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OldAlbert
2007-Nov-18, 09:03 PM
Here I am again,on the same subject, except that in deference to the rules of the forum, I will, this time, reason as from within a static universe.

If a photon is released somewhere, it can generally be detected in any direction from the point of origin. One would expect therefore that in order to intercept the entire photon, one would have to intercept it with a completely encompassing spherical shell. So how can one imagine a whole photon passing through a small slit several feet or several million miles from the point of origin?

Conventional wisdom has the photon moving in all directions at the speed of light, yet that same conventional wisdom does not seem to entertain the idea that the photon is becoming larger (expanding) as it proceeds, so that
both slots of the double slit experiment are passing different segments of the same photon at once.

Would anyone please comment on this?

.

papageno
2007-Nov-18, 10:12 PM
If a photon is released somewhere, it can generally be detected in any direction from the point of origin.


Why would that be?
If the photon is emitted with a well defined momentum, it has a well defined direction of motion.





One would expect therefore that in order to intercept the entire photon, one would have to intercept it with a completely encompassing spherical shell.


And yet the detectors used don't work that way.





So how can one imagine a whole photon passing through a small slit several feet or several million miles from the point of origin?


Since a photon can have a well defined direction of motion, whether a photon passes through a slit depends on its wavelength compared to the width of the slit.





Conventional wisdom has the photon moving in all directions at the speed of light,...


That's not the "conventional wisdom".
If the direction is not defined, then the position of the photon would be well defined, which means that the photon would be localized.

After all, photons follow the Heisenberg principle.





... yet that same conventional wisdom does not seem to entertain the idea that the photon is becoming larger (expanding) as it proceeds, so that both slots of the double slit experiment are passing different segments of the same photon at once.


You are confusing the particle with the wave-function.

Tim Thompson
2007-Nov-19, 01:26 AM
Conventional wisdom has the photon moving in all directions at the speed of light ... One would expect therefore that in order to intercept the entire photon, one would have to intercept it with a completely encompassing spherical shell
Conventional wisdom says no such thing, and therefore one would have no expectation of needing an encompassing spherical shell. The expanding sphere is a wavefront in classical electromagnetism, where there is no concept of "photon" and light is considered entirely as a wave. In the context of the photons, the spherical wavefront becomes an expanding spherical distribution of an extraordinarily large number of photons. The photons are properly treated as particles.


... yet that same conventional wisdom does not seem to entertain the idea that the photon is becoming larger (expanding) as it proceeds, so that both slots of the double slit experiment are passing different segments of the same photon at once.
Not so. Conventional wisdom does in fact explain the double slit experiment exactly that way, for both photons and particles. A single photon, or a single particle (usually an electron), will generate an interference pattern because the photon or electron simultaneously passes through both slits and interferes with itself.

But that only happens when you make no attempt to determine which slit the photon or particle passes through. If you put a counter on the slits, then the photon or particle will pass through one slit only, and there will be no wave interference pattern, despite the double slits. So reality may be stranger than you think.

OldAlbert
2007-Nov-19, 03:07 AM
Thankyou Papa and Tim.

I am currently working on design of a cheap photon detector (with some promise of success, but still some major problems to get past), and I want to reconstruct the double slit experiment. (It didn't seem nearly so interesting years ago in school in the physics lab).

How will I be able to determine whether I am detecting photons as particles or waves? If I DO get photon detectors working, that will become an important problem to solve.

papageno
2007-Nov-19, 12:41 PM
How will I be able to determine whether I am detecting photons as particles or waves?

Usually in the interaction between matter and electromagnetic fields, as it occurs in typical detectors (eye, photographic emulsion, photomultiplier tube, CCD), the photon is detected as a particle.

OldAlbert
2007-Nov-19, 06:40 PM
Papa..:

So would you say that it is appropriate for me to assume that any proton that causes a response in the sensing diode of my photon detector is a particle (as opposed to a wave)?

If so, how might one detect a wave (other than by noting interference fringes)? Perhaps by some sort of resonance efffect? Perhaps a mirror bounded chamber to establish standing waves?

papageno
2007-Nov-20, 01:47 PM
So would you say that it is appropriate for me to assume that any proton that causes a response in the sensing diode of my photon detector is a particle (as opposed to a wave)


If the detector gives a well localized response to the photon (that is, if you can say "the photon hit right here!"), then you are observing the particle-like properties of the photon.




If so, how might one detect a wave (other than by noting interference fringes)? Perhaps by some sort of resonance effect? Perhaps a mirror bounded chamber to establish standing waves?


Interference and diffraction are hallmarks of wave-like properties.
I don't actually know if the are ways to detect a single photon as a wave through interactions with matter.

Tim Thompson
2007-Nov-20, 04:25 PM
If so, how might one detect a wave (other than by noting interference fringes)? Perhaps by some sort of resonance efffect? Perhaps a mirror bounded chamber to establish standing waves?
You cannot detect a single photon as a wave. Solid state detectors detect photons as particles (see George Rieke's book Detection of Light: From the Ultraviolet to the Submillimeter (http://www.amazon.com/Detection-Light-Submillimeter-George-Rieke/dp/052181636X/ref=sr_1_1?ie=UTF8&s=books&qid=1195575684&sr=1-1), an excellent review of the topic). You detect waves with antennae, as in radio & radar, but you cannot distinguish one photon from another with such technology. You either detect continuous waves, or discrete photons, but you will never (and can never) detect an individual photon as a wave. That's what the infamous wave-particle duality is all about. You get whichever one you look for.

OldAlbert
2007-Nov-20, 05:40 PM
I just gotta see some of the wierd behaviour that Angstrom mentioned, myself. Once I get a few photon detectors built up, I will put one at each end of a long slit, to see if they agree, and after that I will so equip both of the parallel slits and play around with it.

I don't know costs yet, but it looks as though Avalanche Photodiodes are the cheapest serious sensor. I will use a common 200 volt power supply and I think a bus logic analyzer on their outputs to detect simultaneity. I don't think the low noise amplifiers will be much of a problem.

MentalAvenger
2007-Nov-21, 01:20 AM
Looking at the situation from a realistic viewpoint, and considering all the information that is currently available (including various theories), it appears to me that light is neither a particle nor a wave. all too often, people are prone to using macro examples to define micro or sub-micro events. In the macro world, we can see particles or waves in action. There does not appear to be any phenomenon in the macro world that exhibits some of the properties of both. So, instead of the obvious answer, we automatically resort to attempting to visualize the micro interactions with macro examples. IMO, the obvious answer is that light is neither a particle nor a wave, but something that does not have a corresponding analogy in the macro world.

Perhaps it would be more productive to attempt to figure out what that micro property is, rather than continuing to attempt to equate it to some observable macro property.

OldAlbert
2007-Nov-22, 01:17 AM
Tim:

Here is my understanding of the last few posts, and the experiment that I think might help solve it.

At some distance from the light source, if I intercept a particle type photon, I will intercept the whole photon's energy. But if I intercept a wave type photon, I will intercept only a small portion of the energy of the expanding wave front (unless I intercept it with an encompassing sphere).

So if I intercept a particle type photon,I can understand that it can have the energy to 'knock' an electron into a higher energy level, so that as it falls back it releases another photon of equal energy. But if I intercept a wave type photon, I assume it would take several (if not hundreds of) cycles to raise an electron to a suffiently higher energy state that it falls back and emits a new photon.

Also, if a pair of photon detectors, operating in 'geiger mode', located at oposite ends of a long slit, both simultaneously detect a photon, then that activity must be the result of an expanding wave-front passing through, whereas if only one of the two detectors 'clicks', then that would indicate the interception of a particle form of photon.

And then there is the following question, that I might be able to 'throw some light on' <<sorry>>:

If I use a 45 degree inclined semi-transparent mirror to intersect a particle type photon, then would I be able to detect a 'reflected' photon (to one side) and at the same time detect a 'passed-through' photon (in line with the original photon path beyont the mirror)? in which event, how could one photon supply energy to release 2 new photons - Or would only one of these events occur? in which case which-one?

I realize that this is not earth shaking stuff, and that the stars will keep turning in the universe, but this is fun stuff, and while I await being able to experiment with it, it is good food for thought.

OldAlbert
2007-Nov-22, 01:28 AM
Mental:

What you say makes sense. I and, I imagine, several others, agree with you. For me, anyway, I am in the early stages of becoming comfortably familiar with the vagaries of photon behaviour, with still lots to learn. I had no idea that this could become so interesting when I was groping my way around in the optics lab at school a few hundrd years ago. I think it is the com****ion of previously unavailable sensors and light sources, and my becoming retired enough to get directly involved with this, that has happened.

OldAlbert
2007-Dec-03, 11:21 PM
A photon may be characterized as having a frequency property. The energy of a photon is considered to vary directly with its frequency (specifically as Plank's constant x frequency). So the photon must have some characteristic that can define frequency, or wavelength.

Option 1
One implication might be that photons are made up of f sub-particles, each of energy equal to h (Plank's constant), so that the energy of a photon of frequency f is h times f.

Option 2
Another possible implication might be that the photon contains some oscillatory feature. For that to be the case, there would have to be an inertial component and a restorative component of force (as with a pendulum). It is possible to imagine such a thing, like a soap bubble whose surface expresses both mass (akin to inductance) and surface tension (akin to capacitance). But it is cumbersome.

I would like to explore the first option. There are some expectations to be had of the first model. One would expect that under the right circumstances, such photons could be "split" into two or more lesser frequency (energy) photons, whose several frequencies add up to that of the original photon. That, of course is the case. Then arises the question of whether that precludes the possibility of option 2. I would like to hear others' opinions on this.

tusenfem
2007-Dec-05, 01:21 PM
A photon may be characterized as having a frequency property. The energy of a photon is considered to vary directly with its frequency (specifically as Plank's constant x frequency). So the photon must have some characteristic that can define frequency, or wavelength.

Option 1
One implication might be that photons are made up of f sub-particles, each of energy equal to h (Plank's constant), so that the energy of a photon of frequency f is h times f.

Option 2
Another possible implication might be that the photon contains some oscillatory feature. For that to be the case, there would have to be an inertial component and a restorative component of force (as with a pendulum). It is possible to imagine such a thing, like a soap bubble whose surface expresses both mass (akin to inductance) and surface tension (akin to capacitance). But it is cumbersome.

I would like to explore the first option. There are some expectations to be had of the first model. One would expect that under the right circumstances, such photons could be "split" into two or more lesser frequency (energy) photons, whose several frequencies add up to that of the original photon. That, of course is the case. Then arises the question of whether that precludes the possibility of option 2. I would like to hear others' opinions on this.

