Chemical potential using Boltzmann equation

[Mikhail_MR]

Homework Statement

I must calculate chemical potential using the Boltzmann equation in relaxation time approximation $$f=f^0-\tau v_z^2 \partial f^0/\partial z,$$ where $f^0$ is given as
$$f^0 = 2(\frac{m}{2\pi\hbar})^3 \frac{1}{\exp{\beta(z)(\frac{mv^2}{2}-\mu(z))}+1}$$
I have to consider only heat current without electrical current. In this case, I can make use of $<v_z>=0$.

Homework Equations

$0=<v_z>=\int d^3 v (v_z f^0 - \tau v_z^2 \partial f^0/\partial z)$. I can also use the limit $\mu \beta \gg 1$ and $I_n=\int_{-\infty}^{+\infty} dx x^n \frac{e^x}{(e^x+1)^2}$ with $I_0=1, ~I_1=0, ~I_2=\pi^2 / 3$

The Attempt at a Solution

Because I have velocity only in $z$ direction, I can calculate the first integral
$$c_0 \int dv_z v_z \frac{1}{\exp{\beta (mv_z^2 / 2 - \mu)}+1} = c_0 \frac{\beta m v_z^2/2 - \ln(\exp{\beta m v_z^2 /2}+\exp{\beta \mu})}{\beta m}$$ Now I can make use of $\beta \mu \gg 1$ and I get $$c_0 (v_z^2/2-\mu/m)$$ with $c_0 = 2(\frac{m}{2\pi\hbar})^3$

To calculate the second integral I need to calculate the derivative first.
$$\partial f^0/ \partial z = - e^x /(e^x+1)^2 dx(v, z)/dz$$ with $x=\beta (m v^2 / 2 - \mu)$
$$\Rightarrow \partial f^0/ \partial z = - e^x /(e^x+1)^2 [\mu d\beta / dz - m v^2/2 \cdot d\beta / dz - \beta d\mu /dz]$$

Now I must somehow get the form of the integrals $I_n$. But I do not see a way I can do it. Have I missed some step or is my attempt at a solution completely wrong?.

Any help would be appreciated

 

[Fred Wright]

I have the following suggestions:
1. For the integral$$\int_{-\infty}^\infty v_z f^0 dv_z$$make the substitution$$x=\beta \left ( \frac{m v_z^2}{2} - \mu \right ),$$express$f^0$ as$$f^0 =\frac{e^{-x}}{1 + e^{-x}},$$and observe that the integral has even symmetry about $v_z = 0$.
2. For the integral$$\int_{-\infty}^\infty \tau v_z^2 \frac{\partial f^0}{\partial z} dv_z$$make the same substitution for x and write$$\frac{\partial f^0}{\partial z}=\frac{\partial f^0}{\partial v_z} \frac{\partial v_z}{\partial z}$$ and observe that $\frac{\partial v_z}{\partial z}\approx \frac{1 }{\tau}$.