PDA

View Full Version : How should we describe the gravitational redshift effect?



Sam5
2005-Apr-27, 07:52 PM
Hi,

Sean has started a thread in which guys are discussing the reason why light bends as it moves through a gravity field. I don’t have much of an opinion about that, but there is a second part to that same Einstein theory, which is generally called the “gravitational redshift” effect.

So I’ve started this thread so we can discuss the gravitational redshift effect that has been observed and that Einstein predicted as early as 1911. I’m talking specifically about the part of the theory where light is being emitted on the surface of a massive astronomical body and is going straight up where it is later received as being “redshifted” at some distance above and away from the surface of the massive body.

Einstein originally explained the cause and the mechanics of the effect one way, which I agree with, but the modern description is completely different and is supposed to be based on the 1959 Pound and Rebka explanation. Unfortunately, the 1959 Pound-Rebka paper is not available unless one subscribes to an expensive physics journal archives, so not many people have ever read it.

My opinion is that the original Einstein description of what happens is correct, while the modern description is not correct. The effect is the same, but the two different descriptions of the reasons why it happens, are not the same. I also have the opinion that most people are not aware of the original Einstein description because there is not much written about it except in his own papers that are available mostly in very expensive books.

So, ironically and unfortunately, the full Einstein description requires a substantial expenditure of cash to read, and so does the Pound-Rebka description.

My interpretation of the two different descriptions is this:

Einstein description: The oscillation rates of atoms slow down in a strong gravity field, causing the atoms to emit a lower frequency of light than atoms in a weaker gravitational field. The light is redshifted to start with, from the very beginning, when emitted from atoms inside a strong gravity field.

Modern description: The frequency of light coming from atoms inside a gravity well changes and redshifts as the light “struggles” to climb up out of the gravity well, and the redshift process takes place while the photons are in transit, because they “lose energy” and they redshift during their struggle to get out of the gravity well.

Ok, I take the original Einstein point of view, so what is your opinion about this?

(I’ve got to go out later today, but I’ll get back to this later.)

SeanF
2005-Apr-27, 09:43 PM
Hi,

Sean has started a thread in which guys are discussing the reason why light bends as it moves through a gravity field.
That ain't why I started the thread.

Why don't you come over (http://www.badastronomy.com/phpBB/viewtopic.php?t=21308&) and answer the question I did ask?

Sam5
2005-Apr-28, 03:08 AM
Sam5, what do you think these guys mean by the word "universal" in the sentence, "That ['simple but nevertheless correct'] way centers on the universal modification of the rate of a clock exposed to a gravitational potential"?


Here is their full paragraph:

“The present article contains little original material; it is primarily pedagogical. The gravitational redshift being, both theoretically and experimentally, one of the cornerstones of General Relativity, it is very important that it always be taught in a simple but nevertheless correct way. That way centers on the universal modification of the rate of a clock exposed to a gravitational potential. An alternative explanation in terms of a (presumed) gravitational mass of a light pulse – and its (presumed) potential energy – is incorrect and misleading. We exhibit its fallacy, and schematically discuss redshift experiments in the framework of the correct approach. We want to stress those experiments in which an atomic clock was flown to, and kept at, high altitude and subsequently compared with its twin that never left the ground. The traveller clock was found to run ahead of its earthbound twin. The blueshift of clocks with height has thus been exhibited as an absolute phenomenon. One sees once over again that the explanation of the gravitational redshift in terms of a naive ‘attraction of the photon by the earth’ is wrong.”

I think they mean what they say on page 1 of their paper:

“Most treatises on GR [11], [12] follow the definitive reasoning of Einstein [2] according to which the gravitational redshift is explained in terms of universal property of standard clocks (atoms, nuclei). The proper time interval between events of emission of two photons as measured by the standard clock at the point of emission is different from the proper time interval between events of absorption of those photons as measured by identical standard clock at the point of absorption (in this way it was first formulated in [13]).”

http://arxiv.org/pdf/physics/9907017

Here’s what the University of Toronto says about it, which is pretty much the same as my position on this matter:

“Both Special and General Relativity predict the slowing down of clocks under certain circumstances, and experiments to test the predictions of either theory must take account of the predictions of the other, since the effects of both theories are often present. Most General Relativity tests use atomic clocks that emit gamma or X-rays at specific frequencies; these frequencies provide the time scale. General Relativity predicts that clocks in greater gravitational fields will run more slowly than those in weaker gravitational fields; thus an atomic clock in a higher gravitational field will emit light of a lower frequency compared to a clock in a lower gravitational field. Since lower frequencies correspond to longer wavelengths and, for visible light, shifts to the red part of the spectrum, the effect predicted by the General Relativity is usually called the Gravitational Redshift.”

www.upscale.utoronto.ca/GeneralInterest/Key/relgen.htm+%22gravitational+redshift%22+atomic+clo cks&hl=en&lr=lang_en&ie=UTF-8]SOURCE: (http://64.233.161.104/search?q=cache:2tfAmnV3U8IJ:[url) scroll down to “gravitational redshift”[/url]

So they don’t say the photons “lose energy” as they “struggle” to “climb” out of the gravity well.

The “clocks”, by the way, in the 1911 theory, were atomic clocks, which were fundamental oscillating atoms. While these atoms are the “clocks,” they are also the things that emit the light. So when the oscillation rates of the atoms slow down in a gravity field, they automatically emit light of a lower frequency, so the light is redshifted as it is being emitted from the slowly oscillating atoms in the gravity field. The atomic clocks that are in the weaker gravity field or are in no noticeable field at all will be running faster (oscillating faster) and will notice that the incoming light waves from the atoms deep in the gravity well are of a lower frequency, i.e. “redshifted”. In the original theory, and in this University of Toronto explanation, and in my opinion, the redshift takes place as the light is being emitted from the slowly oscillating atoms down in the gravity well. It does not redshift while in route from the emitter to the observer.

Sylas
2005-Apr-28, 04:55 AM
Einstein description: The oscillation rates of atoms slow down in a strong gravity field, causing the atoms to emit a lower frequency of light than atoms in a weaker gravitational field. The light is redshifted to start with, from the very beginning, when emitted from atoms inside a strong gravity field.

Modern description: The frequency of light coming from atoms inside a gravity well changes and redshifts as the light “struggles” to climb up out of the gravity well, and the redshift process takes place while the photons are in transit, because they “lose energy” and they redshift during their struggle to get out of the gravity well.

Ok, I take the original Einstein point of view, so what is your opinion about this?

My take is that I also prefer the first description, and this is how I would calculate it as well. I'd use the gravitational time dilation to calculate frequency differences measured at different points in the field.

Given your quotes, at least some modern sources do this as well, so I am not sure why you refer to the modern description.

Some people do describe photons as losing energy as they rise through the field. This almost makes sense in a co-ordinate system corresponding to an observer who moves through the field; but for such an observer the photon frequency depends on the observer's location in the field, not the photon's location in the field! This is a good case for prefering the first explanation; which is that the energy of photons being emitted from a certain point will be different depending on the altitude of the observer making the measurements.

It may be that some non-standard co-ordinate systems have individual photons changing energy; but I'll leave that to GR experts. I don't think it is a good explanation, and it would have some odd implications. My view here is related to my objection to the co-ordinate systems which match up cosmological redshifts to Doppler redshifts. They hinder more than they help.

In the meantime, the first description is the one that works in the co-ordinate systems that work most naturally.

Pound and Rebka performed the famous Harvard Tower experiment, which showed as a matter of simple fact that photons which had a certain energy as measured at one point in the gravitational field had a different energy as measured at another point in the gravitational field.

To say that the photons "lose energy" or "gain energy" implicitly refers to an observer who who is able to see a difference in energy, and the only way an observer can do this is by moving themselves from one place in the field to another. But from the perspective of an observer at a given point in the field, all the photons in a continuous stream from a monochromatic source have a certain frequency, and were emitted and received at that same frequency. The difference in frequency for different observers corresponds to different views of the frequency at which the photons were emitted.

Cheers -- Sylas

Grey
2005-Apr-28, 02:31 PM
The thing I'd criticize in your description is that it makes it sound as though it has something to do with atomic oscillation rates specifically (previous conversations have suggested that you think it does, though you may have changed your mind about that). Remember, though, that you'd also observe a redshift if the radiation were radio waves from a current in an antenna, or a blackbody spectrum, or gamma radiation from nuclear transitions, or any other source. It doesn't just happen to radiation from atomic transitions.

SeanF
2005-Apr-28, 03:15 PM
Sam5, what do you think these guys mean by the word "universal" in the sentence, "That ['simple but nevertheless correct'] way centers on the universal modification of the rate of a clock exposed to a gravitational potential"?


Here is their full paragraph:
I really didn't want a cut and paste, I wanted to know what you thought.

Why do they say "universal modification of the rate of a clock" as opposed to "atomic" or somesuch?

Sam5
2005-Apr-28, 05:44 PM
Sylas, Grey,

Ok, good. I appreciate the information and opinions. The first description is the way I interpret the original theory and the actual effect, and it seems to me that experimental results have proven the original theory to basically be correct. :D

Sylas
2005-Apr-28, 11:23 PM
Sylas, Grey,

Ok, good. I appreciate the information and opinions. The first description is the way I interpret the original theory and the actual effect, and it seems to me that experimental results have proven the original theory to basically be correct. :D

Yes. Note that the original theory does not merely say "atoms slow down". I missed that rather odd expression in your formulation, which could be confused with the notion that the effect is limited to atoms. The original theory says that gravity has an effect on spacetime itself, and that all clocks run more slowly in the gravitational field. It is not an effect on the atoms, so much as an effect on spacetime. That the original and the current theory.

Sam5
2005-Apr-28, 11:56 PM
Sylas, Grey,

Ok, good. I appreciate the information and opinions. The first description is the way I interpret the original theory and the actual effect, and it seems to me that experimental results have proven the original theory to basically be correct. :D

Yes. Note that the original theory does not merely say "atoms slow down". I missed that rather odd expression in your formulation, which could be confused with the notion that the effect is limited to atoms. The original theory says that gravity has an effect on spacetime itself, and that all clocks run more slowly in the gravitational field. It is not an effect on the atoms, so much as an effect on spacetime. That the original and the current theory.

Hi Sylas,

The term I used above was “the oscillation rates of atoms slow down in a strong gravity field.”

The earliest version I can find from Einstein is in one of his 1907 papers (Anna Beck translation):

“There exist ‘clocks’ that are present at locations of different gravitational potentials and whose rates can be controlled with great precision; these are the producers of spectral lines. It can be concluded from the aforesaid that the wave length of light coming from the sun’s surface, which originates from such a producer, is larger by about one part in two millionth than that of light produced by the same substance on earth.”

This generally comes from Maxwell’s suggestion about atoms essentially being ‘clocks’, published in 1873.

Back in the 1920s Charles Steinmetz was one of Einstein’s friends, and he explained the situation quite simply in his 1923 book, “Four Lectures in Relativity and Space”:

”We cannot carry a clock from the earth to Betelgeuse, but we do not need to do this, since every incandescent hydrogen atom, for instance, is an accurate clock, vibrating at rate definitely fixed by the electrical constants of the hydrogen atom and showing us the exact rate of its vibration in the spectroscope by the wave length or frequency of its spectrum lines. Thus in a strong gravitational field the frequency of luminous vibrations of the atoms should be found slowed down; in other words, the spectrum lines should be shifted towards the red end of the spectrum.”

In Einstein’s 1911 paper, he said:

“Es sei vo die Schwingungszahl eines elementaren Lichterzeugers, gemessen mit einer an demselben Orte gemessenen Uru U.”

“vo” is the frequency of the “elementary light producer.” The frequency of the light producer is what changes in the gravity field and the change is what causes the redshift.

worzel
2005-Apr-29, 12:29 AM
I don't understand the problem here. Neither is a theory which contradicts the other. Both are just different ways of describing a feature of GR, aren't they?

Kaptain K
2005-Apr-29, 01:10 AM
I don't understand the problem here. Neither is a theory which contradicts the other. Both are just different ways of describing a feature of GR, aren't they?
Yes, but Sam5 doesn't see it that way.

Sylas
2005-Apr-29, 01:21 AM
Sam5, all your quotes are about observations, and consequences of the theory for observations. Einstein's theory was explicitly and plainly an effect of gravity on spacetime, in such a way that physical laws remain consistent for all observers. All his explanations were consistently based on relativistic effects on spacetime itself. The rate at which time passes is relative to an observer. It is not an effect limited to atomic clocks; and to take Einstein as suggesting any such a thing is a gross failure of comprehension of his original papers and also of all subsequent development in physics.

Worzel, you are correct. The way you describe a phenomenon can depend on the co-ordinates you use in the description. My quibble is that picking the co-ordinates to justify a particular form of explanation can be a bad idea, because such co-ordinates may have other odd consequences that obscure the situation.

Cheers -- Sylas

Sam5
2005-Apr-29, 01:46 AM
You can call the slowdown in the oscillation rate of an atom in a gravity field “an effect of gravity on spacetime” if that’s the way you prefer to describe it. But I think younger students would understand it much better with the Steinmetz type of explanation, which is the way Einstein explained it. The oscillation rate of the atom slows down in a gravity field and thus the atom emits a lower frequency of light. Both Einstein in 1907 and 1911 and Steinmetz in 1923 were speaking specifically about atoms, fundamental “atomic clocks”.

