Given a collision between a high energy cosmic ray proton (~$10^{20}$ eV) and a CMB photon, I want to calculate the final energy of the photon.

I know to proceed, I use the following conservation of four-momentum:

$$P_{p}+P_{\gamma} = P_{p}'+P_{\gamma}'$$

where the primes indicate the final momenta.

I know to subtract off to obtain the proton momentum after the collision, giving:

$$P_{p}' = P_{p} + P_{\gamma}-P_{\gamma}'$$ then to square each of these quantities.

I know $$P_{p}'^2 = -m_p^2c^2$$

but I'm not sure how to handle all of the terms / cross terms on the right.

I know the final answer for the photon energy should be:

$$E'_{\gamma} = [E_{\gamma}(E_p+pc)]/(2E_{\gamma}+E_p-pc)$$

but I'm not sure how to arrive at this point from where I'm at.


Directly squaring the equation you provided

$$(P_{p}+P_{\gamma})^{2} = (P_{p}'+P_{\gamma}')^{2}$$

simplifies it to

$$P_{p}P_{\gamma} = P_{p}'P_{\gamma}'$$

because $P_{p}^{2} = P_{p}'^{2} = m_{p}^{2}$ and $P_{\gamma}^{2} = P_{\gamma}'^{2} = 0$

Now, lets assume photon is traveling in -ve z-axis direction and proton in +ve z-axis direction. Writing the four momentum we have

$P_{p} = (E_{p}, pc)$ and $P_{\gamma} = (E_{\gamma}, -E_{\gamma})$

Noting that $a^{\mu}b_{\mu}=a^{0}b^{0}-a.b$

We simplify the second equation to

$$E'_{\gamma} = \frac{[E_{\gamma}(E_{p}+pc)]}{E'_{p}+P_{p}'}$$

We can find $E'_{p}+P_{p}'$ from applying energy and momentum conservation individually

$$E_{p}+E_{\gamma} = E_{p}'+E_{\gamma}'$$


$$pc-E_{\gamma} = p'c-E_{\gamma}'$$

Combining these two equation we get $E'_{p}+P_{p}'$

Putting everything together we get

$$E'_{\gamma} = [E_{\gamma}(E_p+pc)]/(2E_{\gamma}+E_p-pc)$$


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