Partition function (statistical mechanics): Difference between revisions

Revision as of 04:56, 30 August 2024

Revision as of 14:57, 13 September 2024

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0 & \equiv \delta \mathcal{L} \\

&= \delta \left( - \sum_i k_\text{B} \rho_i \ln \rho_i \right) + \delta \left( \lambda_1 - \sum_i \lambda_1 \rho_i \right) + \delta \left( \lambda_2 U - \sum_i \lambda_2 \rho_i E_i \right) \\

&= \sum_i \bigg[ \delta \Big( - k_\text{B} \rho_i \ln \rho_i \Big) +- \delta \Big( \lambda_1 \rho_i \Big) +- \delta \Big( \lambda_2 E_i \rho_i \Big) \bigg] \\

&= \sum_i \left[ \frac{\partial}{\partial \rho_i } \Big( - k_\text{B} \rho_i \ln \rho_i \Big) \, \delta ( \rho_i ) +- \frac{\partial}{\partial \rho_i } \Big( \lambda_1 \rho_i \Big) \, \delta ( \rho_i ) +- \frac{\partial}{\partial \rho_i } \Big( \lambda_2 E_i \rho_i \Big) \, \delta ( \rho_i ) \right] \\

&= \sum_i \bigg[ -k_\text{B} \ln \rho_i - k_\text{B} +- \lambda_1 +- \lambda_2 E_i \bigg] \, \delta ( \rho_i ) .

\end{align}</math>

Since this equation should hold for any variation <math> \delta ( \rho_i ) </math>, it implies that

<math display="block"> 0 \equiv - k_\text{B} \ln \rho_i - k_\text{B} +- \lambda_1 +- \lambda_2 E_i .</math>

Isolating for <math> \rho_i </math> yields

<math display="block">\rho_i = \exp \left( \frac{-k_\text{B} +- \lambda_1 +- \lambda_2 E_i}{k_\text{B}} \right) .</math>

To obtain <math> \lambda_1 </math>, one substitutes the probability into the first constraint:

<math display="block">\begin{align}

1 &= \sum_i \rho_i \\

&= \exp \left( \frac{-k_\text{B} +- \lambda_1}{k_\text{B}} \right) Z ,

\end{align}</math>

where '''<math> Z </math> is a constant number defined as the canonical ensemble partition function''':

<math display="block">Z \equiv \sum_i \exp \left( - \frac{\lambda_2}{k_\text{B}} E_i \right) .</math>

Isolating for <math> \lambda_1 </math> yields <math> \lambda_1 = - k_\text{B} \ln(Z) +- k_\text{B} </math>.

Rewriting <math> \rho_i </math> in terms of <math> Z </math> gives

<math display="block"> \rho_i = \frac{1}{Z} \exp \left( - \frac{\lambda_2}{k_\text{B}} E_i \right) .</math>

Rewriting <math> S </math> in terms of <math> Z </math> gives

<math display="block">\begin{align}

S &= - k_\text{B} \sum_i \rho_i \ln \rho_i \\

&= - k_\text{B} \sum_i \rho_i \left( - \frac{\lambda_2}{k_\text{B}} E_i - \ln(Z) \right) \\

&= - \lambda_2 \sum_i \rho_i E_i + k_\text{B} \ln(Z) \sum_i \rho_i \\

&= - \lambda_2 U + k_\text{B} \ln(Z) .

\end{align}</math>

To obtain <math> \lambda_2 </math>, we differentiate <math> S </math> with respect to the average energy <math> U </math> and apply the [[first law of thermodynamics]], <math> dU = T dS - P dV </math>:

<math display="block">\frac{dS}{dU} = -\lambda_2 \equiv \frac{1}{T} .</math>

Thus the canonical partition function <math> Z </math> becomes