Canonical Internal Energy

  • Thread starter Abigale
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  • #1
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Hi,
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].



Thx
Abby
 
Last edited:

Answers and Replies

  • #2
1,082
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Try the chain rule.
 
  • #3
vanhees71
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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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