Well that would really get you into trouble, because how are you going to get the number of subparticles of energy h for a photon belonging to light of frequency 223.258746 Hz? Are you going to allow fractional particles?

OldAlbert
2007-Dec-05, 03:52 PM
That WOULD take a lot of sorting and trial fitting together wouldn't it! So much for Option 1.

Cougar
2007-Dec-06, 03:24 AM
I'm not sure of all of the statements made by Nick Herbert, but I think he's pretty close to the mark when he said:



"Electrons [read: quantum particles] cannot really be said to have dynamic attributes [position, momentum, etc.] of their own. What attributes they seem to have depends on how we choose to analyze them... the kind of parts a wave seems to have depends on how we cut it up." [from Quantum Reality, Beyond the New Physics (1985)

OldAlbert
2007-Dec-06, 01:54 PM
Rather than ponder over the nature of a photon, which seems to have gotten me as far as it will, I will try mulling the nature of electromagnetic waves.

An electromagnetic wave, when approaching an observer, appears to have an electric field which is radial, and at right angles to, the wave's direction vector, and which oscillates from expanding outward from the wave's line of travel, to contracting inwards toward the wave's line of travel. There is also a cyclical magnetic field which ebbs and wanes, in coordination with the changing electric field. The magnetic field is at right angles to both the line of travel and to the electric field at all points.

This electric field, taken as a whole in all directions from the light source, forms a spherical shell, which appears to the observer, as an expanding wave-front, approaching at the speed of light. A continuous wave would then consist of an unending set of concentric spherical shells, each shell being a single photon. The observer intersects successive shells, each separated by a single wavelength (lamda).

Each of these photons carries an intrinsic frequency property. One way that this could be encoded into the photon is through physical characteristics of the shell surface. It's mass might be akin to inductance and its surface tension might be akin to capacitance, so that it could vibrate at a resonant frequency that results in the frequency as measured by the observer - like a balloon expanding and contracting slightly at its characteristic frequency.

If the observer could 'run along' with the travel of the wave, he would observe a reference point on the wave to be a radial electric field, cyclically changing at its characteristic frequency. If the observer were 'stationary', he would intercept successive concentric shells of electric field at the characteristic frequency.

I am concluding that photons ARE vibrating electric fields, appearing to expand at the speed of light. How then, do wave like photons cause electrons to take quantum leaps, even though the fields from distant birth points may be very weak? The answer might have to be that each cycle of the wave gives the electron an additional boost until the electron has acquired enough energy to make the quantum jump. That would imply that secondary photon generation is far from instantaneous, but requires many cycles to build up to a release, when the photon stream is very weak.

Why then does a photon seem to cease to exist anywhere else when it is detected (i.e., when an intersected atom's electron makes the jump? Does this 'bubble' burst when a secondary photon is created at its surface? How does the photon get annihilated?

This line of thought leads to the conclusion that electromagnetic radiation is not some kind of relatively pure energy moving in empty space, but is instead, very small amounts of mass (photons) expanding in empty space. This is, to me at least, a more plausible concept of radiation.

stutefish
2007-Dec-06, 08:11 PM
Each shell is made up of a number of photons, spraying out in all directions from the light source.

The interactions of all the individual photons produce a characteristic wavelike behavior when multiple photons are observed as a group.

Observing an individual photon has no effect on the other photons that have been emitted; except perhaps under certain experimenal conditions.

OldAlbert
2007-Dec-06, 08:31 PM
Stutefish:

What if only one photon was produced at the birthspot? Then a single photon would be associated with one shell.

Multiple photons, produced at the same birthspot, would contribute to additional contained concentric shells, unless they were produced exactly simultaneously with the first.

Or so I think.

I want to say here again that it is great to have a forum that critically seriously considers oddball ideas, and often produces excellent knowlegable arguments against - kind of an internationally distributed think tank. It is powerful.

papageno
2007-Dec-06, 08:47 PM
What if only one photon was produced at the birthspot? Then a single photon would be associated with one shell.


Or simply there would be no shell.

I don't think one can get a spherical wave with a single photon.

stutefish
2007-Dec-06, 10:33 PM
Stutefish:

What if only one photon was produced at the birthspot? Then a single photon would be associated with one shell.
There would be no shell. Just a single photon, traveling along a single path in a single direction, until it was intercepted by an observer, which would see the briefest flicker of light--a single photon.

Other observers, not along that photon's path, would see nothing.

stutefish
2007-Dec-06, 10:36 PM
Or simply there would be no shell.

I don't think one can get a spherical wave with a single photon.
OldAlbert seems to be describing something like a single bullet, surrounded by many targets, that grows bigger and bigger until it shoots all the targets simultaneously.

In fact, that single bullet strikes a single target. A gun would need to fire many bullets in many different directions, to strike many different targets.

The same is true for photons.

OldAlbert
2007-Dec-07, 12:00 AM
Don't forget that once a photon is detected in one spot, then that photon ceases to exist at any other spot. It is a bit of a reach that a photon can be annihilated once it is detected, but for a bunch of other photons to disappear in response to one proton being detected, is a harder reach.

