# The GZK bound

1. Dec 31, 2015

### spaghetti3451

1. The problem statement, all variables and given/known data

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. Relevant equations

3. 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?

2. Jan 1, 2016

### Staff: Mentor

That formula is not true.
It might look surprising, but there is a simple analogy in classical mechanics: if you go 20 km/h for 100 km and 100 km/h for 100 km, your average velocity is not 60 km/h, it is ~33 km/h. The average of 1/x is not 1 over the average of x.

To get an average energy, you either have to integrate over the frequency spectrum, or weight the wavelength spectrum with an appropriate factor. I don't think this is necessary, however. You can probably use 3/2 kT for the energy without integrating anything.

3. Jan 1, 2016

### spaghetti3451

Doesn't the equipartition theorem break down for a blackbody?

Are we allowed to use the classical formula $E=\frac{3}{2}k_{B}T$ here?

4. Jan 1, 2016

### Staff: Mentor

The formula is not that classical...
Do the integration if you like, but then do it in the right way.

5. Jan 1, 2016

### spaghetti3451

Is the integration also going to give us a value on the same order as $k_{B}T$?

Do we integrate the spectral radiance $B(\nu,T)$ over the frequency spectrum such that $\bar{B}(\nu,T)=\frac{\int d\nu B(\nu,T)}{\int d\nu}$.

But don't we then get the average spectral radiance, and not the average energy?

6. Jan 1, 2016

### Staff: Mentor

It should, there is no other scale that could be relevant. The prefactor can be different.
Most photons do not have exactly the average energy, and most collisions don't happen head-on. The GZK cutoff is not a precise value where the reaction can happen above but not below - the reaction is just getting more and more likely in some region.
Converting frequency to energy is no problem, that conversion is linear.