“Light is emitted at a lower frequency and longer (or redder) wavelength in a gravitational field than in the absence of a gravitational field.”

http://www.site.uottawa.ca:4321/astronomy/index.html#gravitationalredshift

Sylas
2005-Apr-29, 02:19 AM
You can call the slowdown in the oscillation rate of an atom in a gravity field “an effect of gravity on spacetime” if that’s the way you prefer to describe it. But I think younger students would understand it much better with the Steinmetz type of explanation, which is the way Einstein explained it. The oscillation rate of the atom slows down in a gravity field and thus the atom emits a lower frequency of light. Both Einstein in 1907 and 1911 and Steinmetz in 1923 were speaking specifically about atoms, fundamental “atomic clocks”.

Your explanation does not match the Einsteinian explanation, and is actively wrong. Students will only find your example useful as an exercise for identifying its defects, in order the grasp the difficult notions of relativity of time and space as described by Einstein, and as you have not yet assimilated.

I don't know about Steinmetz, but to say Einstein in 1907 and 1911 was speaking specifically about atoms is a gross failure to comprehend the original papers.

The pheneomenon that light is emitted at lower frequencies in a gravitational field is explained by Einstein as a consequence of relativity of time and space; not as a change to a certain kind of clock. Atomic clocks are useful as fundamental ways to measure time and show those effects, precisely because the characteristic frequencies used are not altered by gravitational forces.

Sound like a contradiction? No; this is the essense of relativity, as Einstein originally conceived it and as it is used today. An observer with such a clock always sees that clock work with same frequency, even in an extremely strong gravitational field, or under a powerful acceleration. Call this observer A.

Another observer B, displaced from the clock, will see everything associated with observer A proceeding more slowly, and they can measure how slowly by observing the frequency photons emitted by the clocks of observer A.

Cheers -- Sylas

Sam5
2005-Apr-29, 03:40 AM
Of course he was talking about atoms when he said: “the frequency of an elementary light-generator” and “There exist ‘clocks’ that are present at locations of different gravitational potentials and whose rates can be controlled with great precision; these are the producers of spectral lines.”

You don’t need “observers” at either of two atomic clocks. The one at sea level will run (“tick”) slightly slower than the one at a high elevation. This is a basic law of nature. There is no need to try to mystify this phenomenon.

Sylas
2005-Apr-29, 04:11 AM
Of course he was talking about atoms when he said: “the frequency of an elementary light-generator” and “There exist ‘clocks’ that are present at locations of different gravitational potentials and whose rates can be controlled with great precision; these are the producers of spectral lines.”

You are cherry picking, and not reading the papers in full, and in particular you are NOT reading the EXPLANATION of the phenomenon. I suspect because it is mathematical, and you skip the maths. In doing so, you have ended up failing to understand the material at all.

Einstein's papers start out by describing the theory. There is lots of reference to "clocks" in general, as ways of measuring time. There is no reference in this explanatory section for the theory itself of anything to do with "atoms" in particular; except the general foundational principle that the laws of physics are the same for all observers.

Then, in the latter portion of the paper, he presents observational consequences. The atoms stand as example of clocks for measuring time; and the explanation that you have entirely missed is the geometry of spacetime and the relative nature of time and space. Atoms are examples of a way of showing consequences of the distortions in spacetime.

The model most emphatically and definitely is NOT "specific" to atoms. It is an egregious error, and diametrically opposed to the fundamental principles Einstein explained in his original papers, to think that Einstein was suggesting some special effect on atoms. The whole point of relativity is that there is no special effect of gravity on the laws of physics. A clock that vibrates at a certain frequency at one place will do the same at another place.... and when it is observed from a distance you can see that the co-ordinates for time and space are dependent on (relative to) the observer.


You don’t need “observers” at either of two atomic clocks. The one at sea level will run (“tick”) slightly slower than the one at a high elevation. This is a basic law of nature. There is no need to try to mystify this phenomenon.

The basic law of nature is precisely the reverse of what you state here. The clocks have a characteristic frequency, and they always vibrate at that frequency. The ONLY way you can detect a change in frequency is at a distance, which is why Einstein makes the notion of observer so central. If you travel up to the mountain, and measure the rate of a clock up there, and then travel back down and measure the rate of a clock down there, you get the SAME ANSWER in both cases. The effect is a change in spacetime itself, and NOT SPECIFIC TO ATOMS.

I'm not mystifying; I'm correcting simple errors of comprehension. The notion is not really that difficult (by comparison with, say, quantum physics!) But you do have to understand that it is all about spacetime.

Cheers -- Sylas

Sam5
2005-Apr-29, 04:40 AM
The ONLY way you can detect a change in frequency is at a distance, which is why Einstein makes the notion of observer so central.

No, you can detect a difference at the location of the clock when you compare it with the rate of another clock at a different elevation in a different gravity field or when you compare it with astronomical time. You don’t need to be “at a distance” from an atomic clock to see that it is getting out ahead or behind astronomical time. It is the gravity field that causes the tick rate change of the atomic clock. This has nothing to do with an observer being in different places. It’s the same with a pendulum clock which speeds up in a gravity field where an atomic clock slows down. They are both obeying the different laws of physics that govern their tick rates.

Regarding the redshift, this has everything to do with atoms. The whole idea of “time dilation” and changing atomic clock rates came from Lorentz in the 1890s and it was directly related to atomic oscillation rates. This came from Maxwell’s comments about atoms being like clocks because he thought their oscillation rates were steady. Lorentz theorized that their rates were not steady all the time under all circumstances and environments. Einstein took that idea and added his own ideas to it. Read his credit to Lorentz about time dilation idea in his 1907 paper.

You are taking this thread off topic in your discussion about clocks.

The point is, the original theory says the light is redshifted as it is being emitted in a gravity field, from the very beginning, from an “elementary light-generator” and “producers of spectral lines”. Certainly he was talking about atoms and not about mechanical alarm clocks or pendulum clocks emitting light on the surface of the sun. As Steinmetz said, an “incandescent hydrogen atom, for instance, is an accurate clock, vibrating at rate definitely fixed by the electrical constants of the hydrogen atom and showing us the exact rate of its vibration in the spectroscope by the wave length or frequency of its spectrum lines. Thus in a strong gravitational field the frequency of luminous vibrations of the atoms should be found slowed down; in other words, the spectrum lines should be shifted towards the red end of the spectrum.” This is very simple, and there is no need to try to mystify it. The light does not redshift while in transit up and out of the field. Not according to the original theory.

Kaptain K
2005-Apr-29, 07:12 AM
I suspect because it is mathematical, and you skip the maths.
Bingo!

Ricimer
2005-Apr-29, 08:13 AM
No, you can detect a difference at the location of the clock when you compare it with the rate of another clock at a different elevation in a different gravity field or when you compare it with astronomical time. You don’t need to be “at a distance” from an atomic clock to see that it is getting out ahead or behind astronomical time. It is the gravity field that causes the tick rate change of the atomic clock. This has nothing to do with an observer being in different places. It’s the same with a pendulum clock which speeds up in a gravity field where an atomic clock slows down. They are both obeying the different laws of physics that govern their tick rates.

okay, a few points: If you are comparing it to a clock at a different elevation...you are checking a clock at a different distance.

And as for the pendulum clock vs atomic clock, those time changes are due to different mechanisms. Also, should you properly acount for how the pendulum clock changes due to a shift in the strength of gravity, you should produce the wrong answer, as the gravitational time dilation will be applied on top of that.

Just trying to make that comparison is a poor experimental setup. If I want to test how an objects electrical properties change with temperature, but I use a measuring device whose readings of those electrical properties is also susceptible to the temperature, I will be unable to determine how much of the observed change is due to the material's properties or the instruments.

To determine it I must take into consideration the behavior of the instrument under those conditions, and remove those effects. For instance, if the measurements have been shown to shift by x amount under the same temperature conditions in the experiment, then I can scale the measurement appropriately to obtain the actual measurement.

So by using a pendulum clock, and not correcting for how the clock measures the passage of time differently due to enviromental factors (even air viscosity can affect it!), you are introducing many complicating and obscuring factors.

The time shift in a pendulum clock due to the different force of gravity is not the same time shift GR predicts due to gravitational time dilation. As such making a comparison will lead you to false conclusions.

worzel
2005-Apr-29, 08:48 AM
Sam5: you appear to be in denail about the reality of spacetime itself. Do you believe that universe is just 3D Eucildean afterall?

SeanF
2005-Apr-29, 01:50 PM
See, this is the discussion I was hoping for when I asked Sam about his cite referring to "the universal modification of the rate of a clock exposed to a gravitational potential."

It's universal, Sam. All clocks - be they atomic, crystal, pendulum, or biological - will experience the effect. Some of them will experience additional effects, of course, but the GR effect is universal. The examples use atomic clocks not because they're the only clocks subject to the effect, but because they are not subject to the other effects.

Sam5
2005-Apr-29, 03:27 PM
Sylas has sidetracked this thread onto a discussion about clocks. The subject of this thread is which is the best way to describe the gravitational redshift effect, by saying the light starts out as redshifted when emitted from atoms located inside a strong gravity field or does it start out at a normal frequency and then redshifts while in route as it “struggles” to “climb out” of the gravity field. The original gravitational redshift theory says the light starts out redshifted in the first place.

Sean, if you want to start a thread about “Do all clocks slow down in a gravity field”, then go ahead.

Ricimer
2005-Apr-29, 03:36 PM
the test for that, is to see if any blueshifting of incoming starlight occurs due to local gravity.

but as I see it, there really are two ways to look at it. It's either a time dilation effect, due to being in different zones of gravitational effect (the it starts redshifted idea).

Or it's an energy loss. But they don't have to be mutually exclusive.

papageno
2005-Apr-29, 03:44 PM
Sylas has sidetracked this thread onto a discussion about clocks. The subject of this thread is which is the best way to describe the gravitational redshift effect, by saying the light starts out as redshifted when emitted from atoms located inside a strong gravity field or does it start out at a normal frequency and then redshifts while in route as it “struggles” to “climb out” of the gravity field. The original gravitational redshift theory says the light starts out redshifted in the first place.
If you are sitting next to the emitter inside a strong gravity field, you do not see the light red-shifted.
You see the red-shift if you sit outside the gravity field.
And this is confirmed experimentally.


Pound and Rebka performed the famous Harvard Tower experiment, which showed as a matter of simple fact that photons which had a certain energy as measured at one point in the gravitational field had a different energy as measured at another point in the gravitational field.
They used gamma radiation (i.e., from nuclear transition) emitted by Fe57 (if I remember correctly), which allowed them to exploit the Mossbauer effect for greater accuracy.

jnik
2005-Apr-29, 03:49 PM
Sylas has sidetracked this thread onto a discussion about clocks. The subject of this thread is which is the best way to describe the gravitational redshift effect, by saying the light starts out as redshifted when emitted from atoms located inside a strong gravity field or does it start out at a normal frequency and then redshifts while in route as it “struggles” to “climb out” of the gravity field. The original gravitational redshift theory says the light starts out redshifted in the first place.
No, it doesn't. What you're referring to as the "original" theory states that the light as observed from a fixed observer outside never changes in frequency. The "tired photon" explanation is from the point of view of an observer travelling with the photon--or, rather, a succession of stationary observers along various points in the path.

By saying "the light starts out redshifted in the first place," you imply that, if one were standing next to a flashlight in a strong gravitational field, it would appear redder than if one were standing next to the same flashlight in the absence of gravitation. That's false--it would be the same colour in both cases.

Sam5
2005-Apr-29, 03:54 PM
the test for that, is to see if any blueshifting of incoming starlight occurs due to local gravity.
That’s an interesting point. Are you aware of any blueshift from any star as the light from it “falls” down our gravity well?

I think the problem in determining this is that stars are more massive than the earth, so their light should always naturally be slightly redshifted when received on earth, since the earth standard for “normal” spectral lines would be measured on earth from earth-based light sources. So, if the earth based sources are in a weak gravity field on earth, it is they that should be “blueshifted” when compared to the “redshifted” light coming from the stars.

There is a long history of studying spectral lines from stars that I’ve been reading about. It goes back to Huggins in the 1860s when astronomers compared the spectral lines of stars to the spectral lines of the same chemicals here on earth, such as hydrogen. They couldn’t see any slight difference due to the gravitational redshift effect, but they could see a difference due to the motion of stars toward or away from the earth. It was Maxwell in the 1870s who first mentioned that glowing atoms and their light could be used as “clocks”. Lorentz in the 1890s first suggested that the oscillation rates of atoms could slow down under certain circumstances. In a 1907 paper Einstein credited Lorentz’s theory as being the origin of his (Einstein’s) time dilation concepts.

While some physics sources describe the gravitational redshift phenomenon in terms of the atoms in a gravity field emitting the light as being redshifted to start off, most sources today tend to explain it in terms of the light “losing energy” and redshifting while it travels out of the gravity field. However, this way of explaining it falsely implies that the light is actually emitted at a normal frequency and wavelength, and then redshifts later, during its travel, which I don’t think is correct.

SeanF
2005-Apr-29, 04:15 PM
Sean, if you want to start a thread about “Do all clocks slow down in a gravity field”, then go ahead.
I did (http://www.badastronomy.com/phpBB/viewtopic.php?t=21308&). You mentioned it in the original post in this thread, and I directly linked to it in the first response (http://www.badastronomy.com/phpBB/viewtopic.php?t=21314&start=1&).

You, so far, have ignored it.