OldAlbert
2007-Dec-07, 12:02 AM
Sorry "to one PHOTON being detected"

stutefish
2007-Dec-07, 12:28 AM
Don't forget that once a photon is detected in one spot, then that photon ceases to exist at any other spot. It is a bit of a reach that a photon can be annihilated once it is detected, but for a bunch of other photons to disappear in response to one proton being detected, is a harder reach.
A photon is a particle. It travels like a particle. It can only be seen in one spot. It is never seen in more than one spot at a time. It does not expand like a shell. It travels like a bullet.

The wavelike properties of light are the result of many photons interacting with each other.

The particle-like properties of light are the result of a single photon examined individually.

People coming out of a building into the town square may form a crowd, but each person in the crowd is still an individual. If you talk to one person in the crowd, you are not having a conversation with the entire crowd. A single person can't create a crowd by walking out of the building and then expanding in every direction until he fills the town square. Photons are like this. They are particles.

Again: Photons are particles. They do not expand and contract. They do not expand. Observers in different places observe different photons, emitted along different trajectories.

Bogie
2007-Dec-07, 01:34 AM
This may not help at all but here is a neat java applet that is supposed to graphically represent the electromagnetic wave and Poynting vector.


It might bring up more questions than it answers but also might help lead you to a better understanding of EMR.

http://www.blazelabs.com/emwaves.htm


http://www.blazelabs.com/pics/empropagation2.gif

MentalAvenger
2007-Dec-07, 06:23 AM
A photon is a particle. That is an assumption. No one has ever seen a “photon”, except perhaps as the result of what-ever-it-is striking an optical receptor in their eye.


It travels like a particle. That does not mean it IS a particle. And it also may depend upon what you consider to be a “particle”. A particle of dust is made up of many different molecules. Those molecules are made up of many atoms, mostly empty space. Those atoms are made up of sub-atomic particles (supposedly). Perhaps those are also made up of largely empty space. No one knows.


The wavelike properties of light are the result of many photons interacting with each other. Can that be proven?


Again: Photons are particles. Perhaps. More likely they are something we are currently unable to sufficiently describe.


Observers in different places observe different photons, emitted along different trajectories.As far as you know. BTW, photons do not always travel at “c”. Or do they?

OldAlbert
2007-Dec-07, 09:59 AM
My post #17 was a mix of straightforward observations and some tentative conclusions.

One of my tentative conclusions was that an expanding (and therefore weakening by square law) wave might spawn many new quantm jumps (and hence photons) by 'pumping up' the energy of an electron cycle-by-cycle until the electron has accumulated enough energy to make the jump. Can anyone comment on that possibility?

Another crux of my posting was to raise the subject of self destruction of a photon, once 'detected'. Of course conservation of momentum requires that it cease to exist as soon as a new secondary photon emerges with new momentum, but just how the photon self destructs might be worthy of discussion. In any event, the fact that a photon does self destruct immediately it is detected implies that it does not HAVE to be of the nature of a bullet. An expanding wave front that self destructed as soon as it spawned a new secondary photon (possibly in order to conserve momentum) would make it appear to an observer that it is like a bullet (that it is here, and not there). This line of thought appears to reconcile the wave-and-particle duality.

Bogie
2007-Dec-07, 01:02 PM
Let's go with the photon as a particle, even with duality it is a particle.

http://www.cpepweb.org/images/chart_2006_4.jpg

OldAlbert
2007-Dec-07, 03:58 PM
Thank you for the chart Bogie, and a genuine thanks for taking an interest in these thoughts.

I suspect that the duality, once understood better, would be neither mysterious nor wierd. For example, if we define a particle as "being HERE and not THERE", and then if we were able to show how a wave-front could seem to fit that definition, then we could brush our hands together and say that there is no duality - just waves!

And I proposed a model to do that, with enough reasonability I think, to warrant further contemplation. My model is incomplete in that there is no explanation of how a weak wave segment can deliver enough energy to an atom to raise an electron's energy enough to instigate a quantum jump, and there is no explanation of the mechanism of the wave self destructing once the secondary photon comes into existence (although conservation of momentum seems to require it).

The answers to these "how"s, if there are any, would go a long way to arriving at a model for the photon.

OldAlbert
2007-Dec-07, 06:04 PM
Here is a little more thought on the self destruction of a wave when it creates a secondary photon.

In order to conserve momentum, it is conceivable that an electron, when it drops back to its normal orbit, creates a new spherical wavefront. Since the new spherical wave-front expands equally in all directions, conservation of momentum is protected. The 'return' facing part of the new wavefront is exactly in the direction of the original wave birth spot, and if it is exactly out of phase with the incoming wave, this may lead to simultaneous extinction of both the incoming wave and the 'return' wave (still maintaining conservation of momentum). This would leave the secondary wave (photon) carrying the exact value and direction of momentum as did the original incoming wave, thus ensuring conservation of momentum.

So when a wave is detected by a photon detector at one spot, it ceases to exist anywhere else, and the "it is HERE and not THERE" test for particle-ness is satisfied.

papageno
2007-Dec-07, 09:15 PM
I suspect that the duality, once understood better, would be neither mysterious nor wierd. For example, if we define a particle as "being HERE and not THERE", and then if we were able to show how a wave-front could seem to fit that definition, then we could brush our hands together and say that there is no duality - just waves!


The duality is there because our minds experience directly a classical world.

And the "just waves" you mention are the wave-functions of the quantum particles.




And I proposed a model to do that, with enough reasonability I think, to warrant further contemplation. My model is incomplete in that there is no explanation of how a weak wave segment can deliver enough energy to an atom to raise an electron's energy enough to instigate a quantum jump, and there is no explanation of the mechanism of the wave self destructing once the secondary photon comes into existence (although conservation of momentum seems to require it).


