Wham!

The reaction is: $p+\gamma=\pi^++n$

First, we need to find when the energy of the proton is minimum.

After collision, the total energy is minimum if the kinetic energy of $pi^++n$ in the center of mass frame is minimum. (Because the rest energy of $pi^++n$ is constant)

The minimum of the kinetic energy of $pi^++n$ in the center of mass frame is 0 when the speed and direction of $pi^++n$ are the same. In other word, $pi^++n$ move like one object with the rest energy: $E_{0n\pi}=139.6+939.6=1079.2 MeV=1.0792*10^9 eV$ .

From the given condition, we have: $E_{\gamma}=kT=2.59*10^{-4} eV$ .

We have: $(E_p+E_{\gamma})^2=E_{n\pi}^2=E_{0n\pi}^2+c^2p_{n\pi}^2$

$=E_{0n\pi}^2+c^2(p_p+p_{\gamma})^2$ .

So, $E_p^2+E_{\gamma}^2+2E_pE_{\gamma}=E_{0n\pi}^2+c^2p_p^2+c^2p_{\gamma}^2+2c^2p_pp_{\gamma}$ .

So, $E_{0p}^2+2E_pE_{\gamma}=E_{0n\pi}^2+2cp_pE_{\gamma}$ .

So, $E_p-cp_p=\frac{E_{0n\pi}^2-E_{0p}^2}{2E_{\gamma}}$ (1).

We have: $E_{0p}^2=E_p^2-c^2p_p^2=(E_p-cp_p)(E_p+cp_p)$ (2).

From (1) and (2), we have: $E_p+cp_p=\frac{2E_{\gamma}E_{0p}^2}{E_{0n\pi}^2-E_{0p}^2}$ (3).

From (1) and (3), we have: $E_p=\frac{E_{0n\pi}^2-E_{0p}^2}{4E_{\gamma}}+\frac{E_{\gamma}E_{0p}^2}{E_{0n\pi}^2-E_{0p}^2}=2.73*10^{20} eV$ .

Energy and momentum are conserved in particle collisions but their conservation is united in the theory of special relativity. Whereas in classical mechanics, the conservation of kinetic/potential energy and the conservation of momentum are treated as separate constraints, relativistic mechanics allows for the exchange of mass for energy as well as the reverse. Because of this, energy that is kinetic before the collision can become mass afterward and likewise. The conservation relation must therefore reflect this interconversion.

This consideration is encapsulated in the idea of the four momentum which is a four entry vector consisting of the total energy and the three components of spatial momentum. For instance, one particle of rest mass $m_0$ traveling along the $x$ axis, with velocity $v$ close to $c$ , would have the following four momentum:

$\begin{aligned} P_4 &= \left(E, p_x, p_y = 0, p_z = 0\right) \\ &= \left(mc^2, m\vec{v}\right) \\ &= \left(mc^2, m_0\gamma v, 0 , 0\right) \\ &= \left(mc^2, p_x, 0, 0 \right) \end{aligned}$

The magnitude of the four vector is given by the square of the energy entry minus the squares of the momentum entries (multiplied by $c^2$ ). I.e. for the four vector above, we have $P_4^2 = m^2c^4 - p_x^2c^2$ .

Applying the famous defining relation for mass-energy equivalence

$E = mc^2 = \sqrt{m_0^2c^4 + p^2c^2}$ ) we see that $P_4^2 = m_0c^2$ .

$P_4^2$ is a number equal to the rest mass of the particle(s) and therefore must be invariant to changes in reference frame (the rest mass is the same in every frame). In addition, the four momentum is itself conserved within the same reference frame, reflecting the conservation of total energy and the conservation of momentum in each dimension. Without knowing much about the subject, we see that the four momentum is a conserved object.

For any four momentum vector $P_4$ , its magnitude $P_4^2$ is equal to $E^2 - \vec{p}\cdot\vec{p}$ and for any two four momentum vectors $P_4^a, P_4^b$ , their dot product is $2P_4^a\cdot P_4^b + P_4^a \cdot P_4^a + P_4^b \cdot P_4^b$ .

So… assuming that we have energy-momentum conservation for the proton-photon collision, we can simply write the four momentum vectors before and after the collision and take their magnitudes. Furthermore, since the magnitude is independent of the reference frame, we can compute the before and after conditions in whatever reference frame is easiest to think about.

We wish to find the minimum energy required for a p to produce a $\pi^+$ and a n after colliding with a CMB $\gamma$ . In the least energy picture, the p and $\gamma$ would meet in a collision and produce a stationary n and stationary $\pi^+$ , i.e. the final four vector only contains rest mass energy terms and the final total energy consists entirely of rest mass. As the classical momentum is zero in the final state, it must be zero in the initial state. This means that we should write down the initial four momentum in the frame in which the momentum is zero, i.e. the center of momentum frame. In this frame, the four momentum of the $\gamma$ and p is given by

$\displaystyle P_4^{\gamma} + P_4^{p} = \left( E_{\gamma}+E_{p}, \vec{p}_{\gamma} + \vec{p}_p, 0, 0 \right)$

The invariant mass of the initial four momentum is then

$\left(\displaystyle P_4^{\gamma} + P_4^{p}\right)^2 = E_{\gamma}^2 + E_{p}^2 + 2 E_{\gamma} E_{p} - p_{\gamma}^2 - p_{p}^2 - 2 \vec{p}_{\gamma} \cdot \vec{p}_p$

The total energy of the photon is entirely due to the momentum so $E_{\gamma}^2 - p_{\gamma}^2 = 0$ . For the high energy proton, the energy is due almost entirely to momentum and we assume $E_p \approx \|\vec{p}_p\|c$ . The $\gamma$ and the p are colliding head-on and so their momenta are aligned in anti-parallel fashion making $2 \vec{p}_{\gamma} \cdot \vec{p}_p = -2E_pE_{\gamma}$ .

In total, the invariant mass of the first frame is given by $4E_pE_{\gamma} + E_{p}^2 - p_{p}^2 = 4E_pE_{\gamma} + \left(m_p^0\right)^2$ .

After the collision, the four momentum is very simple to write down, $P_4 = P_4^{\pi^+} + P_4^n = \left(E_{\pi^+} + E_n, 0,0,0\right)$ and its invariant mass is given by $\left(m_{\pi^+}^0 + m_n^0\right)^2$ .

Equating the invariant mass before and after the collision and solving for the energy of the proton, we find

$\boxed{\displaystyle E_p = \frac{\left(m_{\pi^+}^0 + m_n^0\right)^2 - \left(m_p^0\right)^2}{4E_{\gamma}}}$ .

In the units of our calculation (where we can write the four vector as $\left(E, \vec{p}\right)$ , the rest masses of the particles are given by

$\begin{aligned} m_{\pi^+}^0 &\approx 139.6 \mbox{ MeV/c}^2 \\ m_{p}^0 &\approx 938.3 \mbox{ MeV/c}^2 \\ m_{n}^0 &\approx 939.6 \mbox{ MeV/c}^2 \\ \end{aligned}$

The energy of the CMB photon is equal to $k_BT \approx 2.59 \times 10^{-10} \mbox{ MeV/c}^2$ .

Plugging these values in (all in units of $\mbox{MeV/c}^2$ ) we find $E_p \approx 2.74 \times 10^{20} \mbox{ eV}$ , the minimum energy for the proton and photon to transfer their energy entirely into the rest mass of a charged pion and a neutron.