03:22 AM
Sam, I still don't understand, because an electric field is only measurable (or maybe even existing) when MATTER is involved, so a particle must be IN this field. or doesn't it (in light theory)? I mean, when they say that light can EITHER be regarded as wave OR as particle, how can it be ONLY a wave? Or is that not what they mean then?


I don’t know for sure, but this is the way I look at it:

If a photon is a small single “packet of wavelets”, this “packet” can have an “impact” on, let’s say, a very sensitive scale, and can give the impression that it has “weight”, even though it is only a “wave packet”.

Look at it this way. Get yourself a big bowl of water, then drop a penny in the water. You’ll see the water move in reaction to that “particle” or “group of particles” touching it.

Now, put one of your big hi-fi speakers up near the surface of the water, then turn up your hi-fi full blast, and look at the water. The surface is moving as if you are touching it. So, a “sound wave” is causing the water to physically move, and thus the sound waves are acting like "particles".

Ok, so you say that what is actually causing the water to move is “molecules of air” that are touching and vibrating the water. But, the last I heard about this subject, the “particles” of air don’t actually ever touch anything at all. They don’t even touch each other, ever. What causes a sound wave is not the particles of air bumping into each other, but the “fields” around the air molecules “compressing” each other. I’m not sure what these “fields” are called, but it is they that “compress” and “expand” to form a sound “compression/vacuum” wave in the air. So, a sound “wave” can have a “particle nature” too.

What we are dealing with on the atomic level, the last I heard, aren’t even “physical things”. They aren’t even “particles”, they are “wave forms” or, more properly, “field wave-forms”. So, my view of light is that each photon is a “packet” of tiny wavelets or “wave-forms” that, when they “hit” something, ie, when they rapidly come in contact with the tiny fields that surround other “wave-forms” that we usually call “matter”, the light photon tends to act as if it is a tiny “particle”. But, if I understand this quantum stuff correctly, even the “matter” the photon hits is not “matter” but tiny little “wave forms”.

I hope I haven’t confused anyone. I’m still trying to figure out some of this stuff myself.

I’ve got a 1932 book here, written by an early theoretical atomic physicist, who said, “We have thus been brought to the view that the whole of the material universe is built up of electricity and of nothing but electricity, the atoms of which are aggregated in an infinite number of different ways in the infinite number of different forms in which matter appears.”

So, if this is true, light is a special form of “electricity”, and so are you and I.

Sam5
2004-Jan-04, 03:41 AM
Sam, that also was what I was thinking: I guess it cannot be that 1 photon is really 'something', or a particle, because they say that a radio photon has less energy (longer wave) than a light photon, but all photons have no mass (or not detectable) and all photons travel at the same speed (speed of light), so one photon cannot have more or less energy than another.

But the shorter the wave, the more photons will be fired in 1 time frame, so we can say: the number of photons in 1 time frame is the energy of the 'photon'.


I think that is probably the correct view. The way I understand it, a photon is a long and skinny “thing” and so is a radio wave. Each have a “wave length”, and that determines how long they are. Even if all they are, are two vibrating or oscillating “electric and magnetic fields”, they seem to be long and skinny, with physicists saying the short ones have the most “energy”, because more of them “hit” per second. The physicists seem to ignore the “amplitude” in all this, but that’s their business I guess.

By the way, I finally figured out how radio and TV waves get down our antennas and into our TVs and radios. Actually, they just pass by our antennas, and the ones that “hit” our antennas are “absorbed” by the antennas. This process of “absorption” creates some “energy” inside the antennas, and causes the antennas to “resonate”, if the antennas are the physical length of the “wavelength” of the wave (or a certain percentage of the wavelength, such as a “quarter wave” antenna).

Ok, this “resonation” generates some kind of electron flow or “charge” flow or “current” flow in our antennas. Inside our radios and TVs, there is an electronic thing called a “detector” (or something like that) that detects the “current” flow from the antenna. Other stuff inside our radio and TV “amplifies” that weak current flow by duplicating its “variations” using the higher voltage and amperage coming from our batteries or the AC line that our radios and TVs are plugged in to.

So, the way I understand it, the radio and TV waves never make it past our antennas. It’s the “bump, bump, bump, bump” of the waves against our antennas that cause the antenna “resonation” and that resonation generates the electron or “current” flow inside the antennas.

If any radio expert thinks I’m wrong about any of this, then please point out my errors and tell us what actually happens.

Sam5
2004-Jan-04, 03:54 AM
Metals tend to be really good at absorbing radiation in that range, as are (generally) any materials that have a large number of states for electron excitation.

Do atomic clocks have to be insulated and shielded to keep out unwanted EM waves? If so, if they were not shielded, would some EM waves alter the “tick rates” of the clocks?

Yes, they have to be shielded for a few reasons. EM radiation will cause an AC Stark shift in the clock states, so you either have to ensure they aren't there or measure them if they are. It's not easy, since room temperature blackbody radiation is a significant effect for the best clocks.

Static magnetic field cause Zeeman shifts as well. Even though you use the transition that has zero magnetic moment (no linear shift) there is a second order effect. You put a known bias field in place to split the levels, since you don't want all of the states to mix - they are degenrate in zero field. The key is that it's a known bias, so you know the frequency shift. You shield out the earth's field and the associated fluctuations as best you can.


Great information, thanks!

Now, let me ask you this question:

How much will a temperature change inside the clock change the clock’s rate, as compared with how much the clock’s rate is affected by a change in elevation?

For example. If we note an atomic clock rate in the lowest earth valley, then we take the clock up to the top of Mt. Everest and note the rate change due to the elevation change, then how much of a temperature change would it take for the temperature to affect the “tick rate” of the clock in that same amount?

Sam5
2004-Jan-04, 04:07 AM
[
OK, but can you explain why a light photon has more momentum than a radio photon? The electrical/magnetic wave 'around it' is shorter, ...

Well, Sandor, have you ever heard of E = hf ? (where h is planck's constant & f = frequency)

It's that simple; the greater the frequency, the higher the energy. 8)

So:
momentum p = E/c = hf/c ...........the greater the frequency (shorter wavelenth) means greater the momentum.

G^2 :D

Hmm, this seems to be a space and time related thing. A 1-inch wave will have more “energy” than a 2 inch wave, because the 1 inch wave is carrying all of its energy past a point in space at twice the rate as a 2 inch wave. A hundred-meter wave will take much more time to carry all of its energy past that point in space.

So, with the Doppler effect caused by motion away from an observer, a 1 inch wave can be redshifted into an observed 2 inch wave, and, therefore, physicists will say that the original photon has “lost energy” in the process of being “redshifted”. Is that correct?

Whereas in sound and audio, the “amplitude” of the sound wave is more important and is considered to represent the “energy” of the sound wave, whether it is Doppler shifted or not??

Sam5
2004-Jan-04, 04:11 AM
So:
momentum p = E/c = hf/c ...........the greater the frequency (shorter wavelenth) means greater the momentum.

G^2 :D

Ok, so, when the earth is moving toward a distant star that is “fixed” relative to the sun, and the light of that star is “blueshifted” as a result of the earth’s motion around the sun, the incoming photons “gain energy”, so where does this “energy gain” occur, and where does the blueshift occur? At the earth, in the space between the earth and the star, or at the star?

:D

Sandor
2004-Jan-04, 08:55 AM
Do you know about wave-particle duality. Waves or particles (a photon is the particle aspect of an EM wave) are different aspects of the same thing. An analogy would be vapor and dry ice are different aspect of carbon dioxide

OK, so it's really a matter of 'either' then. There's no photon moving in an EM field. It's either the photon or the wave.


The mass increasing expanation you've read is a simplified explanation (as a matter of fact, mine is also simplified, though less so). The two explanations can be somewhat reconciled if you remember energy and mass are related by c (E=mc^2) I can, if you want, post how relativistic equations modify an objects acceleration thus preventing it from reaching c.

I would be glad to, Tensor.

Sandor
2004-Jan-04, 10:44 AM
Russell & Gsquare, you give me the formulas or the 'knowledge as is', as being very 'simple', and I'm sure they are, but I want to understand HOW it works exactly.

Now it seems that 'momentum of a photon' is really the number of photons or waves per time frame, at least, this is what Sam and I think and it makes sense. Don't you think so Tensor?

It will raise other questions to me, but I still have some questions about 'waves' anyway. Russell and Sam were also mentioning sound waves and amplitude, but these 'waves' are only a graphical picture of what is really a longitudinal wave or movement. We can state the same here as for light: the number of molecules per time frame -or the wave length- determines the pitch, while the 'amplitude' of this graphical wave is really the maximum density of the compressed air compared to the minimum density of the air at a certain point, or the difference between those opposite situations.

This compression of air determines the power with which the molecules bounce in our ear. One would think that a higher compression (louder sound) will speed up the sound, but that's not true, at least as I understand it, because the speed is only depending on the 'matter' that is goes through.

Now, if we use this wave theory for light, we have indeed those 2 different factors:

1. number of photons per time frame
2. momentum or energy of 1 photon

The first one can determine the frequency (color) and the second one...well, what is that? The brightness or intensity of light? I guess the intensity is determined by the 'number of light sources'. Even if you use one lamp, the filament in it will glow at more points when the wattage is higher? So the more photons we see coming from 'one area', which is really a lot of 'points' close to each other, the more intensity we see, while a lot of photons coming from 'one point' will determine the color.

Now I still cannot find a meaning for the energy of 1 photon, but maybe I can for radio signals. Isn't it true that radio transmits at one frequency? So you have to tune at this frequency, which will really determine how many photons will arrive per time frame, you cannot change that. So how can it hold any data, like pitch and loudness? One would think that every single photon should hold different information. So you need an 'amplitude' there.

But what is an amplitude in a light wave? What is even a light wave, I must keep asking. It is only a graphical representation of numbers, but with sound, we can imagine molecules bouncing, getting compressed, speed up and slow down etc. What is the wave line of the electrical wave? Is it positive charge decreasing to zero, changing to negative charge, increasing until the maximun, decreasing again etc? How can it decrease and increase energy? They say the changing electrical charge causes the magnetic wave, simultaneously, or almost, because one thing causing another cannot happen at exactly the same time as the other thing, because it's CAUSING the other thing. The magnetical 'charge' again causes the electrical, but how can it go from positive to negative, or even passing zero and have no charge at all? How can this wave be interpreted?

Sandor
2004-Jan-04, 12:42 PM
Sam, I think you're right about the smallest particles not even being particles, but some sort of energy, of 'moving nothing', just like these light waves. No quark can be 'captured', and by the time we can 'see into' these quarks, we will maybe again find empty space with a much smaller 'part' in it, seemingly moving around etc.

The difference between energy and matter is just the space in which it moves: is it moving at relatively (compared to us or anything else) the same space (matter), or is it moving at greater distances (energy)?

Also I agree on the photon, which, if we have to 'materialize' it, should be a long skinny thing, because it has to 'fit in the whole wave'. One reason for this is that emitting the photons from the source will only send out one photon out if the other one completely left the source. :D

Interesting to read about the antennas. It tells us also why metal is 'shielding' the radio waves: because it absorps them.

About the Doppler effect: when the object is moving away, it sends out photons with the same frequency as it does when it is not moving away, that is, from it's own perspective. Because of relativity, we will see this frequency going down (time of a fast moving clock going slower). If we want to understand this in 'matter', the distance 'between' 2 photons sent out is longer, which in fact means that the photons are longer, or the waves. We then understand that there is not really any loss of energy (or gain of energy in blueshifting), but it is all a matter of perspective (relativity). When we move in the same direction as the emitting object, with the same speed, we will ourselves gain energy in relation to the photons coming our way, or the photons will gain energy, resulting in a 'higher' color.

There is no amplitude as far as I understand. The energy of 1 light ray is the color or wave length. Or maybe there is, but what is it?

Sandor
2004-Jan-04, 01:25 PM
One other interesting point of view in this blueshifting: when the source object is moving towards us, the emitted light fotons will 'want' to go faster. But because of relativity, they can of course not go faster, but instead the photons will shrink (shorter waves).

Sandor
2004-Jan-04, 02:18 PM
Well, the Doppler thing, coming to think of it, clocks go slower when they move away from us; they will however do the same moving towards us. But in the last case, photons will be shorter, so more photons will hit our eye in one time frame, even though there are just as many photons emitted per OTHER time frame. How can the number of photons increase meanwhile? It can not. It just seems that way, because of 'changed' time. Well, in fact, neither of them is THE truth, it's just all relative. The photons from the 'slower' clock coming towards us, will GAIN TIME on their way, so coming in '1 by 1' much quicker eventually, they will speed up 'visible clock time' (as a faster moving clock pointer seen by us).

Sandor
2004-Jan-04, 02:36 PM
One can say: "photons regarded as matter will shrink when at high speed, but how about waves?" Well, it's the same as with photons having no mass but nevertheless being 'attracted' by gravity: you might as well regard this as a massive body not really attracting a photon, but just 'curving' space, or shrinking space, thus decreasing distances.

Gravity is also just 'speed' or acceleration, in fact we'd better speak of ENERGY. Big forces, like the 'speed' of light or gravity, will 'start' the theories of relativity. So light waves speeding towards us will also shrink the space between them and us, making them shorter or higher frequency.

swansont
2004-Jan-04, 02:43 PM
Now, if we use this wave theory for light, we have indeed those 2 different factors:

1. number of photons per time frame
2. momentum or energy of 1 photon

The first one can determine the frequency (color) and the second one...well, what is that? The brightness or intensity of light?

You've got it backwards. The color is determined by the energy per photon (as is the momentum, wavelength and frequency). The number of photons determines the intensity.

Sandor
2004-Jan-04, 02:51 PM
Hahaaa Swansont, that's indeed true but also not true!
It's just how you see it. The number of photons determine the intensity, that's also what I say, but I say that these are DIFFERENT rays, together adding up to the eventual number of photons.

The energy of ONE RAY (wave length/frequency/momentum) is indeed determining the color, but this is not the energy of one photon! All photons have as much energy; there is not something like 1 photon. Only many photons coming in at a certain frequency can determine a color, but one photon doesn't have any frequency of course.

Kaptain K
2004-Jan-04, 02:53 PM
To quote Wolfgang Pauli: "This isn't right, it isn't even wrong".

swansont is 100% right.

Momentum (per photon) = energy = frequency = color.

Number of photons (per second) = intensity. Einstein won the Nobel Prize for explaining this.[/i]

swansont
2004-Jan-04, 02:54 PM
How much will a temperature change inside the clock change the clock’s rate, as compared with how much the clock’s rate is affected by a change in elevation?

For example. If we note an atomic clock rate in the lowest earth valley, then we take the clock up to the top of Mt. Everest and note the rate change due to the elevation change, then how much of a temperature change would it take for the temperature to affect the “tick rate” of the clock in that same amount?

It depends on the clock. Quartz crystals, electronics, and all sorts of components have temperature coefficients of something that affects the overall frequency. For pendulum clocks, there's an obvious correlation between frequency and length of the pendulum, which is temperature dependent unless you've done something to compensate (and really good early 20th century clocks did). That's why you regulate the temperature (and sometime humidity) of good clocks.

A colleague of mine showed me a plot the other day of some Hydrogen maser data showing some fluctuations with strong correlation, noticable above the noise - it turned out to match up really well with barometric pressure changes!

Glom
2004-Jan-04, 02:59 PM
Hahaaa Swansont, that's indeed true but also not true!
It's just how you see it. The number of photons determine the intensity, that's also what I say, but I say that these are DIFFERENT rays, together adding up to the eventual number of photons.

The energy of ONE RAY (wave length/frequency/momentum) is indeed determining the color, but this is not the energy of one photon! All photons have as much energy; there is not something like 1 photon. Only many photons coming in at a certain frequency can determine a color, but one photon doesn't have any frequency of course.

What? Planck's Law: E=hf. The energy of an individual photon determines its frequency. Individual photons most certainly have frequency!

Sandor
2004-Jan-04, 03:07 PM
What? Planck's Law: E=hf. The energy of an individual photon determines its frequency. Individual photons most certainly have frequency!

PLEASE EXPLAIN! :)

What is the differenc between ONE blue photon and ONE red photon? Please don't use just (hollow) terms, but describe the difference in matter/ energy/ space/ time.

Kaptain K
2004-Jan-04, 03:26 PM
E (energy) = h (planck's constant) x f (frequency)

A blue photon has a higher frequency than a red photon, therefore, it has more energy.

Sandor
2004-Jan-04, 03:38 PM
No really, please explain, what is the frequency of ONE photon? The wave/photon of a red photon is longer? So if the wave/photon is longer, with the same speed, how can the energy be LESS?
One photon doesn't make sense, it's a way of saying.

Glom
2004-Jan-04, 03:55 PM
One photon doesn't make sense, it's a way of saying.

Welcome to quantum mechanics!

TriangleMan
2004-Jan-04, 03:59 PM
One photon doesn't make sense, it's a way of saying.
Welcome to quantum mechanics!
Please check all of your common logic in the overhead bins or under the seat in front of you. :lol:

Sam5
2004-Jan-04, 04:27 PM
No really, please explain, what is the frequency of ONE photon? The wave/photon of a red photon is longer? So if the wave/photon is longer, with the same speed, how can the energy be LESS?
One photon doesn't make sense, it's a way of saying.


That’s like saying, “One sound wave doesn’t make sense”.

But I’ve actually seen one sound wave represented graphically on a film soundtrack.

And a single photon can be represented the same way.

A sound wave has a beginning and an end. A “compression” part and a “vacuum” part, and this is often described in terms of a “wavelength”.

A single photon is a long skinny thing. It’s not a moving “dot”. Whatever it is, it’s long. It has at least 2 dimensions, amplitude and length. So, a high energy high-frequency photon is not a “dot”, it’s a long thing, but shorter than a low-frequency photon.

A high-frequency photon is shorter than a low frequency photon. So, the “blip” of a single high frequency photon is a quicker and higher energy “blip” than that of a lower frequency photon.

In sound, the amplitude is usually considered the “energy” of the sound wave, but in physics the frequency or wavelength is considered to be the “energy” of the photon. A high frequency photon carries all its energy in a shorter higher-energy package than a low frequency photon.

If we could transform a photon into a graphical representation, a short high frequency photon would look like this:

<BLIP

A low frequency one would look like this:

<blooooooop

Sam5
2004-Jan-04, 04:47 PM
What? Planck's Law: E=hf. The energy of an individual photon determines its frequency. Individual photons most certainly have frequency!

PLEASE EXPLAIN! :)

What is the differenc between ONE blue photon and ONE red photon? Please don't use just (hollow) terms, but describe the difference in matter/ energy/ space/ time.



One blue photon is physically SHORTER in length than one red photon. A single photon has length, and that is called its “wave length”. It’s not just a dot moving through space. It's a long skinny thing moving through space.

The frequency of any single one photon can be thought of two different ways: It is just "1" passing a point in space. Not one per second, just 1. But that 1 is said to have a certain "frequency" if we imagined a long train of them passing a single point in space over the time of 1 second. Millions of light photons will pass that point in 1 second.

The frequency of a "beam" or a long train of long radio photons, each of 1 mile in wave length, is 186,000 waves (radio photons) passing a point in space in one second. IE, 186,000 radio waves per second, 186,000 radio "photons" per second passing a point in space. So, the frequency of a SINGLE radio wave photon of 1 mile wavelength would be called 186,000 "cyles" or one Hz, because 186,000 of them could pass a point in space in sequence, with all of the rest of them following the one in front, in 1 second.

Sam5
2004-Jan-04, 04:56 PM
PLEASE EXPLAIN! :)

I used to play around with sound when I was a teenager and when I started in the film business. We had optical sound tracks in those days, so we could see the “sound waves” represented on a sound track as a long “graph”.

A single wave could be edited out of a sound track and spliced into a blank track, and when that single wave was run across a sound reader (like in a movie projector), I would hear a “blip”. If I edited a low frequency single wave and a higher frequency single wave, and spliced them about 3 inches apart on a blank sound track, I still heard “blips”, but I could tell that the lower frequency blip had a lower tone than the higher frequency blip.

So, even a single photon can be determined to have a “higher” frequency than a single “lower” frequency photon, by determining how long the two photons take to cross a single point in space. If it’s a short time, then that is a “high frequency” photon. If it’s a longer time, then that is a “low frequency” photon.

Sandor
2004-Jan-04, 05:30 PM
OK Sam, I don't have a problem with you saying that a 'blue photon' has more energy than a red one, because we both understand that only a beam of photons will be visible as light, and that in fact the frequency or NUMBER of photons per time frame will determine the color.

If we only understand that.

Also one 'sound molecule' does not at all have any pitch; the pitch is determined by many molecules reaching our ear in at least one minimum 'recognizable' time frame.

Now, why do you think a light photon has an amplitude and what is it?

About the sound waves that you edited: do you really think that one such a sound wave was presenting just one molecule, being compressed and accelerated and uncompressed again?

Sandor
2004-Jan-04, 06:22 PM
Who knows about the radio waves? If the frequency is a given fact, what else determines all possible sound aspects?

And how on earth is it possible that one light bulb can send out billions of photons that will fill up the entire space for miles, in all 3 dimensions? I mean, one would expect the photons to 'move away' of each other through space, but no matter where you put your eye, you will catch the photons. How is this possible?

Sam5
2004-Jan-04, 07:28 PM
OK Sam, I don't have a problem with you saying that a 'blue photon' has more energy than a red one, because we both understand that only a beam of photons will be visible as light, and that in fact the frequency or NUMBER of photons per time frame will determine the color.

If we only understand that.

Also one 'sound molecule' does not at all have any pitch; the pitch is determined by many molecules reaching our ear in at least one minimum 'recognizable' time frame.

Now, why do you think a light photon has an amplitude and what is it?

About the sound waves that you edited: do you really think that one such a sound wave was presenting just one molecule, being compressed and accelerated and uncompressed again?

It is not a single air “molecule” that is “hitting” your eardrum. It is the compressed and vacuumed “field” of the molecule that is resonating your ear drum, and there is some time factor between the start of the compression and the end of the vacuum. So, it’s not just an instantaneous “click”, it is a “bump” of a certain tone and duration, plus a lot of side by side molecules are doing this at the same time. The old story that many of us learned in school science classes, in the 5th or 6th grade, about the “molecules” physically “hitting” our ear drum, is not true.

If just a single air molecule “hit” your ear drum, all we would hear of “sound” is a series of sharp clicks, more or less per second.

But since a single air molecule carries with it, its compressed field followed by its vacuumed field, or its compressed field density followed by its thinner field density, and so we hear a time difference between the onset of the compression until the end of the vacuum.

There is a time factor involved in this. We don’t hear sound as just a series of instantaneous clicks.

If you throw a single marble at a drum head, you will hear a single “bump”, but that “bump” will take some time for you to hear all of it, from beginning to end, and that single bump will carry the natural resonation frequency of the drum head. If the drum is a snare drum, you will hear a high pitch of that single “bump”. If it is a bass drum, you will hear a low pitch of that single “bump”. If you record the bump and transfer it to an optical sound track, you can see that the snare drum head bump will be shorter than the bass drum head bump. The single “bump” will be a long single waveform, with the higher resonation frequency of the drum head attached to the sides of it (on the optical sound track) as little wavelets or “bump-lets”.

You will see one big “wiggle” on the sound trace, with a bunch of little wiggles attached to it.

The big wiggle will represent the duration of the sound created by the single marble hit, while all the little wiggles will represent the higher frequency of the natural resonation frequency of the drum head.

This will be, in effect, a “compound” musical “note”, and the tone of the note will be different for a snare drum and a bass drum. You will actually hear the snare drum “bump” for an overall shorter time than you hear the base drum “bump”.

So, don’t think of a photon as a single dot moving through space, think of it as a long skinny thing. And don’t think of a single sound wave as being just a single air molecule hitting your eardrum. A single sound wave is actually a compression of a molecule’s fields, followed by a vacuum or a “thinning out” of the molecule’s fields, and this “compression” followed by the “vacuum” produces the single tone of a certain duration, even if you hear only one sound wave. And also, it is not just single molecules that do this. You probably have a million or more molecule fields compressing and vacuuming in unison, all at the same time, all along the surface of your ear drum.

You can’t actually get just one sound wave out of hitting a drum head with a marble. It’s actually not easy to record just one sound wave. But, we can record a continuous sound tone, and then cut just one wave out of an optical sound track. We can therefore look at one sound wave, and we can play it back on a sound reader as one sound wave, but a strange thing happens inside the speaker of the sound reader. The speaker diaphragm doesn’t move out and in just once when we run a single sound wave through the sound reader. It moves out and in several times, just as the drum head vibrates in and out several times when you hit it with a marble.

If you hit a drum head with 100 marbles in one second, you will hear a compound note. One frequency will be 100 hz, and the other frequency will be the natural resonation frequency of the drum head, which I think would be higher than 100 hz.

Here is the formula to calculate the natural resonation frequency of a drum head:

http://hyperphysics.phy-astr.gsu.edu/hbase/music/imgmus/snare.gif

LINK TO SOURCE (http://216.239.57.104/search?q=cache:1JDPZLzH4FkJ:hyperphysics.phy-astr.gsu.edu/hbase/music/snare.html+frequency+snare+drum&hl=en&ie=UTF-8)

We can’t see just one single light photon, since it moves too fast, its energy is too weak, and its physical diameter is too small. So in order to barely be able to see the weakest light, we must see a lot of photons arriving all at once, and a long series of them arriving one after another, with a lot of long wave trains being side by side.

And then we see the weakest light in black and white, not in color. The light has to be brighter for us to see it in color. That means we have to see more side by side photon wave trains per square mm for us to begin to notice the color.

I’m not sure how physicists measure the “amplitude” of light. I don’t know if they think that a single photon can have more or less amplitude or if they think of it in terms of a higher amplitude as being more photons hitting a small area per square mm.

(If anyone notices any errors in what I have said, please offer a suggestion to correct them.)

Sam5
2004-Jan-04, 07:54 PM
Who knows about the radio waves? If the frequency is a given fact, what else determines all possible sound aspects?

And how on earth is it possible that one light bulb can send out billions of photons that will fill up the entire space for miles, in all 3 dimensions? I mean, one would expect the photons to 'move away' of each other through space, but no matter where you put your eye, you will catch the photons. How is this possible?

It is sometimes simpler for teachers to tell us the wrong thing in a science class, than for them to explain the full complex phenomena to us. That’s how I learned in the 9th grade that the electrons of an atom “orbit” the nucleus of an atom “just as the planets orbit the sun”. That was wrong, but the teacher found that simple explanation easier to tell that the real explanation, which is not completely understood yet.

What you said about “no matter where you put your eye, you will catch the photons”, is not true.

If you move a small light bulb 50 miles away from you, you will not notice any of its photons. If you happen to “catch” a few, they will be so small and so far apart per sq mm of your retina, they will be too “weak” for you to notice them.

I live in a place where I can go on top of a mountain and see for more than 50 miles. At night, I can see a bright light for as far away as 30-40 miles. These are usually sodium vapor or mercury vapor lamps, so their output area is wide, and the amplitude of the light is high. I don’t if the amplitude is high because the amplitude of each photon is high or if it’s because I’m seeing more photons per sq mm of my retina.

But, if I tried to see a small 12 volt bulb light at 30 miles, I would not be able to see it, since its putting out fewer photons per second, and they are all spreading out. So, I would be looking at a lot of the empty space in between the spread-out photons, and if I happened to “catch” a few of them, they would be too weak and too spread out per sq mm of my retina for them to register as an electrical “signal” in my brain.

I might be able to see the dim light from a 12 volt bulb at 30 miles with a telescope. The main telescope lens is wider than the lens of my eye, so it is collecting a lot more photons than my eye lens can collect, and then it is focusing those extra photons into a small spot at the back of the telescope, which causes more photons per sq mm to hit my retina. Then I can see the 12 volt light as a small but dim white spot in the darkness. If I use a bigger telescope with a bigger main lens, but if I use the same power of magnification, then I will see the same size of the distant bulb, but I will see it as a brighter spot of light.

Sandor
2004-Jan-04, 08:39 PM
That's quite interesting Sam, about the compound note.

I have this question: when you talk about ONE molecule's field, with a compression and a vacuum state, I guess you mean that the molecule itself is compressed in the beginning (when the other molecule hits it), and it's at a vacuum state when it reaches max. velocity? And then, when it hits the next molecule, it will compress again, so when it's back to the same compression as where it started, we have one wave.
Now, for such ONE wave to reach your ear, you need all the waves from the source to your ear, and all the molecules in between pumping into each other. Finally, when the first molecule hits your eardrum it will compress itself against your eardrum, 'gaining force' etc. I understand this. You can theoretically cut of there, just put something right after this first molecule that hits your eardrum, so the next one will never reach it. You will still hear a sound, even though it's very 'thin', because it's not a compund note.

But now to the photons. Let's assume one photon is intensive enough to be seen. And let's assume it is as long as it's wave. And let's assume it is not a wave, but really a photon. If you want to explain it as a wave instead, be my guest, but you didn't. So it is a photon, a particle, and it starts to fall into the eye at a certain moment, and it ends some moments later. This pops up already a question here, because when many photons, as usual, hit your eye one by one, how can you tell the difference, I mean, how do you know one has ended and the next has started? Well, OK, we're now talking about 1 photon. So from the time it takes to start and end this photon, you know it's lenght. But now this is your conclusion: a long photon has less energy than a short one. How is this possible?

Compare this to what we already guessed, following our commen sense: a higher frequency means more photons to be fired at a time frame, meaning more energy at a time frame.

This is why I say: one photon can never have more or less energy, or a color. This is also common sense when we know that it always has one speed and one mass.

Sandor
2004-Jan-04, 08:56 PM
About the photons spreading around: we don't have ANY hills here, that's why our country is called the Netherlands (down/low or flat land). :)
But of course your answer must be true: the further you are from the source, the less photons will hit your eye, until no photons at all hit it. It's just the extremely small size of the photons that surprises me, because how many photons have to escape from one very small but strong light bulb, let's say 1 mm^3, in how many directions, to be sure that EVERY fraction of a mm in 3-d space around it for maybe 1 MILE away will be 'filled at a certain time with some photons?

1000m.^3 with the light bulb in the middle is only 500 meter away from it, at the maximum. So from 1mm^3 light source for 0,1 second or so, to fill a complete 1000m.^3, this is, I think, 1 : 1.000.000.000, well OK, it's just how it is I guess. Unless this is possible because space shrinks with such a speed.

Sandor
2004-Jan-04, 09:24 PM
I can put this one-photon thing also in another way: a molecule exists of several atoms, which consist of several electrons etc. This is why a molecule can have a certain density or wave length. In fact more electrons will hit our eardrum at a time frame when density is high.
If we see a photon as the smallest particle (in it's 'type'), it cannot be smaller or longer, or having any density or wave by itself. However, a smaller photon can theoretically have more energy because of a higher density.

So, if we succeed in proving that one photon can be recognized as a color (maybe we already did), we have the proof that it consist of even smaller parts or particles.

Sam5
2004-Jan-04, 10:07 PM
That's quite interesting Sam, about the compound note.

Yes, when I was in high school, before I ever saw an optical sound track, I used to wonder how a high frequency was recorded along with a low frequency and what that might “look like”. When I first started working with optical sound tracks, I found that the high frequency wiggles were attached to both sides of a low frequency wiggle. If we flattened out the low frequency wiggle on the optical track, by reducing its amplitude down to zero, then the high frequency wiggles would appear on the optical sound track as a long series of short wiggles.


I have this question: when you talk about ONE molecule's field, with a compression and a vacuum state, I guess you mean that the molecule itself is compressed in the beginning (when the other molecule hits it), and it's at a vacuum state when it reaches max. velocity? And then, when it hits the next molecule, it will compress again, so when it's back to the same compression as where it started, we have one wave.

I’m not sure if the molecule itself is actually compressed or not, but there must be some sort of “field” that surrounds molecules, because I’ve read in several books that they don’t ever touch each other. So, at a certain air pressure, the molecules are a certain distance apart, with their fields acting as a compressed “buffer” zone that keeps them from actually touching.

