How do null coordinates help understand photon's frame of reference?

[6Stang7]

Sorry if this has been asked, but I did a search and didn't find a relating topic.

This thought came to me a few days ago when a friend asked me about time dilation. Time beats slower and distance is shorter for a moving reference frame with respect to a non-moving reference frame, and the amount of each is dependent upon the speed.

So, for a photon that travels at c, would time not pass and would all space be a single point in the photon's moving reference frame with respect to the non-moving reference frame?

 

[ZikZak]

Photons don't have reference frames. By postulate, photons travel at c in all reference frames. Thus there are no reference frames in which photons are at rest.

 

[JesseM]

Some old threads dealing with the question of why light doesn't have its own inertial rest frame:

https://www.physicsforums.com/showthread.php?t=227254
https://www.physicsforums.com/showthread.php?t=212819
https://www.physicsforums.com/showthread.php?t=285875

 

[Phrak]

I've seen a few arguments on PF to the effect that a photon doesn't have a frame of reference where it is at rest. But the inability to provide a map, as these arguments go, from an inertial frame to a hypothetical frame of reference, is only weak evidence of nonexistance.

 

[pervect]

We've really had this discussion several times. See for instance https://www.physicsforums.com/showpost.php?p=1254103&postcount=8

The highlights, since I don't think you'll read the thread otherwise:

It's basically an error to attribute the usual sort of reference frame to an object moving at the speed of light.

If one has some basic knowledge of the concept of the Lorentz interval, it is easy to see why. The Lorentz interval between any two points of the worldline of an ordinary observer is timelike. Thus such an observer experiences time along their worldline. The Lorentz interval along any two points of the wordline of a photon is null. A null interval is neither timelike, nor spacelike. Thus a photon does not experience "time in the same sense that a person or any object made out of matter does.

But while a photon doesn't "experince time", it is possible to mark out regular intervals along its worldline...

So a photon doesn't experience time, but in some abstract sense it "experiences" a null coordinate - at least, one can distinguish regular intervals along a photon's worldline. But in spite of the fact that these intervals are regular in some sense, they should not be confused with time. The intervals are not timelike - they are null intervals, neither timelike nor spacelike.

 

[Phrak]

How does that answer anything, pervect?

The OP asks about the proper time of a photon,

So, for a photon that travels at c, would time not pass

and asks how the measure of space in an inertial frame maps to the measure of space in the comoving frame of a photon,

and would all space be a single point in the photon's moving reference frame with respect to the non-moving reference frame?

 

[ZikZak]

I've seen a few arguments on PF to the effect that a photon doesn't have a frame of reference where it is at rest. But the inability to provide a map, as these arguments go, from an inertial frame to a hypothetical frame of reference, is only weak evidence of nonexistance.

When I claimed above that the photon has no reference frame, I did not argue anything about maps. I pointed out that such a frame would directly contradict the second postulate. That should clearly end any debate, unless it was something other than Relativity you were wanting to talk about.

 

[Fredrik]

The OP asks about the proper time of a photon,

He's also talking about the "photon's frame of reference" as if there's a standard definition of "this object's rest frame" that works for all objects including photons. This is certainly not the case. A good answer should explain the standard definition for massive particles and explain why it doesn't work for photons. I included such an explanation in the thread I mentioned in this quote:

See my posts in this thread about the "photon's point of view". In particular, #8 and #14.

The concept of "proper time" is only defined along those curves in spacetime that represent speeds <c, so the concept of "proper time" isn't really relevant here. We could of course define the proper time of a null curve (a curve that represents speed c) to be =0, but it's a pretty pointless thing to do for several reasons, one of them being that no clocks can travel at that speed anyway.

 

[Phrak]

Fredrik. We've gone over this before, of course. thank you, I've reread your posts. You might also feel free to read my posts at the end of the same thread.

One can spend a great deal of time showing how one cannot make sense of the question using some particular reasoning. But the best we can do, in support of the negative, is say that we cannot make sense of it, rather than sense cannot be made of it.

