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

[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!

 

[Dale]

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.

Sorry if I was unclear. I was not saying that you were wrong, I was just showing how to correctly transform waves and urging caution in interpreting your statement.

Yes, for waves that propagate at v<c it is possible to find an inertial reference frame where they are "at rest". The same is not possible for waves that propagate at v=c as you can see from the simple fact that the wave four-vector is null for the v=c case. This is the caution I was urging: caution about overgeneralizing your correct statement about EM waves in a transparent medium into an incorrect conclusion about EM waves in vacuum.

 

[Phrak]

The question assumes that a photon has such a thing as a proper time - it doesn't.

I'd like to see an informal proof.

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...".

I understand. My interpretation was hasty.

My ongoing objection to the answers that have been received by this category of question that include that proposed by the OP in this thread are the somewhat cavalier responses. There is a significant difference between "I don't know how to make sense of it and I don't have a text that makes sense of it" and "it makes no sense". What's wrong with modesty?

To be fair, to examine all reasonable possibilities may not have exhaustion, but this has never been put forth that I have seen.

 

[Phrak]

The medium is not at rest but the wave is!

That's fascinating Bob. I don't know where one could go with it, but a titillating idea, none the less.

 

[JesseM]

My ongoing objection to the answers that have been received by this category of question that include that proposed by the OP in this thread are the somewhat cavalier responses. There is a significant difference between "I don't know how to make sense of it and I don't have a text that makes sense of it" and "it makes no sense". What's wrong with modesty?

But it's not like there are any open questions anyone isn't sure of here. Everyone agrees that there's no way a coordinate system where a photon was at rest could be an inertial frame, and I assume everyone here also agrees that if we're talking about non-inertial coordinate systems, it's a simple matter to define a coordinate system where a photon is at rest.

 

[Fredrik]

I'd like to see an informal proof.

Proper time is a property of a timelike curve. The concept isn't defined for other types of curves. Nothing more needs to be said.

But I'll say a few more things anyway. Every calculation of the arc length of a curve involves adding upp contributions of the form $$\sqrt g(\dot C,\dot C)$$ along the curve. Here C is the curve, and $$\dot C$$ its tangent vector. But the world line of a massless particle is by definition one that has $$\sqrt g(\dot C,\dot C)=0$$ everywhere. So it would make some sense to define the proper time to be 0. It's just a useless thing to do.

It would also be misleading, because one of the axioms of SR is that what a clock measures is the proper time of the curve in Minkowski space that represents its motion. To define the proper time of a null curve to be 0 seems to suggest that a clock would experience no time along such a curve, but as you know a clock can't be accelerated to that speed, and to imagine a clock made entirely of massless particles would at the very least require a redefinition of what we mean by a "clock". (I think it's more problematic than that actually, but I don't have time to explain right now).

You still seem to think that maybe "the photon's rest frame" can be defined in spite of everything I (and others) have said. Would you also like to see an "informal proof" that it can't? That would be like asking for proof that there can't exist a month called "Octember". Just look at any calendar and you'll see that there's no such month. Now we certainly could make a new calendar that includes the month of Octember, so it doesn't make much sense to ask for proof that it can't be done. The question we should be asking is "Why would we want to, and where would we insert it?".

 

[jtbell]

The question we should be asking is "Why would we want to, [...]?".

Bingo! I think this expresses the uneasiness or puzzlement that has always nagged me when reading these perennial discussions about "what does a photon 'see'?" It may be technically possible to define some kind of weird coordinate system in which a photon is at rest, but what could we do with it except make statements such as "time doesn't pass for a photon?" What kind of quantitative statements could we make that give us new insight into how light behaves, or new ways of solving problems that involve light?

 

[Phrak]

Bingo! I think this expresses the uneasiness or puzzlement that has always nagged me when reading these perennial discussions about "what does a photon 'see'?" It may be technically possible to define some kind of weird coordinate system in which a photon is at rest, but what could we do with it except make statements such as "time doesn't pass for a photon?" What kind of quantitative statements could we make that give us new insight into how light behaves, or new ways of solving problems that involve light?

Statements of the kind made utilizing nonstandard analysis.

 

[Phrak]

We are not unused to working with the infinitesimal bases of tangent spaces nor of finding coordinate singularities in general relativity. It seems infinitesimals should also be useful tools, and singular coordinates equally acceptable in Minkowski space.


