Classical Mechanics Problem 1: Central Potential Solution

Home , Harmonic oscillator

Classical Mechanics Problem 1: Central Potential Solution a) Integrals of motion for a central potential V (r): 2 ˙ Angular Momentum L = rv¡t = r φ ¢ 1 2 2 1 2 Energy per unit mass E = 2 r˙ + vt + V (r) = 2 r˙ + Veff(r) where vt is the tangential velocity and Veff is defined as

L2 V (r) = V (r) + eﬀ 2r2

If the orbit is circular, the distance of the test body from the origin is invariant:r ˙ = 0, which implies that the body is always at the equilibrium-distance:

dV dV L2 v2 eﬀ = 0 ⇒ = = t = rφ˙2 dr dr r3 r

then µ ¶ L 1 dV 1/2 φ˙ = ω = = φ r2 r dr so for the period we get

µ ¶ 2π 1 dV −1/2 Tφ = = 2π ωφ r dr b) Write the orbit as in the statement of the problem:

dV r(t) = r + ²(t) with eﬀ (r ) = 0 and ²2 ¿ r2. 0 dr 0 0

1 2 The energy per unit mass is now E = 2 ²˙ + Veﬀ(r0 + ²), and since ² is small we may Taylor-expand the potential as

2 dVeff 1 d Veff 2 3 Veff(r0 + ²) = Veff(r0) + (r0)² + 2 (r0)² + O(² ) | dr{z } 2 dr =0 so then 1 1 d2V E − V (r ) = ²˙2 + eff (r )²2 + O(²3) = const. eff 0 2 2 dr2 0 In the above equation we readily recognize the equation of the simple harmonic oscillator with µ ¶ d2V 1/2 ω = eff r dr2 r=r0

1 and its general solution is p E − Veﬀ(r0) ²(t) = cos [ωr(t − t0)] ωr

where t0 is an arbitrary constant.

Now return to writing ωr in terms of V (r) instead of Veﬀ(r).

d2V d2V 3L2 d2V d2V 3 dV ω2 = eﬀ = + = + 3ω2 = + r dr2 dr2 r4 dr2 φ dr2 r dr µ ¶ · µ ¶¸ d2V 3 dV 1/2 1 d dV 1/2 ω = + = r3 r dr2 r dr r3 dr dr r=r0 r=r0 And the radial period is 2π Tr = ωr

2 2 c) Stability is determined by the sign of ωr . For stability: ωr > 0, so µ ¶ 1 d dV r3 > 0 r3 dr dr for the Yukawa-potential GM V (r) = − e−kr r so the condition is µ ¶ 1 d dV GM £ ¤ £ ¤ r3 = e−kr 1 + kr − (kr)2 > 0 ⇒ 1 + kr − (kr)2 > 0 r3 dr dr r3 Ã√ !Ã√ ! £ ¤ 5 − 1 5 + 1 1 + kr − (kr)2 = + kr − kr > 0 2 2 which is satisﬁed only if Ã√ ! 5 + 1 kr < 2 Therefore circular orbits are unstable for Ã√ ! 5 + 1 kr > 2

2 d) The outermost stable circular orbit is at Ã√ ! 5 + 1 r = 0 2k

its energy per unit mass is µ ¶ 1 1 dV E = V (r ) + (r ω )2 = V (r ) + r 0 2 0 φ 0 2 dr r=r0 Ã√ ! 1 GM GM 5 − 1 −kr0 −kr0 e (kr0 − 1) = e > 0 2 r0 r0 4

If r0 is decreased only slightly, E > 0 still and the orbit is absolutely stable ¥ The eﬀective potential for the Yukawa-potential has the form shown in Figure 1.

Veff

Figure 1: Eﬀective potential against distance from the origin

3 Classical Mechanics Problem 2: Planar Double Pendulum Solution

q1 l

q2 l

a) L = T − V The moment of inertia for a uniform rod of length l and mass m is 1 I = ml2 about one of the ends 3 and 1 I = ml2 about the rod’s center c 12 The kinetic energy term we can decompose into three parts:

T = T1 + T2,rot + T2,trans

where T1 is the kinetic energy of the ﬁrst rod, T2,trans is the translational energy of the center of mass of the second rod and T2,rot is its rotational energy about its center of mass. Then 1 T = ml2θ˙2 1 6 1 1 T = ml2θ˙2 2,rot 24 2 and 1 ¡ ¢ T = m x˙ 2 +y ˙2 2,trans 2 c c where xc and yc are the coordinates of the second rod’s center of mass, so l x = l sin θ + sin θ c 1 2 2 l y = −l cos θ − cos θ c 1 2 2 from which · ¸ 1 x˙ 2 +y ˙2 = l2 θ˙2 + θ˙2 + θ˙ θ˙ (sin θ sin θ + cos θ cos θ ) c c 1 4 2 1 2 1 2 1 2

