# Rest frame of the photon

1. Dec 21, 2012

### chwie

I know that the rest frame of the photon is a non inertial reference frame. In that sense obviously the physics will not be Lorentz co-variant and so on. I have the following question

Is possible to define a coordinate transformation from an inertial reference frame to the rest frame of a particle traveling at the speed of light?

Obviously the coordinate transformation will not be a Lorentz transforms, but maybe another kind of coordinate transformation can do the trick.

My idea is that we can use any reference frame to describe the physics. The laws of nature will be Lorentz covariant in any inertial reference frame, however if I do a coordinate transformation from an inertial reference frame to a non inertial reference frame then I can see how the laws will be in this coordinate system.

Do you believe that this coordinate transformation can be found?

For example, I can imagine the rest frame of the photon as a non inertial reference frame in which all the massive free particles will travel at the speed of light. My problem is to define a coordinate transformation from this reference frame to an inertial reference frame.

2. Dec 21, 2012

### Mentz114

In GR it is possible to write a metric which describes a spacetime containing massless energy which travels at the speed of light. It is called the 'null dust' solution.
I think there is no static frame field in the spacetime.

Last edited: Dec 21, 2012
3. Dec 21, 2012

### Staff: Mentor

I believe this is correct, except for the trivial case where the density of the null dust is zero (which is just the Schwarzschild vacuum). See here:

http://en.wikipedia.org/wiki/Vaidya_metric

Note that the line elements (6) and (7) are formally identical to the Eddington-Finkelstein forms of the Schwarzschild line element; the only difference is that the metric parameter M is a function of the null coordinate u (for the ingoing form) or v (for the outgoing form), instead of being constant.

4. Dec 21, 2012

### pervect

Staff Emeritus
There isn't any rest "frame" for the photon, but as mentioned, you can find coordinates for space-time such that a constant coordinate is the worldline of a photon. They're called null coordinates.

For the 2d case, the flat Minkowskii space time has the metric
s = dx^2 - dt^2. Let u = x+t and v=x-t. Then in terms of u and v
the metric is just

s = du dv = (dx + dt)(dx-dt) = dx^2 - dt^2

and either u or v is constant depening on the direction of travel as a photon, which is the desired property of represting a photon's world line via a constant coordinate.

5. Dec 22, 2012

### andrien

photon spin component along any chosen z direction can have value either +1 or -1.It does not have any zero.this statement follows from fact that it is not possible to choose a reference frame in which photon polarization state(real) can be increased to 3(a longitudinal component) while it is two for a real photon.this follows rather directly from the fact that photon can not move slowly in any reference frame.The two polarization states can be easily gotten in coulomb gauge in which longitudinal component is already integrated out.Also A0 associated with it does contain any dynamical degree of freedom.so this also can be eliminated.

6. Dec 22, 2012

### chwie

Thanks

I think I can play with light cones coordinates, but I found that it lacks some properties that I would like of the rest frame of a particle traveling at the speed of light. Particularly only the the particle at rest in the original reference frame will "travel" at the speed of light in the light cone coordinates. I believe that 'from the photon point of view' all the inertial observers should travel at the speed of light. Anyway it is a non inertial reference and I can use it as inspiration to play with. Probably it really do not make sense to talk of the rest frame of the photon (inertial or non inertial), but I think it is a fun idea.

Thanks everyone for their contribution.

Andrien

I agree that the polarization of the photon only has two degree of freedom in any inertial reference frame. In a non inertial reference frame non physical degree of freedom can appear or physical degree of freedom can "disappear". For that reason I believe that the argument that you gave is not enough to justified the non existence of a rest frame of the photon.

An example will be a particle that is constrained to move in a two dimension plane in an inertial reference frame, but in a non inertial reference which is oscillating in the perpendicular direction to the plane we can have the impression that the particle is moving in three dimension and not in two dimensions. Obviously this is extra degree of freedom is not a physical degree of freedom.

Maybe I am wrong and your argument is enough to justify that the photon cannot be at rest in any possible reference frame, but then it is not enough to justify that a spinless particle traveling at the speed of light cannot have a non inertial reference frame.

However your argument is important and is a question that I should check if I decide to keep playing with this idea. Thank You.

7. Dec 22, 2012

### robphy

8. Dec 23, 2012

### andrien

photon's mass is zero.if it's mass might not have been zero then it might be possible to go to another reference frame in which it will have a longitudinal component of polarization which will contradict the physical degree of freedoms,which are two for photon.your argument regarding non inertial frame is not valid.
edit:here is a reference related to it-
http://www.damtp.cam.ac.uk/user/tong/qft/six.pdf

Last edited: Dec 23, 2012
9. Dec 23, 2012

### chwie

Which part of the lecture notes of David Tong do you want me to read?

I suppose that you are using some argument based in the gauge symmetry of the photon. Yes I agree that the U (1) gauge symmetry and Lorentz co-variance will implies that the photon does not has a mass term. You probably are thinking in the equation

ε(p)$\bullet$p=0

This equation hold in a inertial reference frame. There is no reason why I cannot define a non inertial reference frame in which it do not hold. In that case I will have something that will appear as a Longitudinal polarization. In general the transformation that will leave this product invariant are Lorentz transformation, then it will hold in any inertial reference frame. This proof that does not exist an inertial reference frame in which the photon does has a longitudinal component. I completely agree. My question is about non inertial reference frame in which that equation does not hold. Probably you are right and it is impossible, but in principle I do not see why it should be true.