The destruction of one photon does not imply the creation of another.

You should try to think of photons as excitations in the electromagnetic fields, like the vibration in a guitar string.

OldAlbert
2007-Dec-08, 01:12 AM
papgeno:

True, but my reasoning does imply that the creation of a secondary photon implies the destruction of the causative photon.

I am trying to rationalize that 'bullet' type photons don't necessarily exist, but that instead ordinary electromagnetic waves can seem to be 'bullet' type photons because once an electromagnetic wave instigates the creation of a secondary electromagnetic wave (which is, I think, a photon also) the act of creating a secondary wavefront does somehow cause the demise of the original electromagnetic wave. And I am suggesting a potential mechanism by which it may happen.

Bogie
2007-Dec-08, 02:08 AM
papgeno:

True, but my reasoning does imply that the creation of a secondary photon implies the destruction of the causative photon.

I am trying to rationalize that 'bullet' type photons don't necessarily exist, but that instead ordinary electromagnetic waves can seem to be 'bullet' type photons because once an electromagnetic wave instigates the creation of a secondary electromagnetic wave (which is, I think, a photon also) the act of creating a secondary wavefront does somehow cause the demise of the original electromagnetic wave. And I am suggesting a potential mechanism by which it may happen.It confuses the issue when you refer to wavefronts. That term could be applicable in the sense that atoms that are energized and emitting photons will emit photons of differing frequencies and in all directions, i.e. one could refer to the radiation from excited atoms as a wavefront of photons.

Planck was the guy who first described photons as discrete buckets of energy and that the atom doesn't emit a photon until the particular bucket is filled. But once filled an individual photon, as quantum packet of energy, is emitted. But given a continuing source of energy, the atom keeps right on filling buckets and emitting photons and they are emitted in all directions depending on the position of the bucket in the electron configuration at the instant they are filled.

The emitted photon is referred to of course as electromagnetic radiation and each photon has frequency and wavelength. The wavefront that you refer to is the ongoing output of the EM that radiates in all directions and includes photons of differing frequencies (and wavelengths) if I interpret your meaning correctly.

These photons each have specific energy based on their frequency and they have a tiny momentum as described by the phenomenon of radiation pressure that I asked about in Q&A.

A photon can be absorbed, reflected, or it can pass right through mass. The radiation pressure coefficient is a number from zero to two, zero being a complete pass through, and 2 being complete reflection. Obviously the radiation pressure will normally fall between 1 and 2 for visible light frequencies. The reflection is absorption and reemission of photons and the re-emitted photons will contain some of the very same energy of the original photons after the buckets have been emptied and refilled in the absorbing atom.

http://www.blazelabs.com/pics/reflabstrans.gif

OldAlbert
2007-Dec-08, 03:27 AM
Bogie:

Thanks for bringing in the concept of building up enough energy in the activated atom to cause an electron to make the quantum jump. I have been thinking that the original electromagnetic wave impinging on the atom might accellerate an electron by, in effect, giving it a small kick to accelerate it as it orbits the nucleus, once per cycle - something like the way an accelerator accelerates electrons around the ring. There are some holes in my model though. But this helps with the concept that even a weak wave from a far-distant source can eventually 'pop up' and electron - it just takes longer for a wealk signal to do it than for a strong one.

I have been thinking of a single photon as being an expanding shell of electric field, hence the wavefront of a photon. A continuing wave would then be a string of such photons, each conscentric with and contained within its predecessor.

An interesting point is that since the activated atom was impinged on by only a segment of the spherical wavefront of the initiating wave, then the impinging wave would have a momentum in the direction of travel. Then after the whole process was completed the secondary radiation, being the entire result, would necessarily have the exact same momentum and direction of momentum, and conservation of momentum would thus be preserved.

Bogie
2007-Dec-08, 03:53 AM
Bogie:

Thanks for bringing in the concept of building up enough energy in the activated atom to cause an electron to make the quantum jump. I have been thinking that the original electromagnetic wave impinging on the atom might accellerate an electron by, in effect, giving it a small kick to accelerate it as it orbits the nucleus, once per cycle - something like the way an accelerator accelerates electrons around the ring. There are some holes in my model though. But this helps with the concept that even a weak wave from a far-distant source can eventually 'pop up' and electron - it just takes longer for a wealk signal to do it than for a strong one.

I have been thinking of a single photon as being an expanding shell of electric field, hence the wavefront of a photon. A continuing wave would then be a string of such photons, each conscentric with and contained within its predecessor.

An interesting point is that since the activated atom was impinged on by only a segment of the spherical wavefront of the initiating wave, then the impinging wave would have a momentum in the direction of travel. Then after the whole process was completed the secondary radiation, being the entire result, would necessarily have the exact same momentum and direction of momentum, and conservation of momentum would thus be preserved.But if the photon spreads out across a wavefront that increases as the photon gains distance from the source, the amount of energy that can be imparted to a distant atom would be a tiny arc of the wavefront and therefore only a tiny amount of the energy in the discrete packet?

So if that is what you are saying, then you must also be saying that the subsequent and consentric photons, each another wavefront, will also contribute a tiny arc of energy and the accumulated energy from the many concentric wavefronts will represent the energy absorbed by that distant atom? Then you also must be saying that all atoms in the path of the photon wavefronts will absorb a tiny arc of energy from a series of photon wavefronts and so the energy of each photon can be disbursed across a circumferance or a sphere of EM?