4 The potential energies are simply Vi = mgyc,i, where yc,i are the vertical coordinates of the rods’ centers of mass. Since both rods are uniform, yc,i are simply the coordinates of the centers. Thus, µ ¶ l l V = −mg cos θ ; V = −mg l cos θ + cos θ 1 2 1 2 1 2 2

The full Lagrangian is then

L = T1 + T2,rot + T2,trans − V1 − V2 · ¸ · ¸ 2 1 3 1 = ml2 θ˙2 + θ˙2 + θ˙ θ˙ cos (θ − θ ) + mgl cos θ + cos θ 3 1 6 2 1 2 1 2 2 1 2 2 b) Expand the Langrangian from part a) for small angles. The only function we have to deal with is 1 cos θ = 1 − θ2 + O(θ4) 2 ˆ ˆ Since we are going to look for normal modes with θj = θj exp (iωt), where the θj ¿ 1, ˙ ˙ we immediately see that in the term θ1θ2 cos (θ1 − θ2), the θ-dependence in the cosine can be dropped, because even the ﬁrst θ-dependent term gives a fourth order correction. Then the approximate Lagrangian is · ¸ · ¸ 2 1 3 1 L = ml2 θ˙2 + θ˙2 + θ˙ θ˙ − mgl θ2 + θ2 + const. 3 1 6 2 1 2 4 1 4 2

The Euler-Lagrange equations are d ∂L ∂L = ˙ dt ∂θj ∂θj so in the speciﬁc case: µ ¶ 4 1 g 3 θ¨ + θ¨ + θ = 0 3 1 2 2 l 2 1 µ ¶ 1 1 g 1 θ¨ + θ¨ + θ = 0 2 1 3 2 l 2 2

if we now look for normal modes, as mentioned, the above set of equations takes the form " ¡ ¢ #" # 4 2 3 g 1 2 ˆ ω − ω θ1 3 2 l ¡ 2 ¢ = 0 1 2 1 2 1 g ˆ 2 ω 3 ω − 2 l θ2 Non-trivial solutions exist if the determinant of the matrix on the left is zero. Denoting ω2 = λg/l, we can write this condition as µ ¶ µ ¶ 4 3 1 1 λ2 λ − λ − − = 0, 3 2 3 2 4

5 that is 7 7 3 λ2 − λ + = 0 36 6 4 whose solutions are 6 λ± = 3 ± √ , 7 so ﬁnally ·µ ¶ ¸ 6 g 1/2 ω± = 3 ± √ 7 l c) To sketch the eigenmodes, ﬁnd eigenvectors of the matrix in part b).

2 • ω = λ−g/l (low-frequency mode) µ ¶ 3 8 1 ³ √ ´ θˆ = − θˆ = 2 7 − 1 θˆ 2 λ 3 1 3 1 √ (2 7 − 1)/3 > 0 and real, therefore the two pendula are in phase; 2 • ω = λ+g/l (high-frequency mode) µ ¶ 3 8 1 ³ √ ´ θˆ = − θˆ = −2 7 − 1 θˆ 2 λ 3 1 3 1 √ (−2 7 − 1)/3 < 0 and real, therefore the two pendula are perfectly out of phase.

a b

Figure 2: The low- (a) and high-frequency (b) normal modes of the planar double pendulum.

6 Electromagnetism Problem 1 Solution a) Normal modes are products of harmonic standing waves in the x, y and z directions. For their frequencies, we have · ¸ q ³πn ´2 ³πn ´2 ³πn ´2 1/2 ω = c k2 + k2 + k2 = c x + y + z ; n , n , n ∈ Z+ x y z a b b x y z

Since a > b, the lowest frequency has nx = 1 and either ny = 1, nz = 0 or ny = 0, nz = 1 (note that ny = 0, nz = 0 does not satisfy the boundary condition Ek,at wall = 0). Since we are told to pick the mode with E~ k yˆ, the boundary conditions require

πx πz E~ (r, t) = E yˆsin sin cos ωt 0 a b

The magnetic induction we can get from Faraday’s Law:

∂B~ ³ ´ ∂E ∂E = −c ∇ × E~ =xc ˆ y − zcˆ y ∂t ∂z ∂x ³ π πx πz π πx πz ´ = xcEˆ sin cos − zcEˆ cos sin cos ωt 0 b a b 0 a a b µ ¶ πcE xˆ πx πz zˆ πx πz B~ (r, t) = − 0 sin cos − cos sin sin ωt ω b a b a a b

where the frequency ω is (by the argument above) µ ¶ 1 1 1/2 ω = πc + a2 b2 b) At a boundary of media, the discontinuity in the normal component of the electric ﬁeld is 4π times the surface charge density σ, so

Ey(x, 0, z) = 4πσ E πx πz σ(x, 0, z) = 0 sin sin cos ωt 4π a b σ(x, b, z) = −σ(x, 0, z)

and σ(0, y, z) = σ(a, y, z) = σ(x, y, 0) = σ(x, y, b) ≡ 0 Similarly, at the boundary of media the discontinuity of the tangential component of the magnetic ﬁeld is given by the surface current ~κ 4π nˆ × B~ = ~κ c

7 wheren ˆ is a unit vector normal to the surface, so µ ¶ c2E zˆ πx πz xˆ πx πz ~κ(x, 0, z) = 0 sin cos + cos sin sin ωt 4ω b a b a a b ~κ(x, b, z) = −~κ(x, 0, z) c2E yˆ πz ~κ(0, y, z) = − 0 sin sin ωt 4ω a b ~κ(a, y, z) = −~κ(0, y, z) c2E yˆ πx ~κ(x, y, 0) = − 0 sin sin ωt 4ω b a ~κ(x, y, b) = −~κ(x, y, 0) c) Since there is no charge on the b × b sides, the force there is purely magnetic and is given by Z 1 ³ ´ F~ (t) = ~κ × B~ d2x 2c b×b Zb Zb E2c2π πz F~ (x = 0, t) = − 0 xˆ dy dz sin2 sin2ωt 8ω2a2 b 0 0 µ ¶ c b 2 = −xˆ E sin ωt 4ω a 0

F~ (x = a, t) = −F~ (x = 0, t) The forces point outwards from the box on both sides (as is indicated by the sign in the equation above). d) Start with the sides where y = const. The magnetic component of the force can be written as above Za Zb µ ¶ E2c2π 1 πx πz 1 πx πz F~ (y = 0, t) = − 0 yˆ dx dz cos2 sin2 + sin2 cos2 sin2ωt mag 8ω2 a2 a b b2 a b 0 0 µ ¶ 1 ³ c ´2 1 b a = −yˆ E sin ωt + 2 2ω 0 4π a b To simplify this result further, use ω from part a) µ ¶ µ ¶ 1 1 −1 b a −1 ω−2 = (πc)−2 + = ab(πc)−2 + a2 b2 a b

Then µ ¶ E 2 ab F~ (y = 0, t) = −yˆ 0 sin ωt mag 4 2π3

8 The electric component of the force can be written as

Z Za Zb 1 E2 πx πz F~ (y = 0, t) = σE~ d2x =y ˆ 0 cos2ωt dx dz sin2 sin2 el 2 8π a b 0 0 µ ¶ 1 E 2 ab =y ˆ 0 cos ωt 2 4 π3

F~tot(y = 0, t) = F~el(y = 0, t) + F~mag(y = 0, t) µ ¶ E 2 ab ¡ ¢ =y ˆ 0 cos2ωt − sin2ωt 4 2π3 µ ¶ E 2 ab = yˆ 0 cos 2ωt 4 2π3

and

F~tot(y = b, t) = −F~tot(y = 0, t) There net force on the top and bottom sides oscillates between the inward and out- ward direction with half the period of the lowest frequency mode. In a time average, therefore, this force cancels. Next, calculate the force on the sides where z = const. Again, there is no charge, therefore no electric component; the force is purely magnetic

Zb Za E2c2π πx F~ (z = 0, t) = − 0 zˆ dy dx sin2 sin2ωt 8ω2b2 a 0 0 a ³ c ´2 = −zˆ E sin ωt b 4ω 0 F~ (z = b, t) = −F~ (z = 0, t)

The magnetic force is pushing the a × b walls outwards, too (sign!). e) From the Maxwell stress tensor, the force per unit surface area is

1 E2 1 B2 f~ = E~ (E~ · nˆ) − nˆ + B~ (B~ · nˆ) − nˆ 4π 8π 4π 8π

On the x = const. wallsn ˆ = ±xˆ, E~ = 0 and B~ · xˆ = 0, so

B2 E2c2π πz f~(x = {0, a}, t) = ∓ xˆ = ∓ 0 xˆ sin2 sin2ωt 8π 8ω2a2 b which is exