Physics Friday 82

One might recall my previous post where I considered a system in contact with a thermal reservoir with which it could exchange energy; from considering the entropy of the system+reservoir combination as a function of the system’s energy to derive the Boltzmann factor. This method of analysing the statistical system, by considering the different probabilities of different states of a system when it is in contact with a thermal reservoir with which it can exchange energy is known as statistical mechanics in “canonical formalism” or in “Helmholtz representation.”
In particular, we noted that for system energy E₁ and total energy E, we could expand the entropy of the system in Taylor series about E and solve for the number of reservoir microstates N₂:
$N_2=e^{\frac{S_2(E)}{k}}e^{-\frac{E_1}{kT}}$
or in terms of the thermodynamic beta $\beta\equiv\frac{1}{kT}$ ,
$N_2={e}^{\frac{S_2(E)}{k}}e^{\beta{E_1}}$
where S₂(E) is the reservoir entropy when the total energy is entirely in the reservoir. The latter exponential factor, the only one which is a function of our system’s energy, is the Boltzmann factor.
We noted that the probability of the system having energy E₁ is proportional to N₂. In particular, if we let N be the total number of microstates of the system+reservoir at total energy E, then the probability that the system is in a state of energy E₁ is
$p(E_1)=\frac{N_2}{N}$
and from the Boltzmann definition of entropy, the total entropy is
$S(E)=k\ln{N}$ ,
so $N=e^{\frac{S(E)}{k}}$
Let U be the average value of the energy of our system. Then S(E)=S₁(U)+S₂(E-U), and expanding our reservoir entropy in Taylor series about E-U (our equilibrium point) instead of E,
$S_2(E-E_1)=S_2(E-U)+\frac{U-E_1}{T}$
so that $N_2=e^{\frac{S_2(E-U)}{k}}e^{\frac{U-E_1}{kT}}$
And so
$p(E_1)=\frac{N_2}{N}=\frac{e^{\frac{S_2(E-U)}{k}}e^{\frac{U-E_1}{kT}}}{e^{\frac{S(E)}{k}}}$
and due to the additivity of the entropy of the system and reservoir:
$S(E)=S_1(U)+S_2(E-U)$ ,
so the probability becomes
$\begin{eqnarray}p(E_1)&=&\frac{e^{\frac{S_2(E-U)}{k}}e^{\frac{U-E_1}{kT}}}{e^{\frac{S(E)}{k}}}\\&=&\frac{e^{\frac{S_2(E-U)}{k}}e^{\frac{U-E_1}{kT}}}{e^{\frac{S_1(U)+S_2(E-U)}{k}}}\\&=&e^{-\frac{S_1(U)}{k}}e^{\frac{U-E_1}{kT}}\\&=&e^{\frac{U-TS_1(U)}{kT}}e^{\frac{-E_1}{kT}}\\p(E_1)&=&e^{\beta{A}}e^{-\beta{E_1}}\end{eqnarray}$ ,
where $A\equiv{U-TS}$ is the Helmholtz free energy of our system, and $e^{-\beta{E_1}}$ is, again, the Boltzmann factor.
We do not yet know the Helmholtz free energy, but we can compute it from the above. We note that the exponential it appears in plays the role of a normalization factor: Summing the probability over all allowed states of our system, where in state s the system has energy E_s, we get:
$\sum_{s}p(E_s)=e^{\beta{A}}\sum_{s}e^{-\beta{E_1}}=1$
so $e^{-\beta{A}}=\sum_{s}e^{-\beta{E_1}}$
we denote this sum, called the “canonical partition function,” by Z. If we consider the sum over energy levels E_i, then we have
$z=\sum_{i}g_i{e^{-\beta{E_i}}}$ ,
where g_i is the degeneracy of the energy level E_i.
[In classical mechanics, the parameters are of a particle are continuous, and we can't actually express the partition function as a sum, having to replace it with an integral with a "coarse graining" procedure (see here). In quantum mechanics, however, energy levels are discrete, and the above summation makes sense.]

