Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

About Relaxation Time Approximation

  1. Sep 8, 2014 #1
    I have seriously stocked in the subject below.

    According to Ashcrift & Mermin (chapter 13):

    If the electrons about [itex]r[/itex] have equilibrium distribution appropriate to local temperature [itex]T(r)[/itex],
    [tex]g_n (r,k,t)=g_n^o (r,k)=\frac {1}{ exp^{(\epsilon_n (k) -\mu (r))/kT} +1} (formula 13.2)[/tex] then collisions will not alter the form of distribution function. We know in the time interval dt a fraction [itex]\frac{dt}{\tau_n(r,k)}[/itex] of electrons in band n with wave vector k near position r will suffer a collision that does alter their band index and/or wave vector. If the above form of distribution function is nevertheless to be unaltered, then the distribution of those electrons that emerge from collisions into band n with wave vector k during the same interval must precisely compensate for this loss. Thus:
    [tex]g_n (r,k,t)=\frac{dt}{\tau_n(r,k)} g_n^o (r,k) (formula 13.3)[/tex]

    I don’t know how is 13.3 obtained.

    In fact I don’t know what the text means by form of distribution function in the expression “collisions will not alter the form of distribution function”? On one hand if the electrons near r which left the point (n,k) due to collision are precisely compensated then at a specific (r,k) the distribution function doesn’t have to change and the expression means that [itex]g_n (r,k,t)=g_n^o (r,k)[/itex] which is 13.2. on the other hand using this interpretation, near position r at time interval dt due to collisions[itex] \frac{dt}{\tau_n(r,k)} [/itex] electrons will leave the point (r,n,k) to another point say [itex](r,n^\prime,k^\prime)[/itex] so that [tex]dg_n (r,k,t)=g^0_n{\prime} (r,k^\prime,t^\prime)- g_n^0 (r,k,t)[/tex] where [itex]t^\prime=t+dt[/itex].

    However if we were to accept this, how can we consider the [itex] dg_n[/itex] as the number of electrons which have left [itex](r,n,k)[/itex] toward [itex](r,n\prime,k\prime)[/itex] due to collision at time interval dt namely[itex] \frac{dt}{\tau_n(r,k)} g_n^o (r,k)[/itex]? If so, we necessarily must have just prior to collision [itex]g^0_n(r,k,t)= g^0_n\prime(r,k\prime,t)[/itex] so that this amount of electrons that enter there, perform a change in distribution function as much as[itex] \frac{dt}{\tau_n(r,k)} g_n^o (r,k)[/itex]!

    Could anyone please help me?
  2. jcsd
  3. Sep 13, 2014 #2
    Hmmm, the definition of thermal equilibrium is that (a) because of the finite temperature collisions occur and (b) in equilibrium the (average) situation is static, i.e. time derivatives of statistical properties vanish.

    For this to be true, every electron scattered away from band n/wave vector k must be replaced by an electron coming from a different band/wave vector. The population of each band/wave vector therefore stays (on average) constant. With that the form of the distribution function remains constant.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook