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Reason behind the definition of Helmholtz free Energy

  1. Aug 30, 2008 #1
    What is the reasoning behind defining the helmholtz free energy as F = -kT ln Z?

    I always wanted to know why it was just defined as the above. Is it as a form of convenience because the macroscopic theromodynamic observables of a system at constant temperature (ie the canonical ensemble) are related to to the partition function as ln Z?


    \bar{E} = -\frac{\partial lnZ}{\partial \beta}


    P = \frac{1}{\beta}\left(\frac{\partial lnZ}{\partial V}\right)_\beta

    So its just convenient to a create a thermodynamic quantity that is related to T ln Z for a system at temperature T?
  2. jcsd
  3. Aug 31, 2008 #2


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    The free energies are defined from classical thermodynamics, without any reference to the partition function of statistical mechanics. Which particular free energy (Enthalpy, Helmholtz, Gibbs etc) is most convenient depends on what variables you keep constant in the experiment.

    The partition function is a quantity that is defined from the canonical ensemble of statistical mechanics. And as you guessed, the identification of F=-kTlnZ is just what turns out to make statistical mechanics and thermodynamics compatible. If you use other ensembles in statistical mechanics, then you make other identifications so that statistical mechanics and thermodynamics are compatible. For example, if you use the microcanonical ensemble, then you use S=kln(number of states). The different ensembles of statistical mechanics yield the same thermodynamical answers in many common situations (magically), but they are not always equivalent: http://arxiv.org/abs/cond-mat/0404655
  4. Aug 31, 2008 #3
    Ah of course, in my haste i forgot the obvious fact that classical thermodynamics was established before the statistical treatment was even investigated.

    So the statistical definitions are just chosen so they correctly correspond to the classical thermo definitions.

    Thanks :)
  5. Aug 31, 2008 #4
    Statistical Mechanics is founded on the entropy-maximum principle, that is, a generalization of a Laplace's indifference principle. Your formal approach is to define a density operator based on informations you have and then find equilibrium state by maximizing the corrispondent entropy. Since entropy is function of extensive quantity (not easly measurable) it is convenient to define another function with same variational principle but different arguments (i.e. intensive variable associate). The mathematically way for doing that is by Legendre transformation.

    In SM view, Helmholtz free energy is Legendre transformation of entropy when your density operator is canonical. You use it for switching between a function of energy (entropy) to a function of temperature (free energy).

  6. Sep 1, 2008 #5
    See here
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