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Lectures on statistical mechanics.

Midterm November 2022

Bosons

1.

Black Body. Use dimensional analysis to find the energy density of a gas of photons at temperature $T$ (Stefan law). Apply the experimental fact that $E/V$ only depends on $T$ ($E$ and $V$ are the thermodynamic energy and volume of the black body).

Solution

We must determine the function

$$\epsilon = E/V = \epsilon(T)$$

for a noninteracting photon gas. A photon is a quantum ($\hbar$ planck constant) of the electromagnetic field ($c$ light speed); therefore we assume that

$$\epsilon=\epsilon(T)=f(T; \hbar, c).$$

The dimensions are:

$$[\epsilon]=EL^{-3}, \; [\hbar c]=E L, \; [k_BT]= E$$

where $E$ and $L$ are dimensions of energy and length, respectively. Noting that to make a volume we need a factor $(\hbar c)^3$, and compensating for the extra powers of $E$, we obtain that

$$\epsilon=\text{const.} \frac{(k_BT)^4}{(\hbar c)^3}$$

has the correct dimensions ($\text{const.}$ is a number). This is the Stefan law.

2.

Debye phonons. Use dimensional analysis to find the maximum energy $\epsilon_M$ a phonon can take in a solid of $N$ atoms and volume $V$; the sound velocity $c_s$ is isotropic. (Hint: assume that the energy is a function of the density $n=N/V$.) Compute $\epsilon_M$ using the exact density of states. Compare the two results.

Solution

A phonon is the quantum of cristal lattice excitations of energy $\epsilon = \hbar \omega$: its frequency $\omega$ is related to its wavenumber $k$ by the sound velocity $c_s$, $\omega=c_s k$. The Debye cutoff frequency is a function of the density:

$$\omega_M=\epsilon_M/\hbar=\omega_M(n)=f(n; c_s).$$

To determine $f$ we analyze the parameter dimensions:

$$[\omega_M]=T^{-1}, \; [c_s] = LT^{-1}, \; [n] = L^{-3}.$$

From these relations we find that

$$\omega_M = \text{const.} c_s n^{1/3},$$

because $n^{-1/3}$ has dimension of length.

To compute the unknown constant, we use the definition of the cutoff frequency:

$$\text{number of modes} = \text{number of dof} = 3N$$

which states that the number of phonons must not exceed the number of degrees of freedom (dof) $3N$ (for $N$ lattice sites). Therefore,

$$3 V \frac{4\pi}{3} \left( \frac{\omega_M}{2\pi c} \right)^3 = 3N,$$

where the first term is $3$ times (for the longitudinal and transversal phonon polarizations) the volume of a sphere of radius $\omega/2\pi c_s$. We finally obtain,

$$\omega_M = (6\pi^2 n)^{1/3} c_s\,.$$

Rigid rotator

We consider a material formed in first approximation, by independent rotators of inertia moment $I$. We are interested in the low temperature limit and assume a Boltzmann distribution.

Write the hamiltonian of the set of \emph{fixed} rotators.
Use dimensional analysis to determine the low temperature condition. (Hint: compare the temperature with a characteristic energy of a rotator.)
Write the exact expression of the one particle partition function and compute its low temperature expansion. (Remark: take into account the degeneracy of the angular momentum.)
Find the specific heat at low temperature.
Plot the specific heat as a function of the temperature. Discuss the result (compare with the classical expectation.)

Solution

The hamiltonian of a rigid rotator of inertia moment $I$ and angular momentum $L$ is

$$H = \frac{L^2}{2I}\,.$$

The eigenspectum is given by the spherical harmonics $\ket{lm}$ of orbital number $l=0,1,\ldots$ and magnetic number $-l\le m \le l$:

$$H\ket{lm} = \frac{\hbar^2}{2I} l(l+1) \ket{lm}.$$

The energy levels are then given by

$$\epsilon_l = \frac{\hbar^2}{2I} l(l+1).$$

This leads to a characteristic energy of the rotator $\hbar^2/I$, which combined with the temperature defines the nondiemnsional parameter:

$$\theta = \frac{\hbar^2}{Ik_BT}\; ;$$

the low temperature regime corresponds to $\theta\gg 1$.

