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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.