I wonder why.

papageno
2005-Apr-29, 04:19 PM
While some physics sources describe the gravitational redshift phenomenon in terms of the atoms in a gravity field emitting the light as being redshifted to start off, most sources today tend to explain it in terms of the light “losing energy” and redshifting while it travels out of the gravity field. However, this way of explaining it falsely implies that the light is actually emitted at a normal frequency and wavelength, and then redshifts later, during its travel, which I don’t think is correct.
Based on what?
The (currently accpeted) theory says that the frequency is the "normal" one. Experimental results support this.
The red-shift is observed if emitter and observer are at different gravitational potentials.
Why do you think this is not correct?

Sam5
2005-Apr-29, 04:20 PM
If you are sitting next to the emitter inside a strong gravity field, you do not see the light red-shifted.
You see the red-shift if you sit outside the gravity field.
And this is confirmed experimentally.


If you are sitting next to the emitter inside a gravity field, such as at sea level on earth, of course you won’t “see” the light redshifted because you’ve got to compare that light (through a spectroscope) to another source of light from the same chemical element which is ALSO located at the same place in the same gravity field, and so BOTH sources will be redshifted exactly the same amount. But that doesn’t mean your biological clock has “slowed down” the same amount at sea level, it just means that both of your two sources of the light – the “emitter” light and the “standard” light that you compare it with – will both be redshifted exactly the same amount because both sources are equally redshifted in the same gravity field.

However, you could do this: Take a modern manufactured atomic clock to a high elevation and use it to determine the rotation rate of a steady rapid pulsar. Determine how many times the star rotates per month, using the atomic clock to determine a “month”, i.e. 30 atomic clock days. Then take the atomic clock down to sea level and again time the rotation of the pulsar. You should notice that the pulsar will rotate more times per atomic clock “month” at sea level. That tells you that your atomic clock has slowed down its “tick” rate at sea level.

The Mossbauer effect allowed Pound and Rebka to make detailed measurements, but it didn’t settle the question as to whether or not the light at the base of the tower was emitted already redshifted at the base, or was emitted at a normal frequency at the base and then redshifted as it “lost energy” as it moved upward toward the top of the tower. The original gravitational redshift theory (and my opinion too) says it was emitted as redshifted in the first place, and that’s what Pound and Rebka were observing with their Mossbauer effect.

papageno
2005-Apr-29, 04:37 PM
If you are sitting next to the emitter inside a gravity field, such as at sea level on earth, of course you won’t “see” the light redshifted because you’ve got to compare that light (through a spectroscope) to another source of light from the same chemical element which is ALSO located at the same place in the same gravity field, and so BOTH sources will be redshifted exactly the same amount.
The fact that the word "redshift" contains the word shift, already implies that there is a reference.
When you say "BOTH sources will be redshifted exactly the same amount", what is the reference?


But that doesn’t mean your biological clock has “slowed down” the same amount at sea level, it just means that both of your two sources of the light – the “emitter” light and the “standard” light that you compare it with – will both be redshifted exactly the same amount because both sources are equally redshifted in the same gravity field.
Except that whatever clock has been used, the results are consistent with time slowing down.



However, you could do this: Take a modern manufactured atomic clock to a high elevation and use it to determine the rotation rate of a steady rapid pulsar. Determine how many times the star rotates per month, using the atomic clock to determine a “month”, i.e. 30 atomic clock days. Then take the atomic clock down to sea level and again time the rotation of the pulsar. You should notice that the pulsar will rotate more times per atomic clock “month” at sea level. That tells you that your atomic clock has slowed down its “tick” rate at sea level.
And this does not happen only with atomic clocks.



The Mossbauer effect allowed Pound and Rebka to make detailed measurements, but it didn’t settle the question as to whether or not the light at the base of the tower was emitted already redshifted at the base, ...
"Redshifted" with respect to what?



...or was emitted at a normal frequency at the base and then redshifted as it “lost energy” as it moved upward toward the top of the tower. The original gravitational redshift theory (and my opinion too) says it was emitted as redshifted in the first place, and that’s what Pound and Rebka were observing with their Mossbauer effect.
They observed the redshift due to a difference in gravitational potential between source and observer.


EDIT to add: in the Doppler effect, you do not observe a shift in frequency if source and observer are at rest with respect to each other.

Sam5
2005-Apr-29, 04:52 PM
The Mossbauer effect allowed Pound and Rebka to make detailed measurements, but it didn’t settle the question as to whether or not the light at the base of the tower was emitted already redshifted at the base, ...
"Redshifted" with respect to what?

I think the atomic clock standard today is based on the gravity field at “sea level”. But you can say it either way regarding the Harvard Tower experiment. You can say the light emitted by atoms at the top of the tower is emitted as “blueshifted” when compared to the light emitted from atoms at the bottom of the tower. Or you could say that the light emitted from the atoms at the bottom of the tower is emitted as “redshifted” when compared to the light from the atoms at the top of the tower. It’s all relative. The original 19th Century standard for “normal” spectral lines of light was simply an earth-surface standard, and their instruments could not tell the difference in altitude. But I think the atomic clock standard today is considered to be at “sea level”. I think that is the world atomic clock time standard, the “sea level” standard. I suppose if the earth heats up and all the polar ice caps melt, the sea level will go up and that’ll throw all our atomic clocks off the current standard. :D

papageno
2005-Apr-29, 05:10 PM
The Mossbauer effect allowed Pound and Rebka to make detailed measurements, but it didn’t settle the question as to whether or not the light at the base of the tower was emitted already redshifted at the base, ...
"Redshifted" with respect to what?
I think the atomic clock standard today is based on the gravity field at “sea level”. But you can say it either way regarding the Harvard Tower experiment. You can say the light emitted by atoms at the top of the tower is emitted as “blueshifted” when compared to the light emitted from atoms at the bottom of the tower. Or you could say that the light emitted from the atoms at the bottom of the tower is emitted as “redshifted” when compared to the light from the atoms at the top of the tower. It’s all relative.
Exactly, that's why you should say where the comparison is done.




The original 19th Century standard for “normal” spectral lines of light was simply an earth-surface standard, and their instruments could not tell the difference in altitude. But I think the atomic clock standard today is considered to be at “sea level”. I think that is the world atomic clock time standard, the “sea level” standard. I suppose if the earth heats up and all the polar ice caps melt, the sea level will go up and that’ll throw all our atomic clocks off the current standard. :D
No, because the standard clock and the other clocks can still be compared at the same gravitational potential (the sea level follows a surface of constant gravitational potential).

Sam5
2005-Apr-29, 05:24 PM
No, because the standard clock and the other clocks can still be compared at the same gravitational potential (the sea level follows a surface of constant gravitational potential).

Hmm, would the strength of the gravity field change at Dallas if the sea level went up by a few hundred feet if the ice caps melted and the new “sea level” moved from Galveston to Dallas?

papageno
2005-Apr-29, 05:36 PM
No, because the standard clock and the other clocks can still be compared at the same gravitational potential (the sea level follows a surface of constant gravitational potential).
Hmm, would the strength of the gravity field change at Dallas if the sea level went up by a few hundred feet if the ice caps melted and the new “sea level” moved from Galveston to Dallas?
Depends on the accuracy you want.
In principle the gravitational field in Dallas will be modified because the distribution of mass on Earth has changed. But I would not know how much.
But this is not the point: two atomic clocks at the same gravitational potential have the same rate.

Kaptain K
2005-Apr-29, 05:43 PM
Hmm, would the strength of the gravity field change at Dallas if the sea level went up by a few hundred feet if the ice caps melted and the new “sea level” moved from Galveston to Dallas?
When backed into a corner...Obfuscate! :roll:

Sam5
2005-Apr-29, 05:59 PM
two atomic clocks at the same gravitational potential have the same rate.

Right. I think we all agree on that. Actually, an atomic clock guy told me that no two atomic clocks “tick” at exactly the same rate at the same place because they can’t be manufactured to do so, so every one has its own unique rate, but theoretically I think we can say that two atoms of the same kind oscillate at the same rate in the same gravity field. The question is, in a strong gravity field, where they oscillate more slowly, do they emit light of a lower frequency than the frequency of light they emit in a weaker gravity field? I say yes they do. And I believe the original gravitational redshift theory says “yes” they do.

SeanF
2005-Apr-29, 06:08 PM
If you are sitting next to the emitter inside a gravity field, such as at sea level on earth, of course you won’t “see” the light redshifted because you’ve got to compare that light (through a spectroscope) to another source of light from the same chemical element which is ALSO located at the same place in the same gravity field, and so BOTH sources will be redshifted exactly the same amount.
Okay, what if, instead of comparing the light to another source of light from the same chemical element, you compare it to a printed spectrogram from the same element that was generated at a different gravitational potential.

Will you see the actual light as being shifted compared to the print-out, or will it match the print-out?

papageno
2005-Apr-29, 06:09 PM
two atomic clocks at the same gravitational potential have the same rate.
Right. I think we all agree on that. Actually, an atomic clock guy told me that no two atomic clocks “tick” at exactly the same rate at the same place because they can’t be manufactured to do so, so every one has its own unique rate, but theoretically I think we can say that two atoms of the same kind oscillate at the same rate in the same gravity field. The question is, in a strong gravity field, where they oscillate more slowly, do they emit light of a lower frequency than the frequency of light they emit in a weaker gravity field? I say yes they do. And I believe the original gravitational redshift theory says “yes” they do.
The light emitted in the strong field has a different frequency compared to the light emitted in the weak field, because the gravitational potential is different.
If you have a set of reference light sources all the way from the strong field to the weak field, to be used a local references, you would observe a gradual shift in frequency.
The shift is not determined at the point of emission, but depends on the point of comparison.

(You have the same with the Doppler effect: observers that have different speeds relative to a source, observe different shifts.)

Tensor
2005-Apr-29, 06:18 PM
The question is, in a strong gravity field, where they oscillate more slowly, do they emit light of a lower frequency than the frequency of light they emit in a weaker gravity field? I say yes they do. And I believe the original gravitational redshift theory says “yes” they do.

Sam5, I really don't see a problem with you looking at it this way, for a single observer. However, the problem with this view is if you have two different observers, at two different heights, each will see a different frequency. The higher the observer (the lower the gravity), the more of a red shift they will see. In this case, it's easier (at least for me) to think of it as a wavelegth getting longer as the light climbs out of the higher gravity, which is the same as losing energy.

Sam5
2005-Apr-29, 06:48 PM
The question is, in a strong gravity field, where they oscillate more slowly, do they emit light of a lower frequency than the frequency of light they emit in a weaker gravity field? I say yes they do. And I believe the original gravitational redshift theory says “yes” they do.

Sam5, I really don't see a problem with you looking at it this way, for a single observer. However, the problem with this view is if you have two different observers, at two different heights, each will see a different frequency. The higher the observer (the lower the gravity), the more of a red shift they will see. In this case, it's easier (at least for me) to think of it as a wavelegth getting longer as the light climbs out of the higher gravity, which is the same as losing energy.

Well, it seems to me that if you say the wavelength gets longer as the light climbs out of the gravity field, that seems like a “tired light” theory, and if we admit to a tired light theory then that leaves open the tired light theory of the redshifts of the distant galaxies. In other words someone could say that the galaxies are fixed relative to us, and are not moving, but the light from them just gradually redshifts as it moves toward us over long distances, and the further the distances the more it redshifts.

And also, your approach would assume that the light is emitted at a normal standard frequency, even though the oscillation rate of the atoms is slowed down in a gravity field. Well, what if the oscillation of the atoms stops under certain conditions, such as inside a black hole. Will it continue to emit light of a “normal” frequency even though it has stopped oscillating?

The best explanation I’ve seen for this is what Steinmetz wrote about it: “every incandescent hydrogen atom, for instance, is an accurate clock, vibrating at rate definitely fixed by the electrical constants of the hydrogen atom and showing us the exact rate of its vibration in the spectroscope by the wave length or frequency of its spectrum lines. Thus in a strong gravitational field the frequency of luminous vibrations of the atoms should be found slowed down; in other words, the spectrum lines should be shifted towards the red end of the spectrum.”

Of course I could be wrong about this, but I don’t think so, and there are a few physicists who have written papers about this and a couple of university physics department websites that agree with my point of view that the light is emitted from a slowly oscillating atom at a lower frequency and at a higher frequency from a more rapidly oscillating atom.

Bob
2005-Apr-29, 07:29 PM
Foolish mortals! You have once again fallen into Sam5's trap. There is no escape.

Normandy6644
2005-Apr-29, 07:42 PM
The question is, in a strong gravity field, where they oscillate more slowly, do they emit light of a lower frequency than the frequency of light they emit in a weaker gravity field? I say yes they do. And I believe the original gravitational redshift theory says “yes” they do.

Sam5, I really don't see a problem with you looking at it this way, for a single observer. However, the problem with this view is if you have two different observers, at two different heights, each will see a different frequency. The higher the observer (the lower the gravity), the more of a red shift they will see. In this case, it's easier (at least for me) to think of it as a wavelegth getting longer as the light climbs out of the higher gravity, which is the same as losing energy.

Well, it seems to me that if you say the wavelength gets longer as the light climbs out of the gravity field, that seems like a “tired light” theory, and if we admit to a tired light theory then that leaves open the tired light theory of the redshifts of the distant galaxies. In other words someone could say that the galaxies are fixed relative to us, and are not moving, but the light from them just gradually redshifts as it moves toward us over long distances, and the further the distances the more it redshifts.