Now, there are a number of thermodynamic variables we can extract from Z. First, as $e^{-\beta{A}}=Z$ , we see $A=-\frac{\ln{Z}}{\beta}=-kT\ln{Z}$ .
Next, consider the average energy U. This is the expectation value of the energy, and will be the internal energy of our system (thus the choice to name it U):
$\begin{eqnarray}U&=&\sum_{s}E_sp(E_s)\\&=&\frac{\sum_{s}E_se^{-\beta{E_s}}}{\sum_{s}e^{-\beta{E_s}}}\\&=&\frac{\sum_{s}E_se^{-\beta{E_s}}}{Z}\end{eqnarray}$
Now, note that
$\frac{\partial}{\partial\beta}Z=\frac{\partial}{\partial\beta}\sum_{s}e^{-\beta{E_s}}=-\sum_{s}E_se^{-\beta{E_s}}$ , so that
$\begin{eqnarray}U&=&\frac{-\frac{\partial{Z}}{\partial\beta}}{Z}\\U&=&-\frac{\partial}{\partial\beta}(\ln{Z})\end{eqnarray}$ ,
or, using the chain rule to rewrite β in terms of T, we have
$U=kT^2\frac{\partial}{\partial{T}}(\ln{Z})$ .
Consider the second derivative of $\ln{Z}$ with respect to β:
$\begin{eqnarray}\frac{\partial^2}{\partial\beta^2}(\ln{Z})&=&\frac{\partial}{\partial\beta}\left(-\frac{\sum_{s}E_se^{-\beta{E_s}}}{Z}\right)\\&=&\left(\frac{\sum_{s}E_s^2e^{-\beta{E_s}}}{Z}\right)-\left(\frac{\sum_{s}E_se^{-\beta{E_s}}}{Z}\right)^2\\&=&\langle{E^2}\rangle-\langle{E}\rangle^2\\&=&\langle(\Delta{E})^2\rangle\end{eqnarray}$ ,
the variance of the energy.
Now, the heat capacity at constant volume is
$\begin{eqnarray}C_v&=&\frac{\partial{U}}{\partial{T}}\\&=&-\frac{1}{kT^2}\frac{\partial{U}}{\partial\beta}\\&=&\frac{1}{kT^2}\frac{\partial^2}{\partial\beta^2}(\ln{Z})\\C_v&=&\frac{\langle(\Delta{E})^2\rangle}{kT^2}\\C_v&=&k\beta^2\frac{\partial^2}{\partial\beta^2}(\ln{Z})\end{eqnarray}$ .
Using the latter expression, we can find the dimensionless specific heat capacity (at constant volume):
$\begin{eqnarray}\hat{c}_V&=&\frac{C_V}{Nk}\\&=&\frac{1}{Nk}k\beta^2\frac{\partial^2}{\partial\beta^2}(\ln{Z})\\&=&\frac{\beta^2}{N}\frac{\partial^2}{\partial\beta^2}(\ln{Z})\end{eqnarray}$ .

Since $A=U-TS$ ,
$\begin{eqnarray}S&=&-\frac{\partial{A}}{\partial{T}}\\&=&\frac{1}{kT^2}\frac{\partial{A}}{\partial\beta}\\&=&-\frac{1}{kT^2}\frac{\partial}{\partial\beta}\left(\frac{\ln{Z}}{\beta}\right)\\&=&k\beta^2\left(\frac{U}{\beta}+\frac{\ln{Z}}{\beta^2}\right)\\&=&k\left(\ln{Z}+\beta{U}\right)\end{eqnarray}$

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Tags: Boltzmann Factor, Canonical Ensemble, Canonical Partition Function, Entropy, Friday Physics, Helmholtz Free Energy, Partition Function, physics, Statistical Mechanics

This entry was posted on July 24, 2009 at 12:08 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

2 Responses to “Physics Friday 82”

Physics Friday 83 « Twisted One 151’s Weblog Says:
July 31, 2009 at 1:09 am | Reply
[...] is done (let εa=0 and εb=ε). Instead, this time let’s compute the canonical partition function . Now, the energy is En=nε, and the degeneracy is . Thus: Note, however, by the binomial [...]
Physics Friday 84 « Twisted One 151’s Weblog Says:
August 7, 2009 at 2:03 am | Reply
[...] of the reservoir and total entropies (as functions of energy and particle number): . As with our canonical formalism, we can expand the entropy as a Taylor series in energy about Et-U and in particle number about the [...]

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