Proof that β = 1/kT

The pressure in the state \(j\) is given by \(p_j = - (\partial E_j/\partial V)\). The average energy is

\[\bar{E}=\frac{\displaystyle \sum_{J} E_{j}(N, V) e^{-\beta E_{f}(N, n)}}{\displaystyle \sum_{j} e^{-\beta E / N, n}} \label{I}\]

The average pressure is

\[\bar{p}=\frac{\displaystyle \sum p_{j}(N, V) e^{-\beta E_{f}(N, n)}}{\displaystyle \sum_{j} e^{-\beta E_{f}(N, n)}} \label{II}\]

According the Gibbs postulate the average energy, average pressure and other average mechanical properties calculated using the partition function are equal to their thermodynamic counterparts.

Note that some authors use \(\bar{E}\) and \(\bar{p}\) bar for the average quantities and elsewhere the angle bracket notation is used. These are equivalent notations.

If we differentiate the expression for the average energy we can treat the denominator, \(Q\) as a function of \(V\) as well since it represents a sum over \(\exp(-\beta E_j(N,V)\). Since \(E_j\) appears both in the exponent and as a function multiplying the exponent we have

\[\left(\frac{\partial E}{\partial V}\right)_{N, \beta}=\frac{\displaystyle \sum\left(\frac{\partial E_{j}}{\partial V}\right) e^{-\beta E / N, n}}{Q}-\frac{\displaystyle \sum-\beta\left(\frac{\partial E_{j}}{\partial V}\right) E_{f} e^{-\beta E / N, n}}{Q} -\frac{\displaystyle \sum_{j} E_{j}^{-\beta E_{j} / N, n} \displaystyle \sum_{j}-\beta\left(\frac{\partial E_{j}}{\partial V}\right) e^{-\beta E_{f}(N, n)}}{Q^{2}}\]

Here we used the quotient rule to take the derivative.

This can written compactly as

\[\left(\frac{\partial E}{\partial V}\right)_{N, \beta}=-\bar{p}+\beta \overline{E p}-\beta \bar{E} \bar{p}\]

We can differentiate Equation \ref{II} with respect to \(\beta\) to obtain

\[\left(\frac{\partial \bar{p}}{\partial \beta}\right)_{N, V}=\bar{E} \bar{p}-\overline{E p}\]

The two derivative expressions can be combined to give

\[\left(\frac{\partial E}{\partial V}\right)_{N, \beta}+\beta\left(\frac{\partial \bar{p}}{\partial \beta}\right)_{N, V}=-p\]

This can be compared to the thermodynamic equation of state

\[\left(\frac{\partial E}{\partial V}\right)_{T}-T\left(\frac{\partial p}{\partial T}\right)_{V}=-p \label{IV}\]

This can be derived as follows from

\[dE = TdS – PdV.\]

First take the derivative of both sides with respect to \(V\) at constant \(T\).

Now, note that

\[\left(\frac{\partial p}{\partial T}\right)_{V}=\left(\frac{\partial S}{\partial V}\right)_{T}\]

This is known as a Maxwell relation. It is obtained from

\[dA = SdT + PdV \label{III}\]

From the fact that \(A\) is a state function (the Helmholtz free energy) we know that the second cross derivatives must be equal. That is:

\[\left(\frac{\partial A}{\partial V \partial T}\right)=\left(\frac{\partial A}{\partial T \partial V}\right)\]

And from inspection of Equation \ref{III} we see that

\[\left(\frac{\partial A}{\partial T}\right)_{V}=S\]

and

\[\left(\frac{\partial A}{\partial V}\right)_{T}=P\]

Finally, using the relation

\[T\left(\frac{\partial}{\partial T}\right)=-\frac{1}{T}\left(\frac{\partial}{\partial(1 / T)}\right)\]

Showing that this is true is a little tricky. For example, we can define \(F = 1/T\). Then

\[\dfrac{\partial F}{\partial T} = \dfrac{-1}{T^2}\]

and

\[\dfrac{\partial F}{\partial F} = 1\]

So we can write

\[\dfrac{\partial F}{\partial T} = \dfrac{-1}{T^2} \left(\dfrac{\partial F}{\partial F}\right)\]

or

\[\dfrac{\partial }{\partial T} = \dfrac{-1}{T^2} \left(\dfrac{\partial }{\partial F}\right)\]

which gives Equation \ref{IV} when both sides are multiplied by \(T\).

\[\left(\frac{\partial E}{\partial V}\right)_{T}+\frac{1}{T}\left(\frac{\partial p}{\partial(1 / T)}\right)_{V}=-p\]

The comparison with the above equation shows that \(\beta \propto 1/T\).

This proves that \(\beta = constant/T\). The constant turns out to be \(k_B\) or Boltzmann’s constant by comparison with expressions for the average energy or average pressure with known thermodynamic equations.