Nuclei of a particular isotope species contained in a crystal have spin \(I=1\), and thus, \(m = \{+1,0,-1\}\). The interaction between the nuclear quadrupole moment and the gradient of the crystalline electric field produces a situation where the nucleus has the same energy, \(E=\varepsilon\), in the state \(m=+1\) and the state \(m=-1\), compared with an energy \(E=0\) in the state \(m=0\), i.e. each nucleus can be in one of 3 states, two of which have energy \(E=\varepsilon\) and one has energy \(E=0\).

1. Find the Helmholtz free energy \(F = U-TS\) for a crystal containing \(N\) nuclei which do not interact with each other.

2. Find an expression for the entropy as a function of temperature for this system. (Hint: use results of part a.)

3. Indicate what your results predict for the entropy at the extremes of very high temperature and very low temperature.

A one-dimensional harmonic oscillator has an infinite series of equally spaced energy states, with \(\varepsilon_n = n\hbar\omega\), where \(n\) is an integer \(\ge 0\), and \(\omega\) is the classical frequency of the oscillator. We have chosen the zero of energy at the state \(n=0\) which we can get away with here, but is not actually the zero of energy! To find the true energy we would have to add a \(\frac12\hbar\omega\) for each oscillator.

1. Show that for a harmonic oscillator the free energy is \begin{equation} F = k_BT\log\left(1 - e^{-\frac{\hbar\omega}{k_BT}}\right) \end{equation} Note that at high temperatures such that \(k_BT\gg\hbar\omega\) we may expand the argument of the logarithm to obtain \(F\approx k_BT\log\left(\frac{\hbar\omega}{kT}\right)\).

2. From the free energy above, show that the entropy is \begin{equation} \frac{S}{k_B} = \frac{\frac{\hbar\omega}{kT}}{e^{\frac{\hbar\omega}{kT}}-1} - \log\left(1-e^{-\frac{\hbar\omega}{kT}}\right) \end{equation}

   

   

   This entropy is shown in the nearby figure, as well as the heat capacity.

1. Find an expression for the free energy as a function of \(T\) of a system with two states, one at energy 0 and one at energy \(\varepsilon\).

2. From the free energy, find expressions for the internal energy \(U\) and entropy \(S\) of the system.

3. Plot the entropy versus \(T\). Explain its asymptotic behavior as the temperature becomes high.

4. Plot the \(S(T)\) versus \(U(T)\). Explain the maximum value of the energy \(U\).

Consider a paramagnet, which is a material with \(n\) spins per unit volume each of which may each be either “up” or “down”. The spins have energy \(\pm mB\) where \(m\) is the magnetic dipole moment of a single spin, and there is no interaction between spins. The magnetization \(M\) is defined as the total magnetic moment divided by the total volume. Hint: each individual spin may be treated as a two-state system, which you have already worked with above.

1. Find the Helmholtz free energy of a paramagnetic system (assume \(N\) total spins) and show that \(\frac{F}{NkT}\) is a function of only the ratio \(x\equiv \frac{mB}{kT}\).

2. Use the canonical ensemble (i.e. partition function and probabilities) to find an exact expression for the total magentization \(M\) (which is the total dipole moment per unit volume) and the susceptibility \begin{align} \chi\equiv\left(\frac{\partial M}{\partial B}\right)_T \end{align} as a function of temperature and magnetic field for the model system of magnetic moments in a magnetic field. The result for the magnetization is \begin{align} M=nm\tanh\left(\frac{mB}{kT}\right) \end{align} where \(n\) is the number of spins per unit volume. The figure shows what this magnetization looks like.

3. Show that the susceptibility is \(\chi=\frac{nm^2}{kT}\) in the limit \(mB\ll kT\).

As discussed in class, we can consider a black body as a large box with a small hole in it. If we treat the large box a metal cube with side length \(L\) and metal walls, the frequency of each normal mode will be given by: \begin{align} \omega_{n_xn_yn_z} &= \frac{\pi c}{L}\sqrt{n_x^2 + n_y^2 + n_z^2} \end{align} where each of \(n_x\), \(n_y\), and \(n_z\) will have positive integer values. This simply comes from the fact that a half wavelength must fit in the box. There is an additional quantum number for polarization, which has two possible values, but does not affect the frequency. Note that in this problem I'm using different boundary conditions from what I use in class. It is worth learning to work with either set of quantum numbers. Each normal mode is a harmonic oscillator, with energy eigenstates \(E_n = n\hbar\omega\) where we will not include the zero-point energy \(\frac12\hbar\omega\), since that energy cannot be extracted from the box. (See the Casimir effect for an example where the zero point energy of photon modes does have an effect.)