When we compress a gas, we aren’t compressing the molecules together surface to surface, but we are compressing their fields together, so the molecules get closer together while their fields are compressed, but the molecules still never touch one another. I don’t know if I’m describing this correctly or not. But think of balloon with a dot like a "bb" in the center of the inside of it that we call a molecule. Then the air inside the balloon would represent the molecule's “field” that surrounds the molecule. We can press two balloons together and that is like compressing two molecules fields together, while the two molecules never actually touch each other. And if we had a long line of balloons, we could send a “wave” down that line without any of the dots of molecules in the center of each balloon touching each other. This is what I think a section of a sound wave is, the compression of the field of a molecule, followed immediately by a "vacuum" in that field. These compressions and vacuums in the fields of the molecules move along through the air as a compression/vacuum "wave" at about 1,100 ft per second, with each molecule itself moving only a very short distance and then it returns to its original position, relative to the earth.

I’ve never been able to find a real explanation of this phenomena, so I’ve had to deduce it myself, based on all the information I have been able to find. They don't teach this stuff in school. They just give us the simple stuff, like "the molecules bump into each other".

I’ll get back to your other questions later.

Sandor
2004-Jan-04, 10:29 PM
This 'field', yes it makes some sense that there is SOMETHING, because what I said about a certain number of electrons hitting the eardrum is very unlikely, because the atoms have a fixed volume, with fixed electron rings, with the outer ones shared by several atoms. I mean, an 'air molecule' is an air molecule, with probably a fixed 'density'.

But density is really more energy or more 'speed' at a certain 'space', so I can imagine 'atoms' moving quicker within the molecule, resulting in a higher density molecule or something like that.

It could also be quite a mess when the outer electrons of one molecule actually hit the electrons of the other molecule, so when we come to think of it, the school stuff we learn is really very, very different from how it actually can be. But this 'field-theory' is, to be honest, about the same level then, very easily 'avoiding' something, but not at all explaining it. :)

The advantage of the way you use it, is that you know that you don't know, and me neither. ;)

Sam5
2004-Jan-04, 10:54 PM
This 'field', yes it makes some sense that there is SOMETHING, because what I said about a certain number of electrons hitting the eardrum is very unlikely, because the atoms have a fixed volume, with fixed electron rings, with the outer ones shared by several atoms. I mean, an 'air molecule' is an air molecule, with probably a fixed 'density'.

Maxwell said in his 1877 book, “Matter and Motion”:

”Air, compressed in the chamber of an air-gun, is capable of propelling a bullet. The energy of compressed air was at one time supposed to arise from the mutual repulsion of its particles. If this explanation were the true one its energy would be potential energy. In more recent times it has been thought that the particles of the air are in a state of motion, and that its pressure is caused by the impact of these particles on the sides of the vessel. According to this theory the energy of compressed air is kinetic energy.”

But I believe it is BOTH, caused by the mutual repulsion of its particles by means of their fields keeping the molecules physically apart, AND by the “impact” of the fields of these moving molecules on the sides of the vessel. With the “compression” causing a compression of the fields, and the rate and strength of their “impact” caused by the heat energy of the gas. Higher temperatures cause more rapid motions of the molecules and deeper compression of their repulsive fields. And higher compression causes more compression of the fields. We can compress the fields with the compression energy alone, or we can compress them with more heat energy supplied to them, or by both means. We don’t have to “pump” up the air gun if we heat its air chamber.

When your soda gets hot in the can, that acts something like a compression effect and the can pops. Or, we can cool the can and put a pressure on its sides with our foot, and it will pop. What we are doing in both cases is pushing the fields of the air molecules inside the can together, while the fields are repelling each other and they want to keep the molecules a certain distance apart.

So, there seems to be a current disagreement about how molecules of a gas work. We’ve got this version (which is the second Maxwell example), where the molecules actually touch and bump into each other:

”In this gas, the atoms and molecules of the elements fly around freely, bumping into each other and everything else.”

LINK TO SOURCE (http://travel.howstuffworks.com/hot-air-balloon5.htm)

And we’ve got this version, which is the same as my version and Maxwell’s first example:

”Note that electrons can "collide" without touching. They repel each other because of alike charges, and this means that they can push against each other. If you throw one electron at another, the first one pushes the second one away, while the second one pushes back upon the first and slows it. But they never touch, and the "pushing" is done by the electrical repulsion forces. It's just like a real collision, but the particles never actually make contact.”

www.madsci.org/posts/archives/nov2000/973264612.Ph.r.html+molecules+never+touch+one+anot her&hl=en&ie=UTF-8]LINK (http://216.239.57.104/search?q=cache:lqcYXiEY36cJ:[url) TO SOURCE[/url]

So, the version I deduced is based on the outermost part of an atom, its electrons, that repel the other electrons of other atoms because of their same negative charge, and two negative charges repel one another. The “compression”, then, in my version, would a compression of these negative fields when they are forced together in a sound wave when the air molecules move closer to other molecules and then rebound, creating a slight compression and “vacuum” wave in the fields, but the molecules never actually touch each other, because of the repulsion forces of their negatively charged electrons.

So while heated air molecules do bounce around, they never actually touch. What interacts is the negative fields of their electrons, and the more rapid the motion of the molecules, the more the fields are “compressed” when they interact.

So I would say that a sound wave is a combination of a mechanical effect and an electrodynamical effect.

And I would say that light is a combination of an electrodynamical effect and a mechanical effect. That is why light behaves like both a "particle" and a "wave", and, in fact, so does sound.

Tensor
2004-Jan-05, 01:48 AM
I’m not sure how physicists measure the “amplitude” of light. I don’t know if they think that a single photon can have more or less amplitude or if they think of it in terms of a higher amplitude as being more photons hitting a small area per square mm.

Sam, for a simplified explanation use the following: using waves to describe EM, the amplitude of the wave is the intensity of the EM radiation and the frequency is it's energy. If using photons, the number of photons per unit time is the intensity and the momentum is it's energy. I noticed you were trying to decribe a photon. In most cases the following way of thinking about it may help. EM radiation propagation is more easily though of in wave terms and more easily thought of as a particle (photon) when interacting with matter (think of Einstein's explanation of the Photelectric effect.) It's not that they change from one to the other, it's just a way to picture it.


(If anyone notices any errors in what I have said, please offer a suggestion to correct them.)

I didn't see any errors (this doesn't mean there aren't any, just I didn't see any :D ) Just a clarification on how physicists describe the phenomena. Hope this helps you sort it out.

Tensor
2004-Jan-05, 02:13 AM
It is sometimes simpler for teachers to tell us the wrong thing in a science class, than for them to explain the full complex phenomena to us. That’s how I learned in the 9th grade that the electrons of an atom “orbit” the nucleus of an atom “just as the planets orbit the sun”. That was wrong, but the teacher found that simple explanation easier to tell that the real explanation, which is not completely understood yet.

Sam, it's not necessarily "wrong", as long as it's valid within its domain. I doubt that most ninth graders have the math or understand (or really care about) electron orbitals, probability clouds or other quantum mechanical effects that constitute the full explanation. When I was working with chemicals, I would often think of atoms (say an hydrogen atom) as having an electron orbiting around the nucleous. Is it completely right, no and I knew it. Did I need the complete explanation, no.

Sam5
2004-Jan-05, 02:22 AM
I’m not sure how physicists measure the “amplitude” of light. I don’t know if they think that a single photon can have more or less amplitude or if they think of it in terms of a higher amplitude as being more photons hitting a small area per square mm.

Sam, for a simplified explanation use the following: using waves to describe EM, the amplitude of the wave is the intensity of the EM radiation and the frequency is it's energy. If using photons, the number of photons per unit time is the intensity and the momentum is it's energy. I noticed you were trying to decribe a photon. In most cases the following way of thinking about it may help. EM radiation propagation is more easily though of in wave terms and more easily thought of as a particle (photon) when interacting with matter (think of Einstein's explanation of the Photelectric effect.) It's not that they change from one to the other, it's just a way to picture it.

Ok, thanks.

Of course you know that I tend to think in a Newtonian/Euclidean “classical” way, so sometimes it takes me a little time to train myself to think in a non-classical way.

If “the number of photons per unit time is the intensity”, does that mean a single photon of one frequency can not be “more intense” than a different single photon of the same frequency?

With sound, a single wave of one frequency can be either of low or high intensity, but I’m not sure if a photon works that way.

Regarding the photoelectric effect, the same results might be obtained by looking at the situation from a classical point of view and saying the photon is not “a particle of a certain energy”, but “a wave of a certain resonation frequency”. The electrons would be driven off the face of the metal not by being “hit by a particle of a certain energy” but by being “resonated by a certain wave-packet frequency”, similar to the way a radio antenna is resonated by a radio wave. Similar to the way some classical physicists say the Tacoma Narrows bridge collapsed not just due to the wind intensity, but due to the natural resonation frequency of the bridge being achieved by the steady wind.

I used to own a big old Victorian house that had window frames that resonated to the frequency of a diesel train switch engine a half a mile away, even though I could not hear that low frequency of the idling engine inside the house. It would appear to someone that big “particles” were hitting the window frame from the outside. I figure it would take about 6 strong guys on ladders to shake that window frame that much, if they acted as big “particles”. But the frame resonated to a low frequency of sound that I could not even feel, see, or hear. I discovered the source by noticing that the resonation started and stopped on a regular basis, and I went out on my front porch and heard the train engine rev up to a higher frequency, then drop back down to a low idle frequency that I could not hear or feel.

But apparently physicists disagree about this, as in these two examples:

www.math.umbc.edu/~gobbert/teaching/math101.s2003/reports/Group1Tacoma.doc+bridge+collapse+wind+resonation&h l=en&ie=UTF-8]LINK (http://216.239.53.104/search?q=cache:Ow_PHFJH2oUJ:

www.geocities.com/marcelleng/chap8.doc+tacoma+collapse+resonation&hl=en&ie=UTF-8]LINK (http://216.239.53.104/search?q=cache:3jqnbVSG59UJ:[url) TO A SOURCE WITH A DIFFERENT OPINION

Sam5
2004-Jan-05, 02:28 AM
Sam, it's not necessarily "wrong", as long as it's valid within its domain. I doubt that most ninth graders have the math or understand (or really care about) electron orbitals, probability clouds or other quantum mechanical effects that constitute the full explanation. When I was working with chemicals, I would often think of atoms (say an hydrogen atom) as having an electron orbiting around the nucleous. Is it completely right, no and I knew it. Did I need the complete explanation, no.

Ok, that makes sense.

Say, I went to a Riemann-shaped library yesterday. What a place! Euclidean books in Euclidean book shelves arranged in a Riemann-shaped library!

The book shelves radiated out from the center of the building like the spokes of that Riemann-gravitational field drawing I posted on the other thread.

I took some pictures inside it. I’ll post them when I get them developed.

Sam5
2004-Jan-05, 03:00 AM
Tensor,

Speaking of pictures, the same classical “resonation/photo-electric effect” might apply to film and photo paper. Under this Sam5 theory, the silver-halide molecules would break apart at certain resonation frequencies.

For example, I disagree with this:

"Everyone knows that the frequency of red light is too low to dissociate silver halide; infrared radiation does not transmit enough energy to excite electrical signals in the optic nerve, and ultraviolet waves are too short to inter-act with the electrons that produce vision."

http://216.239.53.104/search?q=cache:64L7_KBxGOsJ:www.eutopia.no/Pawels.htm+silver-halide+einstein&hl=en&ie=UTF-8

And this:

http://216.239.53.104/search?q=cache:WJy45MCWsPsJ:www-inst.eecs.berkeley.edu/~cs39j/fa03/session07.html+silver-halide+einstein&hl=en&ie=UTF-8

A major reason why I disagree with this is because I learned that the rods and cones in our eyes are about the size of light wavelengths. This indicates that they are EM “antennas” just like a metal radio antenna, and they are “tuned” to the resonation frequencies of visible lightwaves. Not only that, but I think the cones are cone-shaped so that they can resonated at different areas of the cones for the blue, green, and red light. In a TV, we have three electron guns, one for blue, one for green, and one for red. This cuts down on overall sharpness and the pixel count. But in the eye, instead of having three sets of cones of three different resonation wavelengths, which would give us fewer “pixels” per sq mm, we’ve got cone-shaped antennas that can resonate at different areas along the cone, thereby giving us three times the pixel count. LOL, good designer.

In our lifetime, I don’t think it would be a good idea to have our rod and cone molecules “lose” electrons due to the “particle” version of the photo-electric effect. With “resonation”, they don’t “lose electrons”.

Also, this “resonation” idea matches a new way of thinking in botany, with leaves having “antenna” the size of light wavelengths that absorb sunlight.

If the photoelectric effect “particle” theory were correct, we wouldn’t need “resonation” sized rods and cones and “antenna” on plant leaves or full and half-wave length radio antennas.

www.weizmann.org.uk/eng/Enigma/Enigma7/11045_Enigma_8.pdf+plant+leaves+antenna&hl=en&ie=U TF-8]PLANT (http://216.239.53.104/search?q=cache:QhlZsMDx5VgJ:[url) ANTENNA[/url]

Kaptain K
2004-Jan-05, 06:44 AM
A major reason why I disagree with this is because I learned that the rods and cones in our eyes are about the size of light wavelengths.
This is incorrect. The rods and cones are far larger than the wavelength of light. Due to the laws of optics, we cannot image anything that is not larger than the wavelength of the radiation we use.

Sandor
2004-Jan-05, 03:37 PM
I think Sam is right however, about the cones acting as antennas. We can see light or sound as particles as well as waves. Also this window, resonating by the passing train; it is in fact possible to imagine millions of particles per second hitting the frame with exactly the 'own frequency' of the frame, so it will very softly begin to shake, but because it's resonating very naturally as regarded to the material, it will enforce itself in shaking.

Then the 'bouncing', but not hitting of electrons: yes of course, it's the magnetic field (though I didn't think of it). But this is an answer that brings up alike question all over again. We have to see this magnetic force as even smaller particles exchanging the charge, OR as waves, or whatever.

About the amplitude again: in electricity, charge (voltage) is the frequency which is the number of electrons per second. But I would call this serial: it's only one thin line of electrons. Current (Ampere) however, is the 'total amount of electrons ready to take off', but limited by the parallel paths or bandwith, which is the total of parallel thin lines, or just the number of roads accessible. These are the 2 dimensions of all EM waves, bundled together.

Radio does the same. AM uses changing amplitude on a given frequency, which is really the number of parallel waves arriving at a time frame. FM is tuned to a short range of frequencies (haha I might as well quote all of these entire sentences since I'm saying this in amateur language) and catches all of those different frequencies, which together form a constant changing 'total-frequency'. So that must be double quoted. :)

One electron falling back to its ground state, emits one photon. That's it. It only has velocity and mass (not detectable), nothing more. An electron cannot fall back harder or faster. But it can get excited again and fall back again with a certain speed. The electron is doing this together with all of its pals in 3-d around. When more electrons arrive through the wire, more waves are sent out, all with a slightly different angle. But millions of different WAVES can fall in our eye, all coming from one point, just because these waves start with 'in between space of one photon' which is about nothing to us.

Because of quantum laws, only photons with the right energy/frequency will be absorbed by an object; others will be emitted again. This will give the light the color of the object. Of course there is no intensity of one photon here, bumping harder into one electron. The intensity is the number of photons falling in PARALLEL. This is the amplitude.

Now, if one photon is not only undetected, but really cannot have any length or wave, it just cannot exist on its own. This must mean that from the source until the receiver, there must be 'one continous line' of photons and THAT will determine the length of one photon. So the minimum of energy that can be emitted is the distance between the source and receiver divided by the wave length, which can be one wave in theory, but this will have sooo little energy. :)

It is quite obvious that this is the only way to exchange forces, because one can only practice force on something, when this something is returning as much force or resistance. This seems strange, because you might expect nothing to happen then, but every force has its exact opponent, only 'divided' differently in time and space. So the resistance must be 'detected' before any force can be practiced.

But this seems to start the questions also at the beginning, because what particle is sent out for detection? It's the same problem as with the magnetic forces and 'not bouncing' particles. Well, the source force (accelleration) will shrink space in a direction untill source and receiver HAVE NO SPACE in between actually. There is no 'empty space', like there is no 'light' going through 'empty space'. Space is only time, we can see it that way also.

To be honest with you, light is 2-dimensional. :) All EM forces are 2-d. This means that our 3rd space dimension is the time of 2-d. But from our perspective we'd better call it timeless. Timeless is the same as changeless, because time is really change/force. Light photons do not change at all, in fact nothing happens 'on their way'. Only 'when' they hit matter, something happens. In between nothing happens, so there is no in between and no time. But I said space is time. Yes, it takes you time to get on the other side of space, or in another idea, it took time to move matter away from each other. The cumulated time of that is 'now' experienced as space. But for us, 2-d forces are more or less timeless. With the speed of light (immediate accelleration to c and then nothing), a clock stands still. But also, space has shrinked to 2-d.

Sam5
2004-Jan-05, 04:50 PM
A major reason why I disagree with this is because I learned that the rods and cones in our eyes are about the size of light wavelengths.
This is incorrect. The rods and cones are far larger than the wavelength of light. Due to the laws of optics, we cannot image anything that is not larger than the wavelength of the radiation we use.

No, what you are thinking about is the Heisenberg uncertainty principle and some of its ramifications, one of which says that we can’t see the correct image or shape of very small objects (even using special microcopses) because the waves we use to illuminate the object are larger than the object itself.

But that’s not what I’m talking about. I’m talking about eye cones being electro-magnetic wave “antennas” that are as small as light wavelengths and are tuned to the specific wavelength of light that we can see, just like a short-wave radio operator has a long outdoor wire antenna, with the wire being the same length as the wavelength of the radio signal he is receiving. This is classical 19th Century “resonation”, not Einstein’s “photo-electric” effect.

I got that information about the size of the cones out of an optics book and a National Geographic article. I just found a website that says:

“There are 3 types of cones, each with different photo pigment
Short wavelength cones: Max. sensitivity at 440 nm
Medium wavelength cones: Max. sensitivity at 540 nm
Long wavelength cones: Max. sensitivity at 565 nm”

SOURCE (http://216.239.53.104/search?q=cache:b-J8Eev78C8J:arrow.win.ecn.uiowa.edu/56240/eyelecture.ppt+wavelength+eye+cones+size&hl=en&ie= UTF-8)

And this confirms what I said about cones being “antenna” that resonate at certain frequencies:

“three kinds of cones, each "tuned" to absorb light from a portion of the spectrum of visible light
cones that absorb long-wavelength light (red)
cones that absorb middle-wavelength light (green)
cones that absorb short-wavelength light (blue)”

SOURCE (http://216.239.53.104/search?q=cache:ikbA-xgJoysJ:users.rcn.com/jkimball.ma.ultranet/BiologyPages/V/Vision.html+wavelength+eye+cones+size&hl=en&ie=UTF-8)

And that website confirms what I said about all cones being able to see all colors:

“The response of cones is not all-or-none. Light of a given wavelength (color), say 500 nm (green), stimulates all three types of cones, but the green-absorbing cones will be stimulated most strongly. Like rods, the absorption of light does not trigger action potentials but modulates the membrane potential of the cones”

Also see this:

“And....the ratio of their sizes defines the "bandwidth" of the detection system - and the ratio of the size of cones to rods does correspond to the visible band! (see below).”

“Examples that I have come upon, a.) the vision of fish is supposed to lie in the infrared beyond 700 nanometers. Applying this model, retina sensitive to this spectral region should be composed of receptors larger in size than human -the cone receptors of trout are seven microns in diameter (versus the one micron value for human cones), b.) the vision of insects is in the ultraviolet which would indicate smaller receptors - and their are indications that this is the case.”

SOURCE (http://216.239.53.104/search?q=cache:jkKxp3vyJY4J:ghuth.com/A%2520new%2520Model.htm+wavelength+eye+%22size+of+ cones%22+size+diameter&hl=en&ie=UTF-8)

And this:

“(1 micron = 1000 nm)”

“Violet light is electromagnetic radiation with wavelengths of 410 nanometers and red light has a wavelength of 680 nanometers.”

SOURCE (http://216.239.53.104/search?q=cache:fv7vMsAaxU8J:acept.la.asu.edu/PiN/rdg/color/color.shtml+red+light+wavelength+micron&hl=en&ie=U TF-8)

And notice carefully that I said: ”the rods and cones in our eyes are about the size of light wavelengths.”

I did not say they were exactly the same size.

If a cone is 1,000 nm long and it responds to light that has a 680 nm wavelength, then it would seem that 320 nm of the cone length is not designed to resonate along with the rest of the 680 nm of the cone length. So, the cones are about the same size as a light wavelength, and my original statement was correct.

So please, don’t contradict me with your urban legends or bad information you learned in science class in school. I would appreciate it if, when you disagree with me, you would provide some source other than yourself to back up your information so I won’t have to waste my time doing the research required to prove that you are wrong.

If our eye cones were “far larger” than light wavelength, as you claim, then we would not see visible light, but we would see only infrared and microwaves.

Kaptain K
2004-Jan-05, 05:58 PM
Sam5,

No, I am not talking about the Heisenburg principle. I am talking about the laws of optics, that say that you cannot resolve objects that are smaller than the wavelength of the light you are using. The cells (rods and cones) are much larger than the wavelength of light, otherwise we could not see them (much less the stucture within them). The light sensitive pigments are on the order of the wavelength of light, not the cells themselves.

“There are 3 types of cones, each with different photo pigment
Short wavelength cones: Max. sensitivity at 440 nm
Medium wavelength cones: Max. sensitivity at 540 nm
Long wavelength cones: Max. sensitivity at 565 nm”
That is a list of the peak sensitivities, not the size of the cells.

SeanF
2004-Jan-05, 06:12 PM
“Examples that I have come upon, a.) the vision of fish is supposed to lie in the infrared beyond 700 nanometers. Applying this model, retina sensitive to this spectral region should be composed of receptors larger in size than human -the cone receptors of trout are seven microns in diameter (versus the one micron value for human cones), b.) the vision of insects is in the ultraviolet which would indicate smaller receptors - and their are indications that this is the case.”

...

If a cone is 1,000 nm long . . .

Know much about the difference between diameter and length, Sam5?

According to this (http://www.psy.gla.ac.uk/~steve/courses/vision/numbers.html), the cones in a human eye vary from 1-10 microns in diameter and 40-80 microns in length.

That is "much larger" than the wavelength they detect.

Sam5
2004-Jan-05, 08:05 PM
Sam5,

No, I am not talking about the Heisenburg principle. I am talking about the laws of optics, that say that you cannot resolve objects that are smaller than the wavelength of the light you are using. The cells (rods and cones) are much larger than the wavelength of light, otherwise we could not see them (much less the stucture within them). The light sensitive pigments are on the order of the wavelength of light, not the cells themselves.

“There are 3 types of cones, each with different photo pigment
Short wavelength cones: Max. sensitivity at 440 nm
Medium wavelength cones: Max. sensitivity at 540 nm
Long wavelength cones: Max. sensitivity at 565 nm”
That is a list of the peak sensitivities, not the size of the cells.

I see on different websites that there are different estimates for the diameter of rods and cones. Some say they are as small as 1 micron in diameter. Red light is .680 micron. So, whatever the size actually is, they act as “resonating antennas”.

The eye is not “resolving” something as small as a wavelength of light. We can’t see things that small. But the eye rods and cones are receiving EM waves that are small as a wavelength of light, and that’s why the rods and cones are that small, and that’s what causes them to resonate. We don’t “see” anything that small, but the cones must be that small in order to resonate properly when hit by visible light wavelengths.

Some animals have larger rods and cones and can see infra-red light. Many insects have smaller rods and cones and they can see high-blue light. That’s why a bug-zapper machine uses a blue light.

These different lengths of rods and cones can see different frequencies of light:

LINK (http://www.eyedesignbook.com/ch2/fig2-04bBG.jpg)

But this does not mean that we can see or “resolve” small objects that are as small as our rods and cones.

Laser Jock
2004-Jan-05, 09:55 PM
We don’t “see” anything that small, but the cones must be that small in order to resonate properly when hit by visible light wavelengths.

Some animals have larger rods and cones and can see infra-red light. Many insects have smaller rods and cones and they can see high-blue light. That’s why a bug-zapper machine uses a blue light.



This is, in a word, nonsense. Rod and cone size has nothing to do with wavelength sensitivity. They are not "resonating antennas". It is the type of pigment (that turns light into an electrical impulse and fed to the optic nerve) that determines the wavelength response. The size and density of rods and cones may determine how small of object an eye can resolve, but chemistry determines the wavelength response.

BTW, another group (http://www.cvs.rochester.edu/williamslab/) here at the UofR does some great work imaging live human retinas.

Sam5
2004-Jan-06, 12:36 AM
We don’t “see” anything that small, but the cones must be that small in order to resonate properly when hit by visible light wavelengths.

Some animals have larger rods and cones and can see infra-red light. Many insects have smaller rods and cones and they can see high-blue light. That’s why a bug-zapper machine uses a blue light.



This is, in a word, nonsense. Rod and cone size has nothing to do with wavelength sensitivity. They are not "resonating antennas". It is the type of pigment (that turns light into an electrical impulse and fed to the optic nerve) that determines the wavelength response. The size and density of rods and cones may determine how small of object an eye can resolve, but chemistry determines the wavelength response.




It’s probably both. I wouldn’t expect a cone that is more tuned to red to be coated with a chemical that reacts more to green. For example, you don’t plug a full wave CB antenna into a short-wave radio, and you don’t use a VHF antenna for your UHF channels. And you don’t coat a red-tuned cone with a green-sensitive substance. Anyway, color is not on the object you see but only in your brain.

You can find many science references that say the rods and cones are “tuned” to the frequency of the light, and that’s what I’m talking about. A resonating physical thing is a natural amplifier. Why do you think we have a lot of cilia of different length in our ears?

If you don’t understand what I’m talking about, I suggest you go to Radio Shack and buy yourself a book on ham radio and how antennas work.

It is possible that the “pigment” is “tuned” too, of just the right color or chemical makeup to convert the waves into stronger electrical signals, but the rods and cones are “tuned” too, by means of their size. The pigment is probably designed to “absorb” or react to the specific colors. You don’t have two things in nature that are “accidentally” the same size, such as the specific wavelength of light that we see and the diameter of rods and cones that we have in our eyes. If the resonation size didn’t have anything to do with it, bugs wouldn’t see more blue than red and nocturnal mammals wouldn’t see more deep red than blue.

I’ve already posted this, and you can look this stuff up all over the internet and in medical books. Just because you don’t know about it yet, doesn’t mean you are right and the biologists are wrong. “Tuned” in this sense is term of electrodynamics. The electrodynamic effect would apply both to the size of the rods and cones and also to the atomic structure of the chemicals involved. A term related to “resonation”. Why do you think the resonator box on a string bass is much larger than that of a violin? :

“three kinds of cones, each "tuned" to absorb light from a portion of the spectrum of visible light
cones that absorb long-wavelength light (red)
cones that absorb middle-wavelength light (green)
cones that absorb short-wavelength light (blue)”

SOURCE (http://216.239.53.104/search?q=cache:ikbA-xgJoysJ:users.rcn.com/jkimball.ma.ultranet/BiologyPages/V/Vision.html+wavelength+eye+cones+size&hl=en&ie=UTF-8)

“Cones, on the other hand, consist of three different types of cells, each "tuned" to a distinct wavelength peak of response centered at either 430, 535, or 590 nanometers. Often referred to as photopic vision,”

SOURCE (http://216.239.53.104/search?q=cache:iqCe4Wv91EIJ:micro.magnet.fsu.edu/optics/lightandcolor/vision.html+eye+rods+cones+tuned&hl=en&ie=UTF-8)

“Normal humans have three different types of cones with photo-pigments that sense three different portions of the spectrum. Each cone is tuned to perceive mostly either Long wavelengths (reddish), Middle wavelengths (greenish), or Short wavelengths (bluish), referred to as L-, M-, and S- cones. The peak sensitivities occur at light wavelengths that we call red (580 nm), green (540 nm) and blue (450 nm), provided by three different photo-pigments.”

SOURCE (http://216.239.53.104/search?q=cache:jv965p9zWXkJ:webexhibits.org/causesofcolor/1C.html+eye+rods+cones+tuned&hl=en&ie=UTF-8)

Evidently they aren't teaching enough Classical physics and electrodynamics in schools today.

Sandor
2004-Jan-06, 11:22 AM
Guys, I wouldn't make an ego-war out of this.
Sam's basic idea was totally right, and he probably wasn't even wrong about the cones resonating; he only forgot about the pigments.

Not the cones and rods are the primary antenna's, but the pigments on it are. They have 'about' the size of the light waves.

Of course we cannot clearly see things that are smaller than a wave lenght of light, or in other words, smaller than a photon, because photons that are bigger than the 'shapes' they have to communicate back to us, are not able to do that. Even if they feel more pain in their but than in their head, when bumping into the object, they will as a whole be absorbed by the pigment and thus cannot communicate these details.

On the other hand the receiver can be smaller than the wave lenght, which is exactly the case in many antenna's with the size of half a wave lenght. Because the full wave lenght causes electrical 'troubles'. It can also be bigger and it's not true that the 'upper half' of the antenna doesn't resonate in that case, because the antenna as a whole resonates. I'm not sure about the cones with their specific shape, but it seems that the cone structure of

1. a heavy bodypart
2. a light bodypart
3. a thin flexible bodypart in between

will make the light part resonate faster compared to the heavy part, but the total movement of that can only be at a very maximum when the right waves (lengths) are constantly coming through, enhancing the resonance, like in Sam's window frame. A slightly longer or shorter wave will 'bounce' with the cone's natural resonance to exponentially decrease the movements.

So my idea is: the pigments will start to resonate, each picking up one continuous wave (color or frequency). The more pigments resonate, the stronger the resonation of the entire cone (intensity or amplitude). So the cone must also have a shape that is a function of the wave frequency.

About antennas half the size of the wave or a quarter etc, or twice the size etc.: the least energy is needed when the receiver comes as close as possible to the size of the wave, or, data are more detailed. Sometimes radio antennas can be much smaller than the long radio waves, but I guess the magnetic field of the antenna is bigger then, or something like that.

HOWEVER, WE ARE DOING GREAT HERE!
Because the essence here is very right: we might as well see all energy as waves instead of particles and it's just a matter of resonating or not (quantum law for micro), or resonating from just a little bit to exponentially increasing resonance (macro).
So we have our answer to the question which started this topic: why are radio waves going through walls? Because the walls are much smaller than the wave / they don't resonate so much. Quantum laws are not applied on this scale: many waves WILL be absorped, especially in bigger walls, but there are still (enough) waves coming through.

Kaptain K
2004-Jan-06, 12:56 PM
The eye is not “resolving” something as small as a wavelength of light. We can’t see things that small.
Once again, you have misunderstood me. I did not say that we could see (resolve) things that small with the unaided eye. What I said was that we cannot magnify things that are that small to the point where we can see them. As an analogy, imagine a chain-link fence across a pond. Waves will pass through the fence unimpeded because the wires of the fence are so much smaller than the waves. A post in the water will reflect short waves while longer waves will pass around it with minor changes.

Laser Jock
2004-Jan-06, 02:34 PM
OK Sam, I see that you are confused on the word "tune." Yes, in the context of short-wave radio, radar, etc., antennas can be tuned to resonate at a particular frequency. But that is not the sense that they use the word in the sources you cited. For example:


“three kinds of cones, each "tuned" to absorb light from a portion of the spectrum of visible light
cones that absorb long-wavelength light (red)
cones that absorb middle-wavelength light (green)
cones that absorb short-wavelength light (blue)”

Note that "tuned" is in quotes above. In this context, it simply means that different cones are more sensitive to different colors. The reason that they are more sensitive is simply due to the absorption profiles of the different pigments that are used.