Thanks

robphy

Thanks

I agree with your original post. That is the reason I am looking for a non inertial reference frame. My idea is to find a patch of coordinates in which any inertial observer will appear to travel at the speed of light. That is my definition of the photon rest frame which needs to be non inertial. Obviously an inertial reference frame is impossible and some of the arguments that you gave justify it.

Last edited: Dec 23, 2012
10. Dec 23, 2012

### andrien

why do you think that a non inertial frame description would be adequate.do you think that in non inertial frame there will be some extra degree of freedom.

11. Dec 23, 2012

### chwie

My idea is that if I can find a non inertial reference frame in which stuff can move faster than the speed of light. Why I can't find a non inertial reference frame in which particle that move at the speed of light are not moving?

At example of a reference frame in which things can move faster than the speed of light is the earth. During the night I will see that the stars are traveling millions of light years in hours. Obviously the reason is that the earth is a non inertial reference frame. The stars are not moving faster than the speed of light in an inertial reference frame.

The postulates that Einstein used in Special relativity were that the physic will take the same form (covariant) in any inertial reference frame and that the light will travel at the speed of light in any inertial reference frame. If these two postulates define what an inertial reference frame is, then a non inertial reference frame can be defined as a reference frame in which the physics do not take the same form or/and the light does not propagate at the speed of light in the vacuum. I want to find an non inertial reference frame of the second kind in which the speed of the light is zero. (I can define a frame of this kind using the light cone coordinates, but it is not what I am looking for.)

For me the spin and the mass of a particle label the representations of the Poincare group. In an inertial reference frame I can give a satisfactory answer of what mass and spin is. Really it is labeling how the the field will transform under Poincare transformations, but it does not said anything about how will the system will transform under other coordinate transformation.

Finally I believe that the degree of freedom of a particle are defined in a inertial reference frame. In that sense the physical degree of freedom will never change, but in a non inertial reference frame the interpretation will change and the constraints also because the physics will change.

If you do not agree with me it is ok. Probably I will find that I am delusional.

12. Dec 23, 2012

### pervect

Staff Emeritus
I'm not sue what you mean by this. Anyone with u= constant will be travelling at the speed of light, not just u=0 in the metric I gave.

And it's still not right to talk about "inertial observers travelling at light speed" no matter what coordinates you use...

13. Dec 23, 2012

### chwie

Prevect

I used the definition of light cone coordinates (null coordinates) and I found that the speed in this coordinates is

(1-β)/(1+β) such that β is the speed in the inertial reference frame

In the case that β=0 we have a speed of unity. In the case of the photon traveling to the right β=1 we have that the speed is zero and in the case of 0<β<1 the speed will be any value between zero and unity.

Maybe I am doing it wrong.

We can talk about an inertial observer moving at the speed of light relative to a non inertial reference frame. The reason is that there is no reason why we cannot do this. It is just a coordinate transformation. For example a non inertial reference frame which is rotating with a frequency ω relative to a inertial reference frame. In this non inertial reference frame we will have that the inertial observers can have any speed. The speed will be determine by the product ωR.

Obviously we can choose ωR≥c and it will not represent any problem because this reference frame is non inertial.

14. Dec 23, 2012

### pervect

Staff Emeritus
If we take x= βt, and u=x+t and v=x-t, then we have
(u+v) = β(u-v) or (1- β)u = -(1+ β) v

I would guess this is similar (except for sign) as to w hat you are doing because the result looks similar. However, I wouldn't call either du/dv or dv/du a "speed". For one thing, neither u nor v represents distance, or time. "Speed" is a rather ambiguous term in any event. It also has physical implications that don't apply here. So I'd just give the derivative (whichever one you aure using) a symbol, and not call it a "speed".

IT remains true that u=constant represents light moving in one direction, and v=constant represents light moving in the opposite direction in these coordinates. The value of the constant doesn't matter.

15. Dec 23, 2012

### chwie

pervect

The different of sing is because I used v=t-x.

16. Dec 24, 2012

### andrien

ε(p)∙p=0,regarding this relation you are saying that this does not hold.The problem is coulomb gauge which is non-covariant so it might have confused you.You can take lorentz gauge then which in momentum space reads
kμAμ=0,now this relation is covariant.you can write this in expanded form as k0A0-$k.A$=0,now as you can see in the first three pages of reference that A0 is not any degree of freedom.It can be expressed in terms of $∇.\frac{∂A} {∂t}$.It is not dynamical so it can be discarded.it is possible to ellimiate it in other frame also.after it you are left with again
$k.A$=0 so it again implies the transversality condition.However it is tough to see two degree of freedom in lorentz gauge.There are other methods also like bleuler gupta method which introduces indefinite metric(negative probalities) to resolve it.But the point is that even coulomb gauge is non covariant,it's results are covariant in nature.By the way,don't confuse between virtual photons and real photons.Virtual photons can have all 4 polarization states.I hope it will clear some confusions.

17. Dec 24, 2012

### chwie

I perfectly agree with everything that you say. I belief that we are thinking in two different problems.