The one particle partition function $Z_1 = \sum_s \exp(-\epsilon_s/T)$ ($k_B=1$ from now on), where the quantum state here is determined by $s=\{l,m\}$, writes

$$Z_1 = \sum_{l=0}^\infty \sum_{m=-l}^l \E^{-\theta l(l+1)/2}= \sum_{l=0}^\infty (2l+1) \E^{-\hbar^2 l(l+1)/2IT} \,,$$

where we took into account the degeneracy of $\epsilon_l$ (just the sum over $m$). At low temperature each term of the sum is exponentially small (for $l>0$), therfore up the first relevant term we have

$$Z_1 = 1 + 3\E^{-\theta},$$

and, to the same order,

$$F=-3 T \E^{-\hbar^2/IT},$$

for the free energy.

The thermodynamic energy is

$$\frac{E}{N} = \frac{\partial (F/T)}{\partial (1/T)} = \frac{3\hbar^2}{I} \E^{-\hbar^2/IT},$$

and the specific heat per particle,

$$c_v = \frac{\partial E}{N \partial T} = 3 \left( \frac{\hbar^2}{IT} \right)^2 \E^{-\hbar^2/IT}.$$

To plot $c_v(T)$ we note that it tends to zero when $T \rightarrow 0$ and that, for high temperatures, it must match the classical result $c_v = 1$, in units of the boltzmann constant; the change between the two regimes occurs about $\theta=1$.

van der Waals

Find the van der Waals equation of state of parameters $a$ and $b$

\begin{equation} \label{e:vdW} \left( P + N^2a/V^2 \right)(V - Nb) = Nk_B T \end{equation}

where $P$ is the pressure, $V$ the volume, $N$ the number of atoms and $T$ the temperature ($k_B$ is the Boltzmann constant), corresponding to the interaction energy

\begin{equation} \label{e:wr} w(r) = \begin{cases} \infty & r < r_0\\ -\epsilon & r_0<r<r_1 \\ 0 & r>r_1 \end{cases}\,. \end{equation}

Here $r_0,r_1$ have the dimension of a length and $\epsilon > 0$ is an energy.

Plot $w$ and explain the significance of the dimensional parameters appearing in its definition.
Compute the second virial coefficient:
\begin{equation} \label{e:B} B(T) = 2\pi \int_0^\infty r^2 \D r \left[ 1 - \E^{-w(r)/T} \right]\,. \end{equation}

3.Show that one can write

\begin{equation} \label{e:ab} B(T) = v_0 - \frac{a(T)}{T}, \quad v_0 = \frac{2\pi r_0^3}{3}\,, \end{equation}

and demonstrate that $a(T)$ is independent of the temperature at high $T$.

Show how to obtain the excluded volume term and compute the van der Waals parameters $a$ and $b$.

Solution

We :se $k_B=1$. The interaction energy is piecewise constant, repulsive for $r<r_0$, attractive for $r \in (r_0,r_1)$ with a characteristic energy $\epsilon$, and free for $r>r_1$; it can be considered as a crude approximation of the lennard-jones potential.

The second virial coefficient

$$B(T) = 2\pi \int_0^\infty \D r r^2 \left[ 1 - \E^{-w(r)/T} \right] $$

is (the integrals give the volume of a sphere over 2):

$$B(T) = \frac{2\pi}{3}r_0^3 + \frac{2\pi}{3} (r_1^3 - r_0^3)\big(1-\E^{\epsilon/T}\big) \,.$$

It can be written as $B=v_0-a/T$,

$$v_0 = \frac{2\pi}{3}r_0^3, \; a(T) = -\frac{2\pi}{3} (r_1^3 - r_0^3)\big(1-\E^{\epsilon/T}\big)T \,.$$

We note that when $T \gg \epsilon$ (high temperature case) we have $\E^{\epsilon/T}\approx 1 + \epsilon/T$, which leads to

$$a(T) = \frac{2\pi}{3} (r_1^3 - r_0^3)\epsilon = \text{const.} \,,$$

independent of $T$

Putting $B(T)$ in the virial formula $PV = NT (1+ B N/V)$, we obtain

$$PV = NT (1 + \frac{N}{V} v_0) - \frac{N^2}{V} a(T)$$

$$\left[ P + a(T) \frac{N^2}{V^2} \right] V = NT(1 +Nv_0/V) \approx \frac{NTV}{V-Nv_0}$$

where, in the last equality we used that $Nv_0 \ll V$. Therefore, we found the van der Waals equation of state with the microscopic parameters:

$$b = Nv_0, \; \text{and} \; a(T) = a(T) = \frac{2\pi T}{3} (r_1^3 - r_0^3)\big(\E^{\epsilon/T} - 1 \big) \,.$$