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Entropy of fermionic Gas.

  1. Jun 19, 2008 #1
    The problem is to show that for a fermionic gas the entropy is given by:
    [tex]\sigma=-\int d\epsilon D(\epsilon )[f(\epsilon )log(f(\epsilon )-(1-f(\epsilon )log(1-f(\epsilon )][/tex] where D(epsilon) is the derivative operator wrt epsilon, and f(epsilon) is fermi-dirac distribution function.


    Now what I think is that I only need to show that the entropy equals minus the integrand, but I'm not sure where did the minus come from.

    I mean the entropy is defined as logarithm of the number of possible states, the function that counts this number is: (f^f)*((1-f)^(1-f))
    cause f counts the number of possible states there are below the chemical potential and 1-f above it, and we take a power of themselves because there sum equals the number of states of the system.

    but I don't where did the minus sign come from, can you help me on this?

    thanks in advance.
     
  2. jcsd
  3. Jun 20, 2008 #2

    Mute

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    [itex]D(\varepsilon)[/itex] is not an energy derivative, it is the single-particle density of states as a function of energy.

    To derive that expression, you need to calculate the entropy in the grand canonical ensemble, using the relation

    [tex]\sigma = -\left(\frac{\partial \mathcal{F}}{\partial \tau}\right)_{V,\mu}[/tex]

    where [itex]\tau = k_BT[/itex] and [itex]\mathcal{F}[/itex] is the grand free energy. If this is the way you did the problem I would guess you just forgot the minus sign in this relation.
     
  4. Jun 21, 2008 #3
    I don't understand, what is D(epsilon)?
     
  5. Jun 21, 2008 #4
    so because: [tex]-\tau *log(Z_G)=F[/tex]
    I only need to find what is Z_G but it equals:
    [tex]Z_G=1+exp(\beta *(\mu -\epsilon))[/tex]
    but still what is D(epsilon).
    you mean D is the function that counts the number of posiible states, if so then by definition the entropy equals log of this.
     
  6. Jun 21, 2008 #5

    Mute

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    The density of states is precisely what it sounds like: if you plotted the number of particles, [itex]n(\varepsilon)[/itex], in a given state as a function of energy, the density is the number of particles per energy [itex]\varepsilon[/itex]. That is, [itex]d n(\varepsilon)/d\varepsilon = D(\varepsilon)[/itex]. Equivalently,
    [tex]\int_{-\infty}^{\infty}d\varepsilon~D(\varepsilon) = N[/tex]

    where N is the number of particles in the system. You need this density when approximating sums by integrals. If you're summing over a discrete index, say n for example, then

    [tex]\sum_{n} \rightarrow \int dn[/tex]
    when approximating the sum by an integral. In this case, [itex]D(n) = 1[/itex]. However, if you wanted to write that integral as a function of energy instead, then because the states aren't typically equally spaced as a function of energy (and if they were it wouldn't be by "1" -> you need some dimensionful constant), you need the density of states:

    [tex]\sum_{n} \rightarrow \int d\varepsilon~D(\varepsilon)[/tex]

    Now, as for determining the energy, you're almost right about the grand partition function. However, what you've written is the single particle partition function. If you have N particles, then your total grand partition function is going to be the product of the N single particle partition functions, each of which will have a different energy [itex]\varepsilon_n[/itex]:

    [tex]Z_N = \prod_{n}Z(\varepsilon_n) = \prod_n(1 + e^{-\beta(\varepsilon_n - \mu)})[/tex]

    Hence,

    [tex]\sigma = -\left(\frac{\partial \mathcal{F}}{\partial \tau}\right)_{V,\mu} = -\frac{\partial}{\partial \tau} \left(\tau \ln Z_N\right) = -\sum_n \frac{\partial}{\partial \tau} \tau \ln \left(1 + \exp\left[-\beta(\varepsilon_n - \mu\right)\right][/tex]

    In converting that sum to an integral over energy, you introduce the density of states. Some playing around with the summand/integrand of the entropy expression will yield the expression you have in your first post.
     
    Last edited: Jun 21, 2008
  7. Jun 21, 2008 #6
  8. May 19, 2009 #7
    wow, thanks for that, Mute! i'd always wondered why you needed to multiply the integrand by the density of states. fantastic explanation!
     
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