Reference frames for photon collisions

[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.

 

[jtbell]

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)

 

[hartin]

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

 

[hartin]

err I meant beta and gamma

 

[Meir Achuz]

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.

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

 

[pervect]

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.

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?