If I become sufficiently motivated, I might go over my notes to see if I found a sensible answer to this.

 

[A.T.]

I think It is OK to say that proper time doesn't pass for photons. I'm not convinced by the arguments why the concept of proper time is completely inapplicable to photons:

The concept of "proper time" is only defined along those curves in spacetime that represent speeds <c,

Here you a linking the concept of proper time which is a measurable physical quantity with one particular abstract mathematical concept: the Minkowski spacetime. There are other geometrical interpretations than Minkowski's, like the space-popertime diagram, where the proper time of photons is simply zero.

one of them being that no clocks can travel at that speed anyway.

We also cannot place an observer at infinity, but will still use his proper time to define gravitational time dilation. And how do you know that photons are not clocks?

 

[Fredrik]

Fredrik. We've gone over this before, of course. thank you, I've reread your posts. You might also feel free to read my posts at the end of the same thread.

I took a quick look at them, but didn't examine every detail. When you define your new variables a and b, you're really just saying that we take the photon's world line to be the time axis of a new coordinate system, and then you pick another straight line to be the spatial axis. This is of course fine, but you need to be aware of two things: a) the result isn't an inertial frame, and b) after you have chosen the time axis that way, the photon will be stationary in your new coordinate system, no matter what other choices you make in your definition (e.g. what curve to use as the spatial axis). So if you're going to say that a particular coordinate system is "the photon's rest frame" (rather than one of infinitely many rest frames for the photon), you're going to have to come up with a very good reason.

One can spend a great deal of time showing how one cannot make sense of the question using some particular reasoning. But the best we can do, in support of the negative, is say that we cannot make sense of it, rather than sense cannot be made of it.

I think that what's interesting is that neither the theory we're talking about, nor the standard definitions used by people working with it, has made sense of it. The possibility that some weird definitions might give this concept meaning isn't very interesting to me.

 

[Bob_for_short]

Let us take a plane wave Acos(ωt-kx) in a transparent medium with n. The light velocity is c/n. What is the wave solution in a reference frame moving with V=c/n? Isn't it A'cos(k'x') ? And photon energy is not ћω'=0 but ћck'>0?

 

[Fredrik]

I think It is OK to say that proper time doesn't pass for photons. I'm not convinced by the arguments why the concept of proper time is completely inapplicable to photons:

Here you a linking the concept of proper time which is a measurable physical quantity with one particular abstract mathematical concept: the Minkowski spacetime. There are other geometrical interpretations than Minkowski's, like the space-popertime diagram, where the proper time of photons is simply zero.

We also cannot place an observer at infinity, but will still use his proper time to define gravitational time dilation. And how do you know that photons are not clocks?

I can tell that we have pretty different ways of thinking about these things. To me a photon is a mathematical concept defined by a theory, and the definition implies that a photon doesn't have the internal structure it would need to change with time. You're probably talking about "actual photons" in the real world, but I consider that an ill-defined concept, and therefore less interesting to talk about. Also, to me "proper time" is a mathematical property of a curve in Minkowski space, and special relativity is the theory that tells us how to interpret the mathematics of Minkowski space as predictions about results of experiments. If you use another "geometrical interpretation", I would say that you're talking about a different theory. It's an equivalent theory (I hope), but not the same theory, and I tend to interpret questions in the relativity forum as questions about the standard formulations of SR and GR, not as questions about other formulations, or even as questions about the real world (since the concepts we're talking about are defined by the theories we're using).

 

[Count Iblis]

Is it in theory impossible to construct an observer out of zero mass particles?

 

[Ich]

It is possible in principle, as long as they go in different directions. That's equivalent to the system having rest mass.

 

[Dale]

Woohoo, around and around in circles we go over and over and around and around again and again woohoo!

 

[Bob_for_short]

Yes, instead of answering my questions.

 

[nutgeb]

Maybe somebody mentioned this, but another reason why it's not meaningful to talk about a photon's rest frame is that if it were actually at rest, its momentum would be zero in that frame, so its rest mass would be zero. Then its energy content must be zero, and arguably it wouldn't exist at all. A photon is like a shark, it exists only in motion.

