# I would like to see the math of something

1. Nov 21, 2011

### JT73

If someone would be so kind as to show me the math of why a reference frame for a photon wouldn't work?

"Problem #1: In the frame of reference of some photon in that beam, what is the velocity of some other photon in the beam? We have a slight problem here with the second postulate of special relativity, which says that the local speed of light is the exact same value, c, in all reference frames.

Problem #2: Another aspect of special relativity is that one can transform from any one inertial frame to another using the Poincare transform. Try going to/from the photon frame of reference using this transformation. There will be a slight problem with dividing by zero / multiplying by infinity here."

This was quoted from the user D H in a thread I came across. I would like to see the math involed with is, along with any other equations that back up why photons have a frame of reference wouldn't make sense.

Thank you

Last edited: Nov 21, 2011
2. Nov 21, 2011

### Bacle2

You may have better luck posting this in the/a physics forum.

3. Nov 22, 2011

### HallsofIvy

I'm moving this thread to the "Special and General Relativity" forum.

4. Nov 22, 2011

### ZikZak

"Math?" Why does there have to be math for everything?

The speed of light is c in all inertial reference frames, by postulate. An inertial frame in which a photon is at rest violates this postulate. Done.

5. Nov 22, 2011

### JT73

Yes, I know that ZikZak, you're on a science forum, of course we are going to want see math done to back up reasoning...

Now again, I ask if someone would do me the favor of showing me the formula or equations in which 'C' is entered and then the equation comes out with a divison by zero or an answer of infinity or something of the like.

I wish to see this just to better my understanding of the math behind the reasoning. After all math is "the language of physics."

6. Nov 22, 2011

### Matterwave

Reasoning is reasoning and math is math. Math is important in physics, but if you become so dependent on the math that you won't accept physical reasoning, then I think you've gone too far.

But ok, the Lorentz boost in the x-direction is defined by:

$$t'=\gamma (t-\frac{vx}{c^2})$$
$$x'=\gamma (x-vt)$$
$$y'=y$$
$$z'=z$$

We have defined:
$$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

If you plug in v=c (boosting to the frame of the photon), you get gamma is a division by 0 which is a non-nonsensical answer.

7. Nov 22, 2011

### yoron

You know, you put a lot of trust into equations :)

We have countless experiments showing us that 'light' doesn't care about what 'speed' you think you are doing, relative some other frame of reference. It always travel at 'c' measured locally. That's the 'frame of reference' for a 'photon'. To deny this you will need to prove it otherwise, not theoretically but by experiment.

You also will need to explain all other experiments telling us that 'c' is a constant.

(Eh, not you matter wave:)

Last edited: Nov 22, 2011
8. Nov 22, 2011

### JT73

I didn't mean to come off the way I may have. I just always see people talking about light and referencing that a frame of reference for it doesn't work. I understand that reasoning, I do. But for me at least, when I see the math, it becomes even more clear becasue I can picture in my head doing the math instead of picturing in my head a photon traveling at C.

So thank you.

9. Nov 22, 2011

### atyy

The "frame" of a photon can be defined using "light cone" coordinates.

If you have a Lorentz inertial frame whose coordinates are (t,x,y,z), the light cone coordinates are (a,b,c,d), with

a=(t+x)/√2
b=(t-x)/√2
c=y
d=z.

That is not a Lorentz transformation (you can check by using Matterwave's equations in post #6). Since (a,b,c,d) results from a non-Lorentz transformation applied to the Lorentz inertial coordinates (t,x,y,z), the light cone coordinates do not form a Lorentz inertial frame.

10. Nov 22, 2011

### ghwellsjr

The 'c' that is measured locally is always a round-trip "average speed" for light which has nothing to do with any 'frame of reference', only that the measurement takes place under conditions of the apparatus being inertial, that is, not accelerating. This kind of measurement is covered by Einstein's first postulate, the principle of relativity.

But then Einstein has a second postulate which states that the unmeasurable and unknowable one-way speed of light is also 'c' which is a mathematical statement and the basis for his purely mathematical definition for a Frame of Reference which includes the concept of space-time and four-dimensional events. He also derives the mathematical Lorentz Transform to connect events from one Frame of Reference to another FoR moving at some speed, v, with respect to the first one.

