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---
title: "Microcanonical ensemble"
firstLetter: "M"
publishDate: 2021-07-09
categories:
- Physics
- Thermodynamics
- Thermodynamic ensembles
date: 2021-07-08T11:00:59+02:00
draft: false
markup: pandoc
---
# Microcanonical ensemble
The **microcanonical** or **NVE ensemble** is a statistical model
of a theoretical system with constant internal energy $U$,
volume $V$, and particle count $N$.
Consider a box with those properties.
We now put an imaginary rigid wall inside the box,
thus dividing it into two subsystems $A$ and $B$,
which can exchange energy (i.e. heat), but no particles.
At any time, $A$ has energy $U_A$, and $B$ has $U_B$,
so that in total $U = U_A \!+\! U_B$.
The particles in each subsystem are in a certain **microstate** (configuration).
For a given $U$, there is a certain number $c$
of possible whole-box microstates with that energy, given by:
$$\begin{aligned}
c(U)
= \sum_{U_A \le U} c_A(U_A) \: c_B(U - U_A)
\end{aligned}$$
Where $c_A$ and $c_B$ are the number of microstates of
the subsystems at the given energy levels.
The core assumption of the microcanonical ensemble
is that each of these microstates has the same probability $1 / c$.
Consequently, the probability of finding an energy $U_A$ in $A$ is:
$$\begin{aligned}
p_A(U_A)
= \frac{c_A(U_A) \:c_B(U - U_A)}{c(U)}
\end{aligned}$$
If a certain $U_A$ has a higher probability,
then there are more $A$-microstates with that energy,
meaning that $U_A$ is "easier to reach" or "more comfortable" for the system.
Note that $c(U)$ is a constant, because $U$ is given.
After some time, the system will reach equilibrium,
where both $A$ and $B$ have settled into a "comfortable" position.
In other words, the subsystem microstates at equilibrium
must be maxima of their probability distributions $p_A$ and $p_B$.
We only need to look at $p_A$.
Clearly, a maximum of $p_A$ is also a maximum of $\ln p_A$:
$$\begin{aligned}
\ln p_A(U_A)
= \ln{c_A(U_A)} + \ln{c_B(U - U_A)} - \ln{c(U)}
\end{aligned}$$
Here, in the quantity $\ln{c_A}$,
we recognize the definition of
the entropy $S_A \equiv k_B \ln{c_A}$,
where $k_B$ is Boltzmann's constant.
We thus multiply by $k_B$:
$$\begin{aligned}
k_B \ln p_A(U_A)
= S_A(U_A) + S_B(U - U_A) - k_B \ln{c(U)}
\end{aligned}$$
Since entropy is additive over subsystems,
the total is $S = S_A + S_B$.
To reach equilibrium, we are thus
**maximizing the total entropy**,
meaning that $S$ is the [thermodynamic potential](/know/concept/thermodynamic-potential/)
that corresponds to the microcanonical ensemble.
For our example, maximizing gives the following,
more concrete, equilibrium condition:
$$\begin{aligned}
0
= k_B \dv{(\ln{p_A})}{U_A}
= \pdv{S_A}{U_A} + \pdv{S_B}{U_A}
= \pdv{S_A}{U_A} - \pdv{S_B}{U_B}
\end{aligned}$$
By definition, the energy-derivative of the entropy
is the reciprocal temperature $1 / T$.
In other words,
equilibrium is reached when both subsystems
are at the same temperature:
$$\begin{aligned}
\frac{1}{T_A}
= \pdv{S_A}{U_A}
= \pdv{S_B}{U_B}
= \frac{1}{T_B}
\end{aligned}$$
Recall that our partitioning into $A$ and $B$ was arbitrary,
meaning that, in fact, the temperature $T$ must be uniform in the whole box.
We get this specific result because
heat was the only thing that $A$ and $B$ could exchange.
The key point, however,
is that the total entropy $S$ must be maximized.
We also would have reached that conclusion if our imaginary wall
allowed changes in volume $V_A$ and particle count $N_A$.
## References
1. H. Gould, J. Tobochnik,
*Statistical and thermal physics*, 2nd edition,
Princeton.
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