Note

This is a slight approximation, as the boundary conditions for light are a bit more complicated. However, for large \(n\) values this gives the correct result.

1. Show that the free energy is given by \begin{align} F &= 8\pi \frac{V(kT)^4}{h^3c^3} \int_0^\infty \ln\left(1-e^{-\xi}\right)\xi^2d\xi \\ &= -\frac{8\pi^5}{45} \frac{V(kT)^4}{h^3c^3} \\ &= -\frac{\pi^2}{45} \frac{V(kT)^4}{\hbar^3c^3} \end{align} provided the box is big enough that \(\frac{\hbar c}{LkT}\ll 1\). Note that you may end up with a slightly different dimensionless integral that numerically evaluates to the same result, which would be fine. I also do not expect you to solve this definite integral analytically, a numerical confirmation is fine. However, you must manipulate your integral until it is dimensionless and has all the dimensionful quantities removed from it!

2. Show that the entropy of this box full of photons at temperature \(T\) is \begin{align} S &= \frac{32\pi^5}{45} k V \left(\frac{kT}{hc}\right)^3 \\ &= \frac{4\pi^2}{45} k V \left(\frac{kT}{\hbar c}\right)^3 \end{align}

3. Show that the internal energy of this box full of photons at temperature \(T\) is \begin{align} \frac{U}{V} &= \frac{8\pi^5}{15}\frac{(kT)^4}{h^3c^3} \\ &= \frac{\pi^2}{15}\frac{(kT)^4}{\hbar^3c^3} \end{align}

Suppose that a system of \(N\) atoms of type \(A\) is placed in diffusive contact with a system of \(N\) atoms of type \(B\) at the same temperature and volume.

1. Show that after diffusive equilibrium is reached the total entropy is increased by \(2Nk\ln 2\). The entropy increase \(2Nk\ln 2\) is known as the entropy of mixing.

2. If the atoms are identical (\(A=B\)), show that there is no increase in entropy when diffusive contact is established. The difference has been called the Gibbs paradox.

3. Since the Helmholtz free energy is lower for the mixed \(AB\) than for the separated \(A\) and \(B\), it should be possible to extract work from the mixing process. Construct a process that could extract work as the two gasses are mixed at fixed temperature. You will probably need to use walls that are permeable to one gas but not the other.

Note

This course has not yet covered work, but it was covered in Energy and Entropy, so you may need to stretch your memory to finish part (c).

Consider an ideal gas of \(N\) particles, each of mass \(M\), confined to a one-dimensional line of length \(L\). The particles have spin zero (so you can ignore spin) and do not interact with one another. Find the entropy at temperature \(T\). You may assume that the temperature is high enough that \(k_B T\) is much greater than the ground state energy of one particle.

In our week on radiation, we saw that the Helmholtz free energy of a box of radiation at temperature \(T\) is \begin{align} F &= -8\pi \frac{V(kT)^4}{h^3c^3}\frac{\pi^4}{45} \end{align} From this we also found the internal energy and entropy \begin{align} U &= 24\pi \frac{(kT)^4}{h^3c^3}\frac{\pi^4}{45} V \\ S &= 32\pi kV\left(\frac{kT}{hc}\right)^3 \frac{\pi^4}{45} \end{align} Given these results, let us consider a Carnot engine that uses an empty metalic piston (i.e. a photon gas).

1. Given \(T_H\) and \(T_C\), as well as \(V_1\) and \(V_2\) (the two volumes at \(T_H\)), determine \(V_3\) and \(V_4\) (the two volumes at \(T_C\)).

2. What is the heat \(Q_H\) taken up and the work done by the gas during the first isothermal expansion? Are they equal to each other, as for the ideal gas?

3. Does the work done on the two isentropic stages cancel each other, as for the ideal gas?

4. Calculate the total work done by the gas during one cycle. Compare it with the heat taken up at \(T_H\) and show that the energy conversion efficiency is the Carnot efficiency.