I have been trying to solve this problem, and I get a ridiculous answer, but I don't know why. Let's suppose there is a high energy proton in Earth's atmosphere that collides with a cosmic microwave background photon. Let's suppose that this collision results in a pion $\pi^0$ being created, and another proton. I am trying to compute the minimum energy the initial proton should have for this process to be possible. I have worked in natural units, so that $c^2 = 1$, and using the metric described in the Wikipedia article for four-momenta, which I believe is the most common one. This is what I've done:
I have written the four-momenta of all particles involved. The original proton has $p_1 = (E_1,\vec{p_1})$, while the photon has $p_\gamma = (|\vec{p}_\gamma|,\vec{p}_\gamma)$. I imagine the excess energy of the proton is used to generate the pion, and so the products should have zero linear momentum. Therefore, I write for the second proton $p_2 = (E_2,\vec{p}_2) = (m_p,\vec{0})$ and for the pion $p_\pi = (m_\pi,\vec{0})$.
I have decided to use Mandelstam $t$ variable, so that: $$t\equiv(p_1 - p_2)^2 = (p_\gamma - p_\pi)^2$$ Performing the four-momenta products I get: $$m_p^2 + m_p^2 - 2(E_1 m_p - \vec{p1}\cdot\vec{0}) = m_\pi^2 - 2(E_\gamma m_\pi - \vec{p_\gamma}\cdot\vec{0})$$ Which simplifies to: $$2m_p^2 - 2E_1 m_p = m_\pi^2 - 2E_\gamma m_\pi$$ And after solving for $E_1$ (the initial proton's energy) I get: $$E_1 = \frac{2E_\gamma m_\pi + 2m_p^2 - m_\pi^2}{2m_p}$$ Now, a cosmic microwave background photon ($T \approx 2.7$ K) has an energy in the order of $\frac{3}{2}k_B T = 3\times 10^{-4}$ eV, while $m_p \approx 938$ MeV and $m_\pi \approx 135$ MeV, so the photon term is not going influence the outcome much. The thing is, after plugging in the values, I get $E_1 \approx 928$ MeV. But this is ridiculous, since that's not even the proton's mass at rest. So I obviously went wrong somewhere. I checked how this problem was solved in class, and what was done differently is assuming that the products have a total momentum of zero globally, in the center of mass reference frame. I realise this is a weaker requirement that what I did, which was to assume that everything was at rest instead of the net momentum being zero, which still leaves room for particles to move. Nevertheless, my classmates seem to have assumed each particle (the resulting pion and the proton, separately) to have $\vec{p} = \vec{0}$ in their calculations, which baffles me because that's what I did.
Specifically, what they did is write the following four-momentum for the products: $$p_\text{products} = (E_2 + E_\pi, \vec{0})$$ And then, when squaring it, they get $p_\text{products}^2 = (m_p + m_\pi)^2$, which does not make sense to me unless each particle has zero momentum.
Where did I make a mistake? Thanks everyone in advance.