How is that result different from multple discrete packets of energy each with its own momentum and vector and each hitting only one atom, but with the emission of multiple photons in all directions as in the scenario that I described? We get the same result with the same amount of energy, only my scenario agrees with Planck idea of individual discrete packets of energy and it allows the photons to have individual vectors which I think is important.

OldAlbert
2007-Dec-08, 01:02 PM
Bogie:

It is apparent (from your critique) that I AM confusing the issue. I at once describe a photon as a single EM shell, and again as a sequence of EM shells.
I think I am thereby implying that at a distance, a single photon (shell) cannot pump up an electron's energy to cause the quantum jump - that it takes a series of photon segments to do that.

I don't know whether science can discern whether a single photon has been received, or whether a sufficient string of photons has been received!. Our single photon detectors are not really dirctly detecting incoming photons, but are, in fact, detecting secondary photons released within the detector. An avalanche diode probably has the best chance, because, in theory at least, it is reverse biased to such an extent that it may take only a small amount of incoming energy to push the electron over the line. I imagine the situation being like a swing, If the 'pushes' are timed right, and if the swing is already in motion, then the swing amplitude is increased. But if the swing is initially not in motion, can a reasonably sized 'push' establish the swing in full arc motion?

Plank's formula for energy/frequency certainly applies to each emitted photon (shell), but for remote atoms reacting to iminging energy would Plank's formula be applying to the toatal energy received (i.e., carried by a series of partial photons). This is not totally off the wall. Photons DO have a frequency property, and perhaps that is encoded simply into the rate at which partial photons impinge.

THis is all very thought provoking, and I for one will have more to post as I reflect on it over some time.

Thanks for your comment.

papageno
2007-Dec-09, 12:43 PM
I don't know whether science can discern whether a single photon has been received, or whether a sufficient string of photons has been received!.


If you are referring to electronic transitions in atoms and molecules, there are constraints that don't allow more than one photon to produce a "jump". Search for "selection rules".





Our single photon detectors are not really dirctly detecting incoming photons, but are, in fact, detecting secondary photons released within the detector.


Not if the original photon is transformed into an electric current or voltage.

OldAlbert
2007-Dec-09, 10:13 PM
papageno:

Yikes!! The only things I recognise in that reference are wavelength and reduced Plank's constant. Even so I don't recognise any conclusion to the effect you mention.

However, I wonder whether it might be possible that a single photon might be a series of shells each contained and concentric and one wavelength apart, for the purpose of pumping up an electron to make the jump. Since you clearly understand the selection rules, I am hoping that you are able to answer this question.

OldAlbert
2007-Dec-10, 04:57 AM
More specifically, does the Plank formula concern itself with the amount of photon energy required to cause the jump, or with the magnitude of the energy that a single photon would have to carry to do the job?

Is there good evidence that when a photon is released, that it shoots out like a bullet in a specific direction? For example can this be seen in cloud chamber interactions? That would nail it once and for all.

papageno
2007-Dec-11, 01:53 PM
Yikes!! The only things I recognise in that reference are wavelength and reduced Plank's constant. Even so I don't recognise any conclusion to the effect you mention.


If you are referring to the selection rules, these are consequence of the conservation principles.




However, I wonder whether it might be possible that a single photon might be a series of shells each contained and concentric and one wavelength apart, for the purpose of pumping up an electron to make the jump. Since you clearly understand the selection rules, I am hoping that you are able to answer this question.

A photon is not a series of concentric shells, "pumping" an electron.

You really should try to picture a photon as a travelling disturbance in the EM field. If one of the disturbances "hits" an atom and it has the right properties, it can allow an electron to be a bit further away from the nucleus in a stable way. In this case the disturbance will disappear.





More specifically, does the Plank formula concern itself with the amount of photon energy required to cause the jump, or with the magnitude of the energy that a single photon would have to carry to do the job?


The only Planck formula I remember is the one for black-body radiation.

Anyway, the identity of a photon is determined by its momentum (which fixes its energy) and its spin (intrinsic angular momentum). So "amount of photon energy" and "the magnitude of the energy that a single photon would have to carry" are exactly the same. Whether it will cause an electronic transition depends on whether the photon has the right energy and satisfies the selection rules: if it does, then it is absorbed with a certain probability.





Is there good evidence that when a photon is released, that it shoots out like a bullet in a specific direction? For example can this be seen in cloud chamber interactions? That would nail it once and for all.


The evidence exists. But not necessarily in a cloud chamber, because the cloud chamber can trace only charged particles, not photons.

In a teaching lab, for example, we detected the pair of gamma-photons emitted in electron-positron annihilation with photomultiplier tubes which give a well-defined direction.

OldAlbert
2007-Dec-11, 09:22 PM
Summary of seemingly reasonable conclusions to date:

1) When a 3D arc of a spherically expanding wavefront impinges on an atom, the energy of the target electron is increased. After enough such arcs of wavefront have impinged on that atom, the electron has acquired enough energy to 'make the quantum leap'. (papageno's post #38 disputes this). I imagine the energy additions being simillar to synchroton acceleration, so that the photon frequency is significant.

