Description
The definition of the Boltzmann distribution is given by quantities of a system's microstates which can be practically unbounded (temperature \( \htmlClass{sdt-0000000029}{T} \)) or arbitrarily defined (energy \( \htmlClass{sdt-0000000100}{E} \) with multiple possible offsets). This makes the integral over the distribution function different from 1, so a normalization constant is necessary.
\[\htmlClass{sdt-0000000077}{Z} = \htmlClass{sdt-0000000077}{Z}(\htmlClass{sdt-0000000029}{T}) = \int_{\htmlClass{sdt-0000000091}{\mathbf{s}} \in \htmlClass{sdt-0000000026}{S}} \exp\left\{ - \frac{ \htmlClass{sdt-0000000100}{E}(\htmlClass{sdt-0000000091}{\mathbf{s}}) }{ \htmlClass{sdt-0000000029}{T} } \right\} d\htmlClass{sdt-0000000091}{\mathbf{s}}\]
Derivation
- Consider the equation for the unnormalized probability of a system being in a given microstate \( \htmlClass{sdt-0000000091}{\mathbf{s}} \) under the Boltzmann distribution :
\[ p(\htmlClass{sdt-0000000091}{\mathbf{s}}) = \exp\left\{ - \frac{ \htmlClass{sdt-0000000100}{E}(\htmlClass{sdt-0000000091}{\mathbf{s}}) }{ \htmlClass{sdt-0000000029}{T} } \right\} \] - The distribution function must integrate to 1, however this is not the case. Adding some unknown normalization factor \( \frac{1}{\htmlClass{sdt-0000000077}{Z}} \) is needed:
\[p(\htmlClass{sdt-0000000091}{\mathbf{s}}) = \frac{1}{\htmlClass{sdt-0000000077}{Z}} \exp\left\{ - \frac{ \htmlClass{sdt-0000000100}{E}(\htmlClass{sdt-0000000091}{\mathbf{s}}) }{ \htmlClass{sdt-0000000029}{T} } \right\}\] - To find the value of this normalization constant, we can use the fact that the (now normalized) probability distribution must integrate to 1. This gives:
\[ \int_{\htmlClass{sdt-0000000091}{\mathbf{s}} \in \htmlClass{sdt-0000000026}{S}} \frac{1}{\htmlClass{sdt-0000000077}{Z}} \exp\left\{ - \frac{ \htmlClass{sdt-0000000100}{E}(\htmlClass{sdt-0000000091}{\mathbf{s}}) }{ \htmlClass{sdt-0000000029}{T} } \right\} d\htmlClass{sdt-0000000091}{\mathbf{s}} = 1 \] - The constant \( \htmlClass{sdt-0000000077}{Z} \) might depend on \( \htmlClass{sdt-0000000029}{T} \), but importantly not on the state \( \htmlClass{sdt-0000000091}{\mathbf{s}} \) and implicitly neither on \( \htmlClass{sdt-0000000100}{E} \), we have:
\[ \frac{1}{\htmlClass{sdt-0000000077}{Z}} \int_{\htmlClass{sdt-0000000091}{\mathbf{s}} \in \htmlClass{sdt-0000000026}{S}} \exp\left\{ - \frac{ \htmlClass{sdt-0000000100}{E}(\htmlClass{sdt-0000000091}{\mathbf{s}}) }{ \htmlClass{sdt-0000000029}{T} } \right\} d\htmlClass{sdt-0000000091}{\mathbf{s}} = 1 \] - This results in the equality:
\[ \int_{\htmlClass{sdt-0000000091}{\mathbf{s}} \in \htmlClass{sdt-0000000026}{S}} \exp\left\{ - \frac{ \htmlClass{sdt-0000000100}{E}(\htmlClass{sdt-0000000091}{\mathbf{s}}) }{ \htmlClass{sdt-0000000029}{T} } \right\} d\htmlClass{sdt-0000000091}{\mathbf{s}}= \htmlClass{sdt-0000000077}{Z} \]
as required.
Equivalently, the value of \( \htmlClass{sdt-0000000077}{Z} \) can be justified directly as the result of integrating the Boltzmann distribution to some value \( \htmlClass{sdt-0000000077}{Z} \not= 1 \). Reasoning backwards from point 6 to point 2, we find each microstate's probability should be scaled by \( \frac{1}{ \htmlClass{sdt-0000000077}{Z} } \) to make the integration equal to 1.
