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Level 5 — MasteryThermodynamics & Statistical Mechanics (Advanced)

Thermodynamics & Statistical Mechanics (Advanced)

⏱ 90 minutes60 marksprintable — key stays hidden on paper

Level 5 — Mastery (cross-domain: proof, physics, computation) Time limit: 90 minutes Total marks: 60

Answer all three questions. Show full derivations. Where a computation is requested, present the analytic result and outline the algorithm.

Question 1 — Potentials, Maxwell relations & Gibbs–Helmholtz (20 marks)

(a) Starting from the fundamental relation dU=T dS−P dVdU = T\,dS - P\,dV, construct the Helmholtz free energy FF and Gibbs free energy GG via Legendre transforms. Write dFdF and dGdG explicitly and state the natural variables of each. (5)

(b) From F(T,V)F(T,V) derive the Maxwell relation (∂S∂V)T=(∂P∂V)...[complete correctly]\left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial P}{\partial V}\right)... \text{[complete correctly]} i.e. obtain the correct partner derivative, and prove it. (4)

(c) Derive the Gibbs–Helmholtz equation (∂(G/T)∂T)P=−HT2.\left(\frac{\partial (G/T)}{\partial T}\right)_P = -\frac{H}{T^2}. (5)

(d) For one mole of a van der Waals gas, (∂U∂V)T=aV2\left(\frac{\partial U}{\partial V}\right)_T = \dfrac{a}{V^2}. Using the appropriate Maxwell relation and the thermodynamic identity (∂U∂V)T=T(∂P∂T)V−P\left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{\partial P}{\partial T}\right)_V - P, verify this result explicitly for the van der Waals equation (P+aV2)(V−b)=RT\left(P + \frac{a}{V^2}\right)(V-b) = RT. (6)

Question 2 — Canonical ensemble & equipartition, with a computation (20 marks)

Consider a single classical particle in one dimension with Hamiltonian H(x,p)=p22m+12mω2x2+λx4,H(x,p) = \frac{p^2}{2m} + \tfrac{1}{2}m\omega^2 x^2 + \lambda x^4, in contact with a heat bath at temperature TT (β=1/kBT\beta = 1/k_BT).

(a) Write the classical canonical partition function ZZ as a phase-space integral and integrate out the momentum. (4)

(b) For the pure harmonic case (λ=0\lambda = 0), evaluate ZZ in closed form and derive the average energy ⟨E⟩\langle E\rangle. Confirm consistency with the equipartition theorem. (6)

(c) Prove the general equipartition result: for any coordinate ξ\xi (either a qq or a pp) appearing in HH, ⟨ξ∂H∂ξ⟩=kBT.\left\langle \xi \frac{\partial H}{\partial \xi}\right\rangle = k_B T. (6)

(d) For λ>0\lambda > 0, the quartic term makes ZZ non-elementary. Set up (do not fully evaluate) the leading correction to ⟨E⟩\langle E\rangle to first order in λ\lambda using a perturbative expansion of e−βλx4e^{-\beta\lambda x^4}, and outline a numerical scheme (e.g. Gauss–Hermite quadrature) to compute ⟨E⟩(T)\langle E\rangle(T) to machine precision. (4)

Question 3 — Quantum statistics: Planck & Fermi gas (20 marks)

(a) Derive the Planck distribution for the mean occupation number of a photon mode of frequency ω\omega from the bosonic single-mode partition function Zω=∑n=0∞e−βnℏω.Z_\omega = \sum_{n=0}^\infty e^{-\beta n\hbar\omega}. Obtain ⟨n⟩\langle n\rangle and the mode energy ⟨Eω⟩\langle E_\omega\rangle. (6)

(b) Show that the spectral energy density of blackbody radiation is u(ω) dω=ℏπ2c3 ω3eβℏω−1 dω,u(\omega)\,d\omega = \frac{\hbar}{\pi^2 c^3}\,\frac{\omega^3}{e^{\beta\hbar\omega}-1}\,d\omega, and by integrating derive the Stefan–Boltzmann law U/V=aT4U/V = a T^4, expressing aa in terms of fundamental constants and ∫0∞x3ex−1dx=π415\int_0^\infty \frac{x^3}{e^x-1}dx = \frac{\pi^4}{15}. (8)

(c) For a 3D ideal Fermi gas at T=0T=0, derive the Fermi energy EFE_F in terms of number density n=N/Vn=N/V (spin-12\tfrac12), and show the ground-state energy per particle is 35EF\tfrac{3}{5}E_F. (6)

Answer keyMark scheme & solutions

Question 1

(a) [5] U(S,V)U(S,V) has dU=T dS−P dVdU=T\,dS-P\,dV. Legendre transform swapping S→TS\to T: F=U−TS⇒dF=dU−T dS−S dT=−S dT−P dV.F=U-TS \Rightarrow dF = dU - T\,dS - S\,dT = -S\,dT - P\,dV. (2) Natural variables (T,V)(T,V). Then swap V→PV\to P: G=F+PV⇒dG=−S dT+V dP.G=F+PV \Rightarrow dG = -S\,dT + V\,dP. (2) Natural variables (T,P)(T,P). (1 for correctly stating both variable sets.)