It's not tired light at all, since it's not an intrinsic property of the wave. Gravity effects EM waves, we know that.

Tensor
2005-Apr-29, 07:44 PM
Of course I could be wrong about this, but I don’t think so,

Where exactly did I say you were wrong about this?


and there are a few physicists who have written papers about this and a couple of university physics department websites that agree with my point of view that the light is emitted from a slowly oscillating atom at a lower frequency and at a higher frequency from a more rapidly oscillating atom.

Fully agree with the above statement. What your missing is that the energy of the light appears lower (it's been redshifted) to the observer, no matter how you picture it. The exact amount of the redshift can be predicted by comparing the gravitational gradients of the observers and the emitter. It really doesn't matter whether you picture it as the oscillation is slower or the light losing energy climbing out of the higher gravity, what matters is does the predicted effect match observations.

As I said, for multiple observers at different heights above the emitter, it's easier for me to picture it as light losing energy.

Sam5
2005-Apr-29, 07:52 PM
Of course I could be wrong about this, but I don’t think so,

Where exactly did I say you were wrong about this?


and there are a few physicists who have written papers about this and a couple of university physics department websites that agree with my point of view that the light is emitted from a slowly oscillating atom at a lower frequency and at a higher frequency from a more rapidly oscillating atom.

Fully agree with the above statement. What your missing is that the energy of the light appears lower (it's been redshifted) to the observer, no matter how you picture it. The exact amount of the redshift can be predicted by comparing the gravitational gradients of the observers and the emitter. It really doesn't matter whether you picture it as the oscillation is slower or the light losing energy climbing out of the higher gravity, what matters is does the predicted effect match observations.

As I said, for multiple observers at different heights above the emitter, it's easier for me to picture it as light losing energy.


It appears lower because it is lower when it is initially emitted, so of course we will see it lower at some distance from the source. My question is, “Is it lower as it is initially emitted, or is it emitted at a normal frequency and energy level and then lowers while it is in transit.” I think (and the original theory says) it is lower as it is initially emitted. And I think that’s the way the phenomenon should be described.

It is easier for me to picture it as being emitted at a lower frequency to start with.

SeanF
2005-Apr-29, 08:05 PM
Of course I could be wrong about this, but I don’t think so,

Where exactly did I say you were wrong about this?


and there are a few physicists who have written papers about this and a couple of university physics department websites that agree with my point of view that the light is emitted from a slowly oscillating atom at a lower frequency and at a higher frequency from a more rapidly oscillating atom.

Fully agree with the above statement. What your missing is that the energy of the light appears lower (it's been redshifted) to the observer, no matter how you picture it. The exact amount of the redshift can be predicted by comparing the gravitational gradients of the observers and the emitter. It really doesn't matter whether you picture it as the oscillation is slower or the light losing energy climbing out of the higher gravity, what matters is does the predicted effect match observations.

As I said, for multiple observers at different heights above the emitter, it's easier for me to picture it as light losing energy.

It appears lower because it is lower when it is initially emitted, so of course we will see it lower at some distance from the source. My question is, “Is it lower as it is initially emitted, or is it emitted at a normal frequency and energy level and then lowers while it is in transit.” I think (and the original theory says) it is lower as it is initially emitted. And I think that’s the way the phenomenon should be described.

It is easier for me to picture it as being emitted at a lower frequency to start with.
Are you guys really talking about the same thing, though?

If the light is emitted at the bottom of the tower, and there's an observer at the top of the tower, and another observer half-way up the tower, will those two observers measure the same redshift in the light?

I'm thinking Sam would say, "Yes," and Tensor would say, "No."

If that's the case, then you guys aren't just using different terminology to describe the same effect.

Tensor
2005-Apr-29, 08:05 PM
My question is, “Is it lower as it is initially emitted, or is it emitted at a normal frequency and energy level and then lowers while it is in transit.”

Depends on what you mean by normal frequency. An observer next to the emitter will see it as normal. Someone at a lower gravity will see the frequency as lower and someone in an even lower gravity will see the frequency as even lower. So what is the right frequency?


I think (and the original theory says) it is lower as it is initially emitted. And I think that’s the way the phenomenon should be described.

As I said, for a single observer, in a lower gravity than the emitter, this works well. But, for multiple observers, I have trouble picturing how the different observers see different frequencies, from the same emitted frequency.


It is easier for me to picture it as being emitted at a lower frequency to start with.

Not a problem. We all don't have the same picture of how we think it's working. Whatever works for you (as long as your picuture produces the correct result) is fine.

Tensor
2005-Apr-29, 08:11 PM
Are you guys really talking about the same thing, though?

Hmmmmm, not sure.


If the light is emitted at the bottom of the tower, and there's an observer at the top of the tower, and another observer half-way up the tower, will those two observers measure the same redshift in the light?

I'm thinking Sam would say, "Yes," and Tensor would say, "No."

I don't know about Sam5, but you got my answer right, the each would measure a different redshift.


If that's the case, then you guys aren't just using different terminology to describe the same effect.

True. Although I happen to agree with Sam5's picture, if you have one observer.

SeanF
2005-Apr-29, 08:26 PM
True. Although I happen to agree with Sam5's picture, if you have one observer.
Actually, I don't think you do - not entirely. Sam5 predicts that one observer would see a redshift even if the observer were at the same gravitational potential as the light source.

Sam5
2005-Apr-29, 08:27 PM
My question is, “Is it lower as it is initially emitted, or is it emitted at a normal frequency and energy level and then lowers while it is in transit.”

Depends on what you mean by normal frequency. An observer next to the emitter will see it as normal. Someone at a lower gravity will see the frequency as lower and someone in an even lower gravity will see the frequency as even lower. So what is the right frequency?


I think (and the original theory says) it is lower as it is initially emitted. And I think that’s the way the phenomenon should be described.

As I said, for a single observer, in a lower gravity than the emitter, this works well. But, for multiple observers, I have trouble picturing how the different observers see different frequencies, from the same emitted frequency.


It is easier for me to picture it as being emitted at a lower frequency to start with.

Not a problem. We all don't have the same picture of how we think it's working. Whatever works for you (as long as your picuture produces the correct result) is fine.

Think of their atomic clocks in the different locations, ticking more rapidly in the weaker gravity fields and most rapidly in the weakest gravity fields. So, the light emission in the strongest of the fields will be of the lowest frequency of light, an emission of a certain number of photons or waves per your second, where ever you happen to be, based on your atomic clock. If I’m in a weaker gravity field than you, I would measure the frequency by my clock. I would measure the emitted frequency as being a lower frequency by my clock than you would measure it, but that doesn’t mean the light waves/photons have “lost energy” while traveling to me. They started out with the low energy level to start with. The emission frequency does not change for different viewers, but the different atomic clock rates of the different viewers are different in the fields of various strengths where the different viewers are located.

What I call a “normal” frequency would be basically an earth-based “sea level” frequency standard, since this is a world-wide atomic clock standard.

I suppose we could choose some sort of “universal” standard, such as with light being emitted from a small group of atoms in deep space far away from any gravity field. I suppose that that would be the highest frequency of light that could be emitted by those types of atoms.

Tensor
2005-Apr-29, 08:29 PM
True. Although I happen to agree with Sam5's picture, if you have one observer.
Actually, I don't think you do - not entirely. Sam5 predicts that one observer would see a redshift even if the observer were at the same gravitational potential as the light source.

Well, if that is the case, then you are right, I wouldn't agree with that. The one observer would have to be at a different gravitational gradient than the emmiter.

Sam5
2005-Apr-29, 08:31 PM
Are you guys really talking about the same thing, though?

If the light is emitted at the bottom of the tower, and there's an observer at the top of the tower, and another observer half-way up the tower, will those two observers measure the same redshift in the light?

I'm thinking Sam would say, "Yes," and Tensor would say, "No."

I would say “No” because the observer half-way up would be using his own local atomic clock or atomic means of measurement and he would be in a different gravity strength than the guy at the bottom and the guy at the top.

Ricimer
2005-Apr-29, 08:34 PM
but you now see the problem right sam?

If the photon is reshifted as it leaves the emitting atom...and thats the only shift it gets, then the two observers must see the same frequency of photon.

If they do not, then you end up with my two options: It's an effect of gravitational time dilation (both observers see different amounts of time dilation in the atom), or of energy loss...which is the same thing since the time dilation causes the frequency to redden, and be a lower energy.

They're the same picture.

SeanF
2005-Apr-29, 08:35 PM
Are you guys really talking about the same thing, though?

If the light is emitted at the bottom of the tower, and there's an observer at the top of the tower, and another observer half-way up the tower, will those two observers measure the same redshift in the light?

I'm thinking Sam would say, "Yes," and Tensor would say, "No."

I would say “No” because the observer half-way up would be using his own local atomic clock or atomic means of measurement and he would be in a different gravity strength than the guy at the bottom and the guy at the top.
You can measure redshift with a spectrograph, Sam. You don't need an "atomic means of measurement" to compare.

Tensor
2005-Apr-29, 08:38 PM
Think of their atomic clocks in the different locations, ticking more rapidly in the weaker gravity fields and most rapidly in the weakest gravity fields. So, the light emission in the strongest of the fields will be of the lowest frequency of light, an emission of a certain number of photons or waves per your second, where ever you happen to be, based on your atomic clock. If I’m in a weaker gravity field than you, I would measure the frequency by my clock. I would measure the emitted frequency as being a lower frequency by my clock than you would measure it, but that doesn’t mean the light waves/photons have “lost energy” while traveling to me. They started out with the low energy level to start with. The emission frequency does not change for different viewers, but the different atomic clock rates of the different viewers are different in the fields of various strengths where the different viewers are located.

Sam5, I am aware of the above. It's just I find it easier to think of it as the light losing energy as it move againt the gravity. Either way it doesn't matter as we get to the same result.


I suppose we could choose some sort of “universal” standard, such as with light being emitted from a small group of atoms in deep space far away from any gravity field. I suppose that that would be the highest frequency of light that could be emitted by those types of atoms.

Actually, the standard is the emitted frequency in when the emitter and observer are in the same frame of reference.

Van Rijn
2005-Apr-29, 08:47 PM
Ok, let's simplify the picture. Sam, assume there is a tower in an extreme gravity well wth lamps at the bottom and top. The bottom lamp emits blue light and the top lamp emits red light. A hypothetical observer moves up and down the tower. What is he going to see? Forget clocks, what is the observation?

I would say that the observer at the bottom of the tower would see both lamps as blue. As he moved up the tower the light he sees from both lamps would shift towards red. At the top of the tower, both lamps would appear red. At the same time, other observers at different altitudes would disagree.

Do you agree with that?

Sam5
2005-Apr-29, 08:53 PM
Actually, the standard is the emitted frequency in when the emitter and observer are in the same frame of reference.
I don’t think that works, because if the observer always travels up or down with the oscillating atoms, he will always measure them to be oscillating at the same frequency, since he has to use other atoms to measure their oscillation rate. Both sets of atoms will change rates the same amounts in whatever gravitational field they are in.

I’m pretty sure that this is why the world time standard is calculated to be the standard at “sea level” so there will be a stable standard gravitational field in which to measure the tick rates of the clocks.

If we had no world-wide elevation standard, then the atomic clocks of the guys in the Andes and in Switzerland would always tick more rapidly than the atomic clocks of the guys in the swamps of South Louisiana and Florida. And the guys at Boulder would always disagree with the guys at the Naval Observatory in DC about what time it actually is.

Sam5
2005-Apr-29, 09:16 PM
Ok, let's simplify the picture. Sam, assume there is a tower in an extreme gravity well wth lamps at the bottom and top. The bottom lamp emits blue light and the top lamp emits red light. A hypothetical observer moves up and down the tower. What is he going to see? Forget clocks, what is the observation?

I would say that the observer at the bottom of the tower would see both lamps as blue. As he moved up the tower the light he sees from both lamps would shift towards red. At the top of the tower, both lamps would appear red. At the same time, other observers at different altitudes would disagree.

Do you agree with that?

Uhh, not exactly. I don’t think that’s a good thought experiment. You would have to use means other than just gravity to alter the color of your light, and that brings in other complicating factors such as hotter and cooler light sources and maybe filters too. Plus you’ve got to consider what is going on inside the eye cones as the observer changes altitude. They are made up of atoms too, and of course they would oscillate more slowly at the bottom than at the top of the tower. Plus the human eye can’t distinquish any light color change in such a short elevation distance. This is why Pound and Rebka used the Mossbauer effect and iron atoms for their studies.

Ricimer
2005-Apr-29, 09:35 PM
I don’t think that works, because if the observer always travels up or down with the oscillating atoms, he will always measure them to be oscillating at the same frequency, since he has to use other atoms to measure their oscillation rate. Both sets of atoms will change rates the same amounts in whatever gravitational field they are in.

That means he'll never see the ones he travels with give anything but their rest frequency. Correct. However, as he moves around, closer and further from the emiter (which isn't moving with him) the amount of discrepancy between the emitter and the "standards" with the observer varies. This is why they see different redshift values from the emitter.


I’m pretty sure that this is why the world time standard is calculated to be the standard at “sea level” so there will be a stable standard gravitational field in which to measure the tick rates of the clocks. The reason for that is indeed a uniform elevation. The surface of the ocean is, if smooth, along a gravitational equi-potential line. Any clocks sitting on the ocean surface (ignoring waves) will be under the same gravitational influence, and read the same.

