- Entropy Quantum harmonic oscillator Frequency Energy
*assignment*Free energy of a harmonic oscillator*assignment*Homework##### Free energy of a harmonic oscillator

Helmholtz free energy harmonic oscillator Thermal and Statistical Physics 2020A 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}

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

*assignment*One-dimensional gas*assignment*Homework##### One-dimensional gas

Ideal gas Entropy Tempurature 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.*assignment*Free energy of a two state system*assignment*Homework##### Free energy of a two state system

Helmholtz free energy entropy statistical mechanics Thermal and Statistical Physics 2020Find 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 \(\varepsilon\).

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

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

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

*assignment*Entropy and Temperature*assignment*Homework##### Entropy and Temperature

Energy Temperature Ideal gas Entropy Thermal and Statistical Physics 2020Suppose \(g(U) = CU^{3N/2}\), where \(C\) is a constant and \(N\) is the number of particles.

Show that \(U=\frac32 N k_BT\).

Show that \(\left(\frac{\partial^2S}{\partial U^2}\right)_N\) is negative. This form of \(g(U)\) actually applies to a monatomic ideal gas.

*face*Thermal radiation and Planck distribution*face*Lecture120 min.

##### Thermal radiation and Planck distribution

Thermal and Statistical Physics 2020Planck distribution blackbody radiation photon statistical mechanics

These notes from the fourth week of Thermal and Statistical Physics cover blackbody radiation and the Planck distribution. They include a number of small group activities.*face*Introducing entropy*face*Lecture30 min.

##### Introducing entropy

Contemporary Challenges 2021 (4 years)entropy multiplicity heat thermodynamics

This lecture introduces the idea of entropy, including the relationship between entropy and multiplicity as well as the relationship between*changes*in entropy and heat.*assignment*Ideal gas in two dimensions*assignment*Homework##### Ideal gas in two dimensions

Ideal gas Entropy Chemical potential Thermal and Statistical Physics 2020Find the chemical potential of an ideal monatomic gas in two dimensions, with \(N\) atoms confined to a square of area \(A=L^2\). The spin is zero.

Find an expression for the energy \(U\) of the gas.

Find an expression for the entropy \(\sigma\). The temperature is \(kT\).

*assignment*Paramagnetism*assignment*Homework##### Paramagnetism

Energy Temperature Paramagnetism Thermal 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.*assignment*Radiation in an empty box*assignment*Homework##### Radiation in an empty box

Thermal physics Radiation Free energy Thermal and Statistical Physics 2020As 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}

*face*Entropy and Temperature*face*Lecture120 min.

##### Entropy and Temperature

Thermal and Statistical Physics 2020paramagnet entropy temperature statistical mechanics

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
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.