# Reference Frame of a Photon

1. Jan 22, 2010

### jacksonwalter

So if I'm a fast object moving in a sphere, in my frame of reference the sphere begins to contract along the direction I'm moving. As I approach the speed of light, the sphere begins to deform into a disk. If I'm a photon, the sphere is a disk. Rather than requiring requiring 4 coordinates to describe my state, I only need 3. What does this have to do with the holographic principle?

2. Jan 22, 2010

### atyy

Nothing. The reference frame of a photon still has 4 coordinates, but it is not an inertial reference frame.

3. Jan 22, 2010

### Staff: Mentor

...then you have left the domain of relativity theory as we know it.

4. Jan 23, 2010

### ZapperZ

Staff Emeritus
This is false.

I work at a particle accelerator and we routinely accelerates electrons close to c. According to those electrons, *I* am moving very close to c. In fact, these particles also see you as moving close to c. How come everything around you are not deforming into a disk as we speak?

This is another frequent misunderstanding about Special Relativity. The so-called effect of "time dilation" and "length contraction" aren't observed or measured in the moving frame, but in another inertial frame. So in my example above, it is the electrons that are seeing our world as deforming into "disks". You and I see no change. And if those electrons carry "spheres", we'd see them deformed as well.

Zz.

5. Jan 23, 2010

### jacksonwalter

I never said that length contracted in the moving frame of reference... I understand that length contracts in another inertial frame. I was trying to think about how the universe would look I would look if I was travelling at c, which I guess is not addressed in SR as infinities and zeros start to appear everywhere, and thus regarded as a meaningless question. Moving close to the speed of light towards the earth I got that I wouldn't be deformed, but rather the earth. Extrapolating to the speed of light (the reference frame of a photon), the earth would have 0 length along the direction I'm travelling, so would appear as a 2 dimensional disc instead of a deformed sphere.

6. Jan 23, 2010

### bcrowell

Staff Emeritus
The following is cut and pasted from a FAQ I maintain at http://www.lightandmatter.com/cgi-bin/meki?physics/faq .

What does the world look like in a frame of reference moving at the speed of light?

This question has a long and honorable history. As a young student, Einstein tried to imagine what an electromagnetic wave would look like from the point of view of a motorcyclist riding alongside it. But we now know, thanks to Einstein himself, that it really doesn't make sense to talk about such observers.

The most straightforward argument is based on the positivist idea that concepts only mean something if you can define how to measure them operationally. If we accept this philosophical stance (which is by no means compatible with every concept we ever discuss in physics), then we need to be able to physically realize this frame in terms of an observer and measuring devices. But we can't. It would take an infinite amount of energy to accelerate Einstein and his motorcycle to the speed of light.

Since arguments from positivism can often kill off perfectly interesting and reasonable concepts, we might ask whether there are other reasons not to allow such frames. There are. One of the most basic geometrical ideas is intersection. In relativity, we expect that even if different observers disagree about many things, they agree about intersections of world-lines. Either the particles collided or they didn't. The arrow either hit the bull's-eye or it didn't. So although general relativity is far more permissive than Newtonian mechanics about changes of coordinates, there is a restriction that they should be smooth, one-to-one functions. If there was something like a Lorentz transformation for v=c, it wouldn't be one-to-one, so it wouldn't be mathematically compatible with the structure of relativity.

7. Jan 23, 2010

### atyy

Take a look at Figure 1 on p19 pf http://arxiv.org/abs/hep-ph/9705477 .

The coordinates of the "instant form" are inertial coordinates, which are the coordinates of any inertial observer moving at a constant velocity less than the speed of light - as typical of inertial coordinates, the metric matrix is a bunch of "ones" on the diagonal. The "front form" can be considered the reference frame of a photon - there are still 4 coordinates, but the reference frame is not inertial - the metric matrix is not diagonal.

8. Jan 23, 2010

### bcrowell

Staff Emeritus
Those coordinates don't describe a frame of reference, in the sense that there's no way to get from those coordinates to the coordinates of an inertial observer via a Lorentz transformation.

9. Jan 23, 2010

### jacksonwalter

Thanks brcowell, that was what I was looking for. Atyy, that's a little over my head.

One other question: If you're moving close enough the speed of light, could you observe measuring sticks in another reference frame smaller than the Planck length?

$$\frac{l_{0}}{\gamma} < L_{P}$$
$$1-\frac{v^{2}}{c^{2}} < L_{P}^{2}$$
$$v > c\sqrt{1-L_{P}^{2}}$$

Last edited: Jan 23, 2010
10. Jan 23, 2010

### Naty1

11. Jan 24, 2010

### GRDixon

12. Jan 25, 2010

### Demystifier

I would say it is an inertial reference frame, but not a Lorentz one.

13. Jan 25, 2010

### atyy

Interesting! Why?

14. Jan 25, 2010

### Demystifier

Physically, I would call it inertial because a free photon does not need a force to move with a constant velocity. More formally, a transformation from a laboratory Lorentz frame to a proper frame of the photon is a LINEAR transformation. See e.g. Eq. (12) in
http://xxx.lanl.gov/abs/quant-ph/0602024 [Int.J.Mod.Phys.A22:6243-6251,2007]
for beta=1. These are nothing but well-known light-cone coordinates.

Last edited: Jan 25, 2010