There also is a Heisenburg Uncertainty problem, because if the photon's momentum is exactly fixed (at zero in its rest frame), its location must be entirely unbounded.

 

[Bob_for_short]

Maybe somebody mentioned this, but another reason why it's not meaningful to talk about a photon's rest frame is that if it were actually at rest, its momentum would be zero in that frame, so its rest mass would be zero. Then its energy content must be zero, and arguably it wouldn't exist at all. A photon is like a shark, it exists only in motion.

There also is a Heisenburg Uncertainty problem, because if the photon's momentum is exactly fixed (at zero in its rest frame), its location must be entirely unbounded.

And how about my question in post #12?

 

[nutgeb]

And how about my question in post #12?

Agreed.

 

[Dale]

Let us take a plane wave Acos(ωt-kx) in a transparent medium with n. The light velocity is c/n. What is the wave solution in a reference frame moving with V=c/n? Isn't it A'cos(k'x') ? And photon energy is not ћω'=0 but ћck'>0?

Be careful transforming a wave in relativity. The http://en.wikipedia.org/wiki/Four-frequency" is (ω/c,k) which is Lorentz covariant, so it transforms like any other four-vector. And its dot product with the four-position of some event is the phase at that event which is a Lorentz invariant scalar.

Thanks for the brief respite from the usual monotony of this repetitive discussion.

 

[Phrak]

Photons don't have reference frames. By postulate, photons travel at c in all reference frames. Thus there are no reference frames in which photons are at rest.

I've seen a few arguments on PF to the effect that a photon doesn't have a frame of reference where it is at rest. But the inability to provide a map, as these arguments go, from an inertial frame to a hypothetical frame of reference, is only weak evidence of nonexistance.

When I claimed above that the photon has no reference frame, I did not argue anything about maps. I pointed out that such a frame would directly contradict the second postulate. That should clearly end any debate, unless it was something other than Relativity you were wanting to talk about.

Sorry, I missed your last post. Photons have no inertial frame at which they are at rest. That would tend to rule out frames of reference which are inertial. And also comoving. And also those accelerating, I suppose. Is this set exhaustive?

 

[Phrak]

Woohoo, around and around in circles we go over and over and around and around again and again woohoo!

I am not circling.

 

[Phrak]

I took a quick look at them, but didn't examine every detail. When you define your new variables a and b, you're really just saying that we take the photon's world line to be the time axis of a new coordinate system, and then you pick another straight line to be the spatial axis. This is of course fine, but you need to be aware of two things: a) the result isn't an inertial frame, and b) after you have chosen the time axis that way, the photon will be stationary in your new coordinate system, no matter what other choices you make in your definition (e.g. what curve to use as the spatial axis). So if you're going to say that a particular coordinate system is "the photon's rest frame" (rather than one of infinitely many rest frames for the photon), you're going to have to come up with a very good reason.

Thank you. I'll take this all under consideration. Note that neither a nor b are a temporal axis.

I think that what's interesting is that neither the theory we're talking about, nor the standard definitions used by people working with it, has made sense of it. The possibility that some weird definitions might give this concept meaning isn't very interesting to me.

I have not advanced a theory. Personal theories are outlawed under the PF guidlines. I've carefully constrained my analysis to the postulates of special relativity as it should be.

 

[atyy]

Photons have no inertial frame at which they are at rest.

Yes, I think that's the point. You can construct non-inertial frames in which the photon is at rest. http://arxiv.org/abs/hep-ph/9505259

 

[JesseM]

Sorry, I missed your last post. Photons have no inertial frame at which they are at rest. That would tend to rule out frames of reference which are inertial. And also comoving. And also those accelerating, I suppose. Is this set exhaustive?

How do you define "frame"? If you just mean a spacetime coordinate system, then it is trivial to define one where a photon is at rest, but obviously this is not an inertial frame (nor would it really be accurate to describe it as an 'accelerating frame', but you could call it a 'non-inertial frame' if you wanted).

 

[Fredrik]

Thank you. I'll take this all under consideration. Note that neither a nor b are a temporal axis.

When I said "time axis", I just meant the 0th axis.

I have not advanced a theory.

I know, and I didn't mean to suggest that you had.

 

[A.T.]

To me a photon is a mathematical concept defined by a theory, and the definition implies that a photon doesn't have the internal structure it would need to change with time. You're probably talking about "actual photons" in the real world,

Yes that's what I meant, let call it "light".

Also, to me "proper time" is a mathematical property of a curve in Minkowski space,

To me "proper time" is what a clock measures, a real world observable quantity.

and special relativity is the theory that tells us how to interpret the mathematics of Minkowski space

To me it is the other way around: Minkowski space is an geometrical interpretation of the mathematics of Special Relativity.

If you use another "geometrical interpretation", I would say that you're talking about a different theory.

No, just a different type of diagram. The theory gives you formulas which describe the relationship between quantities. The geometrical interpretation is for example which of the quantities you put on axes.

But on the topic I agree that the frame of reference of a photon is not useful / sensible.

 

[Bob_for_short]

Be careful transforming a wave in relativity. The http://en.wikipedia.org/wiki/Four-frequency" is (ω/c,k) which is Lorentz covariant, so it transforms like any other four-vector. And its dot product with the four-position of some event is the phase at that event which is a Lorentz invariant scalar.

In a transparent medium they are not Lorentz four-vectors in an origilnal sense: there the Maxwell equations are different from those in vacuum: they include n so the relationship between ω and k is different. Thus the usual Lorentz tansformations (where n=1) may change the direction of the wave propagation to the opposite: in a medium there may be reference frames faster than light, c/n < V < c. Charged particles with v > c/n emit the Cherenkov's radiation. They leave the radiation behind because they go faster. If in a medium there is a wave propagating in the positive x-direction, a fast charged particle with v > c/n experiences this wave as propagating in the negative x-direction, unlike the vacuum situation.

 

[pervect]

How does that answer anything, pervect?

The OP asks about the proper time of a photon,


and asks how the measure of space in an inertial frame maps to the measure of space in the comoving frame of a photon,

The question assumes that a photon has such a thing as a proper time - it doesn't. And since there isn't any such thing as a "comoving frame", the other question is based on a false assumption as well.

The exchange here is a bit like the old question - "When did you stop beating your wife?" - when the poor SOB not only hasn't beaten his wife, but isn't even married. Of course, people can try to make something out of this response, saying - "Oh - the poor woman - you didn't even marry her...".

 

[Dale]

In a transparent medium they are not Lorentz four-vectors in an origilnal sense: there the Maxwell equations are different from those in vacuum: they include n so the relationship between ω and k is different.

The object (ω/c,k) Lorentz transforms just like a four-vector, what other "original sense" of being a four-vector is there? Even the wave four-vector for a mechanical wave (e.g. sound) Lorentz transforms as a typical four-vector. I don't understand your comment.

Note that c in (ω/c,k) is the speed of light in vacuum, not the speed of the specific wave under consideration. Perhaps that is what confused you?

 

[Count Iblis]

If you perform a lorentz boost in this case then the medium will no longer be at rest.

 

[Dale]

Correct.

 

[Bob_for_short]

...Even the wave four-vector for a mechanical wave (e.g. sound) Lorentz transforms as a typical four-vector. I don't understand your comment...

Ok, for a sound wave there is a reference frame where the wave is still: Acos(ωt-kx) → A'cos(k'x'), isn't there?

I state the same feature for the electromagnetic wave in a medium. There is a rest frame for the EMW in medium where the wave is still: A'cos(k'x'). Your reasoning about Lorentz transformations does not invalidate it.

 

[Bob_for_short]

If you perform a lorentz boost in this case then the medium will no longer be at rest.

The medium is not at rest but the wave is!