- #1
McLaren Rulez
- 292
- 3
I'm nearly at the end of this derivation but totally stuck so I'd appreciate a nudge in the right direction
Consider a set of N identical but distinguishable particles in a system of energy E. These particles are to be placed in energy levels ##E_i## for ##i = 1, 2 .. r##. Assume that we have ##n_i## particles in each energy level. The two constraints we impose are that ##\sum_{i}^{r}n_i = N## and ##\sum_{i}^{r}E_i n_i = E##.
The number of microstates in a given macrostate is given by
\begin{equation}
\Omega = \frac{N!}{\sum_{i}^r n_{i}!}
\end{equation}
We want to maximize this and for ease of notation, we work with ##\ln\Omega## and we use Stirling's approximation (##\ln x! = x\ln x - x##) to obtain
\begin{equation}
\ln\Omega = N\ln N - N - \sum_{i}^{r}n_i\ln n_i - n_i
\end{equation}
Maximizing this function subject to the constraints ##\sum_{i}^{r}n_i = N## and ##\sum_{i}^{r}E_i n_i = E## is a classic Lagrange multiplier problem. We represent the undetermined multipliers to be ##\alpha## and ##\beta## for the two constraints and obtain
\begin{align}
\frac{\partial\ln\Omega}{\partial n_i} &= \alpha\frac{\partial n_i}{\partial n_i} + \beta\frac{\partial E_i n_i}{\partial n_i} \\ \nonumber
\ln n_i &= \alpha + \beta E_i \\ \nonumber
\therefore n_i &= e^{\alpha}e^{\beta E_i}
\end{align}
Now, we use the first constraint equation to determine ##\alpha##. We get
\begin{align}
\sum_i^r n_i &= N \\ \nonumber
\sum_i^r e^{\alpha}e^{\beta E_i} &= N \\ \nonumber
e^\alpha &= \frac{N}{\sum_i^re^{\beta E_i}} \\ \nonumber
e^\alpha &= \frac{N}{Z}
\end{align}
We have introduced the partition function, ##Z=\sum_i^re^{\beta E_i}## in the last line. Next, we have the second constraint equation that determines ##\beta##
\begin{align}
\sum_i^r E_i n_i &= E \\ \nonumber
\frac{\sum_{i}^{r} E_i e^{\beta E_i}}{\sum_i^r e^{\beta E_i}} &= E \\ \nonumber
\end{align}
I'm assuming I should somehow connect ##E## with ##T## so let's say ##E=Nk_B T##. Then we have
\begin{align}
\frac{N}{Z}\frac{\partial Z}{\partial\beta} &= E \\
\frac{\partial\ln Z}{\partial\beta} &= k_B T
\end{align}
How do I get to ##\beta = -\frac{1}{k_B T}## here? Notice that this derivation requires an extra minus sign compared to the usual definition of ##\beta## and this should come out naturally too, shouldn't it?
Consider a set of N identical but distinguishable particles in a system of energy E. These particles are to be placed in energy levels ##E_i## for ##i = 1, 2 .. r##. Assume that we have ##n_i## particles in each energy level. The two constraints we impose are that ##\sum_{i}^{r}n_i = N## and ##\sum_{i}^{r}E_i n_i = E##.
The number of microstates in a given macrostate is given by
\begin{equation}
\Omega = \frac{N!}{\sum_{i}^r n_{i}!}
\end{equation}
We want to maximize this and for ease of notation, we work with ##\ln\Omega## and we use Stirling's approximation (##\ln x! = x\ln x - x##) to obtain
\begin{equation}
\ln\Omega = N\ln N - N - \sum_{i}^{r}n_i\ln n_i - n_i
\end{equation}
Maximizing this function subject to the constraints ##\sum_{i}^{r}n_i = N## and ##\sum_{i}^{r}E_i n_i = E## is a classic Lagrange multiplier problem. We represent the undetermined multipliers to be ##\alpha## and ##\beta## for the two constraints and obtain
\begin{align}
\frac{\partial\ln\Omega}{\partial n_i} &= \alpha\frac{\partial n_i}{\partial n_i} + \beta\frac{\partial E_i n_i}{\partial n_i} \\ \nonumber
\ln n_i &= \alpha + \beta E_i \\ \nonumber
\therefore n_i &= e^{\alpha}e^{\beta E_i}
\end{align}
Now, we use the first constraint equation to determine ##\alpha##. We get
\begin{align}
\sum_i^r n_i &= N \\ \nonumber
\sum_i^r e^{\alpha}e^{\beta E_i} &= N \\ \nonumber
e^\alpha &= \frac{N}{\sum_i^re^{\beta E_i}} \\ \nonumber
e^\alpha &= \frac{N}{Z}
\end{align}
We have introduced the partition function, ##Z=\sum_i^re^{\beta E_i}## in the last line. Next, we have the second constraint equation that determines ##\beta##
\begin{align}
\sum_i^r E_i n_i &= E \\ \nonumber
\frac{\sum_{i}^{r} E_i e^{\beta E_i}}{\sum_i^r e^{\beta E_i}} &= E \\ \nonumber
\end{align}
I'm assuming I should somehow connect ##E## with ##T## so let's say ##E=Nk_B T##. Then we have
\begin{align}
\frac{N}{Z}\frac{\partial Z}{\partial\beta} &= E \\
\frac{\partial\ln Z}{\partial\beta} &= k_B T
\end{align}
How do I get to ##\beta = -\frac{1}{k_B T}## here? Notice that this derivation requires an extra minus sign compared to the usual definition of ##\beta## and this should come out naturally too, shouldn't it?