In my last post I made reference to the group here at the U of R. that can image the live human retina. How do they do it? In order to take any picture of the back of the eye they need to correct for the all the aberrations present in the eye. The cones are just a few microns in diameter and the smallest amount of aberration will blur the image. So what they do is use the same technique that is used by astronomers to correct for atmospheric disturbance (see, a link to astronomy!), namely adaptive optics. They shine a spot on the back of the retina, and use a computer controlled deformable mirror to actively correct any aberrations that might in the eye. This will allow them to take a clear picture of rods and cones in the human eye. Source (http://www.cvs.rochester.edu/williamslab/research/option02.html)

But that's not all. Williams' group can also distinguish between the individual cones. How? Well if we believe Sam, they would just have to look at the size and the large ones would be the red-sensitive, the middle ones would be the green-sensitive and the small ones would be sensitive to blue. But, there is no discernable difference in the size of the cones. Instead, they shine light at a specific wavelength to temporarily bleach cones of certain type. They do this for three different wavelengths and they can now distinguish between the different cones. The results are some very remarkable composite (false color) pictures (http://www.cvs.rochester.edu/williamslab/research/option04.html) of the retina.

Sandor
2004-Jan-06, 04:02 PM
there is no discernable difference in the size of the cones. Instead, they shine light at a specific wavelength to temporarily bleach cones of certain type
If they send out energy of different (specific) wave lenght (always), it's quite assumable that their sizes are different, because it will need less energy if the size is adapted to this frequency. This doesn't have to mean that the size of the entire cone is different. If we imagine all pigments connected to an internal electrical wire, this wire has an optimal thickness or lenght for any given frequency to need as little energy as possible or to be as accurate as possible. So it's assumable that a 'red cone' is SOMEHOW different from a 'blue cone'.

And why shouldn't that be the size? Why does it have the cone shape anyway?
If an average wave is 500mn and the cone diameter is 1000nm, and the max. difference in wave length is 200nm, then a red cone can have a diameter of 1,5*600=900nm and a blue cone can have a diameter of 2,5*400=1000nm, which are both invisibly (?) close to 1 micron. And we are just counting in half wave lengths, just like normal radio antennas. :)

Kaptain K
2004-Jan-06, 06:49 PM
So it's assumable that a 'red cone' is SOMEHOW different from a 'blue cone'.
Yep! They contain different pigments that are sensitive to different wavelengths. It has nothing to do with the size of the cells. It is all in the chemistry.

Sandor
2004-Jan-06, 08:13 PM
Yep! They contain different pigments that are sensitive to different wavelengths. It has nothing to do with the size of the cells. It is all in the chemistry.
You must have read/learned that. Let's just analyze it. I'll bet my *** that red cones are different from blue cones, and NOT only because of the pigments. Why do you think they are cone-shaped? How are the signals from all the pigments in one cone transmitted to elswhere? Are they 'counted up' in one electric wire or are they forming one bigger wave? How about the ideal thickness of the medium?

Let me explain it using wave theory as well as particle theory.

Waves are most easy to imagine. Red cone pigments will resonate at red light frequency, to which red waves contribute most. All red frequency waves (pigments) are counted up in one cone to one red frequency wave with an amplitude as high as the 'sum' of all the fibrating pigments. So this is truely one red cone resonating with a lower frequency than a blue cone. How could it be different?

In particle theory, well, it's the same as with antennas. I really don't know. :) Why are photons which come in at a certain rate, no, let me put it different, because the photons will push the electrons: why are electrons which move with a certain rate (how many pass at a time) through an antenna best moving through an antenna of a certain length?

Laser Jock
2004-Jan-06, 08:32 PM
... it's quite assumable that their sizes are different ...



You can assume whatever you like, but the facts are clear.



I'll bet my [bad word deleted] that red cones are different from blue cones, and NOT only because of the pigments.

Please watch your language. Read the FAQ's.

To clarify things I decided to ask someone in Williams' group who is friend of mine. Here is his response (I did get his permission to print it here).


There are three cone types and they
are classified according to their spectral sensitivities. The
sensitivities are dictated by the type of photopigment that resides
in the inner segments of the cones themselves. It is true that
S-cones have slightly shorter inner segment lengths than the L-- or
M-cones. However, this is not the property that distinguishes S- from
L- and M-cones. It is due to the photopigment.

There has been a theory, first proposed by Edward Land and later
supported by Gerald Huth, that postulates that there is a single
photopigment for the 3 different cone types and they are all
distinguished simply by their length. This theory is not believed in
the vision community. We have very compelling evidence from excised
and living eyes that clearly shows the presence of 3 different types
of photopigment in eyes with normal color vision. Our lab has a done
a fair amount of work in this field, along with the Neitz's at the
Medical College of Wisconsin. There are also several variants in the
L- and M-cone photopigments, with some eyes possessing slightly
different wavelengths where the peak sensitivity occurs in the
absorption spectrum for these two cone classes.



Yep! They contain different pigments that are sensitive to different wavelengths. It has nothing to do with the size of the cells. It is all in the chemistry.


Exactly.

Tensor
2004-Jan-06, 08:51 PM
Let me explain it using wave theory as well as particle theory.



Posted: Sat Jan 03, 2004 10:56 pm Post subject:

--------------------------------------------------------------------------------

OK, so radio waves (fotons) just go through the empty space of most bodies, sometimes bouncing against electrons, but not having enough energy to really move these electrons up to another level? In other words, they do not get absorbed because the energy cannot be USED in any way, something like that?

But I don't understand, because the fotons will at least move the electrons to which they bounce and loose their energy that way? While light fotons, which have more energy, will likely have a bigger chance to survive all the bouncings and have enough energy left to cross the entire body?

Well, obviously I don't understand this at all.

Back to top


Sandor
Bad Newbie


Joined: 03 Jan 2004
Posts: 25
Location: Netherlands
Posted: Sat Jan 03, 2004 10:59 pm Post subject:

--------------------------------------------------------------------------------

The fact is that I don't understand at all what a WAVE is. Yes, it's some sort of electronic or magnetic FIELD, in which a particle (foton) moves, but this foton moves in a straight line.




Amazing, in three days you went from not understanding EM waves and photons, to explaining wave and particle theory.

Sandor
2004-Jan-06, 09:39 PM
Thank you very much Tensor. I just try my best.

Laser Jock, about these facts, I didn't read about it anywhere in this topic. So you say that the 3 types of cones have exactly the same size? How exact is that measured? And what about the inner part? How do the pigments transmit the information elsewhere? What medium is used? If you don't know all this, you actually don't have a clue about the possible differences between cones.

About the words of your friend, I think I can agree to most of it. No-one here said that all pigments are the same, on the contrary. So the second paragraph is not relevant.

Now, in the first paragraph your friend says that there is a difference in cone length between types of cones, but he also says that 'this is not THE property that distinguishes them', but there is no THE, it's just A property that distinguishes them. What is your friend's opinion about this property? What can be the reason that the inner segments are shorter?

But do I understand it clear that you contradict yourself by first saying that all cones have the same size, but later believe your friend who says that there IS a difference in length?

Tensor
2004-Jan-06, 10:10 PM
Thank you very much Tensor. I just try my best.

You obviously missed the sarcasm.

Sandor
2004-Jan-06, 10:16 PM
No, you did. ;)

Sam5
2004-Jan-06, 11:15 PM
OK Sam,

Hey, let's not argue about this.

You tell those guys at Rochester to read Tyndall’s book, and the papers of Hertz and Marconi. Tell them to go down to Radio Shack and pick up a couple of books about antennas, so they can understand why cones are cone shaped and the size that they are.

What scientists of today need to do is study more 19th Century physics, so they can have a better basic understanding of the natural phenomena of nature, so they can learn what words like “tune” really mean.

I didn’t forget the stuff about the pigments. I’ve read a lot about that stuff, about all sorts of different chemicals and stuff in the back of the eye. I just wanted to make a point about the resonation and the size of the cones, and the cones being like “antennas”. If we just had a coating of “pigments” alone, and no cones, I don’t think we would get enough energy out of the light waves for us to be able to see them. That would be like playing guitar strings tied to a 2 x 4 board but without the regular guitar sound-box attached to the board.

Think of it this way: You go to a classical acoustical guitar concert. So, what two things are between you and the strings of the guitar and your ears that help you hear the sound?

The air and the resonation box of the guitar. Remove either, and you hear no sound. As Tyndall cleverly pointed out in his famous book “Sound”, in the 1896 edition, without the resonation box of the guitar, it’s very difficult to hear the vibration of the strings with the observer just a few feet away from the strings.

In his book he says, “Experimenting thus, we learn that there is one particular length of the column of air which, when the fork is placed above it, produces a maximum augmentation of the sound. This reenforcment of the sound is named resonance.”

He also says about the “sound-boards” of stringed musical instruments, “Hence with such vibrating bodies sound-waves may be generated , and loud tones produced, while the thin strings that set them in vibration, acting alone, are quite inaudible.”

This same general rule applies to electrodynamics and the rods and cones in your eye.

So, my 19th Century featured science term of the week is “resonation”.

At the bottom of the folowing webpage is a drawing of Tyndall’s 1867 “mechanical oscilloscope”.

www.uh.edu/engines/epi1670.htm+%22JOHN+TYNDALL%22+SOUND&hl=en&ie=UTF-8]LINK (http://66.102.7.104/search?q=cache:CD9rDLC5IWgJ:[url) TO SOURCE[/url]

Sam5
2004-Jan-06, 11:21 PM
I'll bet my [bottom dollar] that red cones are different from blue cones, and NOT only because of the pigments. Why do you think they are cone-shaped?

You are exactly right, and you have an excellent science teacher.

swansont
2004-Jan-07, 12:38 AM
You tell those guys at Rochester to read Tyndall’s book, and the papers of Hertz and Marconi. Tell them to go down to Radio Shack and pick up a couple of books about antennas, so they can understand why cones are cone shaped and the size that they are.


I think you have to allow for the fact that we've learned a few things in the last 100 or so years, and that rods& cones (AFAIK) are not metal.

Sam5
2004-Jan-07, 01:01 AM
You tell those guys at Rochester to read Tyndall’s book, and the papers of Hertz and Marconi. Tell them to go down to Radio Shack and pick up a couple of books about antennas, so they can understand why cones are cone shaped and the size that they are.


I think you have to allow for the fact that we've learned a few things in the last 100 or so years, and that rods& cones (AFAIK) are not metal.

Lol, neither were my window frames. Neither are guitar and violin sound-boards. Neither was Tyndall’s glass tubes that amplified the sound of his tuning forks. Neither is the pizza that you bake in a microwave, but you are resonating the molecules of your pizza anyway. Why do you think you can cook your pizza with microwaves but not with AM or FM radio waves? Why do you think that some kinds of plastic dishes get hot in a microwave but others, made out of other kinds of material, do not?

Sam5
2004-Jan-07, 01:15 AM
swansont,

Hey, how much would an atomic clock slow down and speed up its “tick” rate, when more and less heat energy is applied to it, assuming it is not insulated for temperature changes?

Sandor
2004-Jan-07, 09:00 AM
how much would an atomic clock slow down and speed up its “tick” rate, when more and less heat energy is applied to it, assuming it is not insulated for temperature changes?

All force is accelleration (changing velocity), or resonance. :) Including matter. More force will slow down time, or, shrink space. These are 2 different perspectives, like particles (matter) and waves (energy) are 2 perspectives of the same. We only 'understand' space and matter, not time and energy. That's why we don't have a clue what waves are, or what time is. We need space and matter to explain those (and when we think we have it, it's gone already). On the other hand we also need time and energy to explain space and matter, but the 'solid mind-grid' from which we comprehend everything is always space and matter.

This comprehension, or consciousness, is the fifth factor, or in fact it's the first. The best way I can explain, is to say that space is changing matter/energy, while time is changing consciousness of that. Well, maybe it's not the best way, because you can exchange many terms, but time is anyhow closer to consciousness, to what we are, than space. And it's about impossible to 'see' your own consciousness. You wouldn't be it, if you could be outside of it.

Back to slowing down time, which is caused by any accelleration or force on it: gravity is accelleration and so is heat. If we take the formula for time and speed:

http://www.lorentz.leidenuniv.nl/vanbaal/SRT/syllabus/chap4-5/img8.gif

we can easily fill in the formula for cosmic gravity and mass (the speed part):

http://www.lorentz.leidenuniv.nl/vanbaal/SRT/syllabus/chap6-7/img63.gif

The same way we can fill in a formula for 'heat-kinetics', like in Joules: 1 Joule = 1 kg m2 s-2. It should be an easy sum. :)

swansont
2004-Jan-07, 01:26 PM
Lol, neither were my window frames. Neither are guitar and violin sound-boards. Neither was Tyndall’s glass tubes that amplified the sound of his tuning forks. Neither is the pizza that you bake in a microwave, but you are resonating the molecules of your pizza anyway. Why do you think you can cook your pizza with microwaves but not with AM or FM radio waves? Why do you think that some kinds of plastic dishes get hot in a microwave but others, made out of other kinds of material, do not?

Microwave absorption in a pizza has nothing to do with antennas. The fact that you note that not all materials get hot show that it's composition, not size - the pizza gets hot because it has water in it, and water readily absorbs radiation at 2.4 GHz. Your contention that cones' sizes matter for absorption would imply that a very small pizza should be cooked with FM and a large pizza with microwaves, but this is obviously not the case.

Sam5
2004-Jan-07, 02:30 PM
Lol, neither were my window frames. Neither are guitar and violin sound-boards. Neither was Tyndall’s glass tubes that amplified the sound of his tuning forks. Neither is the pizza that you bake in a microwave, but you are resonating the molecules of your pizza anyway. Why do you think you can cook your pizza with microwaves but not with AM or FM radio waves? Why do you think that some kinds of plastic dishes get hot in a microwave but others, made out of other kinds of material, do not?

Microwave absorption in a pizza has nothing to do with antennas. The fact that you note that not all materials get hot show that it's composition, not size - the pizza gets hot because it has water in it, and water readily absorbs radiation at 2.4 GHz. Your contention that cones' sizes matter for absorption would imply that a very small pizza should be cooked with FM and a large pizza with microwaves, but this is obviously not the case.


In the pizza, I’m talking about the molecular structure of the water contained therein. That’s what resonates. That has nothing to do with the size of the pizza. There are two different sizes of things that resonate, the “macro” scale, and the “atomic/molecular” scale. We can achieve resonation on both levels. With microwaves and a pizza, we are achieving resonation on the atomic/molecular scale. With resonation in a guitar sound-box, that is resonation on the macro scale. There is no need to make water-molecule-sized pizzas, since the microwaves resonate the water molecules inside pizzas.

It’s possible that for longer waves, such as AM and FM radio, resonation might occur only in metal such as copper and aluminum. A lot of people work at radio and TV stations that have transmitters right there at the satations, but I’ve never heard of any ill-effects on humans because of AM and FM radio waves. However, the waves will resonate tuned aluminum and copper antennas that are held in the hands of the humans. Microwave ovens are of a specific frequency because they are designed to resonate water molecules.

swansont
2004-Jan-07, 02:54 PM
Lol, neither were my window frames. Neither are guitar and violin sound-boards. Neither was Tyndall’s glass tubes that amplified the sound of his tuning forks. Neither is the pizza that you bake in a microwave, but you are resonating the molecules of your pizza anyway. Why do you think you can cook your pizza with microwaves but not with AM or FM radio waves? Why do you think that some kinds of plastic dishes get hot in a microwave but others, made out of other kinds of material, do not?

Microwave absorption in a pizza has nothing to do with antennas. The fact that you note that not all materials get hot show that it's composition, not size - the pizza gets hot because it has water in it, and water readily absorbs radiation at 2.4 GHz. Your contention that cones' sizes matter for absorption would imply that a very small pizza should be cooked with FM and a large pizza with microwaves, but this is obviously not the case.


In the pizza, I’m talking about the molecular structure of the water contained therein. That’s what resonates. That has nothing to do with the size of the pizza. There are two different sizes of things that resonate, the “macro” scale, and the “atomic/molecular” scale. We can achieve resonation on both levels. With microwaves and a pizza, we are achieving resonation on the atomic/molecular scale. With resonation in a guitar sound-box, that is resonation on the macro scale. There is no need to make water-molecule-sized pizzas, since the microwaves resonate the water molecules inside pizzas.

It’s possible that for longer waves, such as AM and FM radio, resonation might occur only in metal such as copper and aluminum. A lot of people work at radio and TV stations that have transmitters right there at the satations, but I’ve never heard of any ill-effects on humans because of AM and FM radio waves. However, the waves will resonate tuned aluminum and copper antennas that are held in the hands of the humans. Microwave ovens are of a specific frequency because they are designed to resonate water molecules.

And what "resonates" in the cones are the molecules comprising the pigment. Not the cones themselves.

Excitation in a molecules is a quantum effect, while antennas are more classical - the electrons are free to move and oscillate at the driving frequency.

Sandor
2004-Jan-07, 03:48 PM
And what "resonates" in the cones are the molecules comprising the pigment. Not the cones themselves.

If we compare the water molecules in a pizza to the pigment molecules, we can say that both resonate because of the waves, but since the water is everywhere in the pizza, this is just about it: there's no need for further 'action'. In the cones however, the signals have to be forwared elsewhere; the cones act as a sort of amplifier.


Excitation in a molecules is a quantum effect, while antennas are more classical - the electrons are free to move and oscillate at the driving frequency.

Electrons moving by photons is also a quantum effect; it's even one of the first quantum effects explained by Einstein.

Sam5
2004-Jan-07, 04:09 PM
And what "resonates" in the cones are the molecules comprising the pigment. Not the cones themselves.

If we compare the water molecules in a pizza to the pigment molecules, we can say that both resonate because of the waves, but since the water is everywhere in the pizza, this is just about it: there's no need for further 'action'. In the cones however, the signals have to be forwared elsewhere; the cones act as a sort of amplifier.

Exactly.

Laser Jock
2004-Jan-07, 04:30 PM
You tell those guys at Rochester to read Tyndall’s book, and the papers of Hertz and Marconi. Tell them to go down to Radio Shack and pick up a couple of books about antennas, so they can understand why cones are cone shaped and the size that they are.

You're really funny. You think you can tell the world's experts on eyes that you understand how eyes work better than they do. :roll:


What scientists of today need to do is study more 19th Century physics, so they can have a better basic understanding of the natural phenomena of nature, so they can learn what words like “tune” really mean.

We did study 19th century physics -- in high school. Now we study 20th and 21st century physics.


I didn’t forget the stuff about the pigments. I’ve read a lot about that stuff, about all sorts of different chemicals and stuff in the back of the eye. I just wanted to make a point about the resonation and the size of the cones, and the cones being like “antennas”. If we just had a coating of “pigments” alone, and no cones, I don’t think we would get enough energy out of the light waves for us to be able to see them. That would be like playing guitar strings tied to a 2 x 4 board but without the regular guitar sound-box attached to the board.

(Emphasis mine)

I put your statement in bold since I think it entirely sums up your position. You don't think that eyes can work without some sort of additional resonance with the structure of the cone. You can't support that statement without hand waving or using guitar sound-box analogies. The fact is that the vision community acknowledges the photopigment is sensitive enough to detect very weak light without resonant amplification.


Think of it this way: You go to a classical acoustical guitar concert. So, what two things are between you and the strings of the guitar and your ears that help you hear the sound?

The air and the resonation box of the guitar. Remove either, and you hear no sound. As Tyndall cleverly pointed out in his famous book “Sound”, in the 1896 edition, without the resonation box of the guitar, it’s very difficult to hear the vibration of the strings with the observer just a few feet away from the strings.

In his book he says, “Experimenting thus, we learn that there is one particular length of the column of air which, when the fork is placed above it, produces a maximum augmentation of the sound. This reenforcment of the sound is named resonance.”

He also says about the “sound-boards” of stringed musical instruments, “Hence with such vibrating bodies sound-waves may be generated , and loud tones produced, while the thin strings that set them in vibration, acting alone, are quite inaudible.”
The resonance of a sound board is well understood, but ...



This same general rule applies to electrodynamics and the rods and cones in your eye.


No, it does not. They are both waves, but the analogy stops there.


In the pizza, I’m talking about the molecular structure of the water contained therein. That’s what resonates. That has nothing to do with the size of the pizza. There are two different sizes of things that resonate, the “macro” scale, and the “atomic/molecular” scale. We can achieve resonation on both levels. With microwaves and a pizza, we are achieving resonation on the atomic/molecular scale. With resonation in a guitar sound-box, that is resonation on the macro scale. There is no need to make water-molecule-sized pizzas, since the microwaves resonate the water molecules inside pizzas.


This is a bit of an over-simplification.

First, consider the simplest case of a bunch of neutral atoms in a dilute gas. These atoms don't interact with each other much; their primary interaction is with electromagnetic waves. If the incident light has exactly the right frequency, a photon will be absorbed from the EM field and an electron in the atom will go (temporarily) to a higher energy state. If you look then at the absorption of the gas as a function frequency, you will see many different narrow peaks. This is known as an electronic transition. The typical frequency band for this type of interaction is near-infrared to optical (and some UV).

If these same atoms are now imbedded in a crystal (eg. sapphire) or amorphous solid (eg. glass), the absorption peaks will be spread out and possibly in different locations due to the interaction with the host. However, the transitions are still purely electronic.

Things become much more complicated when atoms bond to form molecules. Now in addition to the electronic transitions, you can have rotations in three dimensions, compressions, etc. These type of transitions, like electronic transitions, are discrete, and one must use quantum mechanics to describe them. The frequencies where these resonances occur are typically much smaller than electronic transitions, usually in the microwave to mid-infrared region. Just like with electronic transitions, the presence of a host material can also broaden and alter these transitions. This is the type of transition that occurs when you heat pizza in a microwave oven.

Now back to the vision question. In addition, to the above transitions, light can induce a chemical change in a molecule (whereas a water molecule in a microwave is still a water molecule, just in a higher energy state). This (http://palaeo-electronica.org/2000_1/retinal/vision.htm) site provides a detailed description of the process.


When the chromophore absorbs a photon, an electronic rearrangement may occur, which results in a rotation of the constituent atoms around the eleventh carbon bond. This isomerization of the chromophore causes a change in the shape of the opsin, and in its new configuration, the opsin behaves as an enzyme.

The rate at which the reaction catalyzed by the opsin occurs sets the rate at which the photoreceptor releases neurotransmitters to other retinal neurons. Figure 6 shows the absorption probability of a typical vertebrate photopigment as a function of wavelength and, essentially, the absorption spectrum of a photoreceptor as well, since in most cases, all of the opsin molecules within a given photoreceptor are chemically identical.

(Emphasis mine)

Notice they say nothing about the size or shape of the photoreceptor (or cone)! The only thing that will vary the absorption of a cone would be to change the type of photopigment within that cone.

The chemical process is explicitly shown in Figure 5 (http://palaeo-electronica.org/2000_1/retinal/fig_5.htm). Here is Fig. 6 (http://palaeo-electronica.org/2000_1/retinal/fig_6.htm). Also note the absorption profiles of the different types of cones in Fig. 7A (http://palaeo-electronica.org/2000_1/retinal/fig_7.htm).

R.A.F.
2004-Jan-07, 04:45 PM
What scientists of today need to do is study more 19th Century physics...

HUH??!!??

Sam5
2004-Jan-07, 06:10 PM
You're really funny. You think you can tell the world's experts on eyes that you understand how eyes work better than they do. :roll:


I got this information from some of the world’s experts on eyes. It’s in some of their books and papers. I’ve already posted some links to some of them.

You shouldn’t get so upset about this, just because you’ve never heard of this resonation concept.

Sam5
2004-Jan-07, 06:15 PM
What scientists of today need to do is study more 19th Century physics...

HUH??!!??

Seems that with this flood of new and more complex information today, a lot of basic, fundamental, and correct 19th Century concepts have been forgotten, such as the concept of resonation, the physical principles of the basic Doppler Effects, Newton’s use of gravitational fields as a light-speed regulator, his prediction that a light beam would curve when it passed the sun, and a lot of other basic fundamental concepts of physics. That's why some of these guys don't know much about the basics of resonation.

Sam5
2004-Jan-07, 06:28 PM
R.A.F.,

What I like to do – it’s sort of like a hobby – is to read about modern science discoveries, and then I go to my 19th Century books to see if the very same thing was discovered long ago, but was literally forgotten by science and science media.

The story about Louis Leakey and Olduvai Gorge is a classic example. His “new discoveries” were highly promoted by National Geographic in the 1960s as representing a “major breakthrough”. But I found in an old book that the initial discoveries there were actually made there around 1912, and I learned in another book that Leakey learned about them from the writing of the paleontologist who first made the discoveries. So Leakey “discovered” nothing at all. He merely went to the site in Africa where that early paleontologist had made the original discoveries.

And like the “pangaea” thing, supposedly discovered by Wegner early in the 20th Century. Well, that story, including the Greek “pan gaea” term, was already in thousand-year-old books, and the original concept is in a 3,500 year old book. I’ve got 19th Century science books that discuss the same subject.

AstroSmurf
2004-Jan-07, 06:54 PM
Oh boy... this thread is just filled with bad science from beginning to end.

Sandor, Sam5's version of physics, while perhaps intriguing, is not supported by reality. The interaction of visible light with photochemicals is very well understood these days, and is used in (among other areas) laser technology. Tuning is perhaps theoretically possible at frequencies near the visible light, but practically impossible. Cells are too large to interact with the incoming light - normal antennae are one half-wavelength in size, meaning that the cones would have to be around 300 nm in size, about 1/10 of their actual size. The interactions that do take place are of particle nature, not wave nature - the photons are absorbed, not merely used to cause resonation or other wave phenomena.

Nor is 'a single photon' a revolutionary concept - detectors such as photomultiplicators has been in use for quite a long time now, and yes, they can detect a single photon. And it turns out that even at the lowest possible intensity, even when you only catch the photons isolated from each other, all incoming photons in a beam with with a certain frequency have the same energy. A single photon is best thought of as a 'wave packet' - not a single wave period, but a few in sequence. The traditional illustration shows the amplitude rising and falling in a Gaussian curve, with the frequency remaining constant for the whole packet. (The Gaussian curve is IIRC an exp(-x^2) shape)

swansont
2004-Jan-07, 06:56 PM
Excitation in a molecules is a quantum effect, while antennas are more classical - the electrons are free to move and oscillate at the driving frequency.

Electrons moving by photons is also a quantum effect; it's even one of the first quantum effects explained by Einstein.

The photoelectric effect is ionization of an atom or molecule. It's not the same thing as what's going on in an antenna.

Sam5
2004-Jan-07, 07:45 PM
Sandor, Sam5's version of physics, while perhaps intriguing, is not supported by reality. The interaction of visible light with photochemicals is very well understood these days, and is used in (among other areas) laser technology. Tuning is perhaps theoretically possible at frequencies near the visible light, but practically impossible. Cells are too large to interact with the incoming light - normal antennae are one half-wavelength in size, meaning that the cones would have to be around 300 nm in size, about 1/10 of their actual size.

Some of the biology books say that cone cells are about one micron, and the wavelength of red light is about .680 micron. That is too close to be an “accident”. And not all of the length or diameter of the cone cells have to participate in the resonation. Consider the cilia in your ear. Why do you think there are a lot of them and of different lengths? This is exactly the same principle. Resonation.

Normal radio antennas, such as CB radio antenna, are half-wave and quarter-wave and shorter, because the CB radio wavelength is so long. Because you can’t drive around with a 39.34 foot full-wave antenna on your car. One short-wave band is 49 meters, and you can’t put up a 49 meter antenna on your house.

But you can have a full-wave TV antenna on your house, since the wavelengths are shorter. You know, like in VHF and UHF, very high frequency and ultra high frequency. With a wavelength for UHF being about 32.8 cm wavelength, like 12.8 inches. Go to Radio Shack and measure the length of UHF TV antenna elements.

swansont
2004-Jan-07, 09:15 PM
Some of the biology books say that cone cells are about one micron, and the wavelength of red light is about .680 micron. That is too close to be an “accident”. And not all of the length or diameter of the cone cells have to participate in the resonation. Consider the cilia in your ear. Why do you think there are a lot of them and of different lengths? This is exactly the same principle. Resonation.

Normal radio antennas, such as CB radio antenna, are half-wave and quarter-wave and shorter, because the CB radio wavelength is so long. Because you can’t drive around with a 39.34 foot full-wave antenna on your car. One short-wave band is 49 meters, and you can’t put up a 49 meter antenna on your house.

But you can have a full-wave TV antenna on your house, since the wavelengths are shorter. You know, like in VHF and UHF, very high frequency and ultra high frequency. With a wavelength for UHF being about 32.8 cm wavelength, like 12.8 inches. Go to Radio Shack and measure the length of UHF TV antenna elements.

Do you understand how an antenna works, and what is going on? You've already agreed that it's possible for an object can absorb radiation independent of the object's size (within some limits). Why are you insisting that a rod is an antenna?

Sam5
2004-Jan-07, 09:35 PM
swansont,

Would you mind answering the question about the atomic clock rate and the temperature? If you don’t know, just say you don’t know and I’ll stop asking the question.

Laser Jock
2004-Jan-07, 09:43 PM
swansont,

Would you mind answering the question about the atomic clock rate and the temperature? If you don’t know, just say you don’t know and I’ll stop asking the question.

As I said here (http://www.badastronomy.com/phpBB/viewtopic.php?t=9731&start=1175):


Boy, we haven't seen the old "changing-the-topic-since-I'm-losing-the-debate" ploy from Sam5 before have we? :roll:

Sam5
2004-Jan-07, 09:48 PM
Laser Jock,

Why don’t you answer my simple question: Why does a bottle rocket reach a higher altitude when we use both compressed air and water, compared to when we just use compressed air alone?

Tensor
2004-Jan-08, 01:20 AM
[quote]Boy, we haven't seen the old "changing-the-topic-since-I'm-losing-the-debate" ploy from Sam5 before have we? :roll:

Not for a page or two anyway. :roll:

AstroSmurf
2004-Jan-08, 08:50 AM
Some of the biology books say that cone cells are about one micron, and the wavelength of red light is about .680 micron. That is too close to be an “accident”. And not all of the length or diameter of the cone cells have to participate in the resonation. Consider the cilia in your ear. Why do you think there are a lot of them and of different lengths? This is exactly the same principle. Resonation.
You do realise that frequency is much more important than wavelength in resonation, don't you? The propagation speed of the wave phenomenon divided by the frequency yields the actual wavelength. Now, I'm not sure what kind of resonation phenomenon you think is going on inside the cones, but just to take a comparison, I seem to recall that the signal transmission speed in nerves is around 300 km/h! At any rate, the speeds we'd be considering would be vastly lower than c, requiring much smaller sizes than you've ben considering. The size calculations will be left as an exercise to the reader; visible light has frequencies in the range of 400-800 THz.


Normal radio antennas, such as CB radio antenna, are half-wave and quarter-wave and shorter, because the CB radio wavelength is so long. Because you can’t drive around with a 39.34 foot full-wave antenna on your car. One short-wave band is 49 meters, and you can’t put up a 49 meter antenna on your house.
Tell that to my father 8) Anyway, dipole antennae are at the most half-wavelength in size, so it isn't as bad as you think.

The exact length of the antenna elements is determined by desired frequency, the transmission speed in the metal used (we aren't talking the speed of EM radiation in vacuum any more), as well as how the elements are connected - the actual design is determined of a lot of factors, so there's no single "perfect" design. The further you go from the 'ideal' half and quarter-wavelengths, the worse your reception will be, though, which is why you try to match the antenna size to the wavelength as closely as possible. Also, note that in all the cases here, the size of the antenna is smaller than the wavelength. This is not only due to convenience - you wish to make the amplitude response max out for the desired frequency, so a larger size would cause undesired interference.

(I once had the call sign SM6UPS in case you're wondering, though my rig has been gathering dust for about 10 years now)

swansont
2004-Jan-08, 11:22 AM
swansont,

Would you mind answering the question about the atomic clock rate and the temperature? If you don’t know, just say you don’t know and I’ll stop asking the question.

I thought I did, a few days ago (check page 2) - the answer is that it depends on the temperature coefficients of lots of things - your electronics, your cabling, your quartz crystal, etc. Some of the effects will be in opposite directions. There's no simple, universal answer.

Wally
2004-Jan-08, 12:51 PM
Glom, that makes sense! It's the 'quantum amount' that will determine if there is no interaction at all, in an energy-kind of way.