Your comment earlier about how no atomic clock "ticks" at the same rate may be accurate btw. IT doesn't change the accuracy however. The vibration rate for each clock is very stable, you just have to calibrate each clock to a standard first, to know how many of the vibrations equals a second.


If we had no world-wide elevation standard, then the atomic clocks of the guys in the Andes and in Switzerland would always tick more rapidly than the atomic clocks of the guys in the swamps of South Louisiana and Florida. And the guys at Boulder would always disagree with the guys at the Naval Observatory in DC about what time it actually is. indeed they would.[/quote]

Van Rijn
2005-Apr-29, 09:49 PM
Ok, let's simplify the picture. Sam, assume there is a tower in an extreme gravity well wth lamps at the bottom and top. The bottom lamp emits blue light and the top lamp emits red light. A hypothetical observer moves up and down the tower. What is he going to see? Forget clocks, what is the observation?

I would say that the observer at the bottom of the tower would see both lamps as blue. As he moved up the tower the light he sees from both lamps would shift towards red. At the top of the tower, both lamps would appear red. At the same time, other observers at different altitudes would disagree.

Do you agree with that?

Uhh, not exactly. I don’t think that’s a good thought experiment. You would have to use means other than just gravity to alter the color of your light.

Humm? I thought that was exactly what we were discussing. By GR, the color of the light should differ depending on where in the gravity well you are observing it (or more generally, depending on your relative reference frame - for this discussion we are assuming motionless observers and light sources at fixed locations).


and that brings in other complicating factors such as hotter and cooler light sources and maybe filters too.


No, just assume specific output frequencies.



Plus you’ve got to consider what is going on inside the eye cones as the observer changes altitude. They are made up of atoms too, and of course they would oscillate more slowly at the bottom than at the top of the tower. Plus the human eye can’t distinquish any light color change in such a short elevation distance. This is why Pound and Rebka used the Mossbauer effect and iron atoms for their studies.

Now you are sidestepping the issue. As I said: An extreme gravity well, a hypothetical observer. What does he observe?

Sam5
2005-Apr-29, 10:10 PM
By GR, the color of the light should differ depending on where in the gravity well you are observing it (or more generally, depending on your relative reference frame - for this discussion we are assuming motionless observers and light sources at fixed locations).

Ok, I’m still thinking about your thought experiment. In the mean time, let’s see what Tensor, Sean, and Ricimer think about it. I’m trying to work on a solution now.

Ricimer
2005-Apr-29, 10:45 PM
I'm with Van.

Tensor
2005-Apr-29, 11:02 PM
[quote=Van Rijn]




Ok, I’m still thinking about your thought experiment. In the mean time, let’s see what Tensor, Sean, and Ricimer think about it. I’m trying to work on a solution now.

I can wait.

Sam5
2005-Apr-29, 11:05 PM
I'm with Van.

I'm still thinking about it.

Ricimer
2005-Apr-29, 11:06 PM
take your time. I'm quick about it, cause I've been trained to think about such things (and all the physics behind it).

Sam5
2005-Apr-29, 11:08 PM
Ok, I’m still thinking about your thought experiment. In the mean time, let’s see what Tensor, Sean, and Ricimer think about it. I’m trying to work on a solution now.

I can wait.

What's your solution?

Sam5
2005-Apr-29, 11:56 PM
Van,

Ok, I’m tending to lean in the direction of your solution being correct.

worzel
2005-Apr-30, 01:06 AM
Plus you’ve got to consider what is going on inside the eye cones as the observer changes altitude. They are made up of atoms too, and of course they would oscillate more slowly at the bottom than at the top of the tower.
Gosh Sam5, it almost sounds like you're saying that this slowing down of atomic clocks affects everything, almost as if time slows down itself!

You never answered my question about whether or not you think that space is just flat, 3D Euclidean like. The reason I ask this is because, so far, all you've done is argue for a one particular description of gravitational redshift over another with no mention of how the two could be experimentally differentiated. But you have often made remarks about people "mystifying" it with spacetime descriptions. If you really believe that space is flat, then it is more than just a difference in description. You would, presumably, have your own account of Mercury's orbit for instance, given that you don't believe the warped spacetime explanation.

Sam5
2005-Apr-30, 01:18 AM
Plus you’ve got to consider what is going on inside the eye cones as the observer changes altitude. They are made up of atoms too, and of course they would oscillate more slowly at the bottom than at the top of the tower.
Gosh Sam5, it almost sounds like you're saying that this slowing down of atomic clocks affects everything, almost as if time slows down itself!


Hey, get out of here, boy! :D These guys want to talk in term of “time dilation” so I figure I’ll speak their language. If an atom slows down its oscillation rate in a gravity field, that can also be described in plain ol’ classical terms too. If you slow down your photongraph record the sound waves emitted will be lower. I call it tomAto and you call it tomaaahto.


Edit:
*Note to young whippersnappers: A “phonograph” is what we old-timers used before we had CDs. Back in our day, CD meant “Civil Defense”.

Grey
2005-Apr-30, 01:27 AM
Plus you’ve got to consider what is going on inside the eye cones as the observer changes altitude. They are made up of atoms too, and of course they would oscillate more slowly at the bottom than at the top of the tower.
Gosh Sam5, it almost sounds like you're saying that this slowing down of atomic clocks affects everything, almost as if time slows down itself!


Hey, get out of here, boy! :D These guys want to talk in term of “time dilation” so I figure I’ll speak their language. If an atom slows down its oscillation rate in a gravity field, that can also be described in plain ol’ classical terms too. If you slow down your photongraph record the sound waves emitted will be lower. I call it tomAto and you call it tomaaahto.
Well, a serious question is actually whether you think this effect is limited to atoms. For example, when we observe spectral lines in the Sun's light, those are caused by atomic transitions, with the frequency governed by the electromagnetic force between the nucleus and the outer electrons. On the other hand, in Pund and Rebka's experiment, their "clocks" were actually nuclear rather than atomic. That is, they were looking at gamma rays emitted from the nuclei of atoms, so in that case the frequency of the radiation is controlled by the energy levels of the nucleus, which is bound instead by the strong force. So these two types of interactions actually have nothing to do with each other as far as the mechanisms involved are concerned. Yet it's interesting that in both cases, the change in frequency is given by the same relationship from general relativity, so the amount that a nuclear "clock" is slowed is precisely the same as the amount an atomic "clock" is slowed.

worzel
2005-Apr-30, 01:36 AM
Plus you’ve got to consider what is going on inside the eye cones as the observer changes altitude. They are made up of atoms too, and of course they would oscillate more slowly at the bottom than at the top of the tower.
Gosh Sam5, it almost sounds like you're saying that this slowing down of atomic clocks affects everything, almost as if time slows down itself!


Hey, get out of here, boy! :D These guys want to talk in term of “time dilation” so I figure I’ll speak their language. If an atom slows down its oscillation rate in a gravity field, that can also be described in plain ol’ classical terms too. If you slow down your photongraph record the sound waves emitted will be lower. I call it tomAto and you call it tomaaahto.
You are, as ever, as clear as mud. One the one hand it is just a case of tomAtoes verses tomaaahtoes, and yet on the other, time dialation is somehow analogous to paying your 45 at 33. And apologies if it was a stupid one, but you still didn't answer my question.

Sam5
2005-Apr-30, 01:39 AM
Well, a serious question is actually whether you think this effect is limited to atoms. For example, when we observe spectral lines in the Sun's light, those are caused by atomic transitions, with the frequency governed by the electromagnetic force between the nucleus and the outer electrons. On the other hand, in Pund and Rebka's experiment, their "clocks" were actually nuclear rather than atomic. That is, they were looking at gamma rays emitted from the nuclei of atoms, so in that case the frequency of the radiation is controlled by the energy levels of the nucleus, which is bound instead by the strong force. So these two types of interactions actually have nothing to do with each other as far as the mechanisms involved are concerned. Yet it's interesting that in both cases, the change in frequency is given by the same relationship from general relativity, so the amount that a nuclear "clock" is slowed is precisely the same as the amount an atomic "clock" is slowed.


Interesting, thanks for the description.

Sylas
2005-Apr-30, 04:44 AM
Well, a serious question is actually whether you think this effect is limited to atoms.

Another question is whether we think the effect depends on the gravitational forces being experienced by the atom/nucleus. Here is a thought experiment, and we can try the two explanations Sam5 has provided. Unfortunately, I can't think of a good way to conduct the experiment in reality.

Imagine something emitting photons at a constant frequency. It could be from electrons dropping through well defined energy levels in an atom or it could be from well defined excitation states of a nucleus.

Suppose we place such an emitter somewhere in a very high Earth orbit, and we also place an identical emitter at the center of the Earth, or in the center of any shell of matter, with a little hole for the photons to get out.

The total gravitational force experienced at the center of the Earth, or in the high orbit, is zero.

The photons emitted from the center of the Earth are then received in the high orbit, and compared with the photons produced by an identical emitter in orbit.

Question 1. Will the photons from the center of the Earth will be redshifted?

I think they will be redshifted. I guess Sam5 might agree. I know Grey will agree.

There are two explanations for this. The one I prefer goes as follows:

Explanation 1: We can calculate the time dilation of the emitter with respect to the receiver by applying relativity and considering paths from the emitter to the receiver. The emitter is running in exactly the same way as it does anywhere else, producing photons of a fixed characteristic frequency at the emitter. Due to differences in the passage of time at the emitter, and at the receiver, this frequency relative to the receiver is lower: a redshift. The receiver observes the photons with this redshift arising from time dilation.

Some other people here prefer a different explanation, which really corresponds to using a different co-ordinate system in GR.

Explanation 2: The emitter is running in exactly the same way as it does anywhere else, producing photons of a fixed characteristic frequency at the emitter. As the photons travel from the emitter to the receiver, up through the gravitational field, they lose energy and their frequency drops. By the time they arrive at the receiver, they have the redshift that we can calculate by GR.

My preference for the first explanation is subjective. The second explanation is not really wrong. I'm inclined to think it can lead to confusions; but this thread has certainly shown me that the first explanation can lead to confusions as well! In any case, GR comes with the tools and techniques for mapping between co-ordinate systems, and giving a precise transformation from one explanation into the other.

Now as I understand the matter, Sam5 has a third explanation. However, I am not sure that Sam5 would agree, and he is welcome to correct my understanding of his position. The following explanation is my own wording, not that of Sam5.

Explanation 3. The emitter actually behaves differently when at the center of the Earth, and the photons emitted have a different frequency. This frequency corresponds to the redshift seen by the observer.

Sam5 may confirm this, or not. If he confirms it, there is a problem… what physical laws can be used to calculate the emitter frequency? What are the variables involved? The emitter is weightless in both cases, so it can’t simply be the force experienced.

Cheers -- Sylas

papageno
2005-Apr-30, 01:58 PM
The question is, in a strong gravity field, where they oscillate more slowly, do they emit light of a lower frequency than the frequency of light they emit in a weaker gravity field? I say yes they do. And I believe the original gravitational redshift theory says “yes” they do.

Sam5, I really don't see a problem with you looking at it this way, for a single observer. However, the problem with this view is if you have two different observers, at two different heights, each will see a different frequency. The higher the observer (the lower the gravity), the more of a red shift they will see. In this case, it's easier (at least for me) to think of it as a wavelegth getting longer as the light climbs out of the higher gravity, which is the same as losing energy.

Well, it seems to me that if you say the wavelength gets longer as the light climbs out of the gravity field, that seems like a “tired light” theory, and if we admit to a tired light theory then that leaves open the tired light theory of the redshifts of the distant galaxies.
Except that the typical "tired light" theories assume that light is losing energy due to some interaction, not because it was emitted in a gravitational potential different form the observer.



In other words someone could say that the galaxies are fixed relative to us, and are not moving, but the light from them just gradually redshifts as it moves toward us over long distances, and the further the distances the more it redshifts.

And also, your approach would assume that the light is emitted at a normal standard frequency, even though the oscillation rate of the atoms is slowed down in a gravity field.
The emission of light is not due to oscillations.
Are you confusing emission of light with old-type atomic clocks?
Emission of light comes from electrons changing between states with different energy.
Old-type atomic clocks used molecules where one atom oscillates between two positions.

The gravitational red-shift is observed over the whole range of frequencies for EM emission, even those that have nothing to do with "atomic lock oscillation rates" or electronic transitions.
The tower experiment used gamma-radiation, which is produced by nuclear transitions in radioactive Fe nuclei (nothing to do with electrons).




Well, what if the oscillation of the atoms stops under certain conditions, such as inside a black hole. Will it continue to emit light of a “normal” frequency even though it has stopped oscillating?
Yep, you are confused.



The best explanation I’ve seen for this is what Steinmetz wrote about it: “every incandescent hydrogen atom, for instance, is an accurate clock, vibrating at rate definitely fixed by the electrical constants of the hydrogen atom and showing us the exact rate of its vibration in the spectroscope by the wave length or frequency of its spectrum lines. Thus in a strong gravitational field the frequency of luminous vibrations of the atoms should be found slowed down; in other words, the spectrum lines should be shifted towards the red end of the spectrum.”

Of course I could be wrong about this, but I don’t think so, and there are a few physicists who have written papers about this and a couple of university physics department websites that agree with my point of view that the light is emitted from a slowly oscillating atom at a lower frequency and at a higher frequency from a more rapidly oscillating atom.
The light emitted is not the result of pendulum-like oscillations in atoms.
And the effect is not restricted to light.

jnik
2005-May-02, 08:02 PM
But that doesn’t mean your biological clock has “slowed down” the same amount at sea level
I think this is your fundamental misunderstanding. (We *are* talking GR here, right, not some sort of against-the-mainstream modification?)

