# Reference frames for photon collisions

1. Sep 26, 2007

### hartin

Hi,

My question probably has a simple answer, but I've been scratching my
head over it a little too long so I thought I would ask it here. I
have three initial photons involved in a collision with 4-momenta k1,
k2 and k3. I have two reference frames:

frame 1: the centre of mass frame of photons k1 and k2 (so that 3-momenta k1+k2=0)

frame 2: the centre of mass frame of all photons k1, k2 and k3 (sothat 3-momenta k1+k2+k3=0)

I want to find out what the relativistic beta v/c and gamma 1/Sqrt(1-(v/c)^2 are. So there seems no way to directly find the relative velocity of the two frames is as only photons are involved. I thought
to use the expression for the relativistic doppler shift, but that doesn't seem appropriate.

Can anyone help me out here?

cheers.

2. Sep 26, 2007

### Staff: Mentor

Try using the Lorentz transformation for energy and momentum. That is, if an object (including a photon) has energy E and momentum p in one frame, then in another frame with relative velocity v (with respect to the first frame):

$$p^{\prime}c = \gamma (pc - \beta E)$$

$$E^{\prime} = \gamma (E - \beta pc)$$

3. Sep 26, 2007

### hartin

thanks, but $$\beta$$ and $$\delta$$ are what I'm trying to determine

4. Sep 26, 2007

### hartin

err I meant beta and gamma

5. Oct 2, 2007

### clem

Use: P=k3
E=k1+k2+k3
LT with \beta=p/E in the direction of k3.

6. Oct 2, 2007

### pervect

Staff Emeritus
Note that you have only one variable to solve for, $\beta$. By definition,

$$\gamma = \frac{1}{\sqrt{1-\beta^2}}$$

You can (and will have to) invert this to solve for beta as a function of gamma.

If I understand the problem correctly, you have 6 3-vectors as input? I.e. you have the components of k1 in frame1 and frame2, the components of k2 in frame1 and frame2, and the components of k3 in frame 1 and frame 2.

I suppose you could call these k1 and k1', k2 and k2', k3 and k3'?

First step: replace the 3-vectors with 4-vectors. The "missing" component, which is energy, can be found from |k1| = |k2| = |k3| = 0, i.e. E^2 - |p|^2 = 0.

k1 + k2 must be related by a Lorentz transform to (k1'+k2'), because k1' is the Lorentz transform of k1, and k2' is the Lorentz transform of k2.

In geometric units, we know that the components of (k1+k2) are (1,0,0,0). In standard units, this would be (c,0,0,0).

The zeroeth. component of (k1'+k2') will be gamma in geometric units, or c * gamma in standard units.

You apparently are not interested in solving for the direction of the Lorentz boost, just it's absolute value?

Last edited: Oct 2, 2007