Cosmological Boltzmann equation [closed]

Question

Consider the Boltzmann equation:

$$\frac{d \ln{n^c(T)}}{d \ln{T}} = \frac{\Gamma}{H}(1 - \frac{n^c_{eq}(T)}{n^c(T)})$$

We know that the ratio $\Gamma/H$ can be considered constant, let us put it equal to the letter $\alpha$. I have been given that at a certain teperature $T_i$ the system is in thermal equilibrium, so that $n^c(T_i) = n^c_{eq}(T_i)$. How can I mathematically evolve the system , for $T < T_i$ ?

The solution is $$n^c(T) = (\frac{T}{T_i})^\alpha n_{eq}^c(T_i) + \alpha T^\alpha \int_T^{T_i}{\frac{n_{eq}^c(T')}{T'^{1+\alpha}}dT'},$$

but I do not know how to get there: I suppose I have to integrate the Boltzmann equation, but I do not know how exactly.

Answer 1

As you correctly point out, the solution follows by integrating the differential equation. The general idea is to expand the left hand side of your differential equation by realizing that

$$\text{d}\ln n^c(T) = \frac{1}{n^c(T)} \text{d}n^c(T) = \frac{1}{n^c(T)} \frac{\partial n^c(T)}{\partial T}\text{d}T = \frac{T}{n^c(T)} \frac{\partial n^c(T)}{\partial T}\text{d}\ln T$$

which leads to a first-order linear ordinary differential equation

$$ T \frac{\partial n^c(T)}{\partial T} = \alpha (n^c(T) - n_{eq}^c(T)) $$

of the more general form

$$y'+p(x) y = q(x)$$

And I think you are able to solve this one now :) If not, you can refresh your mind using this technique: http://en.wikipedia.org/wiki/Variation_of_parameters

As you can see, you could also keep the logarithmic derivative, solve in terms of this new variable, and substitute in the end.