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Diffstat (limited to 'source/know/concept/canonical-ensemble')
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diff --git a/source/know/concept/canonical-ensemble/index.md b/source/know/concept/canonical-ensemble/index.md index dd7fe90..8a96e91 100644 --- a/source/know/concept/canonical-ensemble/index.md +++ b/source/know/concept/canonical-ensemble/index.md @@ -12,25 +12,25 @@ 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. +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 $N$. +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 $U$, +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 $B$. +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? +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: +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: $$\begin{aligned} p(s_A) @@ -39,12 +39,12 @@ $$\begin{aligned} D \equiv \sum_{s_A} c_B(U - U_A(s_A)) \end{aligned}$$ -In other words, we choose an $s_A$, -and count the number $c_B$ of compatible $B$-microstates. +In other words, we choose an $$s_A$$, +and count the number $$c_B$$ of compatible $$B$$-microstates. -Since the heat bath is large, let us assume that $U_B \gg U_A$. -We thus approximate $\ln{p(s_A)}$ by -Taylor-expanding $\ln{c_B(U_B)}$ around $U_B = U$: +Since the heat bath is large, let us assume that $$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)} @@ -53,8 +53,8 @@ $$\begin{aligned} &\approx - \ln{D} + \ln{c_B(U)} - \bigg( \dv{(\ln{c_B})}{U_B} \bigg) \: U_A(s_A) \end{aligned}$$ -Here, we use the definition of entropy $S_B \equiv k \ln{c_B}$, -and that its $U_B$-derivative is $1/T$: +Here, we use the definition of entropy $$S_B \equiv k \ln{c_B}$$, +and that its $$U_B$$-derivative is $$1/T$$: $$\begin{aligned} \ln{p(s_A)} @@ -63,7 +63,7 @@ $$\begin{aligned} &\approx - \ln{D} + \ln{c_B(U)} - \frac{U_A(s_A)}{k T} \end{aligned}$$ -We now define the **partition function** or **Zustandssumme** $Z$ as follows, +We now define the **partition function** or **Zustandssumme** $$Z$$ as follows, which will act as a normalization factor for the probability: $$\begin{aligned} @@ -74,8 +74,8 @@ $$\begin{aligned} = \frac{D}{c_B(U)} \end{aligned}$$ -Where $\beta \equiv 1/ (k T)$. -The probability of finding $A$ in a microstate $s_A$ is thus given by: +Where $$\beta \equiv 1/ (k T)$$. +The probability of finding $$A$$ in a microstate $$s_A$$ is thus given by: $$\begin{aligned} \boxed{ @@ -84,33 +84,33 @@ $$\begin{aligned} \end{aligned}$$ This is the **Boltzmann distribution**, -which, as it turns out, maximizes the entropy $S_A$ -for a fixed value of the average energy $\Expval{U_A}$, -i.e. a fixed $T$ and set of microstates $s_A$. +which, as it turns out, maximizes the entropy $$S_A$$ +for a fixed value of the average energy $$\Expval{U_A}$$, +i.e. a fixed $$T$$ and set of microstates $$s_A$$. -Because $A\!+\!B$ is a microcanonical ensemble, +Because $$A\!+\!B$$ is a microcanonical ensemble, we know that its [thermodynamic potential](/know/concept/thermodynamic-potential/) -is the entropy $S$. -But what about the canonical ensemble, just $A$? +is the entropy $$S$$. +But what about the canonical ensemble, just $$A$$? The solution is a bit backwards. -Note that the partition function $Z$ is not a constant; -it depends on $T$ (via $\beta$), $V$ and $N$ (via $s_A$). +Note that the partition function $$Z$$ is not a constant; +it depends on $$T$$ (via $$\beta$$), $$V$$ and $$N$$ (via $$s_A$$). Using the same logic as for the microcanonical ensemble, -we define "equilibrium" as the set of microstates $s_A$ -that $A$ is most likely to occupy, -which must be the set (as a function of $T,V,N$) that maximizes $Z$. +we define "equilibrium" as the set of microstates $$s_A$$ +that $$A$$ is most likely to occupy, +which must be the set (as a function of $$T,V,N$$) that maximizes $$Z$$. -However, $T$, $V$ and $N$ are fixed, -so how can we maximize $Z$? +However, $$T$$, $$V$$ and $$N$$ are fixed, +so how can we maximize $$Z$$? Well, as it turns out, the Boltzmann distribution has already done it for us! We will return to this point later. -Still, $Z$ does not have a clear physical interpretation. +Still, $$Z$$ does not have a clear physical interpretation. To find one, we start by showing that the ensemble averages -of the energy $U_A$, pressure $P_A$ and chemical potential $\mu_A$ -can be calculated by differentiating $Z$. +of the energy $$U_A$$, pressure $$P_A$$ and chemical potential $$\mu_A$$ +can be calculated by differentiating $$Z$$. As preparation, note that: $$\begin{aligned} @@ -118,7 +118,7 @@ $$\begin{aligned} \end{aligned}$$ With this, we can find the ensemble averages -$\Expval{U_A}$, $\Expval{P_A}$ and $\Expval{\mu_A}$ of the system: +$$\Expval{U_A}$$, $$\Expval{P_A}$$ and $$\Expval{\mu_A}$$ of the system: $$\begin{aligned} \Expval{U_A} @@ -141,7 +141,7 @@ $$\begin{aligned} = - \frac{1}{Z \beta} \pdv{Z}{N} \end{aligned}$$ -It will turn out more convenient to use derivatives of $\ln{Z}$ instead, +It will turn out more convenient to use derivatives of $$\ln{Z}$$ instead, in which case: $$\begin{aligned} @@ -155,15 +155,15 @@ $$\begin{aligned} = - \frac{1}{\beta} \pdv{\ln{Z}}{N} \end{aligned}$$ -Now, to find a physical interpretation for $Z$. -Consider the quantity $F$, in units of energy, -whose minimum corresponds to a maximum of $Z$: +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: +We rearrange the equation to $$\beta F = - \ln{Z}$$ and take its differential element: $$\begin{aligned} \dd{(\beta F)} @@ -198,7 +198,7 @@ $$\begin{aligned} \end{aligned}$$ As was already suggested by our notation, -$F$ turns out to be the **Helmholtz free energy**: +$$F$$ turns out to be the **Helmholtz free energy**: $$\begin{aligned} \boxed{ @@ -209,7 +209,7 @@ $$\begin{aligned} \end{aligned}$$ We can therefore reinterpret -the partition function $Z$ and the Boltzmann distribution $p(s_A)$ +the partition function $$Z$$ and the Boltzmann distribution $$p(s_A)$$ in the following "more physical" way: $$\begin{aligned} @@ -220,17 +220,17 @@ $$\begin{aligned} = \exp\!\Big( \beta \big( F \!-\! U_A(s_A) \big) \Big) \end{aligned}$$ -Finally, by rearranging the expressions for $F$, -we find the entropy $S_A$ to be: +Finally, by rearranging the expressions for $$F$$, +we find the entropy $$S_A$$ to be: $$\begin{aligned} S_A = k \ln{Z} + \frac{\Expval{U_A}}{T} \end{aligned}$$ -This is why $Z$ is already maximized: -the Boltzmann distribution maximizes $S_A$ for fixed values of $T$ and $\Expval{U_A}$, -leaving $Z$ as the only "variable". +This is why $$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". |