# Question on Relativity.

1. Sep 16, 2007

### zolderick

Hello. I have little knowledge on relativity but I have read the fundamental concepts. What has been confusing me was the following problem:
If a photon A is traveling towards another photon B which is traveling directly into photon A (a head on collision). What will be the speed of photon B relative to photon A?
Since the relative speed cannot be two times the speed of light as nothing can surpass the speed of light, could someone please explain this to me?

Last edited: Sep 16, 2007
2. Sep 16, 2007

### HallsofIvy

The speed of a photon relative to ANY coordinate system is c- the speed of light. That is THE fundamental concept of relativity.

Do you know the formula for combining speeds in relativity? It is NOT u+ v! it is
$$\frac{u+ v}{1+ \frac{uv}{c^2}}[/itex] In particular, if u= v= c, we have [tex]\frac{c+ c}{1+ \frac{c^2}{c^2}}= \frac{2c}{2}= c$$.

3. Sep 16, 2007

### zolderick

Yes I've tried that. What if an observer watches the two photons travel at each other. Will that observer see the two photons traveling towards each other at 2c?

4. Sep 16, 2007

### Staff: Mentor

Yes, that observer will see them come together, covering separation distance (with respect to this observer) at a rate of 2c.

5. Sep 16, 2007

### zolderick

Ah ok thanks for the quick response.

6. Sep 16, 2007

### matteo16

2c? how is it possible? i don't understend this passage

7. Sep 16, 2007

### Staff: Mentor

Don't confuse what can be called the "3rd party separation rate" with relative speed.

Imagine this situation: Spaceship A is 1 light year to the west of me and travels at a speed of 0.9999c towards me; Spaceship B is 1 light year to the east of me and travels at a speed of 0.9999c towards me. What's their distance apart according to me? 2 light-years. When do they reach me? In about 1 year. So the separation distance decreases at a rate of about 2c according to me.

Of course the relative speed of spaceship B as measured by spaceship A is still less than c. (Per the relativistic addition of velocity equation given by Halls' in post #2.)

8. Sep 16, 2007

### jostpuur

I've always found it annoying when some exercises simply ask what is the relative speed of two objects, and don't specify in which frame the result is wanted. The relative speed is by definition $v_1-v_2$ (or with minus sign), and naturally depends on the chosen frame.

Somebody might say that the relative speed should always be calculated in a frame, in which other one of the objects is in rest. It is true, that such frame is in special position, and is the most natural answer to the question, but what do you do with the relative speed of two photons? There is no frame in which other one is in rest, and thus there is no frame where their relative velocity would be c.

9. Sep 16, 2007

### Staff: Mentor

I agree! But it's usually taken for granted that the relative speed of A and B means the speed of A as measured in the frame of B. (But that should be stated clearly.)
That's why I changed the example from photons to spaceships.

10. Sep 16, 2007

### matteo16

so B would see A coming in 1 ligth year instead of 2 ligth year?(sorry for the grammar but i'm an italian 16 boy and i'm self-taught and only this year i start physics at the hight school).

i didn't understend wath's 3rd party separation rate
can you explane it to me with an exemple(so i can traslate it in italian in my mind lol)

11. Sep 16, 2007

### Staff: Mentor

B sees A coming towards him at a speed of about c (a bit less, of course). But B disagrees that A is 2 lightyears away from him at the moment he is 1 lightyear away from me (according to my earth frame): (1) B sees the distance measured by me to be contracted and, (2) B disagrees that A is 1 lightyear away from me at that instant, due to the relativity of simultaneity: B says that A is much closer.

I gave an example in post #7. What I call "3rd party separation rate" is just what jostpuur calls the relative speed of A and B as measured in the earth frame (in this example).

Last edited: Sep 16, 2007
12. Sep 16, 2007

### matteo16

ah ok understood

13. Sep 16, 2007

### robphy

Technically speaking, the "relative speeds" are equal...
they are the magnitudes of the relative velocities of A wrt B and of B wrt A.

What requires distinction is the relative-velocity vectors.
In common usage, they are opposite in direction [as 3-vectors in their respective spatial frames of reference].

This last point is rarely appreciated to mean that:
as spacelike-vectors [orthogonal to their respective [unit] timelike 4-velocity vectors], they are not collinear [i.e. neither parallel or antiparallel] in spacetime.

Last edited: Sep 16, 2007
14. Sep 16, 2007

### robphy

I think your "3rd party separation rate" is a better term than
"relative speed of A and B as measured in the earth frame".

"Relative speed", to me, should be the magnitude of a "relative velocity" vector [i.e., the vector-component of A's 4-velocity that is orthogonal to B's 4-velocity, divided by the temporal component... i.e. essentially, the slope.]

Geometrically, the "relative speed" of any two future-pointing causal worldlines is best described by $|c\tanh\theta|$ (c times the magnitude of the hyperbolic-tangent of the Minkowski-angle intercepted by their worldlines). [Picture the hyperbola with asymptotes along the light cone.] The angle with any lightlike worldine is infinite [even with another distinct lightlike worldline], which corresponds to a relative speed of c.

This is an example where spacetime geometry can really provide definitions and geoemtrical interpretations to clarify physical concepts.

Last edited: Sep 16, 2007
15. Sep 16, 2007

### Staff: Mentor

I agree. That's the term I always use. (I think I made it up years ago--I wonder if there's an "official" term for it?)
Again I agree. But using spacetime geometry requires a bit more sophistication.