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Electrical Current Density

  1. Apr 9, 2013 #1
    1. The problem statement, all variables and given/known data
    Hi.

    I'm having some trouble calculating the electrical current density, which in my case is given by:

    [tex]\sigma =\frac{{{e}^{2}}}{4{{\pi }^{3}}}\int{\left( -\frac{\partial f}{\partial \varepsilon } \right)\tau \cdot \mathbf{v}\cdot \mathbf{v}}\,dk[/tex]


    2. Relevant equations

    Lets assume that [itex]\tau[/itex] is constant, and that:

    [tex]f=\frac{1}{\exp \left( \frac{\varepsilon -\mu }{{{k}_{B}}T} \right)+1},[/tex]
    the Fermi function.
    And that:
    [tex]\varepsilon =\frac{{{\hbar }^{2}}{{k}^{2}}}{2m}[/tex]
    and
    [tex]v=\frac{\partial \varepsilon }{\partial k}[/tex]


    3. The attempt at a solution

    Then, how am I supposed to do this integral ?
    Exponential functions are so annoying if they are not alone.

    I am supposed to end up with a value/number, but I really can't see how this is doable when I have all these exponential functions. When you differentiate the Fermi function, before putting it inside the integral, you get even more exponential functions.

    So yes, I'm kinda lost here. So I was kinda hoping for someone who might be able to give a little hint :)


    Thanks in advance.
     
  2. jcsd
  3. Apr 10, 2013 #2
    Use the chain rule. You have [itex] v = \frac{1}{\hbar} \frac{\partial \epsilon}{\partial k} [/itex], so [tex] v dk = \frac{1}{\hbar} \frac{\partial \epsilon}{\partial k} dk = \frac{1}{\hbar} d \epsilon [/tex] After that, I have no idea. You can try partial integration, which lets you dodge the derivative, although I didn't actually do the calculation so I don't know if it gets easier.
     
  4. Apr 10, 2013 #3
    Hmmm, I've actually narrowed it down to several integral, only one being hard, which is:

    [tex]\int{\exp \left( \frac{\varepsilon +\mu }{{{k}_{B}}T} \right)}\,{{k}^{2}}dk,[/tex]
    where epsilon is the energy given in my first post.

    Any clever way to solve this now ?
     
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