Sam, I still don't understand, because an electric field is only measurable (or maybe even existing) when MATTER is involved, so a particle must be IN this field. or doesn't it (in light theory)? I mean, when they say that light can EITHER be regarded as wave OR as particle, how can it be ONLY a wave? Or is that not what they mean then?

I don't think it's an "either/or" thing necessarily. I always thought of it more and "both". i.e. light travels as a wave which shows properties of a particle.

AstroSmurf
2004-Jan-08, 01:02 PM
It's actually "both" as Wally says - what it will look like depends on how you try to study it. If you try to check wave characteristics, such as frequency, phase, polarization and so forth, it will look like a wave. If you try to check particle characteristics, such as energy or momentum, it will look like a particle. The observations depend on the experiment used to study them.

Sandor
2004-Jan-08, 03:49 PM
“Hence with such vibrating bodies sound-waves may be generated , and loud tones produced, while the thin strings that set them in vibration, acting alone, are quite inaudible.

This same general rule applies to electrodynamics and the rods and cones in your eye.


No, it does not. They are both waves, but the analogy stops there.
It really does Laser Jock. The further I get into this, the more analogy I find between sound and light waves. Try me and name any aspect of either one and I will try to find the similar aspect in the other.


The interaction of visible light with photochemicals is very well understood these days

Well, I 'coincidentally' read these today:

"Much of the neural processing in the retina occurs by graded potentials rather than action potentials, probably because these neurons are so close together that long-range conduction via action potentials is not necessary"

"The neural mechanisms of color perception have still not been completely worked out"

"How the 3 human cone types and the higher-order neurons they contact produce the sensations of color is still unclear"


The interactions that do take place are of particle nature, not wave nature

All forces are as well of a wave nature as of a particle nature. (Which indeed doesn't mean at all that a cone has to resonate)


A single photon is best thought of as a 'wave packet' - not a single wave period, but a few in sequence.

From my perspective you are saying that a photon is several photons. So now we understand each other.


The traditional illustration shows the amplitude rising and falling in a Gaussian curve, with the frequency remaining constant for the whole packet.

You are talking about A photon when stating this, but a photon, even in your understanding as being several wavelengths in ONE wave, can only have either frequency or amplitude.


The photoelectric effect is ionization of an atom or molecule. It's not the same thing as what's going on in an antenna.

I found some clear writings about the working of antennae: (also next pages) (http://www.intuitor.com/resonance/radioTVres.html)

When this one is easy/comprehended, this one (http://www.arrl.org/tis/info/whyantradiates.html) goes deeper.

OK, What I read there is that in an antenna electrons get freed from the atom, what seems to me as ionization. It's true that radio waves don't have enough energy to this on their own. The transmitting antenna is however emitting the photons ('Real Power Flow') by what is called the 'Reactive Power Flow', which is lost in heat by resistance of the antenna material. now they must have found a way to also receive this 'Real Power Flow' by constantly having a charge in the antenna (Reactive Power Voltage or something like that).


Now, I'm not sure what kind of resonation phenomenon you think is going on inside the cones, but just to take a comparison, I seem to recall that the signal transmission speed in nerves is around 300 km/h! At any rate, the speeds we'd be considering would be vastly lower than c, requiring much smaller sizes than you've ben considering. The size calculations will be left as an exercise to the reader; visible light has frequencies in the range of 400-800 THz.

In the same post you mention this: "the transmission speed in the metal used". So you know that this speed in metal is much slower, but still the EM waves 'in it' have the speed of light.



swansont, would you mind answering the question about the atomic clock rate and the temperature? If you don’t know, just say you don’t know and I’ll stop asking the question.

I thought I did, a few days ago (check page 2) - the answer is that it depends on the temperature coefficients of lots of things - your electronics, your cabling, your quartz crystal, etc. Some of the effects will be in opposite directions. There's no simple, universal answer.

I think there is and I gave that answer already. It's in the post with the very nice pictures of formulas.

OK, now you all enjoy laughing about Sam and his resonating cones, but why isn't any of you knowing how it works? First some nice links:

about cones and stuff (http://psych.athabascau.ca/html/Psych402/Biotutorials/23/intro.shtml)
and another one (http://www.psychology.psych.ndsu.nodak.edu/mccourt/website/htdocs/HomePage/Psy486/Organization%20of%20Sensory%20Systems/Organization%20of%20Sensory%20Systems.html)

I learned this: pigments in a cone resonate best at 'their' frequenty. ONE cone transmits all of this information to ONE bipolar cell by graded potentials (instead of action potentials, meaning it 'on' or 'off' and nothing in between). So we have many pigments in let's say one red cone, fibrating AROUND red light frequency (or not fibrating). These energies are all transmitted together through one dendrite, so it has ONE constantly changing frequency (graded potential) to the bipolar cell. Who of you guys can explain this to me NOT using wave theory?

Btw, I'm not saying the entire cone has to resonate, I really don't care what resonates within the cone, but in wave theory it has to be something. What is the particle theory about this?

swansont
2004-Jan-08, 04:26 PM
swansont, would you mind answering the question about the atomic clock rate and the temperature? If you don’t know, just say you don’t know and I’ll stop asking the question.

I thought I did, a few days ago (check page 2) - the answer is that it depends on the temperature coefficients of lots of things - your electronics, your cabling, your quartz crystal, etc. Some of the effects will be in opposite directions. There's no simple, universal answer.

I think there is and I gave that answer already. It's in the post with the very nice pictures of formulas.



Your answer is full of it. Heat is acceleration? :roll:

Sam5
2004-Jan-08, 05:00 PM
Your answer is full of it. Heat is acceleration? :roll:

Heat energy adds accerlative forces to moving atoms and molecules by increasing their RMS velocities, thus adding more inertial/accelerative forces to them during their changes in direction. Thus, a collection of hotter atoms should have lower internal harmonic oscillation rates, thereby causing atomic clocks that are subjected to more heat energy to "tick" more slowly.

Sam5
2004-Jan-08, 05:09 PM
I learned this: pigments in a cone resonate best at 'their' frequenty. ONE cone transmits all of this information to ONE bipolar cell by graded potentials (instead of action potentials, meaning it 'on' or 'off' and nothing in between). So we have many pigments in let's say one red cone, fibrating AROUND red light frequency (or not fibrating). These energies are all transmitted together through one dendrite, so it has ONE constantly changing frequency (graded potential) to the bipolar cell. Who of you guys can explain this to me NOT using wave theory?

Of course, if the atoms inside the chemical coating that surrounds the cones were the only things that “resonated” or “absorbed the energy” of the photons, then our eyes wouldn’t need all the separate little cones that are the size of the light wavelengths, since the coating would do all the absorbing of the light waves all by itself. But all those little cones need to be separate and of the right small light-wavelength size so that they too can “resonate” and help amplify the energy of the light waves and turn it into electrical impulses that go to the brain.

Hey, you might become the next great Hendrik Antoon Lorentz of this modern era. Don’t let these Americans intimidate you.

AstroSmurf
2004-Jan-08, 05:12 PM
The interaction of visible light with photochemicals is very well understood these days
Well, I 'coincidentally' read these today:

"Much of the neural processing in the retina occurs by graded potentials rather than action potentials, probably because these neurons are so close together that long-range conduction via action potentials is not necessary"

"The neural mechanisms of color perception have still not been completely worked out"

"How the 3 human cone types and the higher-order neurons they contact produce the sensations of color is still unclear"
(I shifted the boldfacing to emphasize my point) All this relates to the transmission and interpretation of the information about incoming light, not the actual reception. The reception part, which is what we started talking about, ends after the light has been absorbed by the photopigment. All the rest is neurology, not physics.



The interactions that do take place are of particle nature, not wave nature
All forces are as well of a wave nature as of a particle nature. (Which indeed doesn't mean at all that a cone has to resonate).
Well, yes, but a single phenomenon usually restricts itself to one aspect. When talking about the absorption of light, it seems much more straightforward to think of it in terms of energy levels rather than resonation; the resulting "vibrational frequency" (if this is even possible to determine) might be very different from that of the incoming light.



The traditional illustration shows the amplitude rising and falling in a Gaussian curve, with the frequency remaining constant for the whole packet.
You are talking about A photon when stating this, but a photon, even in your understanding as being several wavelengths in ONE wave, can only have either frequency or amplitude.
Either? Without amplitude, the frequency of a wave isn't all that important, is it :P
I was thinking of an illustration such as in this link (http://hyperphysics.phy-astr.gsu.edu/hbase/optmod/qualig.html).

I'm not certain how (or even IF) the traditional illustration is based in theory. I'm not sure that measuring the amplitude of a single photon is even possible; it might be that it's not a concept that has any physical meaning. I suspect that the illustration is just that - a visualisation of something that is nigh impossible to study.



The photoelectric effect is ionization of an atom or molecule. It's not the same thing as what's going on in an antenna.
OK, What I read there is that in an antenna electrons get freed from the atom, what seems to me as ionization.
But, alas, it isn't. The electrons in a metal have not left the atoms behind (ionization), they're just taking the scenic route through the metal. The physical term for them are "free electrons" - they can move freely, as long as they stay inside the metal body. IIRC, there is a slight potential barrier, which is the cause of electrical resistance.


In the same post you mention this: "the transmission speed in the metal used". So you know that this speed in metal is much slower, but still the EM waves 'in it' have the speed of light.
Not a lot slower, just a little. Speaking of EM fields inside a metal, this is a slightly tricky concept. Since the electrons are free to move, they try to counteract any fields that otherwise would penetrate the metal, so the EM fields are actually close to zero! This causes a lot of interesting (and weird) phenomena - the Hall effect (http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/hall.html) comes to mind. Also, this is what gives us anything to detect when we hook up a radio to an antenna - the electrons oscillate counter to the incoming EM waves, and this oscillation can be amplified and detected. When I speak of the 'transmission' of radio signals in metal, this is actually much more analogous to sound transmission than light in vacuum, though the transmitting particles are the free electrons, not air molecules.


I learned this: pigments in a cone resonate best at 'their' frequenty. ONE cone transmits all of this information to ONE bipolar cell by graded potentials (instead of action potentials, meaning it 'on' or 'off' and nothing in between). So we have many pigments in let's say one red cone, fibrating AROUND red light frequency (or not fibrating). These energies are all transmitted together through one dendrite, so it has ONE constantly changing frequency (graded potential) to the bipolar cell. Who of you guys can explain this to me NOT using wave theory?

Btw, I'm not saying the entire cone has to resonate, I really don't care what resonates within the cone, but in wave theory it has to be something. What is the particle theory about this?
The particle theory is that the photon is absorbed by an electron, exciting it to a higher energy state. I think you're trying the wrong angle here - the energy level of the photochemical (which the sum of the total vibrational, rotational and electronic energies of the molecule) will increase exactly the amount that the incoming photon provides. So, neither the frequencies prior nor after the absorption will match that of the photon, but the difference in energy matches that of the photon.

Sam5
2004-Jan-08, 05:34 PM
Well, yes, but a single phenomenon usually restricts itself to one aspect. When talking about the absorption of light, it seems much more straightforward to think of it in terms of energy levels rather than resonation; the resulting "vibrational frequency" (if this is even possible to determine) might be very different from that of the incoming light.



The resonation term is related to the “energy level” term in some ways, but it’s not exactly the same. For example, the “energy level” that resonated my old large Victorian window frames was a very high energy level, according to my window frames that were tuned to the frequency of that sound wavelength, even though it was a very long wavelength of a very low sound-wave frequency. In the case of the cones, the “energy level” of the light should be considered not to be how high the frequency is but as whatever frequency and wavelength at which the cones are tuned to resonate. For example, as far as the cones are concerned, ultra violet light is of a very low energy level, as are radio waves, since the cones can not resonate to those frequencies. That’s why we don’t “see” those frequencies.

Regarding the radio antenna, it is my understanding that it is the “bump, bump, bump, bump” of the incoming radio waves that causes the current flow inside the antenna. The incoming waves don’t actually enter the antenna, but the “bump, bump, bump, bump” of the waves, if the antenna is of the same length as the wavelength of the waves (or a certain mathematical fraction of it, such as half-wave, quarter wave, etc), causes the metal of the antenna to resonate, and the resonation causes a “current flow” inside the antenna, apparently it induces some kind of electron motion inside the metal, and a “current flow” results inside the antenna, and it is the current flow that is amplified inside the radio. It is not the radio waves that is amplified, but the current flow inside the metal antenna, generated by or caused by the correct frequency of the “bumps” of the radio waves, that is amplified.

Sam5
2004-Jan-08, 05:43 PM
I was thinking of an illustration such as in this link (http://hyperphysics.phy-astr.gsu.edu/hbase/optmod/qualig.html).

Regarding your illustration of the photon of the several waves that are wiggling and traveling together, that is my understanding of what a photon might “look like” from a side view, except that remember it should have a magnetic component and an electrical component, so there should be two groups of waves in the illustration that are perpendicular to each other, traveling side by side, enmeshed within each other.

With a compound and complex “wave packet” like this, then the waves alone, with no mass considered, could act much like a single particle, especially considering their very high speed.

In fact, due to the electrodynamical nature of sound, sound waves can act like “particles” too. For example, place a loud hi-fi speaker up to a bowl of water and you will see the water surface react as if it is being hit by particles.

Laser Jock
2004-Jan-08, 05:52 PM
In fact, due to the electrodynamical nature of sound, ...

:-s

So sound is now something besides pressure waves in air. It also contains electromagntic fields. #-o

You are very confused Sam.

SeanF
2004-Jan-08, 06:05 PM
In fact, due to the electrodynamical nature of sound, sound waves can act like “particles” too. For example, place a loud hi-fi speaker up to a bowl of water and you will see the water surface react as if it is being hit by particles.

Yeah, almost looks like the water's surface is being hit by air molecules, doesn't it?

(Not to mention water molecules, from underneath)

swansont
2004-Jan-08, 06:12 PM
Your answer is full of it. Heat is acceleration? :roll:

Heat energy adds accerlative forces to moving atoms and molecules by increasing their RMS velocities, thus adding more inertial/accelerative forces to them during their changes in direction. Thus, a collection of hotter atoms should have lower internal harmonic oscillation rates, thereby causing atomic clocks that are subjected to more heat energy to "tick" more slowly.

Nope, sorry, it doesn't work that way. An atomic fountain uses atoms that have been cooled to a microKelvin or so in temperature, but one can't say that that clock is inherently running faster than a beam tube clock, where the atoms are above room temperature. There are too many variables involved.

Crystal oscillators have been made by combining materials that have both positive and negative temperature coefficients, and run near the crossover point, so you get something that is temperature independent (to first order), so one crystal, by itself would speed up with an increase in T while the other would slow down. Saying that any oscillator will slow down if heated is just plain wrong.

Sam5
2004-Jan-08, 06:24 PM
Nope, sorry, it doesn't work that way. An atomic fountain uses atoms that have been cooled to a microKelvin or so in temperature, but one can't say that that clock is inherently running faster than a beam tube clock, where the atoms are above room temperature. There are too many variables involved.

A constant temperature is what is important in an atomic clock and in many kinds of electronic clocks. It doesn’t matter what the temperature of the atoms is, as long as it is constant.

In a crystal controlled clock, it doesn’t matter what the frequency of the crystal is, as long as it is constant.

It is the fluctuations that cause the problems in some kinds of clocks.

I worked with this temperature stuff for years in electronics. We even had a machine called a “time-base corrector” that would try to correct in timing fluxuations caused by temperature changes, accelerative forces, frequency drift, etc. This phenomenon is well known throughout the electronics industry.

www.mfelectronics.com/products/tcxo/+crystal+clock+temperature.&hl=en&ie=UTF-8]LINK (http://216.239.57.104/search?q=cache:ohh7K-3YTCcJ:[url) TO SOURCE[/url]

Sam5
2004-Jan-08, 06:31 PM
Crystal oscillators have been made by combining materials that have both positive and negative temperature coefficients, and run near the crossover point, so you get something that is temperature independent (to first order), so one crystal, by itself would speed up with an increase in T while the other would slow down.

LOL, this is based on the 18th Century Harrison clock principle, in which he used two counter-swinging pendulums to cancel out the accelerative effects on rocking ships.

In the 1960s Sony invented a portable audio recorder that used the same principle and that contained counter-rotating flywheels. NASA liked the design so much they let astronauts use them on space missions.

swansont
2004-Jan-08, 07:25 PM
Nope, sorry, it doesn't work that way. An atomic fountain uses atoms that have been cooled to a microKelvin or so in temperature, but one can't say that that clock is inherently running faster than a beam tube clock, where the atoms are above room temperature. There are too many variables involved.

A constant temperature is what is important in an atomic clock and in many kinds of electronic clocks. It doesn’t matter what the temperature of the atoms is, as long as it is constant.

In a crystal controlled clock, it doesn’t matter what the frequency of the crystal is, as long as it is constant.

It is the fluctuations that cause the problems in some kinds of clocks.

I worked with this temperature stuff for years in electronics. We even had a machine called a “time-base corrector” that would try to correct in timing fluxuations caused by temperature changes, accelerative forces, frequency drift, etc. This phenomenon is well known throughout the electronics industry.


So why did you contend that any oscillator heated up will run slow?

The value does matter, though. At some point you need to know your frequency. Removing fluctuations gives you better stability, but the wrong frequency will give the wrong time unless you correct for it.

Sam5
2004-Jan-08, 07:51 PM
So why did you contend that any oscillator heated up will run slow?

The value does matter, though. At some point you need to know your frequency. Removing fluctuations gives you better stability, but the wrong frequency will give the wrong time unless you correct for it.

The wrong frequency in clock, if it’s steady, as been compensated for, during the past 4,000 years. This was done with pendulum clocks 500 years ago by adjusting the bob position on the bob shaft. This is nothing new. Even the modern atomic clocks are checked against the ancient Hebrew sundials. That’s why you add leapseconds to your atomic clocks, to keep pace with the ancient Hebrew sundials. And you don’t want to stop adding those leapseconds or we’ll wind up with midnight at noon, and I'm sure you know what Job said about that. (lol)

You said: “Crystal oscillators have been made by combining materials that have both positive and negative temperature coefficients,”

You are talking about compound oscillators. I’m talking about single fundamental oscillator “materials”.

Of course we can compensate for temperature changes by combining “materials” of different kinds and by making more complex devices that compensate for temperature fluctuations.

I’m talking about the effects on a basic fundamental classical physics level.

And so, fluctuating temperatures in a fundamental atomic clock can slow down and speed up the “tick rate” of the clock. Just as extraneous EM waves can do the same thing. Just as Lorentz Forces can do the same thing.

But the general myth in American schools today is that only “Einstein’s relativity, gravity, and acceleration” can change clock rates. And that is baloney. Einstein “relativity” is Newtonian and Lorentzian relativity. All of Einstein relativity that works, is classical relativity. All of Einstein relativity that doesn’t work, such as “clock slowdowns due only to relative motion”, is bunk, based on Einstein’s own errors, and taught in American schools today as fact, as "amazing science", as "spooky science", but which Einstein later corrected with the GR theory.

You can find the subject of clock slowdowns due to acceleration and gravitational fields in classical physics books of the 19th Century and earlier.

SeanF
2004-Jan-08, 08:14 PM
But the general myth in American schools today is that only “Einstein’s relativity, gravity, and acceleration” can change clock rates.

[Removed because Laser Jock is correct below]

Laser Jock
2004-Jan-08, 08:17 PM
How fun ... another change in topic. Now we all get to watch as this turns into another relativity thread.

#-o

Sandor
2004-Jan-08, 11:01 PM
Heat is acceleration? :roll:

Like Sam says,adding heat is acceleration, it's the delta factor in the formula.


Of course, if the atoms inside the chemical coating that surrounds the cones were the only things that “resonated” or “absorbed the energy” of the photons, then our eyes wouldn’t need all the separate little cones that are the size of the light wavelengths, since the coating would do all the absorbing of the light waves all by itself. But all those little cones need to be separate and of the right small light-wavelength size so that they too can “resonate” and help amplify the energy of the light waves and turn it into electrical impulses that go to the brain.

Almost exactly what I mean: the guys here seem to think that many red pigments can just send their packets through one shared dendrite to the bipolar cell, which last one can understand what those packets mean. First of all, if that was the case, all pigments would indeed be spreaded around the retina, not needing a cone; it would just be a waste of energy. Secondly, of course the bipolar cell doesn't understand anything of the 'untuned' packets of energy that come in. They have to be tuned by the cone. Explanation: if pigment 1 sends out 1000 packets/second through the line and pigment 2 sends 900 and pigment 3 sends 1100, how can the AVERAGE be known by the bipolar cell?
The cone is doing some tuning here, but it doesn't have to be amplifying (also). It's indeed too long to be the most effective resonator. But I guess there is one central nerve in it to which all smaller pigment nerves connect and the cone has to manage that. But indeed the best way to do that, suggested here, is still the cone fibrating itself.

Something about the difference between fibrating and resonating, as I understand: antennae can operate at other lenghts than half or quarter wave etc, we all know that when we have experienced a broken antenna, but it willl not RESONATE anymore, meaning ot doesn't AMPLIFY anymore.


The reception part, which is what we started talking about, ends after the light has been absorbed by the photopigment. All the rest is neurology, not physics.

This discussion is not at all about the reception part, because we all agree that this is done by the pigments. The problems really starts from then.


When talking about the absorption of light, it seems much more straightforward to think of it in terms of energy levels rather than resonation; the resulting "vibrational frequency" (if this is even possible to determine) might be very different from that of the incoming light.

It is not very different, but the average of many red pigments reporting to one bipolar cell. Frequency is just the same as the number of photons per time frame, so it's the same to determine.


Without amplitude, the frequency of a wave isn't all that important, is it :P

Maybe this seems to be so, but in one single wave there is no difference between amplitude and frequency. I tried to explain this many times here. I did the explanation with the particle theory, saying that that one can only send more particles at a time, which is the same as sending them 'harder' (speed is always the same). But I can explain it also with the wave theory: if you hold a rope and move it up and down, to make a wave, changing the amplitude will change the frequency, it's the same. Only with several ropes parallel, you can can change both. That's why radio needs at least 2 waves.


The electrons in a metal have not left the atoms behind (ionization), they're just taking the scenic route through the metal.

I don't have the time now to look this up, but it seemed to me that the electrons are indeed moving up and down the antenna just 'a little', but that this is even a lot more than just exciting to a higher level of the atom, which to my understanding means that they leave the atom for a short while.


In the same post you mention this: "the transmission speed in the metal used". So you know that this speed in metal is much slower, but still the EM waves 'in it' have the speed of light.

Not a lot slower, just a little.

The average SPEED of electrons in metal (the Fermi Speed) is 10^6 m/s, which is 300 times slower than c. Still the speed of electricity is about the same. You say the signal transmission speed in nerves is around 83 m/s, which is 12000 times slower than the electrons in metal. Does this mean that a nerve cannot fibrate at the frequency of light? And what about the pigments? Are they vibrating at the frequency of light and transmitting the energy to the nerves?

I read your next comment:

the energy level of the photochemical (which the sum of the total vibrational, rotational and electronic energies of the molecule) will increase exactly the amount that the incoming photon provides. So, neither the frequencies prior nor after the absorption will match that of the photon, but the difference in energy matches that of the photon.
But I'm curious about how you think about the frequency of the pigment. Do you agree that it should be possible to explain in wave theory, even if we don't know exactly how? Can the pigment have 3 frequencies (vibrational, rotational and electronic) that add up to the light frequency or something like that?

OK, I didn't get beyond the long post of AstroSmurf, just for you to know. I will post this now, so if some answers are already there, I'll find out later.

Sam5
2004-Jan-09, 12:09 AM
Heat is acceleration? :roll:

Like Sam says,adding heat is acceleration, it's the delta factor in the formula.


Of course, if the atoms inside the chemical coating that surrounds the cones were the only things that “resonated” or “absorbed the energy” of the photons, then our eyes wouldn’t need all the separate little cones that are the size of the light wavelengths, since the coating would do all the absorbing of the light waves all by itself. But all those little cones need to be separate and of the right small light-wavelength size so that they too can “resonate” and help amplify the energy of the light waves and turn it into electrical impulses that go to the brain.

Almost exactly what I mean:


Hey! I wrote that second paragraph!

Sandor
2004-Jan-09, 05:55 AM
Ah yes sorry, I changed it now. :oops:

AstroSmurf
2004-Jan-09, 08:59 AM
The reception part, which is what we started talking about, ends after the light has been absorbed by the photopigment. All the rest is neurology, not physics.
This discussion is not at all about the reception part, because we all agree that this is done by the pigments. The problems really starts from then.
Yes, but HOW is the light received by the photopigments? After that, the signal is transmitted with a biochemical reaction, which has absolutely nothing to do with what caused the initial stimulus - it might have been sound, touch, heat or whatever.



When talking about the absorption of light, it seems much more straightforward to think of it in terms of energy levels rather than resonation; the resulting "vibrational frequency" (if this is even possible to determine) might be very different from that of the incoming light.
It is not very different, but the average of many red pigments reporting to one bipolar cell. Frequency is just the same as the number of photons per time frame, so it's the same to determine.
NO! Frequency has nothing to do with the number of photons per time frame - that's the amplitude or intensity of the incoming light! The frequency is the colour of the light.

This distinction is incredibly important - unless this is clear in your mind, no further discussion is possible since our terminology will be totally different. If you consider a normal sine wave, such as ripples on a water surface, the frequency is the number of oscillations per second that pass a certain point, and the amplitude is the height of the waves, compared with a still surface.



The electrons in a metal have not left the atoms behind (ionization), they're just taking the scenic route through the metal.
I don't have the time now to look this up, but it seemed to me that the electrons are indeed moving up and down the antenna just 'a little', but that this is even a lot more than just exciting to a higher level of the atom, which to my understanding means that they leave the atom for a short while.
The trouble is that due to the characteristics of a metal, the locations of the free electrons are not terribly well-defined. The free electrons of a certain atom may indeed be anywhere in the metal body - all we can say is that if there's no electrical field present, they're usually evenly distributed throughout the piece of metal. It might help to think that there's a "gas" of electrons filling the space between/around the atoms, but the electrons still remain inside the metal piece.


In the same post you mention this: "the transmission speed in the metal used". So you know that this speed in metal is much slower, but still the EM waves 'in it' have the speed of light.

Not a lot slower, just a little.
I'll clarify here since there was some remaining confusion. To put it in RF design terms, there's something called a "velocity factor", which simply is the relative speed of EM fields inside a certain metal or electrical component. A normal value of this is around 0.7-0.8 (multiply by c for the actual speed). The electrons don't need to collide to transmit an electrical signal, but they have a certain amount of inertia, which is the cause of this slow-down. An analogue of this for visible light would be the refractive index, though the interactions are rather different.

To return to the reception of light in the eye, one of your links stated this (IMO) rather clearly. The photopigment molecules absorb incoming light, which alters their chemical structure. This reaction is entirely independent of where they are - molecules in a test tube will show the exact same response.

If I read the page correctly, this causes a decrease(!) in the cell's Na+ ion production, which is interpreted by the nervous system as a message that some light has been received by this cell. The Na+ ions then travel through the nerves, transmitting the message to the brain. I suppose the nerve message will contain information on how much light has been received, rather than just a yes/no message, but all of this has nothing to do with what kind of photon has been received - that's an interpretation the brain later makes. The frequency of the incoming light will probably affect the signal strength, since different cells are sensitive in different frequency areas, but for the details of how the brain uses this information to determine the colour of the light that caused all this, you'll have to read for yourself.


I read your next comment:

the energy level of the photochemical (which the sum of the total vibrational, rotational and electronic energies of the molecule) will increase exactly the amount that the incoming photon provides. So, neither the frequencies prior nor after the absorption will match that of the photon, but the difference in energy matches that of the photon.
But I'm curious about how you think about the frequency of the pigment. Do you agree that it should be possible to explain in wave theory, even if we don't know exactly how? Can the pigment have 3 frequencies (vibrational, rotational and electronic) that add up to the light frequency or something like that?
I doubt it - there's no easy way to translate from an energy level to a frequency. For vibrational and rotational energies, it might be possible, but for electronic energies, there simply is no such correlation. And at any rate, there's no way to "add" frequencies like that - adding two signals with different frequencies produces a signal where the two original components are still easily recognisable. When a molecule absorbs a photon, some of the energy might go to the electrons, while the rest changes the vibrational and/or rotational state, so the difference of the sums match up, but the components do not.

Sandor
2004-Jan-09, 11:52 AM
Yes, but HOW is the light received by the photopigments? After that, the signal is transmitted with a biochemical reaction, which has absolutely nothing to do with what caused the initial stimulus - it might have been sound, touch, heat or whatever.

It sure has. This is what Sam and I are stating: all forces can be explained as particles, but also as waves. What you call a biochemical reaction is in fact nothing but many quantumfysical reactions. I get the 'software programmer' association here. There is no programmer in the world that understands MS Windows XP at bit level, but of course this is THE only way to REALLY understand it. Suppose a programmer who works at level 4 is doing some stuff, in which I recognize things at bit level. So I talk to him and he answers: 'no, this has nothing to do with bits, not even with bytes'.
The human body is much more complicated than Windows XP, but I am not discussing the brains or even something that comes close, but just this:

1000 photon-waves - 1000 pigments - 1 cone - 1 dendrite - 1 bipolar cell.
Somewhere from the pigments to the dendrite we are talking about biochemicals, leaving elementary physics, which is OK and even very practical when just interested in biochemicals -it's not possible for 4th level programmers to work and think in bits, because they'll lose overview-, but in this discussion we reach for the deeper/detailed understanding of the cone (as one of the primary subjects).


NO! Frequency has nothing to do with the number of photons per time frame - that's the amplitude or intensity of the incoming light! The frequency is the colour of the light.

I can see that you didn't read this entire topic, because I explained this many times. You and others are rather repeating your 'knowledge' than really thinking yourselves. Science is OK just for 'learning' if you go 'practicing' it, but for you to understand you can't take anything for granted. An active scientist also doesn't; he's constantly questioning and finding problems and looking for answers. An active researcher knows there's much more that he doesn't understand than that he understands, while a science student might think that all the lessons are 'clear knowledge'. There is no value in repeating 'knowledge' without discussing the essence of it.

You are right that we use different terminoligy, and we shouldn't. Well, at least not so much, although it shouldn't make so much of a difference if we would just think a little bit more. I already told you that I was referring to ONE wave. I will not explain this again, you can find it about everywhere in this topic. In one single wave the frequency is really the number of photons passing per time frame. Please imagine it and wipe out the dogma.


The trouble is that due to the characteristics of a metal, the locations of the free electrons are not terribly well-defined. The free electrons of a certain atom may indeed be anywhere in the metal body - all we can say is that if there's no electrical field present, they're usually evenly distributed throughout the piece of metal. It might help to think that there's a "gas" of electrons filling the space between/around the atoms, but the electrons still remain inside the metal piece.

The locations are not well defined, but the distance they travel is well defined. From that we know that they leave the atoms. I'm not saying that the particle perspective is a very realistic one, maybe even not, but that also counts for the other ionizations you were referring to. If we discuss this with particle theory however, let's not make a 'gass' from the electrons to defend that they're not really leaving the atoms.


1. The photopigment molecules absorb incoming light, which alters their chemical structure.
2. If I read the page correctly, this causes a decrease(!) in the cell's Na+ ion production.
3. The Na+ ions then travel through the nerves, transmitting the message to the brain.
4. The frequency of the incoming light will probably affect the signal strength.

1. How?
4. There is one dendrite, coming from many pigments, going to one bipolar cell. Do you agree that an average has to be determined, being this signal strength? If so, is it likely that the cone plays this role? If so, can you describe this better than for the cone just tuning to an average frequency?


And at any rate, there's no way to "add" frequencies like that - adding two signals with different frequencies produces a signal where the two original components are still easily recognisable.