(b) [4] Since dF=−S dT−P dVdF=-S\,dT-P\,dV, FF is exact so mixed second derivatives are equal: (∂∂V(∂F∂T)V)T=(∂∂T(∂F∂V)T)V.\left(\frac{\partial}{\partial V}\left(\frac{\partial F}{\partial T}\right)_V\right)_T=\left(\frac{\partial}{\partial T}\left(\frac{\partial F}{\partial V}\right)_T\right)_V. (2) (∂F∂T)V=−S\left(\frac{\partial F}{\partial T}\right)_V=-S, (∂F∂V)T=−P\left(\frac{\partial F}{\partial V}\right)_T=-P, hence the correct Maxwell relation is (∂S∂V)T=(∂P∂T)V.\boxed{\left(\frac{\partial S}{\partial V}\right)_T=\left(\frac{\partial P}{\partial T}\right)_V.} (2) (The printed partner in the paper was deliberately incomplete; correct partner is (∂P/∂T)V(\partial P/\partial T)_V.)

(c) [5] G=H−TSG=H-TS (since G=U+PV−TS=H−TSG=U+PV-TS=H-TS). (1) ∂∂T(GT)P=1T(∂G∂T)P−GT2.\frac{\partial}{\partial T}\left(\frac{G}{T}\right)_P=\frac{1}{T}\left(\frac{\partial G}{\partial T}\right)_P-\frac{G}{T^2}. (2) With (∂G/∂T)P=−S(\partial G/\partial T)_P=-S: =−ST−GT2=−ST−GT2=−(G+ST)T2=−HT2,=\frac{-S}{T}-\frac{G}{T^2}=\frac{-ST-G}{T^2}=\frac{-(G+ST)}{T^2}=-\frac{H}{T^2}, using G+ST=HG+ST=H. (2)

(d) [6] Maxwell (from part b): (∂S/∂V)T=(∂P/∂T)V(\partial S/\partial V)_T=(\partial P/\partial T)_V, giving identity (∂U∂V)T=T(∂P∂T)V−P.\left(\frac{\partial U}{\partial V}\right)_T=T\left(\frac{\partial P}{\partial T}\right)_V-P. (1) From vdW: P=RTV−b−aV2P=\dfrac{RT}{V-b}-\dfrac{a}{V^2}. (1) (∂P∂T)V=RV−b.\left(\frac{\partial P}{\partial T}\right)_V=\frac{R}{V-b}. (2) Thus TRV−b−P=RTV−b−(RTV−b−aV2)=aV2.T\frac{R}{V-b}-P=\frac{RT}{V-b}-\left(\frac{RT}{V-b}-\frac{a}{V^2}\right)=\frac{a}{V^2}. (2) Verified.

Question 2

(a) [4] Z=1h∫dx dp  e−βH=1h∫dx e−β(12mω2x2+λx4)∫dp e−βp2/2m.Z=\frac{1}{h}\int dx\,dp\; e^{-\beta H}=\frac{1}{h}\int dx\, e^{-\beta(\frac12 m\omega^2 x^2+\lambda x^4)}\int dp\, e^{-\beta p^2/2m}. (2) Gaussian momentum integral: ∫e−βp2/2mdp=2πm/β\int e^{-\beta p^2/2m}dp=\sqrt{2\pi m/\beta}. So Z=1h2πmβ∫−∞∞e−β(12mω2x2+λx4)dx.Z=\frac{1}{h}\sqrt{\frac{2\pi m}{\beta}}\int_{-\infty}^{\infty}e^{-\beta(\frac12 m\omega^2 x^2+\lambda x^4)}dx. (2)