(To justify the concept of several 'hits' of EM energy 'pumping up' and electron, I point to the fact that in an avalanche photon detector working in geiger mode, the device is reverse biased to the point where the internal electric field is almost, but not quite, enough to cause an electron to jump from orbit. The energy of an impinging photon adds enough to cause the jump. This is not a perfect example, because the internal electric field is constant. In the case of partial photons 'pumping up' an electron, the energy accumulated to date must be somehow stored in the electron or its motion).

2) An EM wave type photon would appear to be a 'bullet' type photon, due to conservation of momentum. That is because:

The impinging arc of EM radiation has a momentum in the direction of the arc wavefront travel. When a photon is sreleased from the target atom,it is released as an EM wave, expanding spherically in all directions. That secondary photon, at this point in its life, has no average momentum. It has equal momentum in all directions with no resultant momentum. If the phse of the EM radiation wavefront electric field is in opposite phase to that of the impinging photon wavefront, the arc of the secondary photon wavefront would be directed back along the line of propogation of the impinging photon wave front, and the two would cancel, leaving only the remaining part of the sphere of secondary expanding wavefront which would carry exactly the same momentum as did the impinging photon wavefront, and in the same direction. (My argument now is weak, becasue I don't know how the cancellation takes place, but conservation of momentum requires that it does). The result is that the EM wave model passes the "it is HERE and not THERE test" and so it is behaving as a 'bullet' type photon.

I am arguing that there may not BE particle type photons - that it may be a case of EM forms being mistaken for particle forms because of the above mechanism.

Bogie
2007-Dec-11, 09:34 PM
And if you are correct then there is a background of EM that exerts radiation pressure in all directions, making a case for push gravity. Another aspect of such an EM background is the standing wave idea of particles. Vibrations and resonance play a big role in such an idea, allowing standing waves to combine at the right frequency. Are you thinking along any of those lines?

OldAlbert
2007-Dec-11, 09:49 PM
Bogie:

No but it sounds fascinating. Can you enlarge on that?

Bogie
2007-Dec-11, 10:54 PM
Bogie:

No but it sounds fascinating. Can you enlarge on that?That is some other ATM idea then. Keep your thread going.

I'm still not quite getting it yet. A photon expanding as it travels is where I am having trouble. You're starting from a static universe and defining expansion of the universe as a misinterpretation of red shift maybe, i.e. if the photon is emitted spherically and its energy is spread across a larger and larger sphere as it travels do you think it could affect what we think of as expansion. Then what we think of as red shift might be simply the expansion of the photon sphere maybe?

OldAlbert
2007-Dec-11, 11:07 PM
Bogie:

Nothing quite so esoteric.

Light expands as a spherical wavefront, growing larger and getting weaker (square law) with time. I am simply seeing a photon as a bit of light, expanding as light wavefronts do. However I am suggesting that one photon emission creates one spherical shell of EM field. And that a continuing generation of such photons, each one being a spherical shell of EM force, concentric with and within its predecessor. To an observer fixed in its path, this would appear as a normal EM wave (which it IS). I am declaring this to ba a stream of wave type photons.

To this I have added a rational for the "It is HERE and not THERE" test result that makes a wave type photon seem to be a particle type photon. And I am saying that this process is a necessary consequence of conservation of momentum.

papageno
2007-Dec-12, 01:44 PM
When a 3D arc of a spherically expanding wavefront impinges on an atom, the energy of the target electron is increased. After enough such arcs of wavefront have impinged on that atom, the electron has acquired enough energy to 'make the quantum leap'.


No.
It is not the number of wavefronts: you need a photon with the right energy, momentum and spin, and available states for the electron to "jump" to.




(papageno's post #38 disputes this). I imagine the energy additions being simillar to synchroton acceleration, so that the photon frequency is significant.


We are talking about electrons bound to nuclei, aren't we?




(To justify the concept of several 'hits' of EM energy 'pumping up' and electron, I point to the fact that in an avalanche photon detector working in geiger mode, the device is reverse biased to the point where the internal electric field is almost, but not quite, enough to cause an electron to jump from orbit. The energy of an impinging photon adds enough to cause the jump. This is not a perfect example, because the internal electric field is constant. In the case of partial photons 'pumping up' an electron, the energy accumulated to date must be somehow stored in the electron or its motion).


There is no such thing as "partial photon".

And your "pumping up" is inconsistent with experimental evidence (search for absorption lines).




An EM wave type photon would appear to be a 'bullet' type photon, due to conservation of momentum.


No, a photon is a quantum particle. As such it has properties typical of classical particles and properties typical of classical waves.





The impinging arc of EM radiation has a momentum in the direction of the arc wavefront travel.


That's because a macroscopic EM wave is the result of many photons.





When a photon is sreleased from the target atom,it is released as an EM wave, expanding spherically in all directions. That secondary photon, at this point in its life, has no average momentum. It has equal momentum in all directions with no resultant momentum.

SNIP!


And since the energy of a photon is directly proportional to the momentum, this "secondary photon" would have zero energy. A photon with zero energy is no photon. It does not exist because zero energy means that there is no excitation of the EM field.




I am arguing that there may not BE particle type photons - that it may be a case of EM forms being mistaken for particle forms because of the above mechanism.


Search for quantum particle.