This is not really adding up/taking the average, but just remaining parallel waves. But what if you have to put all waves through the same serial line? If they have about the same frequency, they form, in your words, 'one wave' with the light frequency and a bigger amplitude (which is really many single waves of the same frequency). Since the frequency is a given fact (the dendrite is i.e. a RED dendrite), the amplitude will have to be converted to the frequency of one 'new' wave. This is not something I am making up, although I don't even see a problem with that. Here's an example: a tsunami caused by an earthquake. (When my explanation of a tsunami is not right, it must be easy to think of another example, also with EM waves.) When the quake is several thousands of miles long and 'pointed inwards' (curved), many little waves arise, all pointing to one center point far away (as a wedge of cake). The waves will add up to one big wave eventually.


When a molecule absorbs a photon, some of the energy might go to the electrons, while the rest changes the vibrational and/or rotational state, so the difference of the sums match up, but the components do not.

This is about the best 'programmer' example you can give. Because ALL energy will go to electrons and quarks. Only when these parts form something bigger, you jump to this bigger level. But the 2 levels can never communicate, that's why you can't explain point 1: "The photopigment molecules absorb incoming light, which alters their chemical structure." If you can, you are leveling to the electron level, which is exactly what we need to understand this.

Kaptain K
2004-Jan-09, 11:55 AM
Sandor wrote:

The further I get into this, the more analogy I find between sound and light waves. Try me and name any aspect of either one and I will try to find the similar aspect in the other.
Light is a transverse wave and, as such, it can be polarized. Sound is a pressure wave and cannot be polarized!

swansont
2004-Jan-09, 12:34 PM
You are right that we use different terminoligy, and we shouldn't. Well, at least not so much, although it shouldn't make so much of a difference if we would just think a little bit more. I already told you that I was referring to ONE wave. I will not explain this again, you can find it about everywhere in this topic. In one single wave the frequency is really the number of photons passing per time frame. Please imagine it and wipe out the dogma.



If you are going to discuss science, you need to adopt the accepted terminology. The number of photons gives you the intensity, or the amplitude of the wave. The frequency is independent of that. Your own definitions do not change that.

swansont
2004-Jan-09, 12:39 PM
But I'm curious about how you think about the frequency of the pigment. Do you agree that it should be possible to explain in wave theory, even if we don't know exactly how? Can the pigment have 3 frequencies (vibrational, rotational and electronic) that add up to the light frequency or something like that?


If wave theory were adequate to explain all, quantum theory would never have been developed. There are some things that just can't be explained classically.

AstroSmurf
2004-Jan-09, 01:17 PM
Since this thread is bouncing all over the map, let's concentrate things a bit. I can debate this on a lot of levels - reception of light in the eye, the interaction of EM radiation with a conductor or the absorption of photon energy by a molecule, but I would prefer not to have to do all three at once. Please choose the subject that you feel is the key point and we'll continue with that one.

AstroSmurf
2004-Jan-09, 02:21 PM
I've reread a few pages, and I'd like to clarify one point.

The intensity of light is defined as the amount of energy received per time and area unit. In the wave view, this is the amplitude of the signal, or the magnitude of the EM fields. In the particle view, this is the product of the amount of photons that arrive per time and area unit and the energy per photon! Thus, in the particle view, there's no isolated physical quantity that represents intensity. To put it in another way: It takes fewer blue photons to transfer a certain amount of energy/momentum than red photons.

Sam5
2004-Jan-09, 02:49 PM
NO! Frequency has nothing to do with the number of photons per time frame - that's the amplitude or intensity of the incoming light!

If that’s the case, then why do we see all the separate colors in a rainbow – all of different frequency and wavelength – at the same amplitude and intensity?

I would say intensity is the number of photons per square mm hitting any surface.

The frequency is the color, but you said the number of photons per time frame was the intensity. I think you mean the number of photons hitting one sq mm or cm all at once is the intensity, while the number per second of several individual photons following after the one that is out in front is both the frequency and the color.

Tensor
2004-Jan-09, 03:03 PM
all forces can be explained as particles, but also as waves.
But not at the same time, which you keep trying to do by giving photons frequency. Photons do not have frequency, (For others out there, yes, I do know about the time componet of the 4 vector). You are confusing the wave explanation of frequency (which is wavelenths per unit time) and the Photon explanation of intensity (which is photons per unit time). Just becuase they both have the "per unit time" statement in there, doesn't mean they are the same thing. The two are completely different.


1000 photon-waves - 1000 pigments - 1 cone - 1 dendrite - 1 bipolar cell.

Photons don't wave, so how does this even work?



NO! Frequency has nothing to do with the number of photons per time frame - that's the amplitude or intensity of the incoming light! The frequency is the colour of the light.

Science is OK just for 'learning' if you go 'practicing' it, but for you to understand you can't take anything for granted. An active scientist also doesn't; he's constantly questioning and finding problems and looking for answers. An active researcher knows there's much more that he doesn't understand than that he understands

While this is all true, those scientists and researchers also have a thorough understanding, after years of study, of the current model they are using. As you stated a few days ago, you didn't understand any of this. And while you have learned some of the current model, your statements concerning photons "waving" indicates you really don't have a basic understanding of wave-particle duality .


You are right that we use different terminoligy, and we shouldn't. Well, at least not so much, although it shouldn't make so much of a difference if we would just think a little bit more. [

Or if you understood the model better, there would be no difference either.


I already told you that I was referring to ONE wave. In one single wave the frequency is really the number of photons passing per time frame.


And several people here have told you, you cannot talk about waves and photons at the same time. The two have different properties. If you have a single wave, you can find the frequency by dividing c by the wavelength. Photons have nothing to do with measurements of waves. The number of photons per time is the intensity of the EM radiation. This can be equated to the AMPLITUDE of the wave. The frequency of the wave can be equated to the energy of the photon.


Please imagine it and wipe out the dogma.

Dogma? How would you know it's dogma, if you don't even show a basic understanding of the model?

Sam5
2004-Jan-09, 03:08 PM
[They] and others are rather repeating [their] 'knowledge' than really thinking [them]selves. Science is OK just for 'learning' if [they] go 'practicing' it, but for [them] to understand [they] can't take anything for granted. An active scientist also doesn't; he's constantly questioning and finding problems and looking for answers. An active researcher knows there's much more that he doesn't understand than that he understands, while a science student might think that all the lessons are 'clear knowledge'. There is no value in repeating 'knowledge' without discussing the essence of it.

Exactly. This is what I've been telling some of them on the SR thread. They know the big words but they don't know what they mean, and they question nothing they are told by their American teachers. And if I ask them a question, if they weren't already told the answer by a teacher, they don't know how to figure it out by themselves, and they get mad at me for asking the question. They are the product of the modern American public school system. This is why more of our high technology jobs are going to the Japanese, Chinese, and Indians. For example, more and more PhD physics jobs at Los Alamos are being filled with Asian scientists now.

Sam5
2004-Jan-09, 03:14 PM
While this is all true, those scientists and researchers also have a thorough understanding, after years of study, of the current model they are using. As you stated a few days ago, you didn't understand any of this. And while you have learned some of the current model, your statements concerning photons "waving" indicates you really don't have a basic understanding of wave-particle duality .

Well, why don’t you explain all about it to all of us, and also include the “resonation” concept and the “intensity” in terms of the number of photons falling on one sq mm at the same time? And also explain how a “wave packet” of sound can also act like a “particle”.

swansont
2004-Jan-09, 03:15 PM
NO! Frequency has nothing to do with the number of photons per time frame - that's the amplitude or intensity of the incoming light!

If that’s the case, then why do we see all the separate colors in a rainbow – all of different frequency and wavelength – at the same amplitude and intensity?

I don't know that we do, but also recall that the eye is a nolinear detector.


I would say intensity is the number of photons per square mm hitting any surface.

The frequency is the color, but you said the number of photons per time frame was the intensity. I think you mean the number of photons hitting one sq mm or cm all at once is the intensity, while the number per second of several individual photons following after the one that is out in front is both the frequency and the color.

No, I'm pretty sure he means the number hitting per unit area per unit time, since that's the definition of it. You can have one photon at an energy E1, and you can have lots of them, each at E1. They all can come from a monochromatic source of a given frequency given by E/h. Putting in a filter to lower the intensity doesn't change the energy of the photons or the frequency of the source, even though fewer photons are hitting your detector per unit time.

Sam5
2004-Jan-09, 03:21 PM
No, I'm pretty sure he means the number hitting per unit area per unit time, since that's the definition of it. You can have one photon at an energy E1, and you can have lots of them, each at E1. They all can come from a monochromatic source of a given frequency given by E/h. Putting in a filter to lower the intensity doesn't change the energy of the photons or the frequency of the source, even though fewer photons are hitting your detector per unit time.

My term for intensity = “number of photons per sq mm hitting at the same time” is more accurate. It’s the number “per unit area”, at the same time that’s important for “intensity”. The “number per unit time” should refer to the frequency.

Sandor
2004-Jan-09, 03:29 PM
Kaptain K, thanks for the example of polarization, that's a nice thing to know. I would call this a 'feature', but not really one of the most essential aspects of 'waves'. We can also explain it like this: the waving principle is exactly the same, but since photons go much faster and are even completely flat, some differences occur naturally, that cannot occur to compound particles like molecules. So the particle difference changes the wave, not the wave principle changes the wave or particle.

Swansont, if you still don't understand that one single photon wave doesn't have any amplitude, than we (us 2) might indeed better not discuss science anymore.


It takes fewer blue photons to transfer a certain amount of energy/momentum than red photons.

Sure. As long as you know that 'one blue photon' is a packet with higher energy than a red photon packet, because it holds more wave lengths (more waves arriving at a time).

Laser Jock
2004-Jan-09, 03:50 PM
NO! Frequency has nothing to do with the number of photons per time frame - that's the amplitude or intensity of the incoming light!

If that’s the case, then why do we see all the separate colors in a rainbow – all of different frequency and wavelength – at the same amplitude and intensity?

We don't. As swansont said, our eye is a nonlinear detector.



I would say intensity is the number of photons per square mm hitting any surface.


Close. It is the rate of photons per unit area. When in doubt, look at the units for intensity (SI units): J*s^{-1)*cm^{-2} or Joules (energy) per second per centimeter squared.



The frequency is the color, but you said the number of photons per time frame was the intensity. I think you mean the number of photons hitting one sq mm or cm all at once is the intensity, while the number per second of several individual photons following after the one that is out in front is both the frequency and the color.

No. Frequency (or wavelength) and color are two distinct concepts. Frequency (or wavelength) is a fundamental property of electromagnetic waves. Color is how our eye-brain system interprets some mixture of electromagnetic waves. Check this chromaticity diagram (http://www.cs.rit.edu/~ncs/color/a_chroma.html) to see what I mean. (Aside, from this diagram can anyone tell me why a rainbow on television always looks a little different than it does in real life?)


They know the big words but they don't know what they mean, and they question nothing they are told by their American teachers. And if I ask them a question, if they weren't already told the answer by a teacher, they don't know how to figure it out by themselves, and they get mad at me for asking the question. They are the product of the modern American public school system.

How many times have we caught you using words that you don't know what they mean? You don't need to look further than this post. What you really mean is that since we don't get your answer (when we work it out ourselves), we must be parroting our teachers. You would never consider that you might actually be wrong about something.


This is why more of our high technology jobs are going to the Japanese, Chinese, and Indians. For example, more and more PhD physics jobs at Los Alamos are being filled with Asian scientists now.

Do you have any real numbers to back up this statement? If not, please retract it.

AstroSmurf
2004-Jan-09, 03:57 PM
Another clarification.

What physicists mean by the wave-particle duality is that no single view of photons or other elementary particles explains all the phenomena in nature - you need both. Just as the particle view cannot explain polarization, diffraction and so forth, the wave view cannot explain absorption by electrons and the photoelectric effect. It's not a case of "pick the view you prefer", but "pick the view that explains what you're studying". Neither view is self-sufficient.

Sam5, there are two problems with your term for intensity. Firstly, energy of an equal-"intensity" beam would vary with the frequency of the photons. Higher-frequency photons do carry more energy than lower-frequency ones - this has been seen more times than I care to mention. Secondly, measuring energy/area the way you seem to be attempting does not match the way intensity is used in other areas of physics. The quantity is power/area unit, which is a different beast.

Tensor, it is possible to determine the frequency of a single photon irrespective of its energy, though you need to do probability studies to see it. The frequency is more easily seen when several photons interact, but it's still there regardless.

And finally, Sandor, it simply isn't possible to study EM radiation in smaller quantities than one photon. All we can determine about photons is the result of their interaction with matter, and the wave characteristica won't show up until you have see several photons acting over time. There's no such beast as "one wave" - you never see anything less than one photon. How many "oscillations" there are for each photon - well, your guess is as good as mine; nature doesn't seem to care.

Sandor
2004-Jan-09, 04:33 PM
all forces can be explained as particles, but also as waves.
But not at the same time, which you keep trying to do by giving photons frequency.
No, of course not.


You are confusing the wave explanation of frequency (which is wavelenths per unit time) and the Photon explanation of intensity (which is photons per unit time).
No, not at all.


Photons don't wave, so how does this even work?
This is becoming an absoluty nonsense topic. Don't pretend as if you are consequent in this. You describe the travelling photon as a wave, but the arriving photon as a particle. I really don't care, just describe it as you like, I will do the same. We are not talking about the real stuff anymore, but arguing like old ladies.


While this is all true, those scientists and researchers also have a thorough understanding, after years of study, of the current model they are using. As you stated a few days ago, you didn't understand any of this. And while you have learned some of the current model, your statements concerning photons "waving" indicates you really don't have a basic understanding of wave-particle duality .
No, you still don't understand that not only I, but anybody here doesn't understand about it. But that's no surprise for me by now.


Photons have nothing to do with measurements of waves. The number of photons per time is the intensity of the EM radiation. This can be equated to the AMPLITUDE of the wave. The frequency of the wave can be equated to the energy of the photon.
What are you doing? Describing photons by wave measurements? :)
You don't even understand that your 'wave' is in fact many waves.

Sandor
2004-Jan-09, 05:09 PM
What physicists mean by the wave-particle duality is that no single view of photons or other elementary particles explains all the phenomena in nature - you need both. Just as the particle view cannot explain polarization, diffraction and so forth, the wave view cannot explain absorption by electrons and the photoelectric effect.
Almost true. It can explain, but WE can't explain. When someone here tries to explain anyway, please listen to him carefully and try to understand. Don't just react with the present standard 'knowledge level' as a common conclusion that doesn't need to be questioned or understood.

Although I think that you are getting the point here, since you are talking now about the intensity of a BEAM.
Then Sam5 and I made very clear from the beginning of this topic that WE mean by a photon: one wavelength. I am gladly willing to change this into more wavelengths, or a certain rate. Just tell me the rate or number of wavelengths. You say you don't know. But it can be measured, maybe resulting in an approach (more or less secure), but it can. Science is nothing but assuming and estimating things until someone proves better. So if you cannot even estimate the number of waves in a photon, let's just pick a number. Then we can makes some things clearer here.

Also I would like to repeat the simple rope thing, being waved by one person. Look at this rope as one single wave. Then everyone should be able to understand why frequency and amplitude are directly related in a single wave.

swansont
2004-Jan-09, 06:19 PM
No, I'm pretty sure he means the number hitting per unit area per unit time, since that's the definition of it. You can have one photon at an energy E1, and you can have lots of them, each at E1. They all can come from a monochromatic source of a given frequency given by E/h. Putting in a filter to lower the intensity doesn't change the energy of the photons or the frequency of the source, even though fewer photons are hitting your detector per unit time.

My term for intensity = “number of photons per sq mm hitting at the same time” is more accurate. It’s the number “per unit area”, at the same time that’s important for “intensity”. The “number per unit time” should refer to the frequency.

"Number at the same time isn't more accurate" since, for photons, we're talking about a discrete process and not a continuous one.

"Number per unit time" does not mean frequency, for the reason/example I gave. Can you give a counter to it?

swansont
2004-Jan-09, 06:23 PM
Although I think that you are getting the point here, since you are talking now about the intensity of a BEAM.
Then Sam5 and I made very clear from the beginning of this topic that WE mean by a photon: one wavelength.

The name photon is already taken. It has a particular meaning to those who are familiar with physics. You'll have to use something else if you are going to mean something other than what the scientific community has been calling a photon for the last several decades.

Sandor
2004-Jan-09, 06:36 PM
What does it matter Swansont, if I consider a photon to be 1 wavelength or 1000? How many do you think it is?

Kaptain K
2004-Jan-09, 06:51 PM
What does it matter Swansont, if I consider a photon to be 1 wavelength or 1000? How many do you think it is?
It matters because if you use a nonstandard definition, you cannot tell the whole world it is wrong for using the standard definition. If you are going to discuss physics with physicists, you cannot simply make up your own terms as you go along.

Sandor
2004-Jan-09, 06:56 PM
Yes I know the principle of what's wrong with using different terms for the same things or different things for the same term, but please tell me what is the reasonable difference between a photon of 1 wavelength and a photon of 1000 wavelengths, or are we just making troubles over nothing?

Kaptain K
2004-Jan-09, 07:15 PM
Yes I know the principle of what's wrong with using different terms for the same things or different things for the same term, but please tell me what is the reasonable difference between a photon of 1 wavelength and a photon of 1000 photons, or are we just making troubles over nothing?
The trouble is that you cannot talk about the wavelength of a photon. Again, you must understand the wave-particle duality of EM radiation. If you observe a wave-like atribute of light, it acts like a wave. If you observe a particle-like property of light, it acts like particles.

swansont
2004-Jan-09, 07:18 PM
Yes I know the principle of what's wrong with using different terms for the same things or different things for the same term, but please tell me what is the reasonable difference between a photon of 1 wavelength and a photon of 1000 photons, or are we just making troubles over nothing?

A photon isn't a wave. The phrase has no meaning.

Kaptain K
2004-Jan-09, 07:29 PM
In any field, you have to know and use the correct terminology to be understood.

As an example, a shotgun and a rifle appear (at first glance) to be similar - trigger, barrel, etc. but you don't talk about hunting quail with a rifle.

Sandor
2004-Jan-09, 07:38 PM
Oh guys please, I will ignore first of all the issue that a particle is not a wave, because we simply don't know. In fact energy seems to be both, sometimes better understood as this, sometimes as that. OK, now that I didn't ignore that, let's get on with the real thing:

Do you agree that a certain number of photons per time frame will be fired from a blue ray of light and less photons per time frame will be fired from a red ray of light? In other words, they ARE the ray? Let's use a distance between source and receiver of 1000 m.

And do you also agree that from a wave perspective, there is a certain amount of wavelengths in a blue ray of 1000 m. and another amount in a red ray over the same distance?

Sam5
2004-Jan-10, 12:35 AM
I would say intensity is the number of photons per square mm hitting any surface.


Close. It is the rate of photons per unit area. When in doubt, look at the units for intensity (SI units): J*s^{-1)*cm^{-2} or Joules (energy) per second per centimeter squared.

Well, that’s ok. I’ll let the pros do that, so they can relate it to a standard Newtonian time period of 1 second, but the important part is the number of photos per square centimeter. If they want to standardize that in terms of a second, it’s ok.


Frequency (or wavelength) and color are two distinct concepts. Frequency (or wavelength) is a fundamental property of electromagnetic waves. Color is how our eye-brain system interprets some mixture of electromagnetic waves. Check this chromaticity diagram (http://www.cs.rit.edu/~ncs/color/a_chroma.html) to see what I mean. (Aside, from this diagram can anyone tell me why a rainbow on television always looks a little different than it does in real life?)


Yes, yes, yes, I know that. In fact, I think I already said on this thread that the “color” of an object is not at the object, it is only in our brain. Light frequencies and wavelengths have no inherit “color” in and of themselves, just as a radio wave has no “color”.

When we see “red” car tail lights, the “red” is not at the tail light, it is only in our brain. Our brain makes us think the red is at the tail light.

Regarding the rainbow and TV, a TV screen emits only red, green, and blue light, ie just three frequencies of light, whereas a rainbow emits many frequencies of light.


How many times have we caught you using words that you don't know what they mean? .... You would never consider that you might actually be wrong about something.

Who, me?? Wrong?? lol. I’m occasionally wrong, but I try to keep my error rate down low by not talking about things I don’t already know something about.

But look, this works both ways. I use 19th Century words and Newtonian words that you don’t understand. I know some of the fundamental concepts, such as the rod and cone resonations, that you are never taught and that you don’t understand. You seem to think that it is only the chemical coating that picks up the light waves, and you don’t seem to understand why our eyes have hundreds of thousands of tiny little individual cones that are about the same length or width of a wavelength of light. It doesn’t even occur to you that they might “resonate”. If it wasn’t taught to you, it doesn’t seem to occur to you.

And look, I’m not teaching my ideas in American schools. I’m just talking about them on internet message boards. You guys force wrong ideas into the minds of kids as the “official science point of view of the entire universe, and no independent thinking is allowed” under your system. In fact, your system DISCOURAGES independent thinking. That’s why we have to get the Davis-Lineweaver paper out of New Zealand, many of our computers out of China, our cars and new electronic gadgets out of Japan, and much of our software programmed by guys who were educated in India. That’s why they’ve got so many Asian PhDs working at Los Alamos now.


This is why more of our high technology jobs are going to the Japanese, Chinese, and Indians. For example, more and more PhD physics jobs at Los Alamos are being filled with Asian scientists now.


Do you have any real numbers to back up this statement? If not, please retract it.

Real numbers?? You mean you want me to go to Los Alamos and sit on a streetcorner and actually count the Asians and non-Asians and then post the real numbers? I can’t do that at Los Alamos! Head counting by strangers is not allowed there.

Sam5
2004-Jan-10, 12:51 AM
the wave view cannot explain absorption by electrons and the photoelectric effect.

It can, if the phenomena is actually due to a resonation effect. Just as sound can act like "particles" under certain circumstances.

The "packet" idea can work both with a particle and a "wave packet", especially if a single photon is a "packet of several waves" that travel together. It could be the frequency of the "wave packet" of light that causes the electrons to behave in a certain way.

What is the old story about the breaking of the wine glass with a certain sound wave frequency? The glass breaks as if it is hit by a "particle" but it is actually hit by a certain frequency of sound.

Sam5
2004-Jan-10, 12:56 AM
Resonation: Sound breaking glass by acting as a “particle”:

www.physics.ucla.edu/demoweb/demomanual/acoustics/effects_of_sound/breaking_glass_with_sound.html+wine+glass+break+so und+frequency+&hl=en&ie=UTF-8]LINK (http://216.239.53.104/search?q=cache:1AlGnmVH_GYJ:[url) TO SOURCE[/url]

Sam5
2004-Jan-10, 01:06 AM
No, I'm pretty sure he means the number hitting per unit area per unit time, since that's the definition of it. You can have one photon at an energy E1, and you can have lots of them, each at E1. They all can come from a monochromatic source of a given frequency given by E/h. Putting in a filter to lower the intensity doesn't change the energy of the photons or the frequency of the source, even though fewer photons are hitting your detector per unit time.

My term for intensity = “number of photons per sq mm hitting at the same time” is more accurate. It’s the number “per unit area”, at the same time that’s important for “intensity”. The “number per unit time” should refer to the frequency.

"Number at the same time isn't more accurate" since, for photons, we're talking about a discrete process and not a continuous one.

"Number per unit time" does not mean frequency, for the reason/example I gave. Can you give a counter to it?

I would say that the number of photons per square cm, all hitting at the same time, represents intensity. In other words, 2 photons hitting on 1 sq cm at the same time would represent “low intensity” light, while 50,000 hitting the same 1 sq cm at the same time would represent “high intensity” light.

That’s why we can buy a “low intensity” white lamp and a “high intensity” white lamp, or a high and low intensity red lamp.

Number sequentially per unit time represents frequency. In other words, with one photon following another, 5n arriving per second is a lower frequency than 40n arriving per second.

swansont
2004-Jan-10, 01:17 AM
Number sequentially per unit time represents frequency. In other words, with one photon following another, 5n arriving per second is a lower frequency than 40n arriving per second.

And I reiterate: lowering the number of photons, by putting a filter in place, does not change the frequency of the light. It's a trivial experiment to do (I've done it numerous times, though not to test your hypothesis). Your idea is demonstrably wrong.

You would have a blue light turn red as the brightness is turned down. Doesn't happen. When the facts don't support the hypothesis, you have to abandon the hypothesis.

Sam5
2004-Jan-10, 01:21 AM
Sam5, there are two problems with your term for intensity. Firstly, energy of an equal-"intensity" beam would vary with the frequency of the photons. Higher-frequency photons do carry more energy than lower-frequency ones - this has been seen more times than I care to mention. Secondly, measuring energy/area the way you seem to be attempting does not match the way intensity is used in other areas of physics.

Ok, look at this.

RAINBOW PHOTOS (http://australiasevereweather.com/photography/rainbo01.htm)

The yellow is usually more intense to our eyes, although the blue is of a higher frequency and is supposed to be of a “higher energy”. In rainbow color intensity, I would say it would go from more to less in this order: yellow, red, green, blue, violet, as seen by our eyes and by color film. But violet is generally not available in color film.

Of course these are photos so the deep blue and violet tend to not show up, since color film is not very sensitive to the deep blue or violet. Plus, on the color computer monitor, we see only red, green, and blue pixels. If you look at something that is “yellow” on a computer, use a magnifying glass and you will see the bright green pixels and the darker red ones. The blue ones will be black (totally dark).

Tensor
2004-Jan-10, 01:22 AM
Oh guys please, I will ignore first of all the issue that a particle is not a wave, because we simply don't know. In fact energy seems to be both, sometimes better understood as this, sometimes as that. OK, now that I didn't ignore that, let's get on with the real thing:

Thank you Sandor, since you understand this part, then we can have a meaningful discussion. This is a good statement of the nature of EM radiation.


Do you agree that a certain number of photons per time frame will be fired from a blue ray of light and less photons per time frame will be fired from a red ray of light? In other words, they ARE the ray? Let's use a distance between source and receiver of 1000 m.

No, a bright red light will have more photons than a dim blue light. Contrawise, a bright blue light will have more photons than a dim red light. In the above, distance to the source doesn't matter, as long as the distance is the same for each light.


And do you also agree that from a wave perspective, there is a certain amount of wavelengths in a blue ray of 1000 m. and another amount in a red ray over the same distance?

Yes, The blue light will have more wavelengths than the red light.

swansont
2004-Jan-10, 01:35 AM
But look, this works both ways. I use 19th Century words and Newtonian words that you don’t understand. I know some of the fundamental concepts, such as the rod and cone resonations, that you are never taught and that you don’t understand. You seem to think that it is only the chemical coating that picks up the light waves, and you don’t seem to understand why our eyes have hundreds of thousands of tiny little individual cones that are about the same length or width of a wavelength of light. It doesn’t even occur to you that they might “resonate”. If it wasn’t taught to you, it doesn’t seem to occur to you.


Lots of interesting and useful physics happened in the 20th century. You might try reading up on that sometime.

--

If rods and cones were larger, I would imagine that all that would happen is that our visual acuity would diminish. Few pixels for imaging.

Sam5
2004-Jan-10, 02:13 AM
And I reiterate: lowering the number of photons, by putting a filter in place, does not change the frequency of the light. It's a trivial experiment to do (I've done it numerous times, though not to test your hypothesis). Your idea is demonstrably wrong.

All your filter does is lower the side-by-side photon number per sq cm, because the filter absorbs some of the side-by-side photons, so they never reach your eye. A photo “neutral density” (gray) filter, merely lowers intensity, not frequency. I learned this when I was 14 years old, in 1956.

A welder’s neutral density filter on a welder’s mask cuts down on the side-by-side photon number and density per sq cm, by absorbing many of the photons. There is no color change, unless the filter happens to be colored.


You would have a blue light turn red as the brightness is turned down. Doesn't happen. When the facts don't support the hypothesis, you have to abandon the hypothesis.

No, no, no, no. When the light is turned down in a lamp, you are lowering the side-by-side amount of photons, not the frequency. If you put a neutral density filter in front of the light, you do the same thing.

Sam5
2004-Jan-10, 02:37 AM
If the rods and cones were larger, I would imagine that all that would happen is that our visual acuity would diminish. Few pixels for imaging.

The cones are responsible for pixel count and for resonation amplification of the signal too, and for focusing the light beams, but under your “colored coating” theory, all we would need to do to get a larger pixel count is add more wires (nerve endings) to the backside of the coating, but it doesn’t work that way. What your theory winds up with is one big photo-cell, like an old photo light-meter, sensing only one big round spot of white light. The resonating cones are needed for the pixel count and the focusing of the light beams, and for amplification by means of resonation.

Your theory creates a "photo cell", not an eye retina.

Sam5
2004-Jan-10, 02:44 AM
No, a bright red light will have more photons than a dim blue light.

It could have more side-by-side photons per sq cm, but not more photons-in-a-row. A bright light has more side by side photons per sq mm.

This is why you get dimmer light on a telescope when you change to a stronger lens and a greater magnification power.

With smaller magnification, you are concentrating the side-by-side photon count into a smaller area of your eye. With larger magnification, you are spreading out the same number of photons to a wider area of your eye, ie fewer photons per sq mm.

Tensor
2004-Jan-10, 04:24 AM
No, a bright red light will have more photons than a dim blue light.

It could have more side-by-side photons per sq cm, but not more photons-in-a-row. A bright light has more side by side photons per sq mm.

Nope, all that matters is the number of photons. Doesn't matter where in the measured area they are. And how does your traveling side by side idea explain this single photon experiment (http://copilot.caltech.edu/journal_club/s80h36.pdf) where single photons are emitted and the dector records the intensity? Obviously single photons are not traveling side by side.


This is why you get dimmer light on a telescope when you change to a stronger lens and a greater magnification power.

Care to explain why images in the 15mm eyepiece I borrowed from one of our club members is brighter than the 26 mm I have, when used on the same object in my 6 in refractor? Oh, and what magnification am I getting.


With smaller magnification, you are concentrating the side-by-side photon count into a smaller area of your eye. With larger magnification, you are spreading out the same number of photons to a wider area of your eye, ie fewer photons per sq mm.

Then how would the difference between a smaller or larger FOV eyepiece figure into it?

Sam5
2004-Jan-10, 06:12 AM
Nope, all that matters is the number of photons. Doesn't matter where in the measured area they are. And how does your traveling side by side idea explain this single photon experiment (http://copilot.caltech.edu/journal_club/s80h36.pdf) where single photons are emitted and the dector records the intensity? Obviously single photons are not traveling side by side.

I think what you are talking about is what in photography we would call an “exposure time”. We will get an accumulated higher “intensity” of light over time when our exposure times are longer.

A 1/15 of a second exposure time will give us twice the “accumulated intensity” of a single 1/30 second exposure time.

Here’s what I’m talking about as “side by side” photon intensity: You take three spotlights. Shine one spotlight on a white card. That’s intensity #1. Now, turn on the second spotlight and aim it at the same card. That produces a brighter spot on the card, ie twice intensity. Now add the third spotlight, and you have an even higher intensity, with no frequency change. To get twice the intensity of two spotlights, we need 4 spotlights. This is side-by-side photon intensity. For longer exposure times, we will get an accumulated higher intensity. So we can double the intensity with double the number of photons hitting the same photon-sized spot, or by hitting that spot with more in-line photons over a period of a longer time.

You can do the same thing with three short flashes from strobe lights. One strobe allows you to photograph the card at an f 5.6 lens iris setting. Two strobes allows you to use an f8 iris setting. Three strobes allows you to use a setting between f8 and f11. Add a fourth strobe and you can go up to f11. So with two strobes you double the light intensity. With four you double it again.

In all the examples, we are talking about photographing the total intensity of the light, either by doubling the side-by-side photon intensity by adding more light, or by multiplying the side by side intensity by doubling the exposure time.

We are still talking about the side by side intensity.

With one photo at a time, we can photograph one photon, then double the intensity again by photographing two photons on the same spot. This would be what I would call the “accumulated intensity”. I didn’t see any mention of the frequency of the photon in your article.

But obviously, if the two photons are of the “red” wavelength, then we could accumulate twice the intensity of the red light on the film by exposing the film to a series of two photons (back to back) on the same spot on the film. We could also increase the intensity by aiming two separate photons to hit the same spot at the same time.


