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Canonical Internal Energy

  1. Apr 3, 2013 #1
    I regard an equation for an canonical ensemble.
    I do not understand why both equations should be equal.

    [itex]\bar{E}= -\frac{ \partial \ln{(Z)}}{ \partial \beta} \overset{\text{?}} = k T ^{2 } \frac{\partial \ln{(Z)}}{\partial T}[/itex]

    [itex]Z[/itex] is a canonical partition function.
    [itex]\beta[/itex] = [itex]\frac{1}{kT}[/itex].

    Last edited: Apr 3, 2013
  2. jcsd
  3. Apr 3, 2013 #2
    Try the chain rule.
  4. Apr 3, 2013 #3


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    Do you under stand the first equation? That simply comes from the definition of the partition sum
    [tex]Z=\mathrm{Tr} \exp(-\beta \hat{H}),[/tex]
    where [itex]\hat{H}[/itex] is the Hamiltnian of the system. The Canonical Statistical Operator is
    [tex]\hat{R}=\frac{1}{Z} \exp(-\beta \hat{H}).[/tex]
    The mean energy is
    [tex]\overline{E}=\mathrm{Tr}(\hat{R} \hat{H})=-\frac{1}{Z} \frac{\partial Z}{\partial \beta}=-\frac{\partial \ln Z}{\partial \beta}.[/tex]
    As it turns out by comparing with the first Law of thermodynamics, you have
    [tex]\beta=\frac{1}{k T},[/tex]
    where [itex]T[/itex] is the temperature (contrary what you've written in your posting!).

    Then of course you have for any function of the temperature
    [tex]\frac{\partial f}{\partial \beta}=\frac{\partial f}{\partial T} \frac{\partial T}{\partial \beta}.[/tex]
    From the above formula you find [itex]T=1/(k \beta)[/itex] and thus
    [tex]\frac{\partial T}{\partial \beta}=-\frac{1}{k \beta^2}=-k T^2.[/tex]
    This gives
    [tex]\overline{E}=k T^2\frac{\partial \ln Z}{\partial T}.[/tex]
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