SeanF
2005-May-02, 08:07 PM
But that doesn’t mean your biological clock has “slowed down” the same amount at sea level
I think this is your fundamental misunderstanding. (We *are* talking GR here, right, not some sort of against-the-mainstream modification?)
Most of us are talking about GR. Sam5's talking about some sort of against-the-mainstream modification.

:)

Sam5
2005-May-02, 08:11 PM
But that doesn’t mean your biological clock has “slowed down” the same amount at sea level
I think this is your fundamental misunderstanding. (We *are* talking GR here, right, not some sort of against-the-mainstream modification?)


:D Well, there was a brief discussion about biological time (thermodynamic time) on this board a couple of weeks ago:
HERE (http://www.badastronomy.com/phpBB/viewtopic.php?t=21205).

From the CNN source, (here) (http://www.cnn.com/2005/HEALTH/04/22/medical.hibernation.ap/index.html):

“It works, essentially, like hypothermia. Recall those miraculous cases of people who fall into icy ponds and appear dead but recover after they're warmed up? The extreme cold preserves their brain cells from the certain death that would otherwise quickly follow oxygen deprivation.

---

Within minutes of inhaling the gas, the mice appeared unconscious. Their body temperature plummeted from the normal 98 degrees down to 59 degrees and their respiration slowed to fewer than 10 breaths a minute, down from a normal 120 breaths a minute, Roth reported.

Overall, their metabolic rate dropped by 90 percent -- meaning normal cellular activity slowed to almost a standstill, thus reducing the need for oxygen.”

I brought this very subject up here more than a year ago, and I tried to explain how it works. Some people might remember my “frozen embryo” posts. But the SR and GR relativists here didn’t like the idea since Einstein never mentioned it, so I just dropped the subject. You might want to look it up in biology books.

This basic idea is still ATM in the field of physics, but it has been very mainstream in the field of biology for many decades.

SeanF
2005-May-02, 08:30 PM
Weeks? Seems to me it's been years since you first brought up cryogenics in a relativity-related discussion on this BB.

And after all this time, you still don't understand that it's got nothing to do with it. Absolutely nothing.

Sam5
2005-May-02, 10:14 PM
And after all this time, you still don't understand that it's got nothing to do with it. Absolutely nothing.

Actually, I think it was a “time” related thread. I tried to explain the difference between the physics view of time, being mostly based on atomic time, and the biological view of time, being a thermodynamic phenomenon such as explained in the CNN news story. But nobody here seemed interested because Einstein never said much of anything about biological time.

Remember how I tried to explain to you that Hawking was wrong about this, because he was thinking in terms of physics atomic time instead of biological thermodynamic time:

“Newton’s laws of motion put an end to the idea of absolute position in space. The theory of relativity gets rid of absolute time. Consider a pair of twins. Suppose that one twin goes to live on the top of a mountain while the other stays at sea level. The first twin would age faster than the second.”

LINK (http://64.233.161.104/search?q=cache:APFp6aLhd0kJ:newton.physics.metu.ed u.tr/~fizikt/html/hawking/a.html+hawking+speed+up+on+a+mountain&hl=en&lr=lan g_en&ie=UTF-8)


:D

worzel
2005-May-03, 08:19 AM
Actually, I think it was a “time” related thread. I tried to explain the difference between the physics view of time, being mostly based on atomic time, and the biological view of time, being a thermodynamic phenomenon such as explained in the CNN news story. But nobody here seemed interested because Einstein never said much of anything about biological time.
A slowing down of biological time, as you call it, is a process actually slowing down. That is, the process takes longer. Time dilation in relativity is the slowing of time itself in one frame as viewed from another. You can measure the former in its own frame. You cannot measure the latter in your own frame, indeed the concept of time dilation in your own frame has no meaning.

You obviously don't believe this, and think instead that time dilation is just the slowing down of some atomic process, but that is not what relativity says. If you accept that the speed of light is c in all intertial frames (and not just "appears to be") then how do you explain that without length contraction and time dilation?

papageno
2005-May-03, 10:05 AM
Actually, I think it was a “time” related thread. I tried to explain the difference between the physics view of time, being mostly based on atomic time, ...
Please, do not confuse metrology with physics.
Metrology searches for standards of units, and having good standards does help.




... and the biological view of time, being a thermodynamic phenomenon such as explained in the CNN news story. But nobody here seemed interested because Einstein never said much of anything about biological time.
If biological time is based on thermodynamics, it is already included in physics.
When we say in Relativity that time slows down, it affects every physical process, including thermodynamic ones.

Sam5
2005-May-03, 11:08 AM
When we say in Relativity that time slows down, it affects every physical process, including thermodynamic ones.

So both of the two relatively moving twins in SR see each other freeze?

Sam5
2005-May-03, 11:20 AM
Actually, I think it was a “time” related thread. I tried to explain the difference between the physics view of time, being mostly based on atomic time, and the biological view of time, being a thermodynamic phenomenon such as explained in the CNN news story. But nobody here seemed interested because Einstein never said much of anything about biological time.
A slowing down of biological time, as you call it, is a process actually slowing down. That is, the process takes longer. Time dilation in relativity is the slowing of time itself in one frame as viewed from another. You can measure the former in its own frame. You cannot measure the latter in your own frame, indeed the concept of time dilation in your own frame has no meaning.

Then how can the traveled clock in the “peculiar consequence” thought experiment in the SR theory “lag behind” the “stationary” clock when they unite? Wouldn’t the observer that traveled with the clock that traveled notice, in his own frame, that his own clock “lags behind” the other and therefore ran slow? And wouldn’t he actually notice that the other clock seemed to him to run fast during his travel?

papageno
2005-May-03, 11:35 AM
When we say in Relativity that time slows down, it affects every physical process, including thermodynamic ones.

So both of the two relatively moving twins in SR see each other freeze?

Have a look at this (http://www.mathpages.com/rr/s4-07/4-07.htm).

worzel
2005-May-03, 11:43 AM
Then how can the traveled clock in the “peculiar consequence” thought experiment in the SR theory “lag behind” the “stationary” clock when they unite? Wouldn’t the observer that traveled with the clock that traveled notice, in his own frame, that his own clock “lags behind” the other and therefore ran slow? And wouldn’t he actually notice that the other clock seemed to him to run fast during his travel?
The observer who travelled with the clock would not notice that his clock ran slowly, only that the other clock ran fast. The observer who didn't travel would not notice that his clock ran fast, only that the other clock ran slow.

Sam5
2005-May-03, 11:47 AM
Then how can the traveled clock in the “peculiar consequence” thought experiment in the SR theory “lag behind” the “stationary” clock when they unite? Wouldn’t the observer that traveled with the clock that traveled notice, in his own frame, that his own clock “lags behind” the other and therefore ran slow? And wouldn’t he actually notice that the other clock seemed to him to run fast during his travel?
The observer who travelled with the clock would not notice that his clock ran slowly, only that the other clock ran fast. The observer who didn't travel would not notice that his clock ran fast, only that the other clock ran slow.

No, in SR theory both of two relatively moving observers see each other’s clocks run slow, with no exceptions.

Sam5
2005-May-03, 12:14 PM
Have a look at this (http://www.mathpages.com/rr/s4-07/4-07.htm).
I can sum it up more simply. First, just “relative motion” alone can’t slow down any clock. In the original SR theory, all motion is just “relative”, with no gravity or acceleration. All the SR stories about a traveling twin “aging more slowly” than a twin that “stays at home” is a lot of hokum, fable, and urban legend. It doesn’t happen. In the SR theory, both clocks are always inside “inertial frames” and the motion between the two clocks is just “relative”.

However, in GR theory, atomic clocks really do slow down in a gravity field and when accelerated, but the slow-down is so slight it can’t possibly ever affect human biological time. If an atomic-time slow-down occurred so much it began to affect human biological time, the human would be crushed to death by the extreme gravity/acceleration long before any biological time dilation effect would be noticed.

On the other hand, biological heat-time metabolism slow-downs are caused by temperature factors over just a slight change in temperature.

papageno
2005-May-03, 12:36 PM
Have a look at this (http://www.mathpages.com/rr/s4-07/4-07.htm).
I can sum it up more simply. First, just “relative motion” alone can’t slow down any clock.
It is clear now that you do not grasp the concept of frame of reference and observer.



In the original SR theory, all motion is just “relative”, with no gravity or acceleration.
Motion is "relative" also in the "original" Newtonian mechanics...



All the SR stories about a traveling twin “aging more slowly” than a twin that “stays at home” is a lot of hokum, fable, and urban legend. It doesn’t happen. In the SR theory, both clocks are always inside “inertial frames” and the motion between the two clocks is just “relative”.
But the frames of reference do not coincide, which is why transformations of coordinates are needed..



However, in GR theory, atomic clocks really do slow down in a gravity field and when accelerated, but the slow-down is so slight it can’t possibly ever affect human biological time.
I don't remember anybody claiming that it has been observed in biological process in humans. If somebody did, feel free to provide references.
By the way, it does not work only for atomic clocks.



If an atomic-time slow-down occurred so much it began to affect human biological time, the human would be crushed to death by the extreme gravity/acceleration long before any biological time dilation effect would be noticed.
So, it does not mean that it would not happen in principle.



On the other hand, biological heat-time metabolism slow-downs are caused by temperature factors over just a slight change in temperature.
So, what is the point? What does this have to do with gravitational redshift?
Biological processes depend on chemical and thermodynamic processes, which do depend on temperature (thermo-, as in thermometer).
Why did you bring up "biological time"?

Sam5
2005-May-03, 12:40 PM
Why did you bring up "biological time"?

Because of all the silly thought experiments about “traveling twins” aging more slowly. These are fantasies, sci-fi stories. :D

Celestial Mechanic
2005-May-03, 12:40 PM
[Snip!]I can sum it up more simply. First, just “relative motion” alone can’t slow down any clock. In the original SR theory, all motion is just “relative”, with no gravity or acceleration. All the SR stories about a traveling twin “aging more slowly” than a twin that “stays at home” is a lot of hokum, fable, and urban legend. It doesn’t happen. In the SR theory, both clocks are always inside “inertial frames” and the motion between the two clocks is just “relative”.[Snip!]
Sorry, but there are a lot of people at CERN, Fermilab, Brookhaven, Stanford, Dubna, Serpukhov, Hamburg, etc. who see this every day at their particle accelerators. It is no "urban legend". [-(

Sam5
2005-May-03, 12:44 PM
Sorry, but there are a lot of people at CERN, Fermilab, Brookhaven, Stanford, Dubna, Serpukhov, Hamburg, etc. who see this every day at their particle accelerators. It is no "urban legend". [-(

The keyword in your post is “accelerator”. I’ve already mentioned acceleration effects. :D

worzel
2005-May-03, 12:44 PM
Then how can the traveled clock in the “peculiar consequence” thought experiment in the SR theory “lag behind” the “stationary” clock when they unite? Wouldn’t the observer that traveled with the clock that traveled notice, in his own frame, that his own clock “lags behind” the other and therefore ran slow? And wouldn’t he actually notice that the other clock seemed to him to run fast during his travel?
The observer who travelled with the clock would not notice that his clock ran slowly, only that the other clock ran fast. The observer who didn't travel would not notice that his clock ran fast, only that the other clock ran slow.

No, in SR theory both of two relatively moving observers see each other’s clocks run slow, with no exceptions.
I thought we were talking about the twin paradox with the travelled clock being the one that turned around. Indeed while both clocks are moving relative to each other intertially they will both see each other's clock run slow (after taking into account the doppler effect). It is only when a clock turns around so they can be reunited that it gets interesting - and SR most definitely says that the clock that turned around will lag behind the one that remained inertial throughout - do you disagree with that?

papageno
2005-May-03, 12:44 PM
Why did you bring up "biological time"?

Because of all the silly thought experiments about “traveling twins” aging more slowly. These are fantasies, sci-fi stories. :D

You are obfuscating the issue: the effect happens in principle, although it might not be measurable.
The fact that GPS takes into account SR and GR effects, clearly shows they are not fantasies.

worzel
2005-May-03, 12:46 PM
Sorry, but there are a lot of people at CERN, Fermilab, Brookhaven, Stanford, Dubna, Serpukhov, Hamburg, etc. who see this every day at their particle accelerators. It is no "urban legend". [-(

The keyword in your post is “accelerator”. I’ve already mentioned acceleration effects. :D
What's that supposed to mean? That you don't understand how SR can deal with acceleration?

Sam5
2005-May-03, 12:47 PM
I thought we were talking about the twin paradox with the travelled clock being the one that turned around. In the “peculiar consequence” thought experiment that leads to the clock paradox in SR theory, there is no “turn around.”

Sam5
2005-May-03, 12:49 PM
The fact that GPS takes into account SR and GR effects, clearly shows they are not fantasies.
GPS takes into account LR and GR effects.

Sam5
2005-May-03, 12:53 PM
What's that supposed to mean? That you don't understand how SR can deal with acceleration?

The relatively moving clocks in SR theory do not experience acceleration. That’s why their frames are called “inertial”. Please read the theory.

papageno
2005-May-03, 01:11 PM
The fact that GPS takes into account SR and GR effects, clearly shows they are not fantasies.
GPS takes into account LR and GR effects.
And what would LR be?


