# Center of momentum frame for two dimensions

I have been investigating the scenario of two photons traveling in space (in a lab frame of reference) of different frequencies are about to collide with an angle $$\theta$$ between them. I set their 4-momentum vectors to be $$(hν_1/c , hν_1/c \cos(\theta) ,hν_1/c \sin(\theta), 0)$$ for a gamma photon and $$(hν_2/c , hν_2/c , 0 , 0)$$ for a CMB photon. I am having a hard time figuring out what the energy would be in a zero momentum frame due to the angle. The pair production along with a Lorentz transformation lead me to believe that $$hν_1/c \sin(\theta)$$ would always equal zero which is hard to believe as I can adjust the angle in other frames.

• Do photons interact like this? I’ve never heard of a photon-photon “collision”. Commented Feb 7, 2021 at 17:01
• It is used to create matter, making positron and electron similar to the Breit-Wheeler process. Commented Feb 7, 2021 at 17:09
• Thanks. Didn’t know that was possible. Commented Feb 7, 2021 at 17:14
• As far as i know this is not directly possible, only through some higher-order effects. But I guess this is valid as an approximation or an exercise. Commented Feb 7, 2021 at 18:36
• I am doing it as an exercise. I am thinking of that, but there has to be a way to transition a system of photons to a center of momentum frame. It is the reverse of pair antihalation. Commented Feb 7, 2021 at 21:09

Set c=1, and, so $$h\nu=E=p$$, much less cluttered. Assuming you are serious about the signs you have chosen, the total energy, momentum four vector in the lab frame is $$(p_1+p_2, p_2+ p_1 \cos\theta, p_1\sin\theta, 0).$$ In the center of momentum frame, $$(E,0,0,0)$$. But $$s=E^2-\vec{p}^2$$ is a relativistic invariant, and so the same in both frames, so $$E^2= 2p_1p_2 (1-\cos\theta),$$ with a minimum at 0, and a maximum at 4$$p_1p_2$$. This is the smallest energy squared at which a pair of particles of mass m will be produced at rest, where $$m^2=p_1p_2$$.