2003 Spring Statistical Mechanics

# 1. [35 points total] Entropy of an ideal gas

Throughout this question, clearly state and justify any assumptions or approximations you make. NOTE: Parts A and B are independent.

## A. [15 points]

Problem:

Two samples of an ideal gas (X and Y) are each confined in upright cylinders. Both are initially in equilibrium and are at the same pressure ($P_0$) and temperature ($T_0=300\text{K}$). Each occupies a volume $V_0$. The cylinder containing sample X is sealed at the top by piston of mass M. The piston can move without friction, but no gas can enter of leave the cylinder. The cylinder containing sample Y has a fixed volume.

Each cylinder is placed in thermal contact with a reservoir at $T_{\text{res}}=400\text{K}$. The two samples are allowed t o come to equilibrium with the reservoir. (The system of cylinders and reservoir is isolated from the rest of the universe.)

Greg’s Solution:

# 2. [65 points] An ideal paramagnet

A 2-D system consists of N localized non-interacting spin 1/2 particles in a magnetic field $\textbf{B}=B\hat{\textbf{e}}_z$. The energy associated with each magnetic moment $\textbf{\mu}$ is $\epsilon=-\textbf{\mu}\cdot\textbf{B}$. The system is quantum mechanical — i.e., the z-component of each magnetic moment is given by $\mu_z=s_zg\mu_B$, where g is the g-factor, $\mu_B$ is the Bohr magneton, and $s_z$ can take the values $\pm\frac{1}{2}$.

## A. [5 points] To what value will the total energy of the system tend…

### i. as $T\to 0\text{K}$? Explain.

As the temperature goes to zero, the system will go its lowest energy state, in which the particle magnetic moments are all aligned with the external field. Thus, since there are N particles, the energy of the system in this state will be

(1)
\begin{align} E = -\frac{N g\mu_B B}{2}. \end{align}

### ii. as $T\to \infty$? Explain.

As the temperature becomes infinite, each particle rapidly flips between the two spin states and so on average they have no net magnetic moment; thus the total energy of the system becomes zero.

## B. [15 points] Explain which of the following proposed graphs of entropy as a function of temperature (for some arbitrary value of B) could possible be correct. Then explain how the graph would change, at all, if (i) the field had a greater magnitude and (ii) the field were zero.

The entropy must increase with the temperature, so that immediately rules out I. Furthermore, it asymptotically converges to the logarithm of the number of allowable states in the system, since when the system is at infinite temperature all states are equally likely.

If we allowed the particles to have momentum and position degrees of freedom (as any real system of particles would), then the system would have an infinite number of momentum states available to it and so the entropy would go to infinity. However, this would be inconsistent with what we are supposed to prove in part D, so I assume that the particles only have a spin degree of freedom. In that case, there are a finite number of states, and so the entropy will converge to a finite number, and so II is the correct graph of the entropy.

If the field had greater magnitude then the graph would be stretched out — i.e., the point where the convexity of the graph changes would move to the right — since it takes more energy to make a particle spin-up and so it takes a higher temperature to make the system have no preference for spin-up over spin-down.

## C. [15 points] Derive an expression for the magnetization of the system as a function of temperature and applied field.

We observe that for a small change in the field, the energy — and therefore the free energy, by the law of small increments — changes by

(2)
\begin{align} \delta F = -M \delta B, \end{align}

where M is the magnetization. Thus, we can find the magnetization by computing the partial derivative

(3)
\begin{align} -\frac{\partial F}{\partial B} = M \end{align}

First, though, we need to find the free energy, which means we need the partition function. Since there are N particles which each have energy eigenvalues $\pm \frac{g\mu_B B}{2}$, therefore the partition function takes the form

(4)
\begin{align} Z = \left(e^{+g\mu_B B/2\tau}+e^{-g\mu_B B/2\tau}\right)^N \end{align}

and so the free energy is given by

(5)
\begin{align} F = -\tau ln Z = -N \tau \ln\left(e^{+g\mu_B B/2\tau}+e^{-g\mu_B B/2\tau}\right) \end{align}

and thus the magnetization is given by

(6)
\begin{align} -\frac{\partial F}{\partial B} = +\frac{N g\mu_B B}{2} \frac{e^{+g\mu_B B/2\tau}-e^{-g\mu_B B/2\tau}}{e^{+g\mu_B B/2\tau}+e^{-g\mu_B B/2\tau}} = +\frac{N g\mu_B B}{2}\tanh\left(\frac{g\mu_B B}{2}\right) \end{align}

Observe that this has the correct limits: as $T\to 0$, $M\to +\frac{N g\mu_B B}{2}$ (i.e. completely aligned with the external field), and as $T\to\infty$, $M\to 0$ (i.e. demagnetized).