OldAlbert
2007-Dec-13, 02:39 AM
Thanks papageno - will do. OA

OldAlbert
2007-Dec-13, 04:05 AM
papageno:

<No.
It is not the number of wavefronts: you need a photon with the right energy, momentum and spin, and available states for the electron to "jump" to.>

An avalanche diode in geiger mode uses a high reverse bias to raise electrons to a not-quite-high-enough state to make the jump, then a (partial?) photon adds the needed energy. Without this reverse bias voltage 'priming' the diode, the diode is much less sensetive, It seems that if a photon, to be effective, had to be capable of initiating the 'jump' in an unexcited atom, then reverse biasing the diode to close to breakdown would not be necessary or effective.

How does this fly with your statement re partial photons?


<We are talking about electrons bound to nuclei, aren't we?>

Yes. I should keep the conjecturing to a minimum. I have no idea how an electron can store energy or even be 'pumped up'.


Two photons embarking in opposite directions each have momentum, but the average momentum of the pair is zero. Right? So an EM wave front, expanding spherically in all directions, is without average momentum, but it has energy, I think. I need your comment on this.

papageno
2007-Dec-13, 12:10 PM
An avalanche diode in geiger mode uses a high reverse bias to raise electrons to a not-quite-high-enough state to make the jump, then a (partial?) photon adds the needed energy. Without this reverse bias voltage 'priming' the diode, the diode is much less sensetive, It seems that if a photon, to be effective, had to be capable of initiating the 'jump' in an unexcited atom, then reverse biasing the diode to close to breakdown would not be necessary or effective.

How does this fly with your statement re partial photons?


I told you: there is no such thing as "partial photon".

The voltage is necessary to obtain a measurable signal triggered by a single or few electrons: a photon excites a single electron which is accelerated by the bias voltage. The energy gained by the electron from the electric field is enough to excite other electrons, which are also accelerated by the voltage: an avalanche of electrons has started.





Yes. I should keep the conjecturing to a minimum. I have no idea how an electron can store energy or even be 'pumped up'.


The electrons in question are always part of a system.

For example, for electronic transitions in atoms, the electron is bound to a nucleus, and when a photon is absorbed the whole system (electron+nucleus) gains energy.
Typically one speaks of electron transitions because the electron mass is several orders of magnitude smaller than the mass of the nucleus, and therefore the frame of reference where the nucleus is at rest is most convenient.





Two photons embarking in opposite directions each have momentum, but the average momentum of the pair is zero. Right? So an EM wave front, expanding spherically in all directions, is without average momentum, but it has energy, I think. I need your comment on this.


A macroscopic EM wave is made of many photons. This is different from your single-spherically symmetric photon.
In a macroscopic EM wave, one can define energy and momentum densities, and their flows, which can be non-zero.

OldAlbert
2007-Dec-14, 04:05 AM
Papageno:

<I told you: there is no such thing as "partial photon".>
Got it! Just not ready to buy it yet. It has gotta be a difficult thing to prove either way. Even if it meant abandoning my convictions, you would be doing me a favour if you were to reason me out of thinking that there CAN be a partial photon, if that indeed were the case.



<The voltage is necessary to obtain a measurable signal triggered by a single or few electrons: a photon excites a single electron which is accelerated by the bias voltage. The energy gained by the electron from the electric field is enough to excite other electrons, which are also accelerated by the voltage: an avalanche of electrons has started.>
I think you believe that raising the reverse bias to nearly the breakdown point (or slighly past it with a quenching resistor) just results in a bigger avalanche, and has little to do with whether a partial photon then becomes capable of initiating the avalanche. You might be right. You certainly disqualify my use of geiger mode as a decisive example of the existence of partial photons.


<A macroscopic EM wave is made of many photons. This is different from your single-spherically symmetric photon.
In a macroscopic EM wave, one can define energy and momentum densities, and their flows, which can be non-zero.>
I understand, but how is it that a series of oscillations (transfers of energy between electric and magnetic fields) can express qualities of energy and momentum densities, while a single cycle of the wave cannot? It cannot be that energy and momentum are encoded into the way the individual cycles are organized together, for they are simply consecutive. These attributes MUST be encoded into each cycle of the wave (and so in my model, into each photon.)


No intent to be argumenative Papageno - just enjoying the debate.

papageno
2007-Dec-14, 08:35 AM
Got it! Just not ready to buy it yet. It has gotta be a difficult thing to prove either way. Even if it meant abandoning my convictions, you would be doing me a favour if you were to reason me out of thinking that there CAN be a partial photon, if that indeed were the case.


How did you reason into this conviction?





I think you believe that raising the reverse bias to nearly the breakdown point (or slighly past it with a quenching resistor) just results in a bigger avalanche, and has little to do with whether a partial photon then becomes capable of initiating the avalanche. You might be right.


You don't have t believe me: go to a library and read the books that deal with those systems.





You certainly disqualify my use of geiger mode as a decisive example of the existence of partial photons.


You never qualified it as such in the first place.





I understand, but how is it that a series of oscillations (transfers of energy between electric and magnetic fields) can express qualities of energy and momentum densities, while a single cycle of the wave cannot?


Who said that a single cycle cannot do work on a charged particle?





It cannot be that energy and momentum are encoded into the way the individual cycles are organized together, for they are simply consecutive. These attributes MUST be encoded into each cycle of the wave (and so in my model, into each photon.)


A photon is not a single cycle of wave (otherwise you would not get a single-particle interference, as shown in experiments).

I pointed you several times to think of a photon as a small excitation in the EM field, but you don't seem to follow this lead.