GZK Limit with Sympy

Cosmic Microwave Background Radiation (CMBR) is comprised of photons with energy \(\sim E_\gamma\). Protons moving through the CMBR undergo inelastic scattering with these photons, resulting in the creation of a muon (denoted \(\pi^0\)).

\[\gamma + p \rightarrow \pi^0 + p\]

The particles produced have masses \(m_p\) and \(m_\pi\). This scattering can only occur for protons above a certain energy \(E_\min\). What is \(E_\min\) in terms of the variables given?

Solution:

In my view, the simplest way to solve this problem is with conservation of four-momentum In classical mechanics, what one might call “three-momentum” is conserved. This is your standard \(\frac{d}{dt} \sum_{i} m_i(t) v_i(t) = 0\). But this doesn’t generalize in the relativistic limit. How can photon propulsion work if photons have 0 mass and thus 0 classical momenta? The answer is that three-momentum is a classical limit of a generalized four-momentum \(\vec{P} = (E, p)\) which combines energy and momentum, and is conserved even in relativistic situations (like a very fast proton colliding with a CMBR photon).

We can write the four-momentum of our minimum energy proton as

\[\vec{P}_{\text{proton}} = (E_{\min}, \sqrt{m_p^2 - E_{\min}^2} \ \hat{n})\]

where \(\hat{n}\) is a vector pointing in the direction of the proton’s motion.

Similarly, our CMBR photon four-momentum is

\[\vec{P}_{\gamma} = (E_{\gamma}, E_{\gamma} \ \hat{m})\]

For now, I’ll refrain from explicitly writing down the four-momentum for \(P_\pi\) and the proton after the collision. So we have

\[\vec{P}_{\text{proton}} + \vec{P}_\gamma = \vec{P}_\pi + \vec{P}_{\text{proton}_\text{final}}\]

Dotting both sides with themselvesFour-momentum vectors behave like normal vectors, except they have a metric \(\eta_{\mu \nu}\) that has to be sandwiched in there when taking the dot product. So, \(\vec{P}_\gamma \cdot \vec{P}_\gamma = P_\mu \eta_{\mu \nu} P^{\nu} = P_1^2 + P_2^2 + P_3^2 - P_0^2\). It’s also worth noting that any four-momentum vector dotted with itself is just equal to the mass of the particle. This comes from the Energy-Momentum relation in relativity.

, we get

\[m_{\text{proton}}^2 + 0 + 2 \vec{P}_{\text{proton}} \vec{P}_\gamma = (\vec{P}_\pi + \vec{P}_{\text{pf}})^2\]

We are interested in the minimum energy necessary for this particle conversion to happen. This means that we can make a few relaxing assumptions.

First, at the threshold after the collision, in the center-of-mass frame there shouldn’t be any unnecessary velocity (ie. the products should be at rest). This means that

\[(\vec{P}_\pi + \vec{P}_{\text{pf}})^2 = (m_{\pi} + m_{\text{proton}})^2\]

Second, at the minimum energy collision necessary for this to happen, we want all the energy going into the conversion. No “glancing blows.” This means that \(\hat{n} = -\hat{m}\) - the proton and photon will have a head on collision. Otherwise, you could always have less energy to produce the conversion if there had been a head-on collision.

import sympy