So Einstein's mathematical definition of a FoR and the Lorentz Transform that won't allow for a FoR for anything traveling at v=c with respect to any other FoR which includes photons.

There cannot be any experimental proof for this since the foundation of Einstein's Theory of Special Relativity is purely mathematical. If you want to prove that a photon travels at 'c' or that light propagates at 'c', you're going to need a different theory based on experiments, not mathematics, and I don't think you're going to find one.

11. Nov 22, 2011

### yoron

Don't get you there ghwellsjr?

Light has only one 'speed' locally, in any experiment, as far as I know. The only way you ever will measure anything is locally. And it doesn't matter how you measure your 'motion' relative some other frame for this.

Are you telling me that this is wrong?

(ignoring it travelling through 'mediums' as glass and water etc for this)

12. Nov 22, 2011

### ghwellsjr

When you make a round-trip measurement of the speed of light, you have one timing device located at the source of the light and a mirror some measured distance away. All you know is total time it takes for the light to get from the source to the mirror and back to the source where the timer is located. You cannot know what time the light hit the mirror and therefore you cannot know in that experiment how fast the light was traveling in each direction. Einstein's solution is to make the two time intervals equal (mathematically). That's his second postulate. Now you can define time on a remote clock with respect to a local clock and from that you can build the mathematical concept of a Frame of Reference.

13. Nov 22, 2011

### JT73

What if the final answer in the eqution in post 6 came out to be an actual number with no divison by zero (meaning the object wasn't moving at C), what does that actually mean? Like if the asnwer comes out to .8, what is that telling you?

14. Nov 22, 2011

### robphy

The lorentz (boost) transformation must preserve the square-norms of 4-vectors.
So, a timelike-vector (associated with a typical inertial frame) [with square-norm 1] cannot be transformed to a lightlike-vector [with square-norm 0]... and vice versa.

15. Nov 22, 2011

### Fredrik

Staff Emeritus
You seem to be overlooking the fact that if there is such a thing as an inertial reference frame of the photon, then that photon would have to have both speed 0 and speed c in it. There is an immediate contradiction even if you don't consider any other photons, so the problem is anything but "slight". (The speed must be 0 because it's assumed to be the reference frame in which the photon is at rest, and the speed must be c because it's assumed to be an inertial frame. The contradiction means that if a photon has a reference frame, it's not an inertial frame). Of course, the same thing can be said about any other photon in the beam.

This is what matterwave was showing you. However, rather than concluding that there's a division by zero problem, you should note that Poincaré transformations don't apply to "the reference frame of the photon", because it can't be defined as an inertial frame. That last part is proved by what I said in the reply to problem #1.

It means that v=0.6c. The other equalities tell you that if I'm moving with speed 0.6c relative to you, I would be assigning coordinates (t',x',y',z') to the event you would be assigning coordinates (t,x,y,z). This is assuming that we both use the inertial coordinate systems that are associated with our motions in a standard way. I would say that the reason why "the rest frame of the photon" doesn't make sense (even if we would allow it to be a non-inertial coordinate system) is that this standard way of associating coordinate systems with non-accelerating objects doesn't work for photons. Some of the details are explained in this post.

Last edited: Nov 22, 2011
16. Nov 22, 2011

### Fredrik

Staff Emeritus
You can pick any coordinate system with a time axis that coincides with the photon's world line and call it "the frame of the photon" if you want to. But I don't know a reason why the light cone coordinates should be preferred over any of the others.

17. Nov 22, 2011

### atyy

Yes. The aim was just to pick a reasonable definition (this one that is actually useful) and show that it is not a Lorentz inertial frame.

18. Nov 22, 2011

### JT73

Good information in this thread, thanks. Though I would like to point out, Fredrik, that part of my OP was quoted from the user D H in another thread. It isn't my words.

19. Nov 22, 2011

### JT73

Wait, are you saying in the bolded above that if we were looking for a reference frame of a photon, then we don't neccesarily need to have a speed of 0?

20. Nov 22, 2011

### Fredrik

Staff Emeritus
No, the "reference frame of" something is always a frame or a coordinate system such that the "something" has speed 0. What I'm saying is that such a frame must be non-inertial, because in any inertial frame, the speed of light is c≠0

(I also said that there's no non-inertial frame with properties that singles it out as "the" reference frame of the photon).