Care to explain why images in the 15mm eyepiece I borrowed from one of our club members is brighter than the 26 mm I have, when used on the same object in my 6 in refractor? eyepiece figure into it?


You can get detailed information about the relative brightness of images produced by different types of eyepieces of the same mm. Your 26mm is probably a cheaper eyepiece than the 15mm. Also, the 15 mm probably has wider glass. You can do that research on your own. I don’t have time to do it for you:

“Eyepieces come in several different barrel diameters, .965", 1.25", and 2". The smallest size is found mostly on low-end “department store” telescopes and should be avoided, if possible. Most amateur telescopes are designed to accommodate the 1.25" eyepiece size. The large, 2" models are used mostly with higher-end telescopes, and offer increased field of view and brighter images.”

www.actonastro.com/eyepieces.htm+15mm+brighter+eyepiece+telescope&hl= en&ie=UTF-8]LINK (http://216.239.53.104/search?q=cache:KxYbsh6XXWgJ:[url) TO SOURCE [/url]

“The Siebert was sharper and was a good deal brighter than the Meade Ortho and was running head to head with the UO Ortho. In brightness the Siebert out did the Meade Plossl and was a tad bit brighter than the UO Ortho. The Siebert beat the Plossl in contrast and had just a tad bit more contrast than the UO Ortho and the Meade Ortho (the difference was slight but noticeable among the three). I find myself using this Siebert 12.5mm eyepiece a lot for globular clusters in my 9 1/4" SCT also.
The Siebert 15mm eyepiece ( that I got for deep sky observing ) is everything that the 12.5mm is, but with a slight ghosting on Jupiter. It is brighter than the Meade 4000 series 15mm SP and the sky is much blacker also.”

SOURCE (http://216.239.53.104/search?q=cache:QimAgtnM6J8J:members.tripod.com/irwincur/seibert_eyepieces.htm+15mm+brighter+eyepiece+teles cope&hl=en&ie=UTF-8)

Sandor
2004-Jan-10, 02:42 PM
OK, I now understand the problem here. It's really a nice one. Let's be friends. :)

First of all, before you continue to read this, if you are, please notice that I have only one goal: understanding. Science is a little bit different from that, because it's more about 'controlling', which of course needs a lot of understanding on one hand, but there's a limit to that when 'control' is the result already on the other hand. (It's even a budget thing.)
I only USE science (not very much, as you know ;)) to get a clearer understanding, but I also know that science (the present state of it) sometimes fails. There's a difference between the wisdom of Plato and a present scientist controlling his laser bundle. Science is always RIGHT for it to WORK, but from a deeper understanding point of view we can take a look in the past and see that most 'wisdom' was (partly) WRONG after all. So please know that this present time is abslutely not different from any time in the past, as we will find out in the future. (This is also scientific statistics ;))

I am not saying this to attack science or something like that, but like Sam said, there's a certain way of using the word Science with the capital S here and for those not willing to look beyond the current definitions and understanding (scientist themselves do!), there's no use in continuing reading this. Of course you can still say that I will have to adjust to the 'average' on this forum, but I'm sure some of you want to go a little deeper and this topic still started for me to understand more and you can always leave the topic or ignore me.

These might seem a a lot of pretentious words, while the following maybe doesn't even make sense, but I just want to explain this. Why not use a little energy for this, when we already used so much energy in 'blind fighting'. (Everyone becomes more or less blind when concentrating on a certain 'interest', which is no problem, as long as we can get out every now and then to get the wider view.)

---------------------------------------

About quantumfysics. When we take for example a very big pentagram (five star), and we connect all 5 points of it, the outer shape is that of a pentagon (five sided figure). But the most inner shape is also. So we can connect these 5 points again and thus we can imagine an infinite growing number of smaller pentagrams, but also infinite in the direction of the bigger. Now there's a limit to what we can see of that and in this I am focussing on the limit on the 'smaller' side. This theory can be used I guess for an electron to see (react) or not see a 'passing' wave. We can also use it for the explanation of the rings of an atom, between which no electrons SEEM to be. I'll get back to this last example.

Consider an electron with about the smallest mass we can detect, when it falls back to a lower ring. As it sends out particles which have much more speed, and no mass we can detect, we can only guess how much particles that would be. But when we use the formula of relativity, we find out that at the speed of light particles have a mass as close to no mass as we can detect. The more decimals we calculate (detect) for c, the more zero's we can put in the mass value. But this is no plain theory, this is reality! Because what we can detect is exactly our universe and it is exactly what we are. We might as well say that with 'absolute detection ability' there would be no reality at all.

Well, back to a more practical approach, one electron with a mass that is, from our view, infinitely bigger than that of particles with the speed of light, will send out a not known number of these particles when falling back to a lower ring. Let's call such a particle P. We might not be able to detect such a particle, OR it could be a photon.
But, when it turns out, by the single photon experiment that Tensor came up with, that ONE single photon has as well frequency as amplitude (intensity). What does this mean?

I need to use 2 words to explain, but I'm not sure about the correct English. I would say RAY against BEAM, but correct me if I'm wrong. A ray would be serial (sequential) and a beam would be several rays, which is parallel. I will use these terms until someone tells me better. (I used mostly wave instead of ray untill now.)
A ray, in my definition, can only hold frequency OR amplitude, as you hopefully understand, but it's usually called frequency. Several rays will then add up to a certain amplitude.

So ONE photon consists of several rays (beam), which has frequency and amplitude. We then know for a certain time frame what the energy of such a photon is. But this is really meaning that one beam is actually made up of several rays. If you imagine that several rays combine to ONE bigger ray, this is impossible, because a bigger ray can only be bigger by changing the only one aspect of it, which is frequency. Even when this beam is DETECTED as one 'ray' or one photon, it actually consists of several rays.

When speeding up one electron's emission will change the frequency as well as the intensity in a beam, we can not change them independantly by one electron. (We must understand that only CHANGING the intensity means there is intensity at all. If intensity is not a variable thing, it is not existing. In other words, if only frequency determines energy, there is no other thing as intensity.) So in that case we can change frequency only by speeding up one electron's emission, while we can change intensity by including more electrons in the process, but it also changes automatically when changing the frequency. This means that the interpretation of INTENSITY from the 'single photon experiment' must be different. If you can alter the frequency of light produced by one electron by only changing it's emission speed, and this is the only thing that is changed, you can of course not change anything else INDEPENDANTLY. This should have been clear immediately.

It is however still likely that one electron sends out several rays, containing several 'P's' each. This because the smallest particles with relative infinite masslessness (nice word hey?) are so very much smaller than the energy produced by the change in energy level of an electron with a measurable mass. It is only because of 'quantumreality' that the smallest detectable particle is called a PHOTON, which then consists of several P's.

---------------------------------------


And I reiterate: lowering the number of photons, by putting a filter in place, does not change the frequency of the light.

That's right. Such a filter will make the length of the beam only shorter, because the photons will be absorbed by the filter. The frequency as received by the filter is the same, and as soon as the filter is 'gone' (at a certain coordinate), the beam will continue it's way with the same frequency.

About the single photon experiment: "we can realize a device where a single electron and a single hole are injected into the active region to produce a single photon with well-regulated time interval."
"Such a process can generate a regulated singlephoton stream."

So I was right about being able to at least estimate the number of 'wave lengths of a photon' for a certain frequency.

PS, when I started this post I thought that one photon might really have frequency AND amplitude, because of the 'single photon experiment'. But of course I should have realized right away that one cannot alter these independantly by changing only the emission speed of ONE electron. So if amplitude is mentioned or implied in the experiment, it means something else than a variable of the beam. Bit if I overlooked something there, it is still clear that such a photon has to consist of several rays.
I still liked the friendly introduction so much, that I left it that way. ;)

swansont
2004-Jan-10, 02:47 PM
And I reiterate: lowering the number of photons, by putting a filter in place, does not change the frequency of the light. It's a trivial experiment to do (I've done it numerous times, though not to test your hypothesis). Your idea is demonstrably wrong.

All your filter does is lower the side-by-side photon number per sq cm, because the filter absorbs some of the side-by-side photons, so they never reach your eye. A photo “neutral density” (gray) filter, merely lowers intensity, not frequency. I learned this when I was 14 years old, in 1956.

A welder’s neutral density filter on a welder’s mask cuts down on the side-by-side photon number and density per sq cm, by absorbing many of the photons. There is no color change, unless the filter happens to be colored.


You would have a blue light turn red as the brightness is turned down. Doesn't happen. When the facts don't support the hypothesis, you have to abandon the hypothesis.

No, no, no, no. When the light is turned down in a lamp, you are lowering the side-by-side amount of photons, not the frequency. If you put a neutral density filter in front of the light, you do the same thing.


If a filter is lowering the so-called "side-by-side" photons for any length of time, it also has to lower the so-called "photons-in-a-row"

And since that doesn't happen (by your admission), your model is wrong. Light doesn't work that way. There are quantum effects not explainable by wave descriptions.

swansont
2004-Jan-10, 02:53 PM
Sam5, there are two problems with your term for intensity. Firstly, energy of an equal-"intensity" beam would vary with the frequency of the photons. Higher-frequency photons do carry more energy than lower-frequency ones - this has been seen more times than I care to mention. Secondly, measuring energy/area the way you seem to be attempting does not match the way intensity is used in other areas of physics.

Ok, look at this.

RAINBOW PHOTOS (http://australiasevereweather.com/photography/rainbo01.htm)

The yellow is usually more intense to our eyes, although the blue is of a higher frequency and is supposed to be of a “higher energy”. In rainbow color intensity, I would say it would go from more to less in this order: yellow, red, green, blue, violet, as seen by our eyes and by color film. But violet is generally not available in color film.


Again, our eyes are nonlinear. We are much more sensitive to yellow and green than we are to red and blue. So the same intensity of green will appear to be much brighter than a red or blue source. So more intense to our eyes is not the same as higher intensity.

Sandor
2004-Jan-10, 02:55 PM
By the way, several rays in one photon actually means several 'P's', which then means that there are smaller particles within a photon. Which is very logical, because all matter is nothing but 'moving nothing' (energy), so the smallest level for US stops simply where something (matter) turns out to be nothing (nothing at the highest speed, or photons at c). We can then say that our world or reality CAN exist because or our lack of absolute sense. You may interpret this last concept as you like.

swansont
2004-Jan-10, 02:57 PM
If the rods and cones were larger, I would imagine that all that would happen is that our visual acuity would diminish. Few pixels for imaging.

The cones are responsible for pixel count and for resonation amplification of the signal too, and for focusing the light beams, but under your “colored coating” theory, all we would need to do to get a larger pixel count is add more wires (nerve endings) to the backside of the coating, but it doesn’t work that way. What your theory winds up with is one big photo-cell, like an old photo light-meter, sensing only one big round spot of white light. The resonating cones are needed for the pixel count and the focusing of the light beams, and for amplification by means of resonation.

Your theory creates a "photo cell", not an eye retina.

A larger pixel count won't help us, because then you're making a pixel smaller than a wavelength of light. It's not a photocell, by analogy,it's a CCD. And I'm speculating here, because I'm not a biologist.

Sandor
2004-Jan-10, 03:11 PM
We are much more sensitive to yellow and green than we are to red and blue. So the same intensity of green will appear to be much brighter than a red or blue source. So more intense to our eyes is not the same as higher intensity.

I read that there is most yellow light and the least of blue, coming from the sun or other lightsources which emit WHITE light (all frequencies).

Also we know that blue and yellow are opposite colors, as are red and green. I read somewhere that the 3 types of cones are in fact
1. red/green sensitive
2. black/white
3. blue/yellow

Since there is less of blue in white light, we have less blue/yellow cones.
It seems assumable that our brains correct this. So blue seems as bright as green, which is indeed the reality as I sense it.

Then another thing: the red/green cones would be particularly receptive to lower frequencies, as yellow is. Yellow is red and green added up. So cones number one reacting to light and cones number 3 not reacting to light would mean YELLOW LIGHT? This explains that we are most receptive to yellow light, which then 'is' in fact the first cones reacting.

Sam5
2004-Jan-10, 06:15 PM
If a filter is lowering the so-called "side-by-side" photons for any length of time, it also has to lower the so-called "photons-in-a-row"

No, the number-of-photons-in-a-row intensity is controlled by what photographers call the photographic “exposure time”.

To control this in your thought experiment, you would not use a neutral density filter (ie a gray filter), you would merely turn the light on for a short period of time for some “accumulated intensity”, while you would turn the light on for a longer period of time for “more accumulated intensity”.

Just as a photographer would use a 1 second exposure time for some “accumulated intensity”, while he would use a 2 second exposure time for double the “accumulated intensity”.

But when you or a photographer uses a neutral density filter, you are cutting down on the side-by-side number-per-sq-cm photon/wave intensity by having the filter absorb some of the side by side photons/waves so that they never reach your white card or a piece of photographic film.

Of course you could do both at the same time: 1) increase the side-by-side intensity by turning up the brightness of the light source, by removing the neutral density filter, or by opening up the iris of your camera lens, and 2) by increasing the photons-in-a-row accumulated intensity by leaving the light on for a longer period of time or by increasing the photographic exposure time.


And since that doesn't happen (by your admission), your model is wrong. Light doesn't work that way. There are quantum effects not explainable by wave descriptions.

No. Please excuse my lack of complete elucidation. What I meant by “you do the same thing” was in the photographic sense, whether we think of light as a "photon" or a "wave train". That is, you can cut down on the “intensity” of light hitting a photographic plate either by dimming the light source, or by putting a neutral density filter on your lens, or by “stopping down” your lens iris, OR BY cutting down on the exposure time.

The dimming of the light, the use of the neutral density filter, and the stopping down (reducing the size of the) lens iris cuts down on the side-by-side number of photons hitting the plate, while the lower exposure time reduces the number of photons-in-a-row hitting the plate. In photography both different kinds of phenomena produce the same effect at the photographic plate. This is why we have variable irises on lenses and variable shutter speeds on cameras. And of course we can always add a neutral density filter if we need to.

My model is not “wrong”. It was taught to me by Kodak technical books in 1956 and has proven to be correct throughout all the intervening years. This has been universal knowledge among millions of people for hundreds of thousands of years. I think it was first discovered in reference to photography by Daguerre around 1835. There is some historical evidence that suggests Jan Vermeer also knew about in relation to camera obscura lenses as early as the 1660s. It was probably known by Galileo by 1611 after his experiments with telescopes. It was most likely noticed by the ancient Hebrews, and perhaps even by the earliest cavemen. In fact, many animals know about the two types of phenomena. That’s why on a bright hot sunny day, they either sit in the shade (where there is a reduced side-by-side number of photons/waves) or if they come out of their burrow they sit in the sun for a shorter amount of time (thereby reducing the number of photons/waves-in-a-row they receive). So I don’t understand why these concepts are so difficult for you to comprehend.

Sam5
2004-Jan-10, 06:22 PM
A larger pixel count won't help us, because then you're making a pixel smaller than a wavelength of light. It's not a photocell, by analogy,it's a CCD. And I'm speculating here, because I'm not a biologist.

No, a CCD is a "retina".

I haven’t “made” anything regarding the number of rods and cones in the human eye, nor have I reduced or increased their size. I have merely noted that they are about the size of light wavelengths, and this suggests that visible light waves “resonate” them.

Sam5
2004-Jan-10, 06:33 PM
Again, our eyes are nonlinear. We are much more sensitive to yellow and green than we are to red and blue. So the same intensity of green will appear to be much brighter than a red or blue source. So more intense to our eyes is not the same as higher intensity.

Well, the same effect is noticed by panchromatic photographic film and its sensitivity to different colors does tend to be lenear.

www.photo.net/photo/edscott/pss00030.htm+panchromatic+film&hl=en&ie=UTF-8]CLASSICAL (http://216.239.57.104/search?q=cache:WRT8OfmqR_AJ:

www.kodak.com/US/en/motion/students/handbook/sensitometric4.jhtml+panchromatic+film+COLOR+SENSI TIVITY+SCALE&hl=en&ie=UTF-8]SEE (http://216.239.57.104/search?q=cache:4Jsfg8vDVjkJ:[url) FIGURE 16

swansont
2004-Jan-10, 06:36 PM
If a filter is lowering the so-called "side-by-side" photons for any length of time, it also has to lower the so-called "photons-in-a-row"

No, the number-of-photons-in-a-row intensity is controlled by what photographers call the photographic “exposure time”.


OK, so I put a chopper wheel in place, to physically stop some of the "photons in a row" and change the exposure time. Hmmm. No, still wrong. My frequency is unchanged.

As was pointed out already, this works for a single photon. "Photons in a row" isn't frequency.

Sam5
2004-Jan-10, 10:52 PM
OK, so I put a chopper wheel in place, to physically stop some of the "photons in a row" and change the exposure time. Hmmm. No, still wrong. My frequency is unchanged.

As was pointed out already, this works for a single photon. "Photons in a row" isn't frequency.

Unblocked photons in a row is frequency. You are talking about some blocked photons in a row, blocked on a random basis. That doesn't change the frequency. It just leaves some “photon-less gaps” in the arriving photon beam.

In other words, let’s say your eye or camera is receiving just one long narrow stream of back-to-back photons, a beam that is only one photon wide, then you hold up the neutral density filter and move it around a little. So, you are not blocking that entire stream of photons constantly. You are only blocking, on a random basis, some of the incoming photons in that beam. Let’s say we’ve got a long train of 100 photons. Then your moving filter would be blocking photons number 8, 16, 27, 32, 47, etc. on a random basis. This of course does not change the frequency in the normal sense that we think of light “frequency”, since each photon carries with it the same “frequency” of the entire train and the same frequency of every other monochromatic light photon in that long train.

This would be like a kid throwing 100 marbles through your open window, at the rate of one every second, for 100 seconds, and you use a moving board to block some of them from coming through your window. So you wind up with say 50 marbles inside your house, during the course of 100 seconds, and you’ve managed to block, on a random basis, 50% of the marbles from coming through your window, without changing the frequency of the rate at which the kid threw them.

That’s a one-after-another beam of photons.

Ok, for the side-by side blocking of the photons, we have two kids throwing 50 marbles each through your window for 100 seconds, on the average of one kid throwing a marble every 2 seconds for each kid, which is 1 marble every second for both kids (100 total marbles). If you do nothing, you’ll receive 100 marbles. But, you put up a board blocking half of your window, so that you block 50% of the side-by-side marbles. At the end of 100 seconds you will have 50 marbles inside your house.

These are two different ways of reducing the “received intensity” of the photons.

Regarding the “exposure time” point of view, you simpy do nothing until 50 marbles have been thrown through your window, then you close the window shutters, completely blocking out the final 50.

swansont
2004-Jan-10, 11:06 PM
Let’s say we’ve got a long train of 100 photons. Then your moving filter would be blocking photons number 8, 16, 27, 32, 47, etc. on a random basis. This of course does not change the frequency in the normal sense that we think of light “frequency”, since each photon carries with it the same “frequency” of the entire train and the same frequency of every other monochromatic light photon in that long train.


emphasis added


Right. Each photon carries with it the frequency information. It is independent of the number of photons, or the rate you are sending photons.

Saying that frequency is the number of photons in a row is not the same thing, and is wrong.

Sam5
2004-Jan-11, 12:24 AM
Let’s say we’ve got a long train of 100 photons. Then your moving filter would be blocking photons number 8, 16, 27, 32, 47, etc. on a random basis. This of course does not change the frequency in the normal sense that we think of light “frequency”, since each photon carries with it the same “frequency” of the entire train and the same frequency of every other monochromatic light photon in that long train.


emphasis added


Right. Each photon carries with it the frequency information. It is independent of the number of photons, or the rate you are sending photons.

Saying that frequency is the number of photons in a row is not the same thing, and is wrong.


No it’s not wrong. That’s the standard way of talking about the “frequency” of light, radio waves, and sound waves, also about the cars of a passing train, the marbles thrown at your window, geese flying South for the winter, etc. If you go blocking some of them with the blades of an oscillating fan, then you have to explain to everyone what you are doing. That might block some of them, and that might block their arrival frequency, but it doesn't change their original or emitted frequency.

swansont
2004-Jan-11, 02:05 AM
Let’s say we’ve got a long train of 100 photons. Then your moving filter would be blocking photons number 8, 16, 27, 32, 47, etc. on a random basis. This of course does not change the frequency in the normal sense that we think of light “frequency”, since each photon carries with it the same “frequency” of the entire train and the same frequency of every other monochromatic light photon in that long train.


emphasis added


Right. Each photon carries with it the frequency information. It is independent of the number of photons, or the rate you are sending photons.

Saying that frequency is the number of photons in a row is not the same thing, and is wrong.


No it’s not wrong. That’s the standard way of talking about the “frequency” of light, radio waves, and sound waves, also about the cars of a passing train, the marbles thrown at your window, geese flying South for the winter, etc. If you go blocking some of them with the blades of an oscillating fan, then you have to explain to everyone what you are doing. That might block some of them, and that might block their arrival frequency, but it doesn't change their original or emitted frequency.

No, it's most definitely not standard. The frequency associated with a photon is completely independent of the number of them, or how quickly they are arriving. E=hf. The frequency depends on the energy, and vice-versa. That's all there is to it.

Sam5
2004-Jan-11, 02:46 AM
Let’s say we’ve got a long train of 100 photons. Then your moving filter would be blocking photons number 8, 16, 27, 32, 47, etc. on a random basis. This of course does not change the frequency in the normal sense that we think of light “frequency”, since each photon carries with it the same “frequency” of the entire train and the same frequency of every other monochromatic light photon in that long train.


emphasis added


Right. Each photon carries with it the frequency information. It is independent of the number of photons, or the rate you are sending photons.

Saying that frequency is the number of photons in a row is not the same thing, and is wrong.


No it’s not wrong. That’s the standard way of talking about the “frequency” of light, radio waves, and sound waves, also about the cars of a passing train, the marbles thrown at your window, geese flying South for the winter, etc. If you go blocking some of them with the blades of an oscillating fan, then you have to explain to everyone what you are doing. That might block some of them, and that might block their arrival frequency, but it doesn't change their original or emitted frequency.

No, it's most definitely not standard. The frequency associated with a photon is completely independent of the number of them, or how quickly they are arriving. E=hf. The frequency depends on the energy, and vice-versa. That's all there is to it.


Well, that seems like sort of a backwards way of thinking to me. If photons are emitted from an atom at a certain frequency, at a certain wavelength, and at a certain speed, then it would seem to me that that should determine their “energy” level upon emission. If they are absorbed at a certain frequency, at a certain wavelength, and at a certain speed, then it seems that that should determine their “energy” level upon absorption, but you are certainly free to define their energy level any way you want to.

How do you classify the “energy level” of blueshifted and redshifted photons arriving on earth from a distant star that is stationary relative to the sun, when the earth is moving toward that star and six months later its moving away from that star? Do you say their “energy level” changes due to our “relative motion” toward and away from the star, or what? How do you account for the frequency change without the wavelength change? And if you do say the wavelength changes, about where in space does it change, near the earth or near the star?

Kaptain K
2004-Jan-11, 03:11 AM
... you are certainly free to define their energy level any way you want to.
swansont, SeanF and the rest of us are using the standard, accepted definitions. It is you who feels free to define things as you please. Then you tell us we (and by extension, all of physics) are wrong because we do not use your ad hoc definitions! :-?

Sam5
2004-Jan-11, 03:26 AM
... you are certainly free to define their energy level any way you want to.
swansont, SeanF and the rest of us are using the standard, accepted definitions. It is you who feels free to define things as you please. Then you tell us we (and by extension, all of physics) are wrong because we do not use your ad hoc definitions! :-?

Sorry, but I never noticed you or Sean define the energy level of a photon, so I don’t know what you are getting at, other than trolling me on this thread. Is Sean going to come over here and troll me too?

I’m expressing my classical opinion about photons from a photographer’s point of view. Mine is the very point of view with which astrophotography began. Exposure time, mirror/lens width, the “light gathering ability” of a large lens or mirror, focal length, film sensitivity, etc. In fact, when I visited the Mt. Wilson observatory back in the 1970s, they were still using a 50 year old hand-made wooden camera that was original to the original telescope. They were still using old Edison light bulbs as resistors to control the speed of one of their old telescope motors. If you don’t understand what I’m talking about, then that’s not my problem, so stop trolling me.

Tensor
2004-Jan-11, 03:32 AM
[quote=swansont]

No, it's most definitely not standard. The frequency associated with a photon is completely independent of the number of them, or how quickly they are arriving. E=hf. The frequency depends on the energy, and vice-versa. That's all there is to it.


Well, that seems like sort of a backwards way of thinking to me.

It may seem backward to you, but that's what theory says.


How do you classify the “energy level” of blueshifted and redshifted photons arriving on earth from a distant star that is stationary relative to the sun, when the earth is moving toward that star and six months later its moving away from that star? Do you say their “energy level” changes due to our “relative motion” toward and away from the star?

Yep, exactly.


How do you account for the frequency change without the wavelength change? And if you do say the wavelength changes, about where in space does it change, near the earth or near the star?

In your case above, change is at the earth, due to it motion. It's a straight dopler effect.

Sam5
2004-Jan-11, 03:35 AM
How do you account for the frequency change without the wavelength change? And if you do say the wavelength changes, about where in space does it change, near the earth or near the star?

In your case above, change is at the earth, due to it motion. It's a straight dopler effect.

So then we are moving toward the oncoming light beam at c + v and away from it at c – v? This is what Doppler predicted in 1842.

Tensor
2004-Jan-11, 03:43 AM
So then we are moving toward the oncoming light beam at c + v and away from it at c – v? This is what Doppler predicted in 1842.

Not quite, Sam see this (http://www.mathpages.com/rr/s2-04/2-04.htm)

Sam5
2004-Jan-11, 03:51 AM
So then we are moving toward the oncoming light beam at c + v and away from it at c – v? This is what Doppler predicted in 1842.

Not quite, Sam see this (http://www.mathpages.com/rr/s2-04/2-04.htm)

LOL, although you said “a straight dopler effect”, I didn’t think you really knew what “ a straight Doppler Effect” really is or what he predicted in 1842 or why he predicted it, so I think I should stick around.

Sam5
2004-Jan-11, 03:57 AM
Not quite, Sam see this (http://www.mathpages.com/rr/s2-04/2-04.htm)

Do you happen to know the Φ factor in the 1938 Ives and Stilwell experiment with the hydrogen atoms, which was mentioned in that link you provided?

Sam5
2004-Jan-11, 03:59 AM
Tensor,

I think I have a copy of their 1938 paper around here someplace, I’ll try to look it up tomorrow. You wouldn’t happen to have your copy handy, would you?

Tensor
2004-Jan-11, 04:06 AM
So then we are moving toward the oncoming light beam at c + v and away from it at c – v? This is what Doppler predicted in 1842.

Not quite, Sam see this (http://www.mathpages.com/rr/s2-04/2-04.htm)

LOL, although you said “a straight dopler effect”, I didn’t think you really knew what “ a straight Doppler Effect” really is or what he predicted in 1842 or why he predicted it, so I think I should stick around.

Well, you thought wrong. I don't think you understood the reference I gave you. Or are you claiming the case where they say it will reduce to the classical doppler equation isn't in there?

Kaptain K
2004-Jan-11, 04:10 AM
... you are certainly free to define their energy level any way you want to.
swansont, SeanF and the rest of us are using the standard, accepted definitions. It is you who feels free to define things as you please. Then you tell us we (and by extension, all of physics) are wrong because we do not use your ad hoc definitions! :-?

Sorry, but I never noticed you or Sean define the energy level of a photon, so I don’t know what you are getting at, other than trolling me on this thread. Is Sean going to come over here and troll me too?

I’m expressing my classical opinion about photons from a photographer’s point of view. Mine is the very point of view with which astrophotography began. Exposure time, mirror/lens width, the “light gathering ability” of a large lens or mirror, focal length, film sensitivity, etc. In fact, when I visited the Mt. Wilson observatory back in the 1970s, they were still using a 50 year old hand-made wooden camera that was original to the original telescope. They were still using old Edison light bulbs as resistors to control the speed of one of their old telescope motors. If you don’t understand what I’m talking about, then that’s not my problem, so stop trolling me.
1) I am not trolling you (and neither is anybody else).
2) I do understand what you are talking about and so do the others. We are trying to explain why you and your ad hoc definitions are wrong.
3) The camera at Mt. Wilson is irrelevent to the point of discussion. When I was in college, we used a solar camera that consisted of a wood frame and a wood slide holding a piece of taut aluminum foil with a razor slit. It was powered by 50 stretched rubber bands. We used it on a 12 in. f/15 Brashears refractor that was originally built for the St. Loius worlds fair. When we released the slide, the telescope swung about a meter and rubber bands flew all over the observatory. The exposure time was estimated to be in the 1/100,000th second range. This is also irrelevent to the thread, but no more so than some of the strawman arguments you throw into discussions.

Celestial Mechanic
2004-Jan-11, 04:47 AM
The energy of a photon or massive particle is the time-component of a 4-vector and will differ from frame to frame. One thing that is invariant that all observers will calculate is the mass-squared, E^2 - (p*c)^2 = (m*c^2)^2, which for a photon is zero.

Sam5
2004-Jan-11, 11:06 AM
2) I do understand what you are talking about and so do the others. We are trying to explain why you and your ad hoc definitions are wrong.

No you are not. You are just trying to dominate the conversation by insisting that I use only your non classical terms to explain the same phenomena, and you become insulting when I don’t. I don’t know what your problem is.


3) The camera at Mt. Wilson is irrelevent to the point of discussion. When I was in college, we used a solar camera that consisted of a wood frame and a wood slide holding a piece of taut aluminum foil with a razor slit. It was powered by 50 stretched rubber bands. We used it on a 12 in. f/15 Brashears refractor that was originally built for the St. Loius worlds fair. When we released the slide, the telescope swung about a meter and rubber bands flew all over the observatory. The exposure time was estimated to be in the 1/100,000th second range. This is also irrelevent to the thread, but no more so than some of the strawman arguments you throw into discussions.

I don’t get it. What do you mean that my discussing of an old observatory camera is “irrelevant”, but then it’s ok for you to post a long discussion about an old observatory camera?? Are you trying to say that you want to be the only one who is allowed to talk about old observatory cameras? Please explain.

Sam5
2004-Jan-11, 11:22 AM
Well, you thought wrong. I don't think you understood the reference I gave you. Or are you claiming the case where they say it will reduce to the classical doppler equation isn't in there?

Well, when the earth moves toward the oncoming light from a star that is fixed with the sun, the earth and the light are moving toward each other at c + v. This is why we see the blueshift. Einstein’s own books and papers contain this information. Do you not understand this?

When light from a distant radially moving galaxy is headed toward the earth, the light is moving at c – v relative to the earth while it is still traveling within the galaxy. Do you not understand this?

Do you know what the acceleration factor was in the 1938 Ives and Stilwell experiment? Do you know why that information is important?

Sam5
2004-Jan-11, 12:00 PM
Tensor,

Tell me something. Why is this called “time dilation”?

LINK (http://216.239.57.104/search?q=cache:Vgl5Ks9SJmQJ:www.mpi-hd.mpg.de/ato/rel/+ives+stilwell&hl=en&ie=UTF-8)

And why is this attributed to just “relative motion”? The Lorentz theory was about tick rate slowdowns due to the motion of atoms through fields, and the GR theory was about tick rate slowdowns due to atomic motion through gravitational fields and because of acceleration. So why is this tick rate slowdown attributed only to “relative motion”? What “force” is placed on the hydrogen atoms that could cause them to emit light of a lower frequency? What did Einstein stay about this in 1911?

And why is it called “time dilation”, rather than just a clock tick rate slowdown? When other types of clocks slow down, why is that not called “time dilation”? What makes you and others think that the Doppler-shifted frequencies of an emission line in hydrogen canal rays was due only to “relative motion”, since the canal rays had no way of knowing that the laboratory clock was moving “relative” to them? What did Einstein attribute the shift to in 1911?

