Boltzmann equation for annihilation

[happyparticle]

TL;DR

Trying to understand the right hand side of the Boltzmann equation for annihilation for the rate of change in the abundance of a given particle.

In the Dodelson's textbook, the author introduce the Boltzmann equation for annihilation.

$a^{-3} \frac{d(n_1 a)}{dt} = \int \frac{d^3 p_1}{(2 \pi)^3 2E_1} \int \frac{d^3 p_2}{(2 \pi)^3 2E_2} \int \frac{d^3 p_3}{(2 \pi)^3 2E_3} \int \frac{d^3 p_4}{(2 \pi)^3 2E_4} \times (2 \pi)^4 \delta^3 (p_1 + p_2 - p_3 - p_4) \delta (E_1 + E_2 - E_3 - E_4)|M|^2 \times {f_3 f_4[1 \pm f_1] [1 \pm f_2] - f_1 f_2 [1 \pm f_3] [1 \pm f_4]}$

I don't understand the right hand part of the equation. Where all the part comes from? Why $p_i, f_i$ are outside of the integrals? What |M| means? I can't figure out how all the right hand side is related to the rate of change in the abundance of a given particle.

I'm guessing that $n_i$ is the particle density and $f_i$ is the the expected number of particles in an energy state.

 

[Orodruin]

Why $p_i, f_i$ are outside of the integrals?

They are not.

What |M| means?

That would be the amplitude of the annihilation.

I can't figure out how all the right hand side is related to the rate of change in the abundance of a given particle.

Rate of change = production - annihilation

The delta functions ensure energy and momentum conservation. The distribution functions implement rates - including fermi blocking etc. The integrals integrate over all possible states. You are missing parentheses around the f terms.

 

[happyparticle]

Thank you for the explanation.
However, I'm not sure how exactly the delta functions ensure the energy and moment conservation. Also, I don't see why there is a $(2 \pi)^4$ and $\delta^3$.

I was looking for a full derivation of this equation. Unfortunately, I can't find any. I'm wondering if I'm using the right name for the equation.

 

[Orodruin]

The $\delta^3$ is momentum conservation. Try writing out the conditions for the deltas being nob-zero!

The factors of $2\pi$ are for correct normalization.

 

[happyparticle]

The $\delta^3$ is momentum conservation. Try writing out the conditions for the deltas being nob-zero!

Thus, I have $\delta(0)^3 = \infty$ which give a infinite rate of change. I don't understand.

 

[renormalize]

Thus, I have $\delta(0)^3 = \infty$ which give a infinite rate of change. I don't understand.

This is wrong. Do you know how to evaluate the simple 3D integral $\intop_{\text{All Space}}d^{3}x\,f\left(\vec{x}\right)\delta^{3}\left(\vec{x}-\vec{x}_{0}\right)$?

 

[Orodruin]

Thus, I have $\delta(0)^3 = \infty$ which give a infinite rate of change. I don't understand.

No you don’t. It is inside several integrals over momenta and essentially ensures that the momenta that you integrate over satisfy momentum and energy conservation