In the “peculiar consequence” thought experiment that leads to the clock paradox in SR theory, there is no “turn around.”
Do you mean, the twin does not come back to Earth?


The relatively moving clocks in SR theory do not experience acceleration. That’s why their frames are called “inertial”. Please read the theory.
Even when a clock is accelerating, there is an inertial frame of reference associated with it. But this frame is not same throughout the acceleration.

worzel
2005-May-03, 01:14 PM
I thought we were talking about the twin paradox with the travelled clock being the one that turned around. In the “peculiar consequence” thought experiment that leads to the clock paradox in SR theory, there is no “turn around.”
Then what did you mean by "unite"?


Then how can the traveled clock in the “peculiar consequence” thought experiment in the SR theory “lag behind” the “stationary” clock when they unite?
The clcok could only be said to have lagged behind if the two clocks were compared when together before and after the motion, so at least one must have turned around.



What's that supposed to mean? That you don't understand how SR can deal with acceleration?
The relatively moving clocks in SR theory do not experience acceleration. That’s why their frames are called “inertial”. Please read the theory.
SR can handle acceleration by approximating it with ever more galileon boosts and in limit the infinite lorentz transformations become identical to the GR pseudo gravity well caused by acceleration.

Sylas
2005-May-03, 01:24 PM
The relatively moving clocks in SR theory do not experience acceleration. That’s why their frames are called “inertial”. Please read the theory.

SR can handle a clock in a non-intertial frame. You just can't take a constant dilation factor with respect to an intertial frame. I gave an example of how it is done above. Since you can't handle the maths, it will be a bit difficult to follow, but meaning no offence, until you can follow the maths you don't really know what you are talking about, and your reading of original papers remains inaccurate. I've read the theory, including original papers, and can apply it because I follow the maths.

You can apply GR to the twin "paradox" as well, you know; but it is overkill. SR is sufficient; and gives the same resolution as SR. The accelerated twin has aged less than the one who remains unaccelerated, once they return to a common origin. One of Einstein's insights was a deep equivalence between gravitation and acceleration; but to apply this equivalence you really need to appreciate the mathematical nature of the equivalence.

Cheers -- Sylas

Sam5
2005-May-03, 01:24 PM
The fact that GPS takes into account SR and GR effects, clearly shows they are not fantasies.
GPS takes into account LR and GR effects.
And what would LR be?

Lorentz Relativity, as per ““Versuch Einer Theorie Der Elektrischen Und Optischen Erscheinungen In Bewegten Körpern”, 1895.



In the “peculiar consequence” thought experiment that leads to the clock paradox in SR theory, there is no “turn around.”
Do you mean, the twin does not come back to Earth?

There are only two “systems” in the SR theory. There is no “earth”.

Sam5
2005-May-03, 01:26 PM
The clcok could only be said to have lagged behind if the two clocks were compared when together before and after the motion, so at least one must have turned around.

Nope. Please read the actual paper. There was no turn-around in that thought experiment.

Sam5
2005-May-03, 01:33 PM
The relatively moving clocks in SR theory do not experience acceleration. That’s why their frames are called “inertial”. Please read the theory.

SR can handle a clock in a non-intertial frame. You just can't take a constant dilation factor with respect to an intertial frame. I gave an example of how it is done above.

Thanks for your information. I’ve been reading your other thread. I felt I had nothing to contribute to the discussion, so I just stayed out of it.

Einstein wrote some papers about the very subject you are discussing, such as “The Speed of Light and the Statics of the Gravitational Field,” and “On the Theory of the Static Gravitational Field,” and “Relativity and Gravitation, a Reply to a Comment by Max Abraham,” all translated into English and published in Volume 4 of “The Collected Papers of Albert Einstein.”



You can apply GR to the twin "paradox" as well, you know; but it is overkill. SR is sufficient; and gives the same resolution as SR. The accelerated twin has aged less than the one who remains unaccelerated, once they return to a common origin. One of Einstein's insights was a deep equivalence between gravitation and acceleration; but to apply this equivalence you really need to appreciate the mathematical nature of the equivalence.


Einstein had to add atomic clocks and acceleration to the thought experiments in his 1918 paper in an attempt to try to resolve the SR paradox. He could not clear up the paradox with just SR alone.

I don’t have time to respond to three people at once.

SeanF
2005-May-03, 01:39 PM
The clcok could only be said to have lagged behind if the two clocks were compared when together before and after the motion, so at least one must have turned around.
Nope. Please read the actual paper. There was no turn-around in that thought experiment.
The clock that moves in a "closed curve" is doing nothing but turning around - for the entire duration of the experiment.

Sam5
2005-May-03, 01:54 PM
The clcok could only be said to have lagged behind if the two clocks were compared when together before and after the motion, so at least one must have turned around.
Nope. Please read the actual paper. There was no turn-around in that thought experiment.
The clock that moves in a "closed curve" is doing nothing but turning around - for the entire duration of the experiment.

I figured someone would bring up the “closed curve” thought experiment. Of course nothing can move in a “closed curve” with a “constant velocity”. The German word he used, “Geschwindigkeit,” should have been translated as “speed”, not “velocity.” He was considering an infinite number of straight lines and an infinite number of turns, but without any consideration of the acceleration effects of the turns. That is quite obvious since his clock-related thought experiments in the SR paper don’t consider either gravity or acceleration effects. He was only considering “kinematical” effects. Later he developed GR and did consider gravity and acceleration effects.

I will have some difficulty trying to respond to four poster. Perhaps you could form a comittee and appoint a single spokesperson I could respond to.

Sam5
2005-May-03, 01:58 PM
Sylas, if you don’t have a copy of Volume 4, I think you would enjoy it very much. It contains several papers that cover his transition from SR to GR. If you order a copy, be sure to get the paperback version since that is the English version. The hardback is all in German.

worzel
2005-May-03, 02:12 PM
The clcok could only be said to have lagged behind if the two clocks were compared when together before and after the motion, so at least one must have turned around.

Nope. Please read the actual paper. There was no turn-around in that thought experiment.
You are just switching between two different thought experiments to be contrary. When he talks about two clocks moving relatively and inertially to each other then, from each clock's frame the other clock runs slowly. The twin paradox has a clock that changes its inertial frame, it is not the same. That change in intertial frame is the crucial bit.


I figured someone would bring up the “closed curve” thought experiment. Of course nothing can move in a “closed curve” with a “constant velocity”. The German word he used, “Geschwindigkeit,” should have been translated as “speed”, not “velocity.” He was considering an infinite number of straight lines and an infinite number of turns, but without any consideration of the acceleration effects of the turns. That is quite obvious since his clock-related thought experiments in the SR paper don’t consider either gravity or acceleration effects. He was only considering “kinematical” effects. Later he developed GR and did consider gravity and acceleration effects.
Did it not occur to you when you wrote that that all those infinite number of straight lines and turns equates to an infinite number of lorentz transformations, which is how SR handles acceleration, without considering any extra effects due to acceleration or gravity. Indeed it requires the assumption that nothing extra happens beyond the lorentz transformation when the clock accelerates: the clocks assumption.

SeanF
2005-May-03, 02:19 PM
He was considering an infinite number of straight lines and an infinite number of turns, but without any consideration of the acceleration effects of the turns.
Exactly. With an infinite number of terms, and no consideration of the acceleration effects, you end up with the "infinitely turning" clock lagging behind the other one.

So when you said there was no turn-around in that thought experiment, you were wrong. There is. And the "acceleration effects" are nothing more or less than the integrated effects of the relative motion.

So what's the problem?

Sam5
2005-May-03, 02:30 PM
So when you said there was no turn-around in that thought experiment, you were wrong.

I was talking about the clock paradox in the first “peculiar consequence” thought experiment in Section 4.

There are absolutely no “turn arounds” in it.

Sam5
2005-May-03, 02:35 PM
The clcok could only be said to have lagged behind if the two clocks were compared when together before and after the motion, so at least one must have turned around.

Nope. Please read the actual paper. There was no turn-around in that thought experiment.
You are just switching between two different thought experiments to be contrary. When he talks about two clocks moving relatively and inertially to each other then, from each clock's frame the other clock runs slowly. The twin paradox has a clock that changes its inertial frame, it is not the same. That change in intertial frame is the crucial bit.

No it’s not. Only relative motion is considered. Both systems are moving relative to each other. The K observers see the K’ system move, and the K’ observers see the K system move. There are only two systems in the thought experiment. No acceleration is considered. The motion is only relative. Observers in both systems see themselves as being “stationary” and the other system as “moving”. Thus the paradox.

SeanF
2005-May-03, 02:36 PM
So when you said there was no turn-around in that thought experiment, you were wrong.
I was talking about the clock paradox in the first “peculiar consequence” thought experiment in Section 4.
Why didn't you say so before?


There are absolutely no “turn arounds” in it.
But there is a change in inertial frames, which amounts to the same thing. The clock that is moved from A to B has motion now relative to its motion then. The clock that remains at B does not have motion now relative to its motion then. Thus, reciprocity is broken.

worzel
2005-May-03, 02:39 PM
So when you said there was no turn-around in that thought experiment, you were wrong.

I was talking about the clock paradox in the first “peculiar consequence” thought experiment in Section 4.

There are absolutely no “turn arounds” in it.
Then I ask you again, what did you mean by this?

Then how can the traveled clock in the “peculiar consequence” thought experiment in the SR theory “lag behind” the “stationary” clock when they unite?

SeanF
2005-May-03, 02:43 PM
So when you said there was no turn-around in that thought experiment, you were wrong.

I was talking about the clock paradox in the first “peculiar consequence” thought experiment in Section 4.

There are absolutely no “turn arounds” in it.
Then I ask you again, what did you mean by this?

Then how can the traveled clock in the “peculiar consequence” thought experiment in the SR theory “lag behind” the “stationary” clock when they unite?
Worzel, the thought experiment he's referring to goes like this:


From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by 1/2(tv^2/c^2) (up to magnitudes of fourth and higher order), t being the time occupied in the journey from A to B.
The clocks start out physically separated but in the same inertial frame. Then one clock is shifted into another inertial frame so that the two clocks approach each other.

Sam5 is confused by the distinction between

1) Either inertial frame can be considered to be "stationary"; and
2) Either clock can be considered to have changed inertial frames.

The first is true, the second is not.

Edited to Add: Actually, I don't think I've really got Sam5's problem down with that simplistic description.

Take two inertial frames, call them K and K', that are in relative motion to each other. There are differences in how observers in those two frames would measure space and time, right?

Now, what Sam5 thinks, is that in the peculiar consequence I quoted above, the two clocks start out in their own, separate, inertial frames that have no relative motion to each other. Then, when the motion starts, those same two inertial frames now have motion relative to each other.

The truth, of course, is that the two clocks start out in the same, single, inertial frame and end up in two different inertial frames.

It is that failure to understand what an "inertial frame" actually is that is at the heart of Sam5's confusion.

Sam5
2005-May-03, 02:49 PM
So when you said there was no turn-around in that thought experiment, you were wrong.
I was talking about the clock paradox in the first “peculiar consequence” thought experiment in Section 4.
Why didn't you say so before?

I did, several posts above.

Anyway, the “closed curve” thought experiment leads to a paradox too, but that is a little more difficult for average people to grasp. In fact, people have trouble grasping the paradox in the first “peculiar consequence” thought experiment.

SeanF
2005-May-03, 02:56 PM
Anyway, the “closed curve” thought experiment leads to a paradox too, but that is a little more difficult for average people to grasp. In fact, people have trouble grasping the paradox in the first “peculiar consequence” thought experiment.
The "paradox" is easy to grasp, Sam5. It's grasping that there isn't really one that is difficult for some people.

Sam5
2005-May-03, 03:10 PM
So when you said there was no turn-around in that thought experiment, you were wrong.

I was talking about the clock paradox in the first “peculiar consequence” thought experiment in Section 4.

There are absolutely no “turn arounds” in it.
Then I ask you again, what did you mean by this?

Then how can the traveled clock in the “peculiar consequence” thought experiment in the SR theory “lag behind” the “stationary” clock when they unite?
Worzel, the thought experiment he's referring to goes like this:


From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by 1/2(tv^2/c^2) (up to magnitudes of fourth and higher order), t being the time occupied in the journey from A to B.
The clocks start out physically separated but in the same inertial frame. Then one clock is shifted into another inertial frame so that the two clocks approach each other.

Sam5 is confused by the distinction between

1) Either inertial frame can be considered to be "stationary"; and
2) Either clock can be considered to have changed inertial frames.

The first is true, the second is not.



No, you don’t get it. Einstein doesn’t use the term “inertial frames” and he doesn’t count them as “frame 1, frame 2, frame 3, etc.” He uses the term “system” and there are two systems. Both systems are always in what you call an inertial frame all the time, meaning they never experience acceleration and they can’t “switch frames”. Both systems are always “inertial” no matter in which direction the relative motion is taking place, or if there is no relative motion taking place.

He says: “Let us take a system of co-ordinates in which the equations of Newtonian mechanics hold good. In order to render our presentation more precise and to distinguish this system of co-ordinates verbally from others which will be introduced hereafter, we call it the ‘stationary system.’”

This choice is totally arbitrary. To observers fixed with each system, their own system is the “stationary” one. He’s talking about kinematics, not accelerations.

