USAPhO 2017 Problem A4

Past Contests USAPhO 2017
1

Relativistic particles obey the mass energy relation

E^{2}=(p c)^{2}+\left(m c^{2}\right)^{2}

where E is the relativistic energy of the particle, p is the relativistic momentum, m is the mass, and c is the speed of light.

A proton with mass m_{p} and energy E_{p} collides head on with a photon which is massless and has energy E_{b}. The two combine and form a new particle with mass m_{\Delta} called \Delta, or “delta”. It is a one dimensional collision that conserves both relativistic energy and relativistic momentum.

a. Determine E_{p} in terms of m_{p}, m_{\Delta}, and E_{b}. You may assume that E_{b} is small.

b. In this case, the photon energy E_{b} is that of the cosmic background radiation, which is an EM wave with wavelength 1.06 \mathrm{~mm}. Determine the energy of the photons, writing your answer in electron volts.

c. Assuming this value for E_{b}, what is the energy of the proton, in electron volts, that will allow the above reaction? This sets an upper limit on the energy of cosmic rays. The mass of the proton is given by m_{p} c^{2}=938 \mathrm{MeV} and the mass of the \Delta is given by m_{\Delta} c^{2}=1232 \mathrm{MeV}.

The following relationships may be useful in solving this problem:

\begin{array}{ll} \text { velocity parameter } & \beta=\frac{v}{c} \\ \text { Lorentz factor } & \gamma=\frac{1}{\sqrt{1-\beta^{2}}} \\ \text { relativistic momentum } & p=\gamma \beta m c \\ \text { relativistic energy } & E=\gamma m c^{2} \\ \text { relativistic doppler shift } & \frac{f}{f_{0}}=\sqrt{\frac{1-\beta}{1+\beta}} \end{array}