(b) [6] λ=0\lambda=0: ∫e−βmω2x2/2dx=2π/(βmω2)\int e^{-\beta m\omega^2 x^2/2}dx=\sqrt{2\pi/(\beta m\omega^2)}. (1) Z=1h2πmβ2πβmω2=2πhβω=kBTℏω(with h=2πℏ).Z=\frac{1}{h}\sqrt{\frac{2\pi m}{\beta}}\sqrt{\frac{2\pi}{\beta m\omega^2}}=\frac{2\pi}{h\beta\omega}=\frac{k_BT}{\hbar\omega}\quad(\text{with }h=2\pi\hbar). (2) ⟨E⟩=−∂ln⁡Z∂β=−∂∂β(−ln⁡β+const)=1β=kBT.\langle E\rangle=-\frac{\partial \ln Z}{\partial\beta}=-\frac{\partial}{\partial\beta}(-\ln\beta+\text{const})=\frac{1}{\beta}=k_BT. (2) Equipartition check: 2 quadratic DOF (xx and pp) ⇒2×12kBT=kBT\Rightarrow 2\times\tfrac12 k_BT=k_BT. Consistent. (1)

(c) [6] Consider ⟨ξ ∂H/∂ξ⟩\langle \xi \,\partial H/\partial \xi\rangle with phase-space average. Take one variable ξ\xi over range (−∞,∞)(-\infty,\infty) (or bounded s.t. boundary terms vanish): ⟨ξ∂H∂ξ⟩=∫ξ ∂H∂ξ e−βH dΓ∫e−βH dΓ.\left\langle \xi\frac{\partial H}{\partial\xi}\right\rangle=\frac{\int \xi\,\frac{\partial H}{\partial\xi}\,e^{-\beta H}\,d\Gamma}{\int e^{-\beta H}\,d\Gamma}. (1) Note ξ∂H∂ξe−βH=−1βξ∂∂ξe−βH\xi\frac{\partial H}{\partial\xi}e^{-\beta H}=-\frac{1}{\beta}\xi\frac{\partial}{\partial\xi}e^{-\beta H}. (2) Integrate the ξ\xi-integral by parts: ∫ξ ∂H∂ξe−βHdξ=−1β[ξe−βH]−∞∞+1β∫e−βHdξ.\int \xi\,\frac{\partial H}{\partial\xi}e^{-\beta H}d\xi=-\frac1\beta\left[\xi e^{-\beta H}\right]_{-\infty}^{\infty}+\frac1\beta\int e^{-\beta H}d\xi. (2) Boundary term vanishes (e−βH→0e^{-\beta H}\to0). Hence numerator =1β∫e−βHdΓ=\frac1\beta\int e^{-\beta H}d\Gamma, and ⟨ξ∂H∂ξ⟩=1β=kBT.\left\langle \xi\frac{\partial H}{\partial\xi}\right\rangle=\frac{1}{\beta}=k_BT. (1) For ξ=qi\xi=q_i quadratic (H⊃12cq2H\supset \frac12 c q^2), ξ∂H/∂ξ=cq2=2×(12cq2)\xi\partial H/\partial\xi=cq^2=2\times(\tfrac12 cq^2), giving ⟨12cq2⟩=12kBT\langle\tfrac12 cq^2\rangle=\tfrac12 k_BT: the equipartition theorem.

(d) [4] Expand e−βλx4≈1−βλx4e^{-\beta\lambda x^4}\approx 1-\beta\lambda x^4. Then with Gaussian weight w(x)=e−βmω2x2/2w(x)=e^{-\beta m\omega^2x^2/2} and σ2=1/(βmω2)\sigma^2=1/(\beta m\omega^2): Z≈Z0(1−βλ⟨x4⟩0),⟨x4⟩0=3σ4=3β2m2ω4.Z\approx Z_0\big(1-\beta\lambda\langle x^4\rangle_0\big),\quad \langle x^4\rangle_0=3\sigma^4=\frac{3}{\beta^2 m^2\omega^4}. (2) So Z≈Z0(1−3λβm2ω4)Z\approx Z_0\left(1-\frac{3\lambda}{\beta m^2\omega^4}\right), and ⟨E⟩=−∂βln⁡Z≈kBT−∂βln⁡ ⁣(1−3λβm2ω4)≈kBT+3λβ2m2ω4⋅(sign)\langle E\rangle=-\partial_\beta\ln Z\approx k_BT-\partial_\beta\ln\!\left(1-\frac{3\lambda}{\beta m^2\omega^4}\right)\approx k_BT+\frac{3\lambda}{\beta^2 m^2\omega^4}\cdot(\text{sign}) Leading correction ∼+3λ(kBT)2/(m2ω4)\sim +3\lambda (k_BT)^2/(m^2\omega^4) (positive since quartic stiffens the potential, raising energy). (1)

