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Chemical potential using Boltzmann equation

  1. Jun 22, 2017 #1
    1. The problem statement, all variables and given/known data
    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##.

    2. Relevant 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##

    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
     
  2. jcsd
  3. Jun 27, 2017 #2
    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}##.
     
    Last edited: Jun 27, 2017
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