Kaptain K
2004-Jan-11, 12:25 PM
I don’t get it. What do you mean that my discussing of an old observatory camera is “irrelevant”, but then it’s ok for you to post a long discussion about an old observatory camera?? Are you trying to say that you want to be the only one who is allowed to talk about old observatory cameras? Please explain.
Here it is folks. Proof that Sam5 does not actually read posts, but picks and chooses what he wishes to reply to. My response was:

3) The camera at Mt. Wilson is irrelevent to the point of discussion. When I was in college, we used a solar camera that consisted of a wood frame and a wood slide holding a piece of taut aluminum foil with a razor slit. It was powered by 50 stretched rubber bands. We used it on a 12 in. f/15 Brashears refractor that was originally built for the St. Loius worlds fair. When we released the slide, the telescope swung about a meter and rubber bands flew all over the observatory. The exposure time was estimated to be in the 1/100,000th second range. This is also irrelevent to the thread, but no more so than some of the strawman arguments you throw into discussions.
Emphasis added. As I said, my post was just as irrelevent as his.

Maybe I should add a point to my post above:

4) Never wrestle with a pig - you just get dirty and the pig enjoys it.

I'm outta here! I'm tired of ](*,)

Sam5
2004-Jan-11, 12:45 PM
Emphasis added. As I said, my post was just as irrelevent as his.


This is an astronomy board, so I hardly think that my or your stories about old observatory cameras is “irrelevant” to a discussion “about photons and light and things”.


4) Never wrestle with a pig - you just get dirty and the pig enjoys it.

What a rude guy you are. I would never call anyone on this board or any other board a “pig”.

Sam5
2004-Jan-11, 12:57 PM
So then we are moving toward the oncoming light beam at c + v and away from it at c – v? This is what Doppler predicted in 1842.

Not quite, Sam see this (http://www.mathpages.com/rr/s2-04/2-04.htm)

Ok, I found my copy of the 1938 Ives-Stilwell paper that you referenced to. They attributed the Doppler shift to a "Larmor-Lorentz" effect, not an Einstein effect. So the link you provided is inaccurate and does not agree with the actual Ives-Stilwell paper.

Here also is something Ives wrote in 1940 that was published in Science magazine:

“Thus if we measure the Doppler shift of a line in the spectrum of a star in the east in the early evening, and again in the west in the early morning, the change in observed wavelength gives the change in the velocity of light relative to the earth’s surface, due to the earth’s axial velocity. This is the only interpretation, unless we are so hopelessly geocentric that we believe the whole stellar universe oscillates with a 24-hour period in order to maintain the velocity constant with respect to the earth!”

This is what I’ve explained here several times.

Sandor
2004-Jan-11, 01:54 PM
Sam, the width of a cone is about a wave length, but the length is (much) bigger. Resonation is always waving along the length, isn't it? And this is at best when close to a wave or half wave. It takes more energy when the resonating medium is smaller or bigger than that. So why would our body make the cones so big if they had to resonate?

Could it be that the cones have to fibrate, but just not resonate? A cone shape is perfect for fibration in the way that it will keep fibrating for a while, but it's just not good for resonating, because it's too soft, too flexible, whereas resonating media are typically rigid.

Sandor
2004-Jan-11, 01:56 PM
About this one photon experiment: can anybody here explain how ONE electron can emit light, on which you can alter the frequency as well as the amplitude?

swansont
2004-Jan-11, 02:28 PM
The frequency associated with a photon is completely independent of the number of them, or how quickly they are arriving. E=hf. The frequency depends on the energy, and vice-versa. That's all there is to it.


Well, that seems like sort of a backwards way of thinking to me. If photons are emitted from an atom at a certain frequency, at a certain wavelength, and at a certain speed, then it would seem to me that that should determine their “energy” level upon emission. If they are absorbed at a certain frequency, at a certain wavelength, and at a certain speed, then it seems that that should determine their “energy” level upon absorption, but you are certainly free to define their energy level any way you want to.


How is it backwards, when you say exactly the same thing - the frequency determines the energy, and vice-versa, at emission. Not the number of photons emitted. For a single photon.

swansont
2004-Jan-11, 02:32 PM
Sorry, but I never noticed you or Sean define the energy level of a photon, so I don’t know what you are getting at, other than trolling me on this thread. Is Sean going to come over here and troll me too?

I’m expressing my classical opinion about photons from a photographer’s point of view. Mine is the very point of view with which astrophotography began.

E=hf has been mentioned numerous times in this thread. No need for each poster to state it.

"Classical opinion about photons" is an oxymoron. Photons are not a classical phenomenon. They are a quantum phenomenon.

swansont
2004-Jan-11, 02:34 PM
And why is it called “time dilation”, rather than just a clock tick rate slowdown? When other types of clocks slow down, why is that not called “time dilation”?

It's called time dilation, in part, so that we can differentiate the effects - it's only due to relative motion, and not due to any other phenomenon that might affect the clocks.

swansont
2004-Jan-11, 02:45 PM
About this one photon experiment: can anybody here explain how ONE electron can emit light, on which you can alter the frequency as well as the amplitude?

Are you asking about an electron transition? The frequency is determined by the energy difference of the two levels in the atom. You can't adjust the amplitude - it's one photon. If you want more photons, you need to have more atoms undergoing the transition. If you want a different frequency, you have to use a different energy level(s).

Sandor
2004-Jan-11, 02:52 PM
You can't adjust the amplitude - it's one photon. If you want more photons, you need to have more atoms undergoing the transition. If you want a different frequency, you have to use a different energy level(s).

Exactly Swansont, that's exactly what I mean. One photon can be seen as a part of a single (serial) ray, where only the frequency determines the color and therefore the energy.
Only when more rays contribute to a (paralllel) beam, we also have amplitude, which also contributes to the total energy, by intensity. So you agree to that?

Sam5
2004-Jan-11, 03:59 PM
Could it be that the cones have to fibrate, but just not resonate?

What is "fibrate"?

Sam5
2004-Jan-11, 04:02 PM
"Classical opinion about photons" is an oxymoron. Photons are not a classical phenomenon. They are a quantum phenomenon.

Well, Doppler predicted redshifts and blueshifts in starlight, and also he is why we have Doppler radar today. That is why we have "Doppler shifts". Photons are classical things.

Newton described photons as “particles”. He was the first to predict that light rays would bend when they pass near the sun. He was the first to say that matter could be turned into light energy and light energy turned into matter. This is all old-time classical stuff.

Sam5
2004-Jan-11, 04:10 PM
And why is it called “time dilation”, rather than just a clock tick rate slowdown? When other types of clocks slow down, why is that not called “time dilation”?

It's called time dilation, in part, so that we can differentiate the effects - it's only due to relative motion, and not due to any other phenomenon that might affect the clocks.

It’s not due to “relative motion”. It’s due to the Lorentz effect and acceleration. A clock can not slow down due only to “reltive motion” since it doesn’t know that anything is moving relative to it. It feels no force due only to relative motion. With the Lorentz effect and acceleration, the clock feels a physical force.

swansont
2004-Jan-11, 05:14 PM
And why is it called “time dilation”, rather than just a clock tick rate slowdown? When other types of clocks slow down, why is that not called “time dilation”?

It's called time dilation, in part, so that we can differentiate the effects - it's only due to relative motion, and not due to any other phenomenon that might affect the clocks.

It’s not due to “relative motion”. It’s due to the Lorentz effect and acceleration. A clock can not slow down due only to “reltive motion” since it doesn’t know that anything is moving relative to it. It feels no force due only to relative motion. With the Lorentz effect and acceleration, the clock feels a physical force.

Well, clocks do run slow due to relative motion, and it's been measured. No acceleration needed - effects due to that can also be measured, and they have been.

I see no point in discussing further. ](*,)

Sam5
2004-Jan-11, 05:20 PM
Well, clocks do run slow due to relative motion, and it's been measured. No acceleration needed - effects due to that can also be measured, and they have been.

What physical “force” is placed on the clocks that are moving relatively that causes their rates to change? A spooky force at a distance? If you don’t want to use the term “force”, then what did cause the clocks to change rate? What physical reason caused them to change rates?

Where were these “relative motion” tests conducted? At the surface of the earth, inside a gravitational field?

If you conduct electrodynamical experiments at the surface of a large massive geoid, then you are going to get geocentrical results. You are making the same fundamental mistake Ptolemy made 1852 years ago.

Sandor
2004-Jan-11, 06:18 PM
Sam, I mean vibrate instead of fibrate.

Swansont, Sam is right. Just motion isn't doing anything. It is one of the first learned principles in relativity that a train passenger is only undergoing these special forces when accelerating (up and down). Just travelling with constant speed isn't doing anything to a clock in the train.

Sam5
2004-Jan-11, 06:53 PM
Sam, I mean vibrate instead of fibrate.

Hi,

Well, “vibration” is sort of a weaker form of “resonation”. Perfect “resonation” tends to “amplify” a signal and/or makes the maximum use of wave-energy transferal.


Swansont, Sam is right. Just motion isn't doing anything. It is one of the first learned principles in relativity that a train passenger is only undergoing these special forces when accelerating (up and down). Just travelling with constant speed isn't doing anything to a clock in the train.


These guys are taught in American schools that “relative motion” causes “time dilation” and they are taught that they are “crackpots” if they try to disagree with the teacher about that. They are indoctrinated with this incorrect concept. They are taught to always agree with the teacher and not think for themselves or to question that “law” of “science” about “relative motion”, which is actually a mistake Einstein made in 1905 and he corrected it in 1918, but these guys don’t know about his correction.

Are you taught much about Lorentz theory in schools in the Netherlands?

Sandor
2004-Jan-11, 07:16 PM
OK Sam, I will ask my question again using the right word VIBRATION.


Sam, the width of a cone is about a wave length, but the length is (much) bigger. Resonation is always waving along the length, isn't it? And this is at best when close to a wave or half wave. It takes more energy when the resonating medium is smaller or bigger than that. So why would our body make the cones so big if they had to resonate?

Could it be that the cones have to vibrate, but just not resonate? A cone shape is perfect for vibration in the way that it will keep vibrating for a while, but it's just not good for resonating, because it's too soft, too flexible, whereas resonating media are typically rigid.

Sandor
2004-Jan-11, 07:27 PM
Are you taught much about Lorentz theory in schools in the Netherlands?

Usually only at the technical universities, but some physics teachers at highschools might teach some principles of it, or it might even be in some school programs as an introduction to university subjects.

I have my own thoughts about it, my own way of understanding and I was about to react to the clock-stuff discussed here. Now please brace yourselves, here I come. :)


Crystal oscillators have been made by combining materials that have both positive and negative temperature coefficients, and run near the crossover point, so you get something that is temperature independent (to first order), so one crystal, by itself would speed up with an increase in T while the other would slow down. Saying that any oscillator will slow down if heated is just plain wrong.

You can see relativity theory completely apart from crystals or whichever clock mechanism. Because TIME is relatively changing, no matter what mechanism you use. All energy you put in, will simply slow down time. Although it's not always so simple to see the 'relations'. Sam already said that it's not the velocity, but the acceleration. In fact this is the most difficult part, because the difference between those is about the relations. An object moving through the cosmos with constant speed, doesn't have any absolute speed at all, there's only speed compared to something else, but this only has effect/can be measured/is really different when forces are exchanged. These forces are by definition acceleration, ALL forces are acceleration. A hot clock in a cold environment is going slower in relation to the colder environment. If the clock is permanently hotter, it is not 'once heated', but constantly heated, otherwise it couldn't stay hot. This constant heating is constant acceleration, because it will constantly lose heat also. Difference in velocity is like potential difference in time, which will come 'true' when velocities meet and 'tune' together (in acceleration) to the same reality.

Let me explain clearer by using the example of a big planet with more gravity and thus more acceleration. From our perspective, space over there is smaller (compressed) and/or time is bigger (slower). When we exchange power, the space-time frame will be the same at the 'exchange point'. But since we have a solid 'space grid', we see that time in the other frame is different. A clock that was sent from earth to this big planet and comes back again, will be more behind than one that was sent to a smaller planet, but not only because of the bigger forces during traveling. Although time isn't absolutely going slower on the big planet, it seems that way for us, because of the different spacetime frame. When we insist on saying time was (is) really going slower (using our reality), we also have to state that the clock was smaller, because space over there is smaller (using our reality). Let's use a pendulum clock as an example. When we literally watch this clock on a big planet, from our planet, we see 2 things: the pendulum is going faster (more force) than the same pendulum here, but the time on it is going slower. This is logical, because a bigger swing of the pendulum means slower ticking. But what we don't see is the 3rd thing changing, which is the space, including the clock, being smaller than the same clock here. Well, if we measure it using the known distance, we know it, but we do not 'sense' it, because it's too far away. (Things speeding by with high speed are also shrinking from our perspective, but we cannot sense it either, because it goes too fast.) Even if we could sense the object being smaller, we just cannot sense SPACE (around) being smaller, because we have this solid space grid. But if we reason it out, we decide that the pendulum only seems to go faster, because space is smaller. And smaller space mean bigger time. I will explain now differently.

Places with much force, will suck up space, or make it smaller, more compressed. I prefer to call it just 'smaller'. We do not sense this, because space is our solid grid from which to understand the world. But there is difference between one space and the other and this is time. It's just that. When another place (big planet) decreases space, it will increase time, or time will be bigger. Space/time = speed. It's easier to talk about difference in speed or energy. When 10 Joules here is 10 spaces high, it will be 2 spaces high on the big planet. The difference, 8 spaces, will show up as extra (bigger) time. If we put our 10 spaces at the left and the 2 spaces of bigplanet at the right, and we connect bottoms and tops, our 10 spaces become our time while their 2 spaces remain their space. Here's a picture of it:

http://members.home.nl/sandeman/timespace.gif

Their blue space is not only relating to our blue space, but we can add the red 'space' also. Crossing their blue space with a certain speed is reflected on our 'scale' as a much bigger speed. But we don't see space differently, so the left scale becomes a time scale on which the space is fastly going by.
When we look at their time on the right, this is much bigger than ours. This is exactly how we should see it. Time cannot go 'fast'. Only speed can be fast and speed = space/time. Bigger space and or smaller time is more speed. But we compare only their speed to ours. We see space the same everywhere and time is just the difference which is not visible at all. So when we look through one space frame of ours to their world, theirs is holding more matter (of in fact total energy), because our red line holds the same space as their blue line, although we consider their red line the same as ours (equal space).
So crossing their bigger time (red) at the right with a certain speed, is reflected on our timeline as much slower.

If we want to see this from the perspective of Bigplanet, we just have to change the colors.
Space and time together are always equally big everywhere. This is the global spacetime frame. When space gets bigger, time gets smaller and the opposite. It only means relative motion or relative difference in speed, which will get 'real' in acceleration when the different speeds meet (in force).

What we feel is not time, but acceleration (force). The total force 'on' a planet is called time. More force in a clock will slow down time, but only because the space of this clock is shrinking compared to any other spaceframe which is then not shrinking. When our clock comes back from Bigplanet we notice the bigger speed, but at the crossover point, spaces are tuned, resulting in slower time. For those who really want to imagine how every different clock mechanism is in fact SHOWING slower time with more force in it; it's interesting to find that this is a logical reaction. But we also have to know that TIME is not what we usually feel: force or change. Because on a bigger planet as well as on a speeding rocket, much more actually happens compared to here. So stating that a twin brother ages less on his accelerations, is really only meaning that time was indeed bigger during his acceleration, but this has to be compared to the smaller space, all resulting in more forces compared to our world. It has principally nothing to do with aging, although it's not hard to imagine that one will probably sooner die with more forces working, than with less.

Note: when using only relative space differences to describe accelerations (we can even do this at any level), the movement is just a 'direction', nothing more. If you don't believe space is really getting smaller on a very small level, i.e. when you lift up a cup of coffee, just think of a truck passing an atom clock, this slowing it, or a little heat getting in the clock.

I can still think of a paradox here: people on the moon were also going slower from their own perspective/feeling. They WERE bigger however. That might seem crazy, but you can just visit the 'mystery spot' in Santa Cruz and even get bigger yourself. Anyway, the fact that they also felt they were going slower is because of the 'own' (cumulated earthly) time of their bodies. These bodies are not only the felt memory of greater gravity, but also the 'showing' memory of that, as 'presenting the reality of their cumulated past'. If they would stay a little longer on the moon, their bodies would also start to grow bigger (in mass), but fall apart eventually. These bodies were born and raised in earthly spacetime.

Sam5
2004-Jan-11, 07:40 PM
OK Sam, I will ask my question again using the right word VIBRATION.


Sam, the width of a cone is about a wave length, but the length is (much) bigger. Resonation is always waving along the length, isn't it? And this is at best when close to a wave or half wave. It takes more energy when the resonating medium is smaller or bigger than that. So why would our body make the cones so big if they had to resonate?

Could it be that the cones have to vibrate, but just not resonate? A cone shape is perfect for vibration in the way that it will keep vibrating for a while, but it's just not good for resonating, because it's too soft, too flexible, whereas resonating media are typically rigid.


I’ve got a CB radio antenna that is a circle. The FCC and the FBI uses circular antennas to track down clandestine transmitters. A radio transmitter can be horizontal or vertical, as long as the antenna matches the orientation of the transmitter antenna. The “length” of the resonating part of the cone might be horizontal rather than vertical. It might be a horizontal circle around the outside of the cone. This could explain why they are “cones” rather than “rods”, with different-diameter circles on one single cone that resonate at different frequencies.

Different parts and areas of the cone might serve various functions. Only a part of the cone might resonate. It might be some stuff inside or outside the cone that resonates, with the cone just being a “carrier” or a “holder” of that stuff so it will be circle shaped. The hard stuff might be the molecules in the coating on the cones.

Sandor
2004-Jan-11, 07:51 PM
I’ve got a CB radio antenna that is a circle. The FCC and the FBI uses circular antennas to track down clandestine transmitters. A radio transmitter can be horizontal or vertical, as long as the antenna matches the orientation of the transmitter antenna. The “length” of the resonating part of the cone might be horizontal rather than vertical. It might be a horizontal circle around the outside of the cone. This could explain why they are “cones” rather than “rods”, with different-diameter circles on one single cone that resonate at different frequencies.

Different parts and areas of the cone might serve various functions. Only a part of the cone might resonate. It might be some stuff inside or outside the cone that resonates, with the cone just being a “carrier” or a “holder” of that stuff so it will be circle shaped. The hard stuff might be the molecules in the coating on the cones.

But even if a circle shaped medium is resonating, it is still waving along the length of it, isn't it?
So I can imagine circles in the cone, as slices, but if the inside of the cone is not hollow, and also these slices are sandwiched by the slices above and underneath, it seems quite hard for them to resonate.

Sam5
2004-Jan-11, 08:33 PM
I’ve got a CB radio antenna that is a circle. The FCC and the FBI uses circular antennas to track down clandestine transmitters. A radio transmitter can be horizontal or vertical, as long as the antenna matches the orientation of the transmitter antenna. The “length” of the resonating part of the cone might be horizontal rather than vertical. It might be a horizontal circle around the outside of the cone. This could explain why they are “cones” rather than “rods”, with different-diameter circles on one single cone that resonate at different frequencies.

Different parts and areas of the cone might serve various functions. Only a part of the cone might resonate. It might be some stuff inside or outside the cone that resonates, with the cone just being a “carrier” or a “holder” of that stuff so it will be circle shaped. The hard stuff might be the molecules in the coating on the cones.

But even if a circle shaped medium is resonating, it is still waving along the length of it, isn't it?
So I can imagine circles in the cone, as slices, but if the inside of the cone is not hollow, and also these slices are sandwiched by the slices above and underneath, it seems quite hard for them to resonate.



A circle antenna is on the end of a stick. You hold up the stick and aim the whole circle at the transmitter. To zero in on where the transmitter is located, you slowly turn the circle antenna until you get a “null” with the edge of the antenna pointing at the transmitter, then you turn it again and the loudest signal comes when you point the full circle at the transmitter.

That kind of antenna will take a signal from the front or the back, but of course in the eye, a circle antenna would take it only from the front, so the circle would be around the cone, and the back of the cone is where the wiring to the brain is.

When you stand up, your cones point out toward the front of your eyes. A resonating circle around the cones would be a circle around the cones so that would be like a round antenna aiming at the transmitter, out the front of your eyes, with the full circle aiming out the front of your eyes.

If you are looking at me, one circle around one of your cones would be aimed at me like this, with that cone pointing toward me:

O

The shorter-wave and higher frequency circles would be in the forward part of the cone, so looking directly toward the cone, an illustration would show a circle within a circle within a circle for three different frequencies.

Not “slices” but circles around the outside of the cone, as if you wrapped a fine wire around the outside of a cone.

Sam5
2004-Jan-11, 08:54 PM
But even....

HA HA! Seems that microwave antennas are cones:

www.flann.com/Products_Home/Antennas/Lens_Horn/lens_horn.html+microwave+cone+antenna&hl=en&ie=UTF-8]CONE (http://216.239.57.104/search?q=cache:kVl6jb4gUtIJ:

www.seaveyantenna.com/catalog/p36/pg37.htm+cone+antenna&hl=en&ie=UTF-8]CONE (http://216.239.57.104/search?q=cache:6J3I_2_vhsMJ:[url) ANTENNA

Sandor
2004-Jan-11, 09:05 PM
They're cone shaped, but backwards and also for a different reason.

I understand the o-thing, but these rings 'around' the cones, they vibrate backward and forward don't they? Still standing up I mean. So the short wave will make a small ring resonate AGAINST the bigger ring behind it. So it will be slowed down in resonation by the bigger ring. But it will also be slowed down by the inner tissue to which it is attached. Do you understand what I mean?

Sam5
2004-Jan-11, 09:47 PM
They're cone shaped, but backwards and also for a different reason.

I understand the o-thing, but these rings 'around' the cones, they vibrate backward and forward don't they? Still standing up I mean. So the short wave will make a small ring resonate AGAINST the bigger ring behind it. So it will be slowed down in resonation by the bigger ring. But it will also be slowed down by the inner tissue to which it is attached. Do you understand what I mean?


4 PAGE PDF FILE, SEE DRAWING ON PAGE 3 (http://www.pv.unsw.edu.au/Osaka_Conf/Corkish.pdf)

“The upper frequency limit is determined by the
feed point at the vertex, where the spacing should be
smaller than quarter of the upper frequency. The cone
outer radius determines the lower frequency limit. A 10:1
bandwidth has been demonstrated.”

Tensor
2004-Jan-12, 01:12 AM
But even....

HA HA! Seems that microwave antennas are cones:....

So you're saying the cones in our eyes pick up microwaves?

Sam5
2004-Jan-12, 01:42 AM
But even....

HA HA! Seems that microwave antennas are cones:....

So you're saying the cones in our eyes pick up microwaves?

No. Microwaves are short radio waves. Light waves are shorter, thus our cones are smaller.

Tensor
2004-Jan-12, 02:49 AM
Tensor,

Tell me something. Why is this called “time dilation”?

The answer is in the paper, but he's a short version of why. Because the shift of frequencies in the Li ions is caused by time dilation.


LINK (http://216.239.57.104/search?q=cache:Vgl5Ks9SJmQJ:www.mpi-hd.mpg.de/ato/rel/+ives+stilwell&hl=en&ie=UTF-8)

Sam, I used that link in the "Time Dilation" thread and here is the part of that post (it's on page 44 of the "Time Dilation" thread) with the questions you still haven't answered:





the Kinematical part and the Electrodynamical part. The Kinematical part contains the “relative motion” error. The Electrodynamical part contains the Lorentz Force.

So, the muons “time dilate” not due to “relative motion” but due to acceleration and due to motion through the earth’s fields.

Interesting. How do you explain that the amount of time dilation from relative motion (which you claim is wrong) matches the amount of dilation due to acceleration and movement through the earth's fields (which you claim is why time dilation occurs)?

Could you post the equations showing how much dilation is caused by the acceleration and how much is caused by the motion through earths fields ? I would like to compare them to the SR dilation equation.

Also, how does this experiment, which confirms the relative motion equations, fit into the acceleration and motion through fields, since the acceleration and movement would be different for* mu-meson's, but somehow still matches the relative motion dilation?



People should stop attributing the “muon” time dilation to “relative motion”. They should attribute to Lorentz type forces


How does the Lorentz force apply to time dilation of a mu-meson moving in a gravitational field since it is the force (F) exerted on a charged particle (charge = q) moving with velocity v in a magnetic field B. It is computed F=vq X B where X is the vector cross product. It mentions nothing about time dilation.

I'm still waiting for your answer on these questions. Could you please answer them?

And now to your questions.



And why is this attributed to just “relative motion”?

Because that is what causes the slowdown. And now you have experimental proof that SR has been verified to a high degree of accuracy.


The Lorentz theory was about tick rate slowdowns due to the motion of atoms through fields,

That is your interpratation, not that of almost all of those who understand the theory. It's not only clocks, it's also time itself that slows down and it's caused by relative motion. What term in the Lorentz contraction equations represent the fields that cause the slowdown or what equations do we need to put the Lorentz contraction into to show the fields cause the slowdown by the amount in the Lorentz contraction?


So why is this tick rate slowdown attributed only to “relative motion”? What “force” is placed on the hydrogen atoms that could cause them to emit light of a lower frequency? What did Einstein stay about this in 1911?

Again, it's relative motion that causes the slowdown. And what Einstein said in 1911 doesn't matter, the link is talking about verifiying the SR theory from 1905 .


And why is it called “time dilation”, rather than just a clock tick rate slowdown?

Because that is the term used in Relativity for the contraction of the time term. The clock tick rate slowdown is the term you use.


When other types of clocks slow down, why is that not called “time dilation”?

Well, using your pendulum example, there are two effects going on. The mechanical and relativistic. What don't you understand about that?


What makes you and others think that the Doppler-shifted frequencies of an emission line in hydrogen canal rays was due only to “relative motion”, since the canal rays had no way of knowing that the laboratory clock was moving “relative” to them?

Well, the emission lines in the canal rays were from the Ives and Stilwell experiment. Or are you saying the canal rays were used in the experiment in the link? What part of the experimental setup would you claim won't show relative motion time dilation?


What did Einstein attribute the shift to in 1911?

Why would what the 1911 paper matter, the experiment in the link was trying to verify the 1905 theory, not what was said in 1911.

*Changed the word from "from" to "for" to clarify the question.

Sam5
2004-Jan-12, 03:27 AM
Interesting. How do you explain that the amount of time dilation from relative motion (which you claim is wrong) matches the amount of dilation due to acceleration and movement through the earth's fields (which you claim is why time dilation occurs)?

LOL, because Lorentz developed this theory in 1895 and the LORENTZ TRANSFORMATION equation is HIS, based on his "motion through fields" theory. His Lorentz Transformation equation WAS DESIGNED for his "motion through fields" theory. HE invented "time dilation", Length Contraction, and Mass Increase due to motion through fields. NASA sent up their tether, and caused electrons to flow through it, as it moved THROUGH the earth's magnetic field, based on Lorentz's 1895 theory.

Einstein borrowed the equation in 1905 and messed it up. He corrected his mess-up in 1918.

Sorry I missed your questions earlier. I’ll try to answer them. This is one of the most unusual situations in all of physics.

First, Einstein got the idea of the “clock rate slowdown” directly from the Lorentz relativity theory and his theory of electrodynamics, with his principal papers being published in 1892, 1895, and 1904.

Einstein credits Lorentz more in one of Einstein’s 1907 papers. He notes that in an 1895 paper, Lorentz called dilated time “local time”, and in Einstein’s 1905 paper he just renamed it “time”, which is the “time” of his local “frames” or “systems” in the 1905 paper.

So, “time dilation” comes from Lorentz, not Einstein.

Einstein tried to explain it in the Kinematical part of his 1905 paper, and he messed up big time. That’s why he had to correct the paper with a “patch” in 1918. He added a gravitational field to the “moving” k frame.

This info has been “lost” for years, mentioned only in a reprint of Pauli’s 1921 book, and mentioned in just a couple of other books. That’s why you don’t know about. But, if you will go to a big university library, look up Volume 7 of “The Collected Papers of Albert Einstein”, you will find his 1918 paper that corrects the big “relative motion” error in the 1905 paper.

So, based on the Lorentz "motion through fields" theory and Einstein’s GR theory: Atomic clocks slow down when they move through fields, when they are resting inside a gravitational field, and when they are accelerating.

So, motion of atomic clocks around and near the surface of the earth will slow them down, due to Lorentz “forces”, but NOT due to just “relative motion” alone.

All of this information is available in books, in Lorentz papers, and in Einstein papers. Get off the internet and go out to a big university library and start reading. And yes, you’ll get Lorentz “time dilation” out of his old theory. It’s also mentioned in his 1904 paper, reprinted in “The Principle of Relativity”, Dover press.

So the idea of motion-related atomic clock slow-downs was Lorentz’s idea. The idea of atomic clock slow-downs due to resting inside gravitational fields and acceleration, that was Einstein’s idea.

I'm a retired journalist. I used to get paid to do these types of investigations. I've got no emotional ties to any of these people or their theories, so I can investigate them objectively.

Sam5
2004-Jan-12, 03:51 AM
Tell me something.



You need to get a copy of “The Principle of Relativity”, Dover press. It costs about $10 in paperback, and a mall book store can order it. It’s got all of Lorentz’s major equations and two of his relativity/electrodynamics papers, and several Einstein papers, including his 1911 gravitational redshift theory.

Lorentz uses the basic Lorentz transformation equation and variations of it, and the title of his 1904 paper is, “Electromagnetic Phenomena in a System Moving with any Velocity Less than that of Light.”

So he is the one who first suggested the “limiting speed” of “c”.

Based on what I’ve read, he and Einstein got along well, so he seems to have thought of Einstein as improving and promoting the Lorentz theory through Einstein’s General Relativity theory. And of course, Einstein always used the Lorentz Transformation name when he used that famous equation, and that made Lorentz happy.

If all physics professors had just gotten copies of that 1918 “patch” paper, none of this controversy stuff would have ever come up, and there would be no “twins paradox” today. Einstein straightened it out by adding the gravitational field to the 1905 paper in 1918, and that should have ended all the controversy.

In fact Ives and Stilwell should have attributed their observed canal ray frequency shifts to: 1) Lorentz motion-through-fields theory and 2) Einstein’s acceleration GR theory, since the canal rays both accelerated and moved through the earth’s fields.

But Ives hated Einstein theory because of that “clock paradox”. So I suppose Ives never read that 1918 correction. Einstein never should have repeated his error in his 1916 book, because he knew better by then. He knew better by 1911. In his 1916 book, he should have attributed the “relative motion” slowdown to Lorentz’s “motion through fields” theory. I’m not sure why he didn’t, but I’m still working on that part of my investigation.

I have some very interesting background information about why Einstein was so coy about admitting to that particular 1905 "relative motion" error. This is a very interesting historical story, and I might try to write something about it later.

He admitted to the “cosmological constant” error, although it was not really an error. He based the constant on what astronomers told him about a “fixed” universe filled with “fixed” stars, so that’s why he added it. It was the astronomers who were wrong, not him.

Musashi
2004-Jan-12, 04:17 AM
I was at the bookstore today and I picked up (and then put back down) a bunch of Dover books. They make decent books usually for relatively cheaply. if I hadn't already been spending a bunch of money on books, I would have picked up 5 or 10 Dover science books (but this shop didn't have the Theory of Relativity. They did have stuff on optics, electromagnetic theory, and a few others that made me think of you Sam.).

Sam5
2004-Jan-12, 04:26 AM
I was at the bookstore today and I picked up (and then put back down) a bunch of Dover books. They make decent books usually for relatively cheaply. if I hadn't already been spending a bunch of money on books, I would have picked up 5 or 10 Dover science books (but this shop didn't have the Theory of Relativity. They did have stuff on optics, electromagnetic theory, and a few others that made me think of you Sam.).

lol, I've spent so much money on books. I've got a good collection of old 19h Century and reprinted books. That's where I get all the classical science info from.

Today, atomic and light stuff is called "quantum mechanics". In the late 19th and early 20th Centuries, it was called "electrodynamics".

Oh, and look at this. A few days ago I found an introduction to the THIRD Einstein theory, the one that never worked out. So, we’ve go SR, and GR, but he never finished......

