- #1

spaghetti3451

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## Homework Statement

In 1966 Greisen, Zatsepin and Kuzmin argued that we should not see cosmic rays (high-energy protons hitting the atmosphere from outer space) above a certain energy, due to interactions of these rays with the cosmic microwave background.

(a) The universe is a blackbody at ##2.73\ \text{K}##. What is the average energy of the photons in outer space (in electronvolts)?

(b) How much energy would a proton ##(p^{+})## need to collide with a photon ##(\gamma)## in outer space to convert it to a ##135\ \text{MeV}## pion ##(\pi^{0})##? That is, what is the energy threshold for ##p^{+}+\gamma\rightarrow p^{+}+\pi^{0}##

*?*

(c) How much energy does the outgoing proton have after this reaction?

This GZK bound was finally confirmed experimentally 40 years after it was conjectured [Abbasi et ai., 2008]

2. Homework Equations

2. Homework Equations

## The Attempt at a Solution

(a) The spectral radiance ##B(\lambda,T)## of a blackbody is given by ##B(\lambda,T) = \frac{2hc^{2}}{\lambda^{5}}\frac{1}{e^{\frac{hc}{\lambda k_{B}T}}-1}##, where the symbols are self-explanatory.

Using this formula, I would like to calculate the average wavelength ##\lambda_{av}=\frac{\int_{0}^{\infty} d\lambda\ \lambda\ B(\lambda,T)}{\int_{0}^{\infty} d\lambda\ B(\lambda,T)}## and hence the average energy ##E_{av}=\frac{hc}{\lambda_{av}}## of the photons.

So, ##\int_{0}^{\infty} d\lambda\ \lambda\ B(\lambda,T) = \int_{0}^{\infty} d\lambda\ \frac{2hc^{2}}{\lambda^{4}}\frac{1}{e^{\frac{hc}{\lambda k_{B}T}}-1} = (2hc^{2})\Big(\frac{k_{B}T}{hc}\Big)^{3}\int_{0}^{\infty}\ dx\ \frac{x^{2}}{e^{x}-1}## under change of variables ##x=\frac{hc}{\lambda k_{B}T}## and

##\int_{0}^{\infty} \frac{x^{2}}{e^{x}-1} = \Gamma(3)\zeta(3) = (2!)(1.202)=2.4##

Also, ##\int_{0}^{\infty} d\lambda\ B(\lambda,T) = \int_{0}^{\infty} d\lambda\ \frac{2hc^{2}}{\lambda^{5}}\frac{1}{e^{\frac{hc}{\lambda k_{B}T}}-1} = (2hc^{2})\Big(\frac{k_{B}T}{hc}\Big)^{4}\int_{0}^{\infty}\ dx\ \frac{x^{3}}{e^{x}-1}## under change of variables ##x=\frac{hc}{\lambda k_{B}T}## and

##\int_{0}^{\infty} \frac{x^{3}}{e^{x}-1} = \Gamma(4)\zeta(4) = (3!)(1.0823)=6.4938##

So, ##\lambda_{av}=\frac{hc}{k_{B}T}\frac{2.4}{6.4938}##.

Therefore, ##E_{av} = k_{B}T\frac{6.4938}{2.4} = 6.3 \times 10^{-4}\ \text{eV}##.

Am I correct so far?