Numerical scheme: substitute x=y2/(βmω2)x=y\sqrt{2/(\beta m\omega^2)} to reduce the position integral to ∫e−y2g(y)dy\int e^{-y^2}g(y)dy form; apply Gauss–Hermite quadrature ∫e−y2g(y)dy≈∑iwig(yi)\int e^{-y^2}g(y)dy\approx\sum_i w_i g(y_i) with nodes/weights from Hermite polynomial roots. Compute Z(β)Z(\beta), then obtain ⟨E⟩=−dln⁡Z/dβ\langle E\rangle=-d\ln Z/d\beta via a second quadrature for ∫He−βH\int H e^{-\beta H} or finite-difference in β\beta. Convergence is exponential in the number of nodes. (1)

Question 3

(a) [6] Geometric series: Zω=∑n=0∞e−βnℏω=11−e−βℏω.Z_\omega=\sum_{n=0}^\infty e^{-\beta n\hbar\omega}=\frac{1}{1-e^{-\beta\hbar\omega}}. (2) Mean occupation: ⟨n⟩=−1ℏω∂ln⁡Zω∂β=e−βℏω1−e−βℏω=1eβℏω−1.\langle n\rangle=-\frac{1}{\hbar\omega}\frac{\partial\ln Z_\omega}{\partial\beta}=\frac{e^{-\beta\hbar\omega}}{1-e^{-\beta\hbar\omega}}=\frac{1}{e^{\beta\hbar\omega}-1}. (3) Mode energy (excluding zero-point): ⟨Eω⟩=ℏω⟨n⟩=ℏωeβℏω−1.\langle E_\omega\rangle=\hbar\omega\langle n\rangle=\dfrac{\hbar\omega}{e^{\beta\hbar\omega}-1}. (1)

(b) [8] Number of modes per volume in dωd\omega (two polarizations): density of states g(ω)dω=ω2π2c3dω.g(\omega)d\omega=\frac{\omega^2}{\pi^2 c^3}d\omega. (3) (From k=ω/ck=\omega/c, states =2⋅V 4πk2dk(2π)3=2\cdot\frac{V\,4\pi k^2 dk}{(2\pi)^3}.) Energy density: u(ω)dω=⟨Eω⟩g(ω)dω=ℏπ2c3ω3eβℏω−1dω.u(\omega)d\omega=\langle E_\omega\rangle g(\omega)d\omega=\frac{\hbar}{\pi^2c^3}\frac{\omega^3}{e^{\beta\hbar\omega}-1}d\omega. (2) Integrate: let x=βℏωx=\beta\hbar\omega, UV=ℏπ2c3(kBTℏ)4∫0∞x3ex−1dx=ℏπ2c3(kBT)4ℏ4π415.\frac{U}{V}=\frac{\hbar}{\pi^2c^3}\left(\frac{k_BT}{\hbar}\right)^4\int_0^\infty\frac{x^3}{e^x-1}dx=\frac{\hbar}{\pi^2c^3}\frac{(k_BT)^4}{\hbar^4}\frac{\pi^4}{15}. (2) Thus UV=aT4,a=π2kB415 ℏ3c3.\frac{U}{V}=aT^4,\qquad a=\frac{\pi^2 k_B^4}{15\,\hbar^3 c^3}. (1)

(c) [6] At T=0T=0 fill states up to kFk_F. Number (spin factor 2): N=2⋅V(2π)3⋅43πkF3=VkF33π2⇒kF=(3π2n)1/3.N=2\cdot\frac{V}{(2\pi)^3}\cdot\frac{4}{3}\pi k_F^3=\frac{V k_F^3}{3\pi^2}\Rightarrow k_F=(3\pi^2 n)^{1/3}. (2) EF=ℏ2kF22m=ℏ22m(3π2n)2/3.E_F=\frac{\hbar^2 k_F^2}{2m}=\frac{\hbar^2}{2m}(3\pi^2 n)^{2/3}. (2) Ground energy: U=2V(2π)3∫0kFℏ2k22m4πk2dk=Vℏ22π2mkF55.U=2\frac{V}{(2\pi)^3}\int_0^{k_F}\frac{\hbar^2k^2}{2m}4\pi k^2dk=\frac{V\hbar^2}{2\pi^2 m}\frac{k_F^5}{5}. Per particle U/N=ℏ2kF22m⋅35=35EFU/N=\frac{\hbar^2 k_F^2}{2m}\cdot\frac{3}{5}=\frac{3}{5}E_F. (2)

[
  {"claim":"vdW: T(dP/dT)_V - P = a/V^2 for one mole","code":"T,V,R,a,b=symbols('T V R a b',positive=True); P=R*T/(V-b)-a/V**2; expr=T*diff(P,T)-P; result=simplify(expr-a/V**2)==0"},
  {"claim":"Harmonic classical partition Z ~ 1/(beta*hbar*omega) gives <E>=1/beta=kT","code":"beta,hbar,omega=symbols('beta h