Here’s an easier way to think of it:

Put a camera on each of two space shuttles that are close together. Then view the scene on TV from EITHER camera but from only one of the cameras at a time. Then move one of the shuttles. What you will see in the TV view is the OTHER shuttle appearing to be the one that moves, no matter which one actually fires its rockets. That is an effect of kinematics and relative motion.

Celestial Mechanic
2005-May-03, 03:18 PM
All right! Special Relativistic Food Fight!
Watch out for any bleu-shifted cheese coming your way! (Red-shifted cheese is OK, it's moving away from you.) :lol:

I will have some difficulty trying to respond to four poster[s]. Perhaps you could form a com[m]ittee and appoint a single spokesperson I could respond to.
What did I say elsewhere about Sam5's MO? :)

papageno
2005-May-03, 03:19 PM
No, you don’t get it. Einstein doesn’t use the term “inertial frames” ...
Semantic nitpicking.



...and he doesn’t count them as “frame 1, frame 2, frame 3,
etc.” He uses the term “system” and there are two systems. Both systems are always in what you call an inertial frame all the time, meaning they never experience acceleration and they can’t “switch frames”. Both systems are always “inertial” no matter in which direction the relative motion is taking place, or if there is no relative motion taking place.

He says: “Let us take a system of co-ordinates in which the equations of Newtonian mechanics hold good. ...
Which means, to anybody who understands classical mechanics, inertial frames of reference.
And please note "system of co-ordinates", which means frame of reference.




...In order to render our presentation more precise and to distinguish this system of co-ordinates verbally from others which will be introduced hereafter, we call it the ‘stationary system.’”

This choice is totally arbitrary.
Yes. He just puts some labels to be clear.




To observers fixed with each system, their own system is the “stationary” one. He’s talking about kinematics, not accelerations.
They would be stationary even when they accelerate.




Here’s an easier way to think of it:

Put a camera on each of two space shuttles that are close together. Then view the scene on TV from EITHER camera but from only one of the cameras at a time. Then move one of the shuttles. What you will see in the TV view is the OTHER shuttle appearing to be the one that moves, no matter which one actually fires its rockets. That is an effect of kinematics and relative motion.
Why not trains (which are closer to everyday experience than space shuttles)?

Sam5
2005-May-03, 03:26 PM
I will have some difficulty trying to respond to four poster[s]. Perhaps you could form a com[m]ittee and appoint a single spokesperson I could respond to.
What did I say elsewhere about Sam5's MO? :)


I can generally handle 2 at a time, but 4 is difficult and now you are #5. :D

worzel
2005-May-03, 03:33 PM
Sam5, you make no sense whatsoever. You keep on talking about that section from Einstein that is simply showing how a moving clock runs slowly, and insisting that because of that, there is no change of frame of reference in the twin paradox and no acceleration in SR.

Two systems of coordinates, frames of reference, intertial frames, whatever you want to call them, that are moving with respect to each other see clocks in the other frame run slowly. Apply the lorentz transformation to events in either frame shows you how those events appear in the other frame. That was the point of the section you are now quoting. And it holds for either frame. But the clocks could only ever coincide in space and time once in this case - so when you say "reunited" you are not talking about this thought experiment.

If two clocks start out synchronized and at the same point in space and time, move apart and come together, while one clock remains in one intertial frame, then the other clock must have changed its intertial frame. Although both clocks see the other running slow while they are both moving intertally, the one that turns around must change its frame of reference, that is, it must accelerate. SR deals with this by using one, many, or infinite lorentz transformations with the result being that when the clocks are reunited the one that turned around laggs behind the one that remained in an intertial frame throughout. That is the twin paradox.

What I can't work out is what bit of that you don't get. If you can't even distinguish between the two thought experiments then I don't think you're likely to know where your confustion lies either.

SeanF
2005-May-03, 03:54 PM
No, you don’t get it. Einstein doesn’t use the term “inertial frames” and he doesn’t count them as “frame 1, frame 2, frame 3, etc.” He uses the term “system” and there are two systems.
What do you mean "there are two systems"? He only ever refers to two, K and K' (or K and k, depending on your translation), but there are an infinite number of systems. The "closed curve" clock uses quite a few of them during it's trip.


Both systems are always in what you call an inertial frame all the time, meaning they never experience acceleration and they can’t “switch frames”. Both systems are always “inertial” no matter in which direction the relative motion is taking place, or if there is no relative motion taking place.
Yes, systems can't "switch frames" because systems are frames. Clocks (and observers) can, however, switch systems.


He says: “Let us take a system of co-ordinates in which the equations of Newtonian mechanics hold good. In order to render our presentation more precise and to distinguish this system of co-ordinates verbally from others which will be introduced hereafter, we call it the ‘stationary system.’”

This choice is totally arbitrary. To observers fixed with each system, their own system is the “stationary” one. He’s talking about kinematics, not accelerations.
Yes. Any "system of co-ordinates in which the equations of Newtonian mechanics hold good" can be considered stationary. Doesn't change the thought experiments.


Here’s an easier way to think of it:

Put a camera on each of two space shuttles that are close together. Then view the scene on TV from EITHER camera but from only one of the cameras at a time. Then move one of the shuttles. What you will see in the TV view is the OTHER shuttle appearing to be the one that moves, no matter which one actually fires its rockets. That is an effect of kinematics and relative motion.
And if the two shuttles were motionless relative to each other (and in a "system of co-ordinates in which the equations of Newtonian mechanics hold good"), and their clocks were synchronized before the firing of the rockets, and the shuttles then move together, the clock on the shuttle whose rockets fired will lag behind, because it changed inertial frames (or systems, if you prefer) - after the rockets are done firing, the shuttle is now in a different inertial frame. Nonetheless, during the travel, both shuttles will observer the other's clock as running slow.

And - this is important - how far behind the clock lags when they meet up depends on the final relative velocity and the amount of time it takes for them to meet up, which are only tangentially related to the magnitude of the acceleration.

And this remains true even if we consider that the two shuttles were both moving before the rockets firing, and the rockets served to stop one shuttle and allow the other to catch up to it.

Tensor
2005-May-03, 04:09 PM
I'm abstaining for now. This has all been explained to Sam5 several times here (http://www.badastronomy.com/phpBB/viewtopic.php?t=9731), here (http://www.badastronomy.com/phpBB/viewtopic.php?t=14284) and here (http://www.badastronomy.com/phpBB/viewtopic.php?t=12739) . Either he just doesn't understand, doesn't want to understand, or is pulling our chain again. I think the BA post in this thread (http://www.badastronomy.com/phpBB/viewtopic.php?p=298391) may be applicable here.

Sam5
2005-May-03, 06:02 PM
No, you don’t get it. Einstein doesn’t use the term “inertial frames” and he doesn’t count them as “frame 1, frame 2, frame 3, etc.” He uses the term “system” and there are two systems.
What do you mean "there are two systems"? He only ever refers to two, K and K' (or K and k, depending on your translation), but there are an infinite number of systems. The "closed curve" clock uses quite a few of them during it's trip.

In the paper he called them “rigid bodies (systems of co-ordinates)” or just “systems”. Since he was dealing with kinematics only, he never dealt with mass, gravity, inertia, or acceleration.

Look, there is no use in continuing this, since you can’t understand what I’m talking about. You are getting his original theory confused with stories about it.

Sam5
2005-May-03, 06:08 PM
Anyway, Tensor doesn’t want me to say anything else. He’s giving me the warning to “Just Shut Up.” Dat's Relativity! :D

papageno
2005-May-03, 06:11 PM
No, you don’t get it. Einstein doesn’t use the term “inertial frames” and he doesn’t count them as “frame 1, frame 2, frame 3, etc.” He uses the term “system” and there are two systems.
What do you mean "there are two systems"? He only ever refers to two, K and K' (or K and k, depending on your translation), but there are an infinite number of systems. The "closed curve" clock uses quite a few of them during it's trip.

In the paper he called them “rigid bodies (systems of co-ordinates)” or just “systems”. Since he was dealing with kinematics only, he never dealt with mass, gravity, inertia, or acceleration.
Acceleration is part of kinematics.
Also, you don't seem to grasp that "system of coordinates" and "frame of reference" are the same.
Einstein uses "rigid rods" and "clock" just to give something more tangible to the reader.



Look, there is no use in continuing this, since you can’t understand what I’m talking about. You are getting his original theory confused with stories about it.
You have shown that your understanding of Classical Mechanics is lacking.
You cannot expect to understand Einstein's "original" theory, because you misinterpret the terms and concepts he uses.



Anyway, Tensor doesn’t want me to say anything else. He’s giving me the warning to “Just Shut Up.” Dat's Relativity!
The BA's warning in the other thread was clear: back up your claims.

SeanF
2005-May-03, 06:21 PM
No, you don’t get it. Einstein doesn’t use the term “inertial frames” and he doesn’t count them as “frame 1, frame 2, frame 3, etc.” He uses the term “system” and there are two systems.
What do you mean "there are two systems"? He only ever refers to two, K and K' (or K and k, depending on your translation), but there are an infinite number of systems. The "closed curve" clock uses quite a few of them during it's trip.
In the paper he called them “rigid bodies (systems of co-ordinates)” or just “systems”. Since he was dealing with kinematics only, he never dealt with mass, gravity, inertia, or acceleration.
Sure, why not? But there's still no paradox.


Look, there is no use in continuing this, since you can’t understand what I’m talking about. You are getting his original theory confused with stories about it.
I understand exactly what you're talking about. You don't understand what Einstein was talking about. As I pointed out above (and you've verified with your later comments), you don't understand what a "system of coordinates," as Einstein described it, is.


Anyway, Tensor doesn’t want me to say anything else. He’s giving me the warning to “Just Shut Up.” Dat's Relativity! :D
[-X

If you bail on this, it's because you chose to, not because Tensor made you. Besides, Tensor's dropping out just means one less person you have to try to respond to, right?

Tensor
2005-May-03, 06:23 PM
Anyway, Tensor doesn’t want me to say anything else. He’s giving me the warning to “Just Shut Up.” Dat's Relativity! :D

Nope, not at all. Your statements have been shown to be wrong before and you continue to make them. Just asking you provide some sort of proof for your statements, not your mis-interpretation of an analogy or a misunderstanding of the explanation of the math.

Sam5
2005-May-03, 06:27 PM
Anyway, Tensor doesn’t want me to say anything else. He’s giving me the warning to “Just Shut Up.” Dat's Relativity! :D

Nope, not at all. Your statements have been shown to be wrong before and you continue to make them. Just asking you provide some sort of proof for your statements, not your mis-interpretation of an analogy or a misunderstanding of the explanation of the math.

Thank you for your opinion. :D

Bob
2005-May-03, 07:31 PM
I think the BA post in this thread (http://www.badastronomy.com/phpBB/viewtopic.php?p=298391) may be applicable here.

BA can always do what he wants on his BB, but Sam5 didn't hijack this thread, he started it.

Tensor
2005-May-03, 07:40 PM
I think the BA post in this thread (http://www.badastronomy.com/phpBB/viewtopic.php?p=298391) may be applicable here.

BA can always do what he wants on his BB, but Sam5 didn't hijack this thread, he started it.

Didn't claim he hijacked it, just requesting he provide some evidence for his ATM claims, that's all.

SeanF
2005-May-03, 07:54 PM
Since Sam's got no one else to respond to but me, maybe he can find the time to answer a question now.

Sam, you've said that you accept the GR theory that acceleration causes time changes in atomic clocks. I'm wondering if you can tell me where, in all of Einstein's papers, he quantified this effect.

In other words, what's the equation used to calculate how much a given acceleration will affect a clock, and what do the different variables in the equation mean?

If you do understand GR, this shouldn't be too difficult.

Sylas
2005-May-03, 10:05 PM
Sylas, if you don’t have a copy of Volume 4, I think you would enjoy it very much. It contains several papers that cover his transition from SR to GR. If you order a copy, be sure to get the paperback version since that is the English version. The hardback is all in German.

Sam5, if you ever actually learned the maths of this, I think you would enjoy it very much. Einstein was particularly struck with the mathematical beauty of the theory he discovered, and so far you are missing that entirely. Your descriptions are completely incorrect, and misunderstand Einstein completely.

Your own enjoyment of these papers will increase immeasurably when you learn enough about them to stop making so many trivial errors in understanding them.

Best wishes -- Sylas

Sam5
2005-May-03, 11:16 PM
Sorry, Sylas, but Tensor has given me a warning to shut up, so I can’t say any more.

Tensor
2005-May-04, 03:15 AM
Sorry, Sylas, but Tensor has given me a warning to shut up, so I can’t say any more.

I did no such thing, please retract this statement. Point out exactly where I told you to shut up or not to post. All I did was to provide a link to where the BA requested you provide support for your claims.

Normandy6644
2005-May-04, 03:45 AM
*waiting patiently for some science*

:D

Fram
2005-May-04, 07:46 AM
Sorry, Sylas, but Tensor has given me a warning to shut up, so I can’t say any more.

I did no such thing, please retract this statement. Point out exactly where I told you to shut up or not to post. All I did was to provide a link to where the BA requested you provide support for your claims.

Well, Tensor, to be fair, as Sam5 cannot support his own claims, he has no choice but to shut up. :lol:

SeanF
2005-May-04, 01:21 PM
Sorry, Sylas, but Tensor has given me a warning to shut up, so I can’t say any more.
Nice try, Sam. If you can't say more it's because you've realized you've got nothing to say. Like I said, bailing is your choice, not Tensor's.