# D. [15 points]

Problem:

Show that the entropy of the system is given by:

(7)
\begin{align} S = Nk\left[\ln \left\{ 2\cosh \left(\frac{\beta g\mu_B B}{2}\right)\right\}-\frac{\beta g \mu_B B}{2}\tanh\left(\frac{\beta g \mu_B B}{2}\right)\right] \end{align}

where $\beta:=1/kT$ and k is the Boltzmann factor.

Greg’s Solution:

Life will be easier if I define the constant $\eta := \frac{g \mu_B B}{2}$ so that

(8)
\begin{align} F = -N\tau \ln \left(e^{+\eta/\tau}+e^{-\eta/\tau}\right) \equiv -N\tau \ln \left\{2\cosh \left(\frac{\eta}{\tau}\right)\right} \end{align}

The entropy is just the negative of the partial derivative of the free energy with respect to temperature,

(9)
\begin{aligned} S &= -\frac{\partial F}{\partial \tau} \\ &= N\left[\ln \left\{2\cosh \left(\frac{\eta}{\tau}\right)\right} + \tau \frac{2\sinh \left(\frac{\eta}{\tau}\right)}{2\cosh \left(\frac{\eta}{\tau}\right)}\frac{-\eta}{\tau^2}\right] \\ &= N\left[\ln \left\{2\cosh x \right)\right} + x\tanh(x)\right], \\ \end{aligned}

where

(10)
\begin{align} x = \frac{\beta g\mu_B B}{2} \end{align}

and

(11)
\begin{align} \beta = \frac{1}{\tau}. \end{align}

Of course, if you had insisted in doing everything with $T$ (Kelvin) as your temperature variable rather than $\tau$ (Joules/energy), then you’d have to use

(12)
\begin{align} \tilde S(T) = k S(kT). \end{align}

## E. [5 points] Describe in a few sentences (using equations if necessary) how one could determine experimentally how the entropy of a system depends on temperature.

Greg’s Solution:

The thermodynamic equation is

(13)
\begin{align} dE = T dS - P dV + \mu dN \end{align}

If we take the derivative with respect to temperature, holding volume and particle number constant, then we obtain

(14)
\begin{align} \frac{1}{T} \frac{\partial E}{\partial T} = \frac{\partial S}{\partial T} \end{align}

Thus, we see that the total entropy can computed by integrating,

(15)
\begin{align} S(T) = \int_0^T \frac{1}{T} \frac{\partial E}{\partial T} \, dT \end{align}

So experimentally, what we could do is set up a system which has a fixed number of particles and a fixed volume, and then take measurements of how the heat capacity under constant volume changes from zero temperature (as cold as you can get it) to the desired temperature. Then we integrate our heat capacity function (or, more accurately, compute a sum which approximates the integral above since its unlikely that we’ve taken uncountably many measurements) to find the entropy.

## F. [10 points] Consider an instant at which the magnetization is near its maximum value. Imagine that at this instant, the magnetic field is reversed in direction over a time interval much smaller than the relaxation time of the magnetic moments. Describe in a few sentences (using equations if necessary) the relationship between the energy and the entropy during this time interval. What does this relationship imply about the sign of the temperature?

Greg’s Solution:

Restating the question, we have that our system was near the ground state so that it was almost completely magnetized in the same direction as the external field, and thus had the least possible energy. When the field flips, it suddenly has the greatest possible energy. The energy can only go down as some of the particles start to flip to match the new field. But as they start to flip, the entropy increases since now not all particle have the same spin. Thus, we have that $dE/dS < 0$, and so we have a negative temperature.

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