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.
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 \(F\approx k_BT\log\left(\frac{\hbar\omega}{kT}\right)\).
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}
Entropy of a simple harmonic oscillatorHeat capacity of a simple harmonic oscillator
This entropy is shown in the nearby figure, as well
as the heat capacity.
Find the entropy of a set of \(N\) oscillators of frequency
\(\omega\) as a function of the total quantum number \(n\). Use the
multiplicity function: \begin{equation}
g(N,n) = \frac{(N+n-1)!}{n!(N-1)!}
\end{equation} and assume that \(N\gg 1\). This means you can
make the Sitrling approximation that
\(\log N! \approx N\log N - N\). It also means that
\(N-1 \approx N\).
Let \(U\) denote the total energy \(n\hbar\omega\) of the
oscillators. Express the entropy as \(S(U,N)\). Show that the total
energy at temperature \(T\) is \begin{equation}
U = \frac{N\hbar\omega}{e^{\frac{\hbar\omega}{kT}}-1}
\end{equation} This is the Planck result found the hard
way. We will get to the easy way soon, and you will never again need
to work with a multiplicity function like this.
EnergyTemperatureParamagnetismThermal and Statistical Physics 2020
Find the equilibrium value at temperature \(T\)
of the fractional magnetization \begin{equation}
\frac{\mu_{tot}}{Nm} \equiv \frac{2\langle s\rangle}{N}
\end{equation} of a system of \(N\) spins each of magnetic moment
\(m\) in a magnetic field \(B\). The spin excess is \(2s\). The energy
of this system is given by \begin{align}
U &= -\mu_{tot}B
\end{align} where \(\mu_{tot}\) is the total magnetization. Take the
entropy as the logarithm of the multiplicity \(g(N,s)\) as given in
(1.35 in the text): \begin{equation}
S(s) \approx k_B\log g(N,0) - k_B\frac{2s^2}{N}
\end{equation} for \(|s|\ll N\), where \(s\) is the spin excess, which
is related to the magnetization by \(\mu_{tot} = 2sm\). Hint:
Show that in this approximation \begin{equation}
S(U) = S_0 - k_B\frac{U^2}{2m^2B^2N},
\end{equation} with \(S_0=k_B\log g(N,0)\). Further, show that
\(\frac1{kT} = -\frac{U}{m^2B^2N}\), where \(U\) denotes
\(\langle U\rangle\), the thermal average energy.
These notes, from the third week of Thermal and Statistical Physics cover the canonical ensemble and Helmholtz free energy. They include a number of small group activities.
Task:Draw a right triangle. Put a circle around the right angle, that is, the angle that is \(\frac\pi2\) radians.
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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.
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!
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}
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}
These lecture notes for the second week of Thermal and Statistical Physics involve relating entropy and temperature in the microcanonical ensemble, using a paramagnet as an example.
These notes include a few small group activities.
Thermal and Statistical Physics 2020
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.
