Special relativity with particles

[tskuzzy]

Homework Statement

Consider the event $p + \gamma \to p + \pi$, where $p$ is a cosmic ray of a proton, $\gamma$ is a microwave background photon, and $\pi$ is a generated meson. What is the minimum energy of the proton for such an event to happen? Proton has a rest mass of 1 GeV/c2, π particle has a rest mass of 100 MeV/c2, and microwave background photon has an energy of 2.5*10^-4 eV.

Homework Equations

Conservation of energy and momentum equations.
$$ E = mc^2 $$

The Attempt at a Solution

From conservation of energy, we get the equation:
$$
\begin{align*}
E_i &= E_f \\
E_{p_i} + E_{\gamma} &= E_{p_f} + E_{\pi} \\
E_{p_i} + 2.5 \times 10^{-4} &= \gamma_{p_f}(1 \times 10^6) + \gamma_{\pi}(1 \times 10^5)
\end{align*}
$$

In order to minimize the energy of the initial proton, is it reasonable to simply set the gammas on the right side equal to 1 (i.e. let them be at rest)?

 

[vela]

Nope, that would violate the conservation of momentum.

Try solving the problem in the center-of-mass frame. In that frame, the resulting proton and pion will be at rest. Then transform the results back to the lab frame.

 

[tskuzzy]

Nope, that would violate the conservation of momentum.

Try solving the problem in the center-of-mass frame. In that frame, the resulting proton and pion will be at rest. Then transform the results back to the lab frame.

Thanks for the speedy response!

Could you explain why the resulting proton and pion would be at rest in the COM frame? Wouldn't that imply that the two are traveling together with the same velocity? Why couldn't they fly off in different directions wrt the COM?

 

[vela]

Because you're looking for the minimum energy. Some of the energy goes into creating the pion. Any extra ends up as the kinetic energy of the resulting particles, so to find the minimum, you want the kinetic energy to be as small as possible.

 

[asteriatic]

If so, then we can solve for the minimum energy of the proton using the equation:
$$
E_{p_i} = \gamma_{p_f}(1 \times 10^6) + \gamma_{\pi}(1 \times 10^5) - 2.5 \times 10^{-4}
$$
where $\gamma_{p_f}$ and $\gamma_{\pi}$ are the Lorentz factors for the final proton and pion, respectively. This would give us a minimum energy of approximately 1.001 GeV for the initial proton in order for the event to occur.

However, if we want to consider the actual physical processes involved in the event, we would need to take into account the momentum of the particles as well. This would require using the conservation of momentum equation:
$$
\vec{p_i} = \vec{p_f} + \vec{p_{\gamma}} + \vec{p_{\pi}}
$$
where $\vec{p_i}$ is the initial momentum of the proton, and $\vec{p_f}$, $\vec{p_{\gamma}}$, and $\vec{p_{\pi}}$ are the final momenta of the proton, photon, and pion, respectively. This would give us a more accurate minimum energy for the initial proton, taking into account the momentum of the particles and their interactions.

In conclusion, the minimum energy of the proton for the given event to occur would depend on whether we are considering only energy conservation or both energy and momentum conservation. In either case, the rest masses of the particles involved and the energy of the microwave background photon would play a significant role in determining the minimum energy of the initial proton.