# I Deriving the thermodynamic beta from Lagrange Multipliers

Tags:
1. Aug 22, 2016

### McLaren Rulez

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

\Omega = \frac{N!}{\sum_{i}^r n_{i}!}

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

\ln\Omega = N\ln N - N - \sum_{i}^{r}n_i\ln n_i - n_i

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?

2. Aug 23, 2016

### vanhees71

Of course, it's easier to flip the sign of $\beta$ from the very beginning, because then you have $\beta>0$ for the usual case where the Hamiltonian of the system is bounded from below but can become infinite (e.g., for an ideal gas you have $H=\vec{p}^2/(2m) \geq 0$. Then you have
$$Z(\beta,\alpha)=\sum_i \exp(-\beta E_i + \alpha n_i).$$
Now to get the interpretation of $\beta$ and $\alpha$ in terms of the thermodynamical quantities note that
$$U=\langle E \rangle=-\partial_{\beta} \ln Z, \quad \mathcal{N}=\langle N \rangle=\partial_{\alpha} \ln Z.$$
Further the probability distribution is
$$P_i=\frac{1}{Z} \exp(-\beta E_i+\alpha \mathcal{N}).$$
This implies that the entropy is given by
$$S=-k_B \sum_i P_i \ln P_i \rangle=k_B (\ln Z+\beta U-\alpha \mathcal{N}).$$
With the above relations you get
$$\frac{1}{k_B} \mathrm{d} S = \beta \mathrm{d} U - \alpha \mathrm{d} \mathcal{N}.$$
Then comparing this to the 1st Law
$$\mathrm{d} U=T \mathrm{d} S +\mu \mathrm{d} \mathcal{N}$$
you'll get
$$\beta=\frac{1}{k_B T}, \quad \alpha=\frac{\mu}{k_B T}.$$