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Clasical statistical mechanics

  1. Oct 17, 2012 #1
    Hello,I'm studing the fundamental principles of statistical mechanics and I don't understand the "principle of equal a priory probabities" for this reason: the principle states that all states in the representative emsemble have the same probability which means that the density [itex]\rho(p,q)[/itex] should be constant in the corresponding region of phase space, this is ok for the microcanonical emsemble however for the canonical emsemble this is no longer true if the corresponding region in phase space includes points with different energies.
    In any case, according to that principle it seems that the only function compatible with it is[itex]\rho(p,q)=const.[/itex]
    Can some one please tell what is wrong with my interpretation of the principle?
     
  2. jcsd
  3. Oct 17, 2012 #2

    Mute

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    I'm not entirely sure if this will answer your question: In the usual formulation of equilibrium statistical mechanics, the microcanonical ensemble is a closed system in which every state is equally probable (corresponding to ##\rho(q,p) = \mbox{const}## in your formulation). To develop the canonical ensemble, one images a very large microcanonical ensemble of which we only look at a subset that can exchange energy with the rest of the system, which we refer to as the "heat resevoir".

    The subset+reservoir taken together form a microcanonical ensemble, so overall the probability is the same for every state, but if we just look at a restricted region of the system then not all states in this restricted system are equally likely. This is due the flow of energy in and out of the system. Lower energy states are more likely to be sampled than higher energy states, so in a canonical system ##\rho(q,p)## is not constant and is in fact ##\rho(q,p) = \exp(E(q,p)/k_BT)/Z##, where ##Z## is the partition function.
     
  4. Oct 18, 2012 #3
    Thanks for your answer Mute. In your example the problem is that our system of interest is the subset and in setting up the representative ensemble of that system we allow in the ensemble states with different energies, so far so good, but next we need to set up the function [itex]\rho(q,p)[/itex] and we see that it is not constant so it seems to contratic the principle of equal a priori probabilities as it is stated in the clasic book by Tolman that says that in setting up any emsemble we must consider uniform probabilities distribution but, it seems to me, that uniform distribution must be applied only when E=const for the states in the ensemble otherwise it makes no sense
     
  5. Oct 18, 2012 #4

    DrDu

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    When speaking of a priori probability this includes the ignorance of the value of E.
     
  6. Oct 18, 2012 #5
    Tolman says that the representative ensemble for the system of interest is set up by whatever partial knowledge we have of the system, this knowledge may include a specific energy value or a range of values of energy or no restriction of energy at all. May be I need to study more and work out specfic applications
     
  7. Oct 18, 2012 #6

    DrDu

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    This is the correct a priori probability. You obtain it by maximizing entropy without any restriction.
    If you have restrictions, e.g. on E, you can calculate the kanonical distribution. However, the canonical distribution is not an a priory probabilty.
     
  8. Oct 18, 2012 #7
    The microcanonical ensemble holds for an isolated system whose energy is conserved. The canonical ensemble, on the other hand, represents a sample in contact with a heat bath that is held at a certain temperature.

    Although their distributions look quite quite different, they essentially give the same ensemble averages for macroscopic samples.

    Consider the integral in phase space:
    [tex]
    \langle f \rangle_{\mathrm{micro}} = \frac{1}{A(E)} \, \int{dq dp \, f(q, p) \, \delta(H(q, p) - E)}
    [/tex]
    of a quantity f(q, p) that is a function of the generalized coordinates q, and generalized momenta p. The delta function ensures that we take into account only points that lie on a constant energy E hypersurface with equal weight. These kinds of integrals are essentially what is evaluated for the average in a microcanonical ensemble. It is not hard to see that this integral is a parametric function of E. The normalization constant A(E) is chosen so that [itex]\langle 1 \rangle = 1[/itex].

    Now, represent the delta function by an inverse Laplace transform:
    [tex]
    \delta(x - E) = \frac{1}{2 \pi i} \, \int_{\eta - i \infty}^{\eta + i \infty}{e^{s (x - E)} \, ds}
    [/tex]

    Then, the above integral may be rewritten as:
    [tex]
    I_{\mathrm{micro}}(E) = \frac{1}{2 \pi i} \, \int_{\eta - i \infty}^{\eta + i \infty}{ds \, e^{-s \, E} \, \int{dq dp f(q, p) \, e^{s H(q, p)}}}
    [/tex]
    Notice that the integral over phase space is essentially what is evaluated in the canonical ensemble, albeit with a complex variable s.

    We want to evaluate this integral within the stationary phase approximation. We will do it first for the normalization constant A(E). Let us introduce:
    [tex]
    \tau^D \, e^{s \, F(s)} \equiv \int{dq dp e^{s H(q, p)}}
    [/tex]
    where [itex]\tau[/itex] is a quantity with the dimension of action ([itex]\int dq_{j} dp_{j}[/itex]). Then
    [tex]
    A(E) = \tau^D \, \frac{1}{2 \pi \, i} \, \int_{\eta - i \infty}^{\eta + i \, \infty}{ds \, e^{s \, (F(s) - E)}}
    [/tex]
    Now, we choose [itex]\eta = \beta[/itex] such that:
    [tex]
    \frac{d}{ds} \left. s \, (F(s) - E) \right\vert_{s = \beta} = 0
    [/tex]
    In other words, if we know F(s), then E is a function of [itex]\beta[/itex]:
    [tex]
    E = \frac{d}{d\beta} \left( \beta \, F(\beta) \right)
    [/tex]
    The method of stationary approximation then gives
    [tex]
    A \sim \tau^D \, \frac{1}{\sqrt{2\pi \, \vert F''(\beta)\vert}} \, e^{\beta \, (F(\beta) - E)}
    [/tex]

    When evaluating the integral we treat the same approximation, and we essentially get:
    [tex]
    \langle f \rangle_{\mathrm{micro}}(E) \sim \int{\frac{dq dp}{\tau^D} \, e^{\beta \, (F(\beta) - H(q, p))} \equiv \langle f \rangle_{\mathrm{canon}}(T = \frac{1}{\beta}})
    [/tex]
    which is the canonical ensemble average if we accept F as the free energy of the system. Notice that the energy is then expressed as:
    [tex]
    \frac{d}{d\beta} = \frac{dT}{d\beta} \, \frac{d}{dT} = -T^2 \, \frac{d}{dT}
    [/tex]
    [tex]
    E = -T^{2} \, \frac{d}{dT} \left( \frac{F}{T} \right) = F - T \, F'(T) = F + T \, S
    [/tex]
    which is the correct thermodynamic relation.
     
    Last edited: Oct 18, 2012
  9. Oct 18, 2012 #8

    atyy

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    In the situation Mute is describing, E is constant for the entire system, not just the subsystem you are interested in, ie. the microcanonical ensemble applies to the entire system. Then for the subsystem you are interested in, E is not constant, and the microcanonical ensemble does not apply.
     
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