The Unitary Field Theory (http://www.rain.org/~karpeles/einsteindis.html)

Sandor
2004-Jan-12, 09:07 AM
It's not only clocks, it's also time itself that slows down and it's caused by relative motion.
That's right Tensor, but these relative motions have to MEET first, and that's acceleration/force.

Sandor
2004-Jan-12, 09:32 AM
Sam, I was mistaken about the direction of where the light comes from, hitting the cones. Light indeed comes in just like in these conical antennae. But they are hollow and I still didn't read about them resonating. Isn't a cone one massive fleshy thing? How can an 'outer ring' of that resonate when it's attached from 3 directions (up, down and inner, leaving only the outer part free)? Antennae only have a very small attachment point so they can resonate.

AstroSmurf
2004-Jan-12, 02:58 PM
A few points:
The frequency of light can be determined mainly by two means. The first is to measure the photon energy, seeing how much energy is contained within each photon. This isn't all that convenient - it's usually quite a hassle to get a precise enough measurement for the small quantities that a single photon can carry, and also requires that the light is monochromatic (all photons have the same energy/frequency). The second, and more common, is to use wave effects. The method I used most frequently was using a diffraction grating or some other optical component to deflect the light - the deflection angle is directly dependant on the frequency, and this makes it easy to measure the frequency value directly. The validity of such measurement methods naturally have to be verified, but this is an entirely different topic of discussion. Suffice to say that such verifications have been made, repeatedly, and that it's possible to determine the frequency of light as a quantity isolated from all other factors.

Blue/redshifted photons change their energy as well as their frequency. This is very easy to verify as a 'real' effect - all electron interactions follow the new, changed energy value of the light; frequency-dependant effects also follow the new frequency.

The one-photon experiment has two independent variables - the frequency or energy of the light, which is controlled by the chosen electron reaction, and the number of such electron reactions per time frame, the 'reaction rate'. The reaction rate does indeed not equal the light frequency in a physical sense - the frequencies are usually in completely different orders of magnitude, such as light emitted with a frequency in the 10^14 Hz region, at a 'reaction rate' chosen to be whatever is convenient, probably something like a few thousand photons per second. Since the experiment makes it possible to control reaction rate directly, verifying that the light frequency remains constant would be trivially easy.
Sandor, your view makes sense, of a sort. The QM view of your definition of rays/beams would be that the 'width' (actually, area) of the beam is quantisized; for a certain frequency of light, the beam width can't be reduced below a certain size, or the light stops arriving completely. However, this view of the light gets slightly awkward, since the quantisized quantity is actually the energy of the light, which would be beam area x ray length, or the 'volume' of light. Also, the width would not be an actual space measurement; the intensity of a light beam isn't only related to its width. (Let a lightsource transmit light through an fixed-size aperture, and measure how the intensity varies with how strong the lightsource is). The conclusion is that you need a separate quantity for intensity; just measuring beam with and ray length/timespan doesn't suffice.

To put in mathematical terms, the energy imparted by a monochromatic light beam is:
E = beam area * intensity * timespan, or beam area * intensity * ray length / c
The particle view becomes simpler, since you can count the incoming photons directly, and get E = number of photons * energy per photon = N * h * f

Since the retina topic has come up again, let me remind you that light is a transverse wave, so any electron resonance motions would have to be transverse as well. In other words, the length of the cones is irrelevant; the widths would have to match up for resonance to occur. To take the RF antenna example, the induced amplitude in an antenna is at its maximum when the antenna element(s) are perpendicular to the direction of the EM beam. Additionally, the electrons in the cone have to be free to move, so the cone would have to be a fairly good conductor, or the signal would dissipate. This doesn't seem to be the case.

And before you call me an indoctrinated-by-my-evil-US-teachers-drone, check my location... 8)

Sandor
2004-Jan-12, 03:37 PM
And before you call me an indoctrinated-by-my-evil-US-teachers-drone, check my location
I'm sure nobody will cal you that, since your post makes a lot of sense. Actually you really use your independant intelligence.


The one-photon experiment has two independent variables - the frequency or energy of the light, which is controlled by the chosen electron reaction, and the number of such electron reactions per time frame, the 'reaction rate'.
'The chosen electron reaction', do you mean the atoms of the emission medium used? But since this is a given fact, you cannot change it then. But I didn't think of that. Could it mean that an electron falling back to a ground state in a certain atom type will emit more photons parallel (intensity) than an atom of another material?


the intensity of a light beam isn't only related to its width. (Let a lightsource transmit light through an fixed-size aperture, and measure how the intensity varies with how strong the lightsource is).
Yes, but when you increase the 'power' on the emission side, thus increasing the intensity (with the same frequency), the light beam wil have more density (parallel photons are travelling closer to each other).


E = beam area * intensity * timespan, or beam area * intensity * ray length / c
I guess you only mean 'intensity' once? In a one photon experience, you don't need to care about the beam area do you? What's the difference with intensity? As long as you catch all the photons, what does it matter how big the area is? It could even be bigger if you stand further away, isn't it?

Sam5
2004-Jan-12, 03:45 PM
Since the retina topic has come up again, let me remind you that light is a transverse wave, so any electron resonance motions would have to be transverse as well. In other words, the length of the cones is irrelevant; the widths would have to match up for resonance to occur. To take the RF antenna example, the induced amplitude in an antenna is at its maximum when the antenna element(s) are perpendicular to the direction of the EM beam. Additionally, the electrons in the cone have to be free to move, so the cone would have to be a fairly good conductor, or the signal would dissipate. This doesn't seem to be the case.

And before you call me an indoctrinated-by-my-evil-US-teachers-drone, check my location... 8)


Ahh, Smurffy, I agree with you. That’s why I said the actual resonance might occur in a ring around the cones, because the resonating antenna needs to be perpendicular to the incoming waves. This is what I explained in my “round antenna loop” example. A “null” signal is obtained when the loop antenna is aimed at the transmitter in a parallel manner. That’s how the FBI tracked down spies using short wave radios during WW II.

AstroSmurf
2004-Jan-12, 04:12 PM
And before you call me an indoctrinated-by-my-evil-US-teachers-drone, check my location
I'm sure nobody will cal you that, since your post makes a lot of sense. Actually you really use your independant intelligence.
The depends on how indoctrinated you (or actually Sam5) think I am... I do have a fair bit of schooling in these subjects, so it all comes down to whether or not you trust my mind to have remained unwarped through all of that :o


'The chosen electron reaction', do you mean the atoms of the emission medium used? But since this is a given fact, you cannot change it then.
It is possible to alter, though not continously; you need to pick a material, initial and final energy state and so forth. So with setup A, you get light of frequency f1, and with setup B, you get frequency f2. With the same reaction rate, the power of the emitted light will be proportional to the frequencies. (All of this discussion is similar to the stuff I was taught during the first year btw - how to design an experiment, how to analyse the data to extract the actual physical relationships, avoid irrelevant relationships to creep in and so forth)


But I didn't think of that. Could it mean that an electron falling back to a ground state in a certain atom type will emit more photons parallel (intensity) than an atom of another material?
Or, put differently, how can we tell the difference between one photon with a lot of energy, or several with an equally large total energy. The short answer is that we can't - there's no way to tell what the 'truth' is. However, there are some indications that there's always only one single photon at work, or that the photons always work as a group.

1. The photon or photons always travel together, never in different directions. If you try to measure photons emitted only in a certain direction, you'll always get the whole bundle together, not separate parts.
2. The photon or photons can interact with other electrons only when the total energy equals the reaction energy we're looking at, never anything where a hypothetical 'component photon' might initiate the reaction.
3. All frequency-dependant phenomena show a single frequency value for the photon/photons, not several, lower, values.

There are probably other experiments to be made, but the conclusion is that it doesn't matter - we can't separate the photon into smaller parts, so we might as well consider it a single one. This is the whole basis of the 'photon' view of light.


Yes, but when you increase the 'power' on the emission side, thus increasing the intensity (with the same frequency), the light beam wil have more density (parallel photons are travelling closer to each other).
Yes, that would be the case. So call the density 'intensity', and our views are the same.



E = beam area * intensity * timespan, or beam area * intensity * ray length / c
I guess you only mean 'intensity' once? In a one photon experience, you don't need to care about the beam area do you? What's the difference with intensity? As long as you catch all the photons, what does it matter how big the area is? It could even be bigger if you stand further away, isn't it?
I meant that the formulae are equivalent, since ray length = timespan * c.
Intensity only have a physical meaning in the wave view. The reason I include beam area is that it's part of the definition of intensity, in order to work with the classical optics from where the term comes. With your view, you might think of it that beam area * intensity is the 'number of rays'.

Kaptain K
2004-Jan-12, 05:07 PM
Could it mean that an electron falling back to a ground state in a certain atom type will emit more photons parallel (intensity) than an atom of another material?
One electron falling back to a ground state emits one photon.

Sandor
2004-Jan-12, 07:49 PM
it all comes down to whether or not you trust my mind to have remained unwarped through all of that :o
I don't trust or believe. Plain knowledge is worthless to me. My opinion is that only they who can explain (in relative simple words) why their knowledge makes sense, prove their value.


The photon or photons always travel together, never in different directions. If you try to measure photons emitted only in a certain direction, you'll always get the whole bundle together, not separate parts.
You mean if they come from one electron? Because a light bulb will emit in all directions, increasing space in between the several beams when the light gets further away. So by beam area, you mean by beam: one sequential train of photons?

Sandor
2004-Jan-12, 09:03 PM
Kaptain K, do you like ZZ Top's 70's music?

Sandor
2004-Jan-12, 09:57 PM
I would like to sum a view things here, as I understood:

1. One electron emits one photon when falling back to a ground state

2. With a given emission material, a given temperature etc, the reaction rate will determine only the frequency of emitted light. Intensity is then this given constant when the material is constant.

3. Keeping the reaction rate and thus the frequency the same, more power in the emitting source will increase intensity by
a. more electrons to involve in the process, so more photons emitted;
b. changes in the material can increase the intensity of one photon

4. it's not possible to tell the difference between one photon with a lot of energy, or several with an equally large total energy

5. The photon or photons can interact with other electrons only when the total energy equals the reaction energy we're looking at


Questions:

1 + 2. So the 'smallest' (least energy) possible photon with a given color is emitted from one electron in a certain material under certain circumstances which determine the photon to have as little intensity as possible? Can we say that this is for instance an aliminium electron at -273 degrees C. falling back from the 5th ring to the 4th or something like that? Can it also be excited like 2 rings up and fall back 2 rings? And how many other methods are there to create photons? One electron and one positron reacting will also change into one photon? Is this photon usually smaller than the smallest photon from an electron falling back?

2 + 3b. I guess that the use of certain material and the conditions of it will not be able to alter the intensity of i.e. a blue beam very much (like 10 times)? If that is true, blue photons will not be 'very' various in intensity. If you want to multiply intensity many times, you'll need more electrons participating, meaning more photons emitted.

1 + 4 + 5. We can easily detect the color. So when we know the frequency, we only need to know the intensity. But if we detect how many electrons are reacting, don't we just know that? In other words, can several photons together react with one electron?

daver
2004-Jan-12, 10:12 PM
4. it's not possible to tell the difference between one photon with a lot of energy, or several with an equally large total energy

No, this isn't the case. It's very simple to tell the difference between one photon at a given energy level and a bunch at a lower level--they'll refract differently, for one.

Sandor
2004-Jan-12, 10:17 PM
I also had my question about that yes, although differently.

swansont
2004-Jan-12, 11:10 PM
Questions:

1 + 2. So the 'smallest' (least energy) possible photon with a given color is emitted from one electron in a certain material under certain circumstances which determine the photon to have as little intensity as possible? Can we say that this is for instance an aliminium electron at -273 degrees C. falling back from the 5th ring to the 4th or something like that? Can it also be excited like 2 rings up and fall back 2 rings? And how many other methods are there to create photons? One electron and one positron reacting will also change into one photon? Is this photon usually smaller than the smallest photon from an electron falling back?


Energy of a photon and color are synonymous. Energy, wavelength and frequency are all inter-related.

Yes, an electron can make different transitions within an atom, but not all transitions are possible because of conservation laws (e.g. S state to S state is forbidden, because the photon has 1 unit of angular momentum, given by Planck's constant, so the electron state's angular momentum muct change by 1 unit)

Photons can be also created by accelerating a free charge (this is how incandescent bulbs work, thermally agitated electrons colliding), in nuclear transitions similar to electronic ones, and annihilation of matter with antimatter.

Electron-positron annihilation requires 2 photons (or more) to conserve momentum. The two photons would have .511 MeV of energy, which is more than in your typical outermost-electron atomic transition, which are typically an eV or so. You need to have knocked out an electron from an inner shell to get keV-ish energies (and we call these X-rays). Note that an X-ray machine typically produces radiation consisting of a continuous spectrum of photons from electron acceleration (Bremsstrahlung) as well as discrete peaks from electron level changes.

AstroSmurf
2004-Jan-13, 10:40 AM
1. One electron emits one photon when falling back to a ground state

5. The photon or photons can interact with other electrons only when the total energy equals the reaction energy we're looking at

It's not necessarily the ground state, any lower energy state will do. Also, the exact reverse is true, e.g, if an electron going from state A->B emits a photon of frequency f, any electron in state B can then absorb that photon, and will go up to state A. Only a single photon is ever involved in these transitions; see below for more on this.


2. With a given emission material, a given temperature etc, the reaction rate will determine only the frequency of emitted light. Intensity is then this given constant when the material is constant.
Put it this way: The material, temperature and so forth determines the frequency of the photons, or the "energy per photon". The total emitted energy will then be the reaction rate times the energy per photon. So the energy can only be varied in increments of the "energy per photon" - this is the very definition of quantisation.


4. it's not possible to tell the difference between one photon with a lot of energy, or several with an equally large total energy
What I meant was this: While it might be possible that a photon consists of several, smaller parts, we've never been able to find any physical effects from this. So for all practical purposes, a photon can be considered to be a single unit. The points I listed in my previous post is the kind of stuff you see if there actually are several photons being emitted - they won't all be going the same direction (that is, even when there's only a single reaction happening), you'll detect different, lower frequency maxima than the energy sum would indicate, and they'll interact with electrons differently.

Now, to your questions:


1 + 2. So the 'smallest' (least energy) possible photon with a given color is emitted from one electron in a certain material under certain circumstances which determine the photon to have as little intensity as possible? Can we say that this is for instance an aliminium electron at -273 degrees C. falling back from the 5th ring to the 4th or something like that? Can it also be excited like 2 rings up and fall back 2 rings? And how many other methods are there to create photons? One electron and one positron reacting will also change into one photon? Is this photon usually smaller than the smallest photon from an electron falling back?
1) All photons with a certain colour (frequency) have the same energy.
2) A specific electron transition always causes the emitted photon to have the same frequency - this is in fact how we examine the electron energy levels.
3) The process of going from the frequencies of emitted light to an energy spectrum is a long and tedious detective work, which keeps quite a few physicists occupied. It's interesting though - I used to work as a lab assistant in one of those places for a while.
4) The transitions aren't restricted to just one layer difference, which lets just a few energy levels give rise to a lot of possible photon energies. There are some conservation laws which restrict the possible transitions, but the total number is still fairly huge.

Photons can be produced in a lot of other ways. What's interesting about electron layer transitions is that the photons involved are for the most part in the visible spectrum. Other reactions include particle-particle collisions (X-rays), nuclear reactions (gamma rays) and chemical reactions (IR).


1 + 4 + 5. We can easily detect the color. So when we know the frequency, we only need to know the intensity. But if we detect how many electrons are reacting, don't we just know that? In other words, can several photons together react with one electron?
Theoretically perhaps, but it seems to be highly unusual. The reason for this is that all the participating particles/photons have to be in the same place at the same time, which is fiendishly difficult to arrange, both in experiments and in nature. Those reactions which involve more than two participants split up into several, less complex sub-reactions once you start to scrutinise them.

Kaptain K
2004-Jan-13, 10:53 AM
Kaptain K, do you like ZZ Top's 70's music?
Yes. Why? How is this relevant to the discussion?

Sandor
2004-Jan-13, 01:04 PM
Thanks a lot Swansont and Astrosmurf, the science part is now quite clear to me (partly of course). I will now tell you something about my understanding of it.

Excuse me for comparing waves with photons, but these are properties of the same thing. And maybe we need one vision to explain this and the other to explain that; we might still want to exchange the visions and regard the traveling waves as photons and the electron-reacting photons as waves. When waves enter our eye with a certain frequency, our detection system has to determine the frequency. It can do that only by determining the wavelength, which it can only do by detecting a wave length and count the time for that to 'pass by' or to start and stop. Or, from a photon vision, it will have to count the incoming particles per time frame. No matter which vision we use, time (and/or space) is determining this frequency. So if our spacetime on earth changes, we see different colors. This is exactly what happens with red- and blueshifting. Sam asked where this would happen exactly. I already stated by i.e. regarding our planet as moving towards the star/light, that relative motion (and also i.e. gravity from the sun) is changing space, but instead of space changing, we might consider time changing, which will have exactly the same result of course (spacetime is just one). But we all see the same colors always. We would also see red-shifting the same, all of us. So this proves that we all have an 'objective' time-sensing. By objective I mean that it's the same for all persons on earth. In other words: we can feel of course all forces of (and around) the earth working in our bodies and we just compare other forces relative to it. So, although we all feel time differently in one way, we also have (nearly) exactly the same time-sensing on a physical level on the other way. (But this was really an extra thing that I didn't want to talk about, but I just understood it.)

I will pick out this phrase of mine: which it can only do by detecting a wave length and count the time for that.
If you consider light as waves when traveling and as photons when reacting, the detector (eye) has to detect the waves and turn them into photons. ;) Or they were already photons during traveling. Anyway, from the wave vision, the detector has to detect and distinct every single wave length. Otherwise it couldn't sense the frequency.
From the photon vision, it can detect every single photon coming in at a certain rate. But it cannot do that plainly by energy of one photon, because intensities might vary. So a photon has to have 2 dimension at least. Which is exactly in line with a wave having frequency and amplitude. The thickness of a photon is the intensity and the length is the frequency.

I already knew that EM force is 2-dimensional, traveling at c. We cannot 'see' any 2-d thing; we can only detect it when it reacts with 3-d, thus becoming 3-d. We also know that traveling at c will shrink space, in fact, anyhting with MASS cannot reach c, because all space would disappear. What we forget in this formula is that 'our' mass is 3-d. Traveling at c makes you 2-d, thus eating up one dimension. This is also why a photon KNOWS it can exchange force with the other end of universe. It HAS to know, because it cannot find out after 2 years of traveling that it CANNOT exchange force because there isn't any, and go back very ashamed. :) It cannot not even leave the source when it's not confirmed in a anti-force, which is always needed in any force exchange: 2 opposite forces of exactly the same 'momentum'. Lorentz and Einstein knew that an accelerating ruler would shrink space very locally, like a big planet with acceleration gravity will shrink space globally. A photon traveling at c will shrink not only space locally (on it's own path), but space is shrinking in only 1 dimension. So the photon is really at source and detector at the same time, in time of the photon. Because the photon eats one dimension, making our 3rd spacial dimension it's time dimension. Our 4th time dimension is then totally dissolved. When the photon makes the 3rd dimension infinitely small, it makes it's time infinitely big (the same 3rd dimension), standing still. For a photon, there is no difference in our past or future. For you to enter this timeless zone, which is as well past as future, you will indeed have to travel at c, but you will become 2-d.

You see, every motion in 3-d has 3 components or dimensions. If we consider them as pure single dimensions, we could simply add them up, resulting in the total distance. That's very logical: if you travel in a box from the left lower corner to the upper right corner, you are in fact traveling all 3 ribs TOTALLY. But if we count the 3 ribs up, we get more and that because every single rib is also made up of 3 dimensions. We cannot see pure single dimensions. We cannot even see 2; but always 3. But, with an absolute view, we would be able to see them and thus we could see also the line from left lower to upper right as one pure single dimension, which is limited by c. When one dimension has disappeared, there is REALLY no third (spacial) dimension left, by which I mean that the entire universe is in 2-d. So you don't have to bother from which angle you theoretically have to look at something 2-d. Well, in fact you might best see it as you see it: put the surface in line with your (inner) view, so you see nothing. You cannot see something flying at c over the 'horizon', from left to right in your view; it will 'turn away' from your view. While it gets thinner, it turns more away from you, and when it is dissolved, it will continue it's way straight ahead (when it gets > c).

http://www.lorentz.leidenuniv.nl/vanbaal/SRT/syllabus/chap6-7/img40.gif

It seemed to take 50 years 'after' Lorentz before someone wondered HOW we would see an object speeding by at 'close' to c. When we take an imaginairy distance from which all light is coming almost straight at us, light from the back side (from our view, perpendicular to the motion direction) of a moving box is getting much later in our eye than light from the front side. This will a bigger back-side (now from the motion direction). Calculations prove that the proportions, including the contraction in the direction of the motion, will be exactly as if the object is turning away from us. The funny thing is, that even when we're doing relativity theory here, and we already know that things are often 'really' how they seem, they didn't seem to believe or understand that the object was in fact turning away from them (or that space was twisting into 2-d).

My last line from the former paragraph was: "when it gets > c". I saw Sam writing about blue****ing and "c+v". That's correct. As c is our max. speed, it is the min. speed in 2-d. So the light may go faster (from then it's also traveling in the world of 2-d, meaning 'moving at all', through their time) over there. When one dimension is eaten up by the light, it can 'continue' with more speed. When we looked at the box, we saw that the line from down left to upper right could be seen as a pure single dimension, reaching max. speed (c). But from our view, it is still made up of 3 dimensions, from which only 1 reached c. It will literally continue speeding on the next dimension (when blueshifted of course). This is the most difficult part to understand. Because we cannot think in 2-d. We MUST visualize directions. But I think this will help: all light we see, we can only see from straight away. (When you think you have a vision of several angles high, it's really your view traveling up and down very fast. When you really look at something, you have to look at it straight.) By this I mean that the world of light REALLY comes straight away at us. Even if we think a light from the lamp at our left is shining on the couch at the right, all of this light we see is coming straight at us (we see no light going from the left to the right). Of course this is not new to any of you, but you have to THINK that way also. The world might be as well 2-d from a light perspective. Just think of it as you yourself turning your head, with a sheet of paper in front of, against, your 2 eyes, so you hardly see the paper at all. Whichever way you turn your head, upside down, 2-d will move with you in a way that everything comes to you from this paper only. The past comes toward us in a straight line. We can only observe old stars correctly when the light has not changed on the way. And light NEVER changes, because it's timeless. Only by reacting with objects, it gets absorbed and emitted again, which means totally new, different light. Back to the blueshifting: when light is going faster than light, it gains in frequency, by shrinking the next dimension.

By the way, ever heard of holography? That's a nice way to show how 3-d information can be 'stored' 2-d.

Sandor
2004-Jan-13, 01:10 PM
Kap, I sent you a PM.

Sam5
2004-Jan-13, 06:33 PM
Excuse me for comparing waves with photons, but these are properties of the same thing. And maybe we need one vision to explain this and the other to explain that; we might still want to exchange the visions and regard the traveling waves as photons and the electron-reacting photons as waves. When waves enter our eye with a certain frequency, our detection system has to determine the frequency. It can do that only by determining the wavelength, which it can only do by detecting a wave length and count the time for that to 'pass by' or to start and stop. Or, from a photon vision, it will have to count the incoming particles per time frame. No matter which vision we use, time (and/or space) is determining this frequency.


I would explain it this way:

Our cones merely resonate to the proper frequency of the light, the proper “bump, bump, bump, bump” sequence, in time.

So I would say that the cones don’t consider the “wavelength” directly, but indirectly. The size of the cone “antenna”, if it matches the light wavelength, tends to resonate more when it is just the right size, and so the “bump, bump, bump, bump” causes it to resonate at its maximum rate and efficiency, and this generates the maximum electrical signal going out the back of the cone, through the nerve “wires” to the brain.

It’s like, we can pick up a radio signal using the wrong length antenna, but we get the maximum signal going into the radio when the length of the antenna perfectly matches the wavelength of the incoming radio waves. Even though the antenna doesn’t measure the wavelength directly, it measures it indirectly when it receives the certain frequency, in the form of just the right rate of “bump, bump, bump” signals.

If the antenna is of the correct length to resonate the maximum when it receives a “bump, bump, bump” frequency, then when it receives a lower frequency, like “bump.....bump.....bump”, it would not resonate properly.

If the signal is redshifted, that is, if the radio waves are being sent from a spacecraft that is moving away from the earth, the “bump, bump, bump” frequency is turned into a “bump.....bump.....bump” frequency, so the radio receiver must be tuned to the lower frequency, and the antenna must be longer in order to resonate the maximum to the redshifted radio waves.

The "tuner" inside the radio also works on the "resonation" principle.

Years ago, when radio engineers tried to explain to me why a radio antenna of 1 meter in length resonated best to a wave of 1 meter in length, I couldn’t understand this since the antenna was vertical and the radio wavelength was horizontal. Finally, I figured out that the radio antenna DOES NOT KNOW the wavelength of the incoming signal. All the antenna feels is the “bump, bump, bump”, and the bumps are spread out in time by the wavelength of the signal, so when the vertical length of the antenna matches the horizontal wavelength of the signal, then the antenna resonates at the maximum because the “bump, bump, bump” is of the correct frequency and the horizontal wavelength of the radio waves is of the correct length to match the vertical length of the antenna.

Sandor
2004-Jan-13, 10:33 PM
Our cones merely resonate to the proper frequency of the light, the proper “bump, bump, bump, bump” sequence, in time.
That's very interesting to think about. A 'bumping feeling', that's what you and me can imagine. But when we analyze such of our feelings, and we get down to the most elementary level, can we then still say an electron just feels a photon and it's intensity? Is that different from 'knowing' it's width? Is an electron who feels the difference between one photon and the other something else than an electron 'measuring' the length of photons? Is the 'power' of 'hardness' of a photon lower at the points where the EM wave is going from plus to minus? Is this also where the photon 'ends'? It's easy to say it just feels a bump, but does this mean it doesn't need at least 2 dimensions to feel that? I will have to think about this. How do an alectron and a photon shake hands to agree that they will exchange forces?

Sam5
2004-Jan-14, 12:09 AM
That's very interesting to think about. A 'bumping feeling', that's what you and me can imagine. But when we analyze such of our feelings, and we get down to the most elementary level, can we then still say an electron just feels a photon and it's intensity? Is that different from 'knowing' it's width? Is an electron who feels the difference between one photon and the other something else than an electron 'measuring' the length of photons? Is the 'power' of 'hardness' of a photon lower at the points where the EM wave is going from plus to minus? Is this also where the photon 'ends'? It's easy to say it just feels a bump, but does this mean it doesn't need at least 2 dimensions to feel that? I will have to think about this. How do an alectron and a photon shake hands to agree that they will exchange forces?

When I say an electron or a cone “feels” a photon’s frequency, I’m basically saying that a certain small amount of physical pressure is being put on the electron or cone when they encounter a photon. They don’t sit around and say, “Hmm, I think I ‘feel’ something.” They just react to it right away.

With a series of photons, I think they react quicker and with more energy, by means of “resonation”.

It’s like when a pendulum is swinging. All you’ve got to do is add just a tiny amount of more energy to keep it swinging. But, the first time you make it swing, and if you try to make it swing in a non-resonating way, that takes a lot more energy to get the swing accomplished. But once you already have it swinging harmonically, just a tiny amount of more energy keeps it swinging.

The electrons and cones might care about the wavelength, since that wavelength also represents the duration and gradual power increase and decrease of each “bump”. The beginning of the bump (the wave-front) starts the motion or the resonation, and it continues on through to the end of the bump (the wave-end, ie the last of that wave).

On an optical sound track a “bump” is a little longer than a “click”. A click is a very short sharp sound, while a bump is a longer one.

A gunshot recorded with the gun near a microphone will record not as a shot sound but as a big short “click”. You can actually have the gun so close to the microphone so that you record no gunshot “sound” at all, just a big click, which is the powerful compression wave that pushes the microphone diaphragm in so far, it only vibrates in and out basically once.

A snare drum sound records as a short click, while a bass drum records more as a “bump”.

If you turned a large gear wheel and held a stick or piece of metal down on the gears to make a “clicking” sound, you would record a lot of clicks. If you turn the gear faster, you will notice that the clicks begin to turn into a low frequency hum. Turn it faster and they turn into a high frequency hum. A hum made up of a series of short clicks.

A single click looks like this on an optical sound track: ^

A series of clicks look like this: ^^^^^^^^^^

A series of bumps look like this: ~~~~~~~~

I would say that a photon would be more like a bump than a click.

Here is one animated interpretation of the Maxwell-Hertz magnetic/electric oscillation theory of light:

LINK (http://www.sciencemaster.com/jump/physical/magnetic_waves.php)

Sam5
2004-Jan-14, 12:17 AM
Sandor,


SEE 1ST ILLUSTRATION (http://www.u.arizona.edu/~molinero/)

Sandor
2004-Jan-14, 03:00 PM
Haha Bumping Sam, OK this is a nice one. I now understand what is the biggest 'problem' of mankind. It's people assuming they understand about things. Even when you explain to them that they don't, they cannot hear.
Of course we have several levels here. One is that you ask about 'what waves are' and people come up with formulas. Even when you tell them 10 times that you're looking for understanding, they will be coming back with this formula. There's no bad spirit in that, it's just where thinking stops. But even on another level we find 'bumping' very normal. It's the thing you first experience when you're a baby and from then it gets so familiair that you don't even ask yourself what the heck is going on. It's just bumping! Okayyyy.

'Well OK, bumping is when you put a force in a direction where another force (matter or energy) is already working, resisting your force.'

'What's then the difference between matter and energy?'

'That's just relative. Energy is going quicker related to matter, which is more or less staying at one place.'

'But it's still moving?'

'Yes, everything moves all the time when you look inside. It just the perspective you choose.'

'So from a certain perspective, everything is energy?'

'Yep, that's right.'

'So there is no smallest particle?'

'Well, at a certain level we cannot catch the particles anymore to hold them; we can only see them. But then comes the level where we cannot see them; we can then only detect them. But there's also a limit to detecting particles.'

'Yes, but that's just a matter of time before we have equipment to detect those too, isn't that true?'

'Not really. Quantum physics and Heisenberg tell us that our laws of nature round off values, so that we can only see the 'integers' of which are not really integers. So there seems to be a smaller world that is more or less outside our world, although it's in it also.'

'So if we follow this line down, it looks like energy is 'moving nothing' in the end?'

'Maybe. The smallest particles we know, we actually don't know at all. That makes some sense too, because we can only understand something by knowing what's inside, but when nothing smaller is inside, it really is nothing hey? Now you're confusing me.'

'I feel the same man. What is left for us eventually?'

'I guess it's space, time and directions. I can't think of anything else.'

'Ah I see. Space shrinking in a certain direction, within a time frame means velocity, like in a one dimensional world. You can call it matter as well as energy, but it's non of those, because there is no difference. In a 2-d world space can also 'spin', it can shrink staying at one place, for it to become matter.'

'Yes, and when it spins and it's going, as a whole, in another direction, it's moving matter, so this is energy. At least, compared to something bigger that is spinning.'

'Hmm, but isn't time just another spacial dimension?'

'Yes, sort of, you might see it as spacetime for a better understanding. It's to get a grip of the world on one hand (space), and for not ruining 'the plot' on the other hand (time). It's really no fun when you already know everything. Life would not be possible.'

'So these are several of the same dimensions, only different in perspective from where you are?'

'Yes, in that case we only know dimensions and, well, what was the other thing, direction.'

'But when be bring everything down to the smallest level, we only have 2 directions, isn't it? In daily life we know compound directions, but at elementary these can only be opposite or, or what?'

'Well, only opposite yes. When 2 'things' have the same direction and the same velocity, they do not sense each other at the lowest level. When velocities are difference, they'll bump. So there's only opposite directions eventually. I think you're right.'

'So when spacetime shrinks in opposite directions, it bumps?'

'That must be the definition of bumping, I cannot think of anything else. Well, I can. You could just say that A dimension is shrinking, because that's always opposite directions.'

'Thanks a lot! Now I know what bumping means, it's one shrinking dimension.' :)