We may specify Minkowski space in http://en.wikipedia.org/wiki/Light_cone_coordinates" .

$$a = \frac {1} {\sqrt{2}} (x+ct) \ \ \ \ \ \ b = \frac {1} {\sqrt{2}} (x-ct)$$

y and z remain unchanged.

For a Lorentz transformation in the x direction, $\mu$ is defined as

$$\mu = \sqrt {\frac{1-\beta}{1+\beta} }$$

where $\beta=$ v/c .

Coordinates a and b undergo simple dilation and contraction.

$$a' = \mu a \ \ \ \ \ \ b' = \mu^{-1}b$$


As v=v_x approaches c from below, using nonstandard analysis,

v = c-dv or

$$\beta = 1 - \frac{dv}{c} \ .$$

dv is the infinitesimal of velocity. Then

$$\mu = \sqrt{\frac{dv}{2c-dv}} \ .$$

The transformation of coordinates as v approaches c to an infinitesimal difference are

$$a' = \sqrt{\frac{dv}{2c-dv}}a$$

$$b' = \sqrt{\frac{2c-dv}{dv}}b$$

Immediately we have the conserved quantity ab = a'b'.

We may make a change to null coordinates, perform the Lorentz transform, reduce the nonstandard form, and change back to null coordinates to obtain

[tex]x' = \frac{1}{2} \sqrt{\frac{2c}{dv}}(x-ct)[/itex]<br /> <br /> $$ct' = \frac{1}{2} \sqrt{\frac{dv}{2c}}(x-ct)$$<br /> <br /> These expressions are reduced and irreversible. Leaving out the reduction, the transformation is reversible as boost velocity is decreased from c back to zero.[/tex]

 

[Fredrik]

It's only reversible because you don't actually take the limit v→c. What you're doing is equivalent to saying that v-c is small but non-zero. I don't see what you're accomplishing by calling it an "infinitesimal" and claiming that you're doing "non-standard analysis".

I also don't see why you bother to take the detour over the variables a and b. Why not just introduce your new variable dv=1-v in a Lorentz transformation? (I'm using units such that c=1).

$$\begin{pmatrix}t'\\ x'\end{pmatrix}=\gamma\begin{pmatrix}1 & -v\\ -v & 1\end{pmatrix}\begin{pmatrix}t\\ x\end{pmatrix} =\frac{1}{\sqrt{1-(1-dv)^2}}\begin{pmatrix}1 & -1+dv\\ -1+dv & 1\end{pmatrix}\begin{pmatrix}t\\ x\end{pmatrix}\approx\frac{1}{\sqrt{2dv}}\begin{pmatrix}t-x\\ -t+x\end{pmatrix}$$

 

[Phrak]

It's really not a detour. It's nice to see what happens in transforming (x,t) to (x',t'), but the essence of what is going on is best seen in null coordinates. We would really want to use the infinitessimal [itex]d\mu[/tex], as $\mu$ becomes an infinitesimal when v = c - dv.<br /> $$d\mu = \sqrt{\frac{dv}{2c-dv}}$$<br /> <br /> With the additional clutter removed,<br /> <br /> $$a' = d\mu \ a$$<br /> <br /> $$b' = \frac{1}{d\mu} \: b$$<br /> <br /> a' is seen for what it is, an infinitesimal. b' is an infinity.<br /> <br /> By definition, c-v = dv is smaller than any number you can name that is greater than zero, but it's not all together satisfying as you noticed, but this is the best you can do with http://en.wikipedia.org/wiki/Hyperreal_number" . There is also an extension of the hyperreals that I don't sufficiently understand.[/itex]

 

[pervect]

Kruskal-Szerkes coordinates, used for black holes, are one example of commonly used null coordinates in GR.

Conceptually, null coordinates aren't too hard to visualize.

On the x-t plane, you'd assign a constant, u, to left-propagating light rays, and another constant, v, to right propagating light rays. So left-propagating light rays would have constant u, and a variable v. Right-propagating light rays would have a constant v, and variable u.

Any point in the x-t plane can be described by giving the u and v coordinate of it.

For the black hole case, you'd assign u and v to ingoing and outgoing null geodesics.

Null tetrads are another example of where the "trick" of using null coordinates can be useful.