1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Homework Helper

    [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


    User Avatar
    Homework Helper

    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]


    [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!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Entropy of fermionic Gas.
  1. Molar Entropy Of A Gas (Replies: 1)

  2. Fermion gas problems. (Replies: 0)