Why do we ignore higher-order derivatives of entropy with energy in deriving the Boltzmann distribution?

Question

I am taking my first course in statistical mechanics, one point that I don't really get is the justification for ignoring higher-order derivatives of entropy w.r.t energy.

We began the course by postulating the microcanonical distribution using the principle of equal a priori probabilities for microstates of an isolated system at equilibrium and then arrived at the canonical distribution by recognizing,

\begin{equation} P_r \propto \mathcal{N_{env}}(E_{total} - E_r) \end{equation}

Where,
$r\, :$ microstate of the system interacting with an environment at temperature T (only allowing exchange of energy).
$E_{r}\, :$ Energy associated with the microstate of the system.
$\mathcal{N_{env}}(E_{total} - E_r) :$ Number of microstates of the environment that satisfy the constraint $E_{total} = E_{r} + E_{env}$

Next, we used $S_{env} =k_{B} \ln (\mathcal{N_{env}}(E_{env}) ) \implies \mathcal{N_{env}}(E_{env}) = e^{\frac{S_{env}(E)}{k_B}}$ and Taylor expanded $S_{env}$ about $E_{total}$ to get,

\begin{equation} \mathcal{N_{env}}(E_{env}) \approx \,e^{\frac{1}{k_B} \left[S_{env}(E_{total}) - \frac{\partial S_{env}}{\partial E_{env}}E_r \right]} \end{equation}

Recognizing $\frac{\partial S_{env}}{\partial E_{env}} = \frac{1}{T}$ we get,

\begin{equation} P_r \,\propto \,e^{-\frac{E_r}{k_BT}}. \end{equation}

The justification given for neglecting higher-order terms say for example $\frac{\partial^2S_{env}}{\partial E_{env}^2} E_r^2$ was that since these are all extensive quantities the scale with number of particles i.e.

\begin{equation} \frac{\partial^2S_{env}}{\partial E_{env}^2} E_r^2 \rightarrow \frac{\partial^2(N_{env}S_{env})}{\partial (N_{env}E_{env})^2} (N_rE_r)^2 = \frac{\partial^2S_{env}}{\partial E_{env}^2} E_r^2 \left( \frac{N_r}{N_{env}}\right). \end{equation} And since $N_r << N_{env}$ this is valid, however, I'm having a hard time convincing myself that this is true since we are dealing with derivatives and we do encounter discontinuities in derivatives quite often in this subject. I faced pretty much the same issue with the Gibbs distribution too.

I'd be grateful for any insights on why this works, and some alternate approach apart from the above-mentioned scaling. I am new to this subject so I'd be fine with a non-rigorous explanation too.

Answer 1

I've probably defeated the purpose of asking a question by answering it myself after a while. Here's something I just came up with that seems like an alternative approach to "scaling" (open to comments that may point out any errors in the reasoning).

\begin{equation} S_{env}(E_{tot}-E_{r}) = S_{env}(E_{tot}) - \frac{\partial S_{env}}{\partial E_{env}}E_r + \frac{1}{2} \frac{\partial^2 S_{env}}{\partial E_{env}^2} E_r^2 \,\,- \,\,... \end{equation}

Now $\frac{1}{T} = \frac{\partial S_{env}}{\partial E_{env}}$ (by definition).

\begin{equation} \frac{\partial^2 S_{env}}{\partial E_{env}^2} = \frac{\partial}{\partial E_{env}} \left( \frac{\partial S_{env}}{\partial E_{env}}\right) \end{equation}

\begin{equation} \;\;\;\;\;\;= \frac{\partial}{\partial E_{env}} \left( \frac{1}{T}\right) \end{equation}

\begin{equation} \;\;= \frac{-1}{T^2}\frac{\partial \,T}{\partial E_{env}} \end{equation}

The environment at temperature T is essentially a thermal reservoir. This is because we assume that the temperature of the environment is fixed even if energy is gained from the system. So the energy required to increase the temperature (heat capacity) is practically infinite.

\begin{equation} \frac{\partial E_{env}}{\partial T} \rightarrow \infty \implies \frac{\partial \;T}{\partial E_{env}} \rightarrow 0 . \end{equation}

Hence, $\frac{\partial^2 S_{env}}{\partial E_{env}^2} = \frac{-1}{T^2}\frac{\partial \,T}{\partial E_{env}} \rightarrow 0$, and so do all higher derivatives.