The GZK Bound and Cosmic Rays: Calculating Photonic Energy in Outer Space

[spaghetti3451]

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

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?

 

[mfb]

$E_{av}=\frac{hc}{\lambda_{av}}$

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.

 

[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?

 

[mfb]

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

 

[spaghetti3451]

The formula is not that classical...

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

Do the integration if you like, but then do it in the right way.

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?

 

[mfb]

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

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.

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?

Converting frequency to energy is no problem, that conversion is linear.

 

[Matt King]

I don't think the OPs approach is quite right because it doesn't contain any information about numbers of photons, but it does seem to give a numerically similar answer.

The average energy of a photon in a cloud must be equal to the total energy of the cloud divided by the total number of photons in the cloud.

Where the density is uniform (as in the case of the cosmic background), we can just divide total energy density L by total number of photons density N to get the same number. Density here means per unit volume. The formulae for Planck spectral density are usually given per unit solid angle but the volume spectral density formulae are proportional to those so as we are dividing any constants of proportionality will cancel.

The total energy density is given by integrating the spectral density B(λ, T) over the whole wavelength range (or B(ν, T) over the entire frequency range - they must give the same answer), and is given by

L = ( 2 (π k T) ³ ) / (15 c² h³) - see https://en.wikipedia.org/wiki/Black-body_radiation#Planck's_law_of_black-body_radiation, so that's the top of the fraction sorted.

For the bottom of the fraction, N, the total number of photons (density) is found by integrating the frequency-dependendent number of photons (density), so we need to work out what that is.

The frequency-dependent number of photons (density) is found by dividing the frequency-dependendent energy (density) by the energy of a single photon at that energy, h ν, or h c / λ.

So we are looking for integral [ 0 - infinity ] (1 / h ν) * B (ν, T) wrt ν = ( 2 (k T)³ / (c² h³) ) * ∫ [0 - infinity] x² / e^x -1 wrt x, where x = (h ν) / (k T)

Integrals from 0 to infinity of x^{y -1} / e^x -1 are always equal to RiemanZeta(y) * Gamma (y) - it's a standard result but don't ask me to prove it, so the bottom part of our original fraction, the total number of photons (density) evaluates to

N = 4 (k T )³ * RiemannZeta(3) / c² h³ as Gamma(3) = 2.

Dividing L by N then gives the average energy of photons in black body radiation as π⁴ k T / (30 * RiemannZeta(3)).

RiemannZeta(3) has the numerical value 1.20205..., so this works out at

Mean E(photon) = 2.701 k T.

Plugging in a temperature of 2.73 Kelvin gives E = 1.018 exp -22 Joules, which is 6.3539 exp -4 electron volts.

Wikipedia quotes the peak energy of photons in the CMB as 6.6 exp -4 electron volts at https://en.wikipedia.org/wiki/Cosmic_microwave_background, and the distribution graph on that page does look like the mean energy would be a little less than the peak.

This source also has it at 6.4 electron volts https://web.stanford.edu/~oas/SI/SRGR/notes/SRGRLect11Prob_2015.pdf

I found this source helpful in solving this problem too: http://www.astro.wisc.edu/~townsend/resource/teaching/astro-310-F11/solutions-2011-10-21.pdf

That was harder than I thought!