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---
title: "Microcanonical ensemble"
sort_title: "Microcanonical ensemble"
date: 2021-07-09
categories:
- Physics
- Thermodynamics
- Thermodynamic ensembles
layout: "concept"
---
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 numbers of subsystem microstates
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,
so, statistically, for an *ensemble* of many boxes,
we expect that $$U_A$$ is more common.
The maximum of $$p_A$$ will be the most common in the ensemble.
Assuming that we have given the boxes enough time to settle,
we go one step further,
and refer to this maximum as "equilibrium".
In other words, the subsystem microstates at equilibrium
are maxima of $$p_A$$ 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 $$S_A \equiv k \ln{c_A}$$,
where $$k$$ is Boltzmann's constant.
We thus multiply by $$k$$:
$$\begin{aligned}
k \ln p_A(U_A)
= S_A(U_A) + S_B(U - U_A) - k \ln{c(U)}
\end{aligned}$$
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:
$$\begin{aligned}
0
= k \dv{(\ln{p_A})}{U_A}
= \pdv{S_A}{U_A} + \pdv{S_B}{U_A}
= \pdv{S_A}{U_A} - \pdv{S_B}{U_B}
\end{aligned}$$
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 point is that the most likely state of the box
maximizes the total entropy $$S$$.
We also would have reached that conclusion
if our imaginary wall was permeable and flexible,
i.e if it 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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