From 6ce0bb9a8f9fd7d169cbb414a9537d68c5290aae Mon Sep 17 00:00:00 2001 From: Prefetch Date: Fri, 14 Oct 2022 23:25:28 +0200 Subject: Initial commit after migration from Hugo --- source/know/concept/canonical-ensemble/index.md | 239 ++++++++++++++++++++++++ 1 file changed, 239 insertions(+) create mode 100644 source/know/concept/canonical-ensemble/index.md (limited to 'source/know/concept/canonical-ensemble/index.md') diff --git a/source/know/concept/canonical-ensemble/index.md b/source/know/concept/canonical-ensemble/index.md new file mode 100644 index 0000000..aca2a33 --- /dev/null +++ b/source/know/concept/canonical-ensemble/index.md @@ -0,0 +1,239 @@ +--- +title: "Canonical ensemble" +date: 2021-07-10 +categories: +- Physics +- Thermodynamics +- Thermodynamic ensembles +layout: "concept" +--- + +The **canonical ensemble** or **NVT ensemble** builds on +the [microcanonical ensemble](/know/concept/microcanonical-ensemble/), +by allowing the system to exchange energy with a very large heat bath, +such that its temperature $T$ remains constant, +but internal energy $U$ does not. +The conserved state functions are +the temperature $T$, the volume $V$, and the particle count $N$. + +We refer to the system of interest as $A$, and the heat bath as $B$. +The combination $A\!+\!B$ forms a microcanonical ensemble, +i.e. it has a fixed total energy $U$, +and eventually reaches an equilibrium +with a uniform temperature $T$ in both $A$ and $B$. + +Assuming that this equilibrium has been reached, +we want to know which microstates $A$ prefers in that case. +Specifically, if $A$ has energy $U_A$, and $B$ has $U_B$, +which $U_A$ does $A$ prefer? + +Let $c_B(U_B)$ be the number of $B$-microstates with energy $U_B$. +Then the probability that $A$ is in a specific microstate $s_A$ is as follows, +where $U_A(s_A)$ is the resulting energy: + +$$\begin{aligned} + p(s_A) + = \frac{c_B(U - U_A(s_A))}{D} + \qquad \quad + D \equiv \sum_{s_A} c_B(U - U_A(s_A)) +\end{aligned}$$ + +In other words, we choose an $s_A$, +and count the number $c_B$ of compatible $B$-microstates. + +Since the heat bath is large, let us assume that $U_B \gg U_A$. +We thus approximate $\ln{p(s_A)}$ by +Taylor-expanding $\ln{c_B(U_B)}$ around $U_B = U$: + +$$\begin{aligned} + \ln{p(s_A)} + &= -\ln{D} + \ln\!\big(c_B(U - U_A(s_A))\big) + \\ + &\approx - \ln{D} + \ln{c_B(U)} - \bigg( \dv{(\ln{c_B})}{U_B} \bigg) \: U_A(s_A) +\end{aligned}$$ + +Here, we use the definition of entropy $S_B \equiv k \ln{c_B}$, +and that its $U_B$-derivative is $1/T$: + +$$\begin{aligned} + \ln{p(s_A)} + &\approx - \ln{D} + \ln{c_B(U)} - \frac{U_A(s_A)}{k} \Big( \pdv{S_B}{U_B} \Big) + \\ + &\approx - \ln{D} + \ln{c_B(U)} - \frac{U_A(s_A)}{k T} +\end{aligned}$$ + +We now define the **partition function** or **Zustandssumme** $Z$ as follows, +which will act as a normalization factor for the probability: + +$$\begin{aligned} + \boxed{ + Z + \equiv \sum_{s_A}^{} \exp(- \beta U_A(s_A)) + } + = \frac{D}{c_B(U)} +\end{aligned}$$ + +Where $\beta \equiv 1/ (k T)$. +The probability of finding $A$ in a microstate $s_A$ is thus given by: + +$$\begin{aligned} + \boxed{ + p(s_A) = \frac{1}{Z} \exp(- \beta U_A(s_A)) + } +\end{aligned}$$ + +This is the **Boltzmann distribution**, +which, as it turns out, maximizes the entropy $S_A$ +for a fixed value of the average energy $\Expval{U_A}$, +i.e. a fixed $T$ and set of microstates $s_A$. + +Because $A\!+\!B$ is a microcanonical ensemble, +we know that its [thermodynamic potential](/know/concept/thermodynamic-potential/) +is the entropy $S$. +But what about the canonical ensemble, just $A$? + +The solution is a bit backwards. +Note that the partition function $Z$ is not a constant; +it depends on $T$ (via $\beta$), $V$ and $N$ (via $s_A$). +Using the same logic as for the microcanonical ensemble, +we define "equilibrium" as the set of microstates $s_A$ +that $A$ is most likely to occupy, +which must be the set (as a function of $T,V,N$) that maximizes $Z$. + +However, $T$, $V$ and $N$ are fixed, +so how can we maximize $Z$? +Well, as it turns out, +the Boltzmann distribution has already done it for us! +We will return to this point later. + +Still, $Z$ does not have a clear physical interpretation. +To find one, we start by showing that the ensemble averages +of the energy $U_A$, pressure $P_A$ and chemical potential $\mu_A$ +can be calculated by differentiating $Z$. +As preparation, note that: + +$$\begin{aligned} + \pdv{Z}{\beta} = - \sum_{s_A} U_A \exp(- \beta U_A) +\end{aligned}$$ + +With this, we can find the ensemble averages +$\Expval{U_A}$, $\Expval{P_A}$ and $\Expval{\mu_A}$ of the system: + +$$\begin{aligned} + \Expval{U_A} + &= \sum_{s_A} p(s_A) \: U_A + = \frac{1}{Z} \sum_{s_A} U_A \exp(- \beta U_A) + = - \frac{1}{Z} \pdv{Z}{\beta} + \\ + \Expval{P_A} + &= - \sum_{s_A} p(s_A) \pdv{U_A}{V} + = - \frac{1}{Z} \sum_{s_A} \exp(- \beta U_A) \pdv{U_A}{V} + \\ + &= \frac{1}{Z \beta} \pdv{}{V}\sum_{s_A} \exp(- \beta U_A) + = \frac{1}{Z \beta} \pdv{Z}{V} + \\ + \Expval{\mu_A} + &= \sum_{s_A} p(s_A) \pdv{U_A}{N} + = \frac{1}{Z} \sum_{s_A} \exp(- \beta U_A) \pdv{U_A}{N} + \\ + &= - \frac{1}{Z \beta} \pdv{}{N}\sum_{s_A} \exp(- \beta U_A) + = - \frac{1}{Z \beta} \pdv{Z}{N} +\end{aligned}$$ + +It will turn out more convenient to use derivatives of $\ln{Z}$ instead, +in which case: + +$$\begin{aligned} + \Expval{U_A} + = - \pdv{\ln{Z}}{\beta} + \qquad \quad + \Expval{P_A} + = \frac{1}{\beta} \pdv{\ln{Z}}{V} + \qquad \quad + \Expval{\mu_A} + = - \frac{1}{\beta} \pdv{\ln{Z}}{N} +\end{aligned}$$ + +Now, to find a physical interpretation for $Z$. +Consider the quantity $F$, in units of energy, +whose minimum corresponds to a maximum of $Z$: + +$$\begin{aligned} + F \equiv - k T \ln{Z} +\end{aligned}$$ + +We rearrange the equation to $\beta F = - \ln{Z}$ and take its differential element: + +$$\begin{aligned} + \dd{(\beta F)} + = - \dd{(\ln{Z})} + &= - \pdv{\ln{Z}}{\beta} \dd{\beta} - \pdv{\ln{Z}}{V} \dd{V} - \pdv{\ln{Z}}{N} \dd{N} + \\ + &= \Expval{U_A} \dd{\beta} - \beta \Expval{P_A} \dd{V} + \beta \Expval{\mu_A} \dd{N} + \\ + &= \Expval{U_A} \dd{\beta} + \beta \dd{\Expval{U_A}} - \beta \dd{\Expval{U_A}} - \beta \Expval{P_A} \dd{V} + \beta \Expval{\mu_A} \dd{N} + \\ + &= \dd{(\beta \Expval{U_A})} - \beta \: \big( \dd{\Expval{U_A}} + \Expval{P_A} \dd{V} - \Expval{\mu_A} \dd{N} \big) +\end{aligned}$$ + +Rearranging and substituting +the [fundamental thermodynamic relation](/know/concept/fundamental-thermodynamic-relation/) +then gives: + +$$\begin{aligned} + \dd{(\beta F - \beta \Expval{U_A})} + &= - \beta \: \big( \dd{\Expval{U_A}} + \Expval{P_A} \dd{V} - \Expval{\mu_A} \dd{N} \big) + = - \beta T \dd{S_A} +\end{aligned}$$ + +We integrate this and ignore the integration constant, +leading us to the desired result: + +$$\begin{aligned} + - \beta T S_A + &= \beta F - \beta \Expval{U_A} + \quad \implies \quad + F = \Expval{U_A} - T S_A +\end{aligned}$$ + +As was already suggested by our notation, +$F$ turns out to be the **Helmholtz free energy**: + +$$\begin{aligned} + \boxed{ + F + \equiv - k T \ln{Z} + = \Expval{U_A} - T S_A + } +\end{aligned}$$ + +We can therefore reinterpret +the partition function $Z$ and the Boltzmann distribution $p(s_A)$ +in the following "more physical" way: + +$$\begin{aligned} + Z + = \exp(- \beta F) + \qquad \quad + p(s_A) + = \exp\!\Big( \beta \big( F \!-\! U_A(s_A) \big) \Big) +\end{aligned}$$ + +Finally, by rearranging the expressions for $F$, +we find the entropy $S_A$ to be: + +$$\begin{aligned} + S_A + = k \ln{Z} + \frac{\Expval{U_A}}{T} +\end{aligned}$$ + +This is why $Z$ is already maximized: +the Boltzmann distribution maximizes $S_A$ for fixed values of $T$ and $\Expval{U_A}$, +leaving $Z$ as the only "variable". + + + +## References +1. H. Gould, J. Tobochnik, + *Statistical and thermal physics*, 2nd edition, + Princeton. -- cgit v1.2.3