Question

I have been given a lengthy question regarding the angle that two equal energy photons will emit from an axis if an antiparticle moving with speed v collides with a stationary (to the lab) corresponding particle. It first asks to find the energy of each photon which I found as: 1/2m₀c²(γ+1)

. I added the two initial energies of the particles and set one γ=1
then took half because there are two emitted photons.

The question then asks to prove that the angel θ
that the photons emit at is given by
cos(θ)=√(γ−1/γ+1)

I tried using my energy for the photons in the relativistic energy equation to find the velocity from the momentum given that v=cos(θ)c which gave me a strange γ polynomial. I then tried to check my photon energy by reversing the given equation which gave me m₀c²√(2γ/γ+1).

My working for these are very long so I can't include them in the question. I would like to know if this approach to showing θ
is correct and if someone could provide a framework for the algebra that would be amazing, Thank you.

Answer 1

To prove the equation for the angle θ that two equal energy photons will emit, we can use the relationship between photon momentum and energy and the relativistic total energy equation. By simplifying the equation and considering the total energy before and after the collision, we can derive the equation for cos(θ).

Step-by-step explanation:

In order to prove the equation cos(θ) = √(γ-1/γ+1) for the angle θ that two equal energy photons will emit, we need to use the relationship between photon momentum p and photon energy E, which is consistent with the relativistic total energy equation E² = (pc)² + (mc)². Since the rest mass of a photon is zero, we can simplify this equation to E² = (pc)². Using this equation and the fact that the total energy of the particles before the collision is equal to the sum of the energies of the photons after the collision, we can derive the equation for the angle θ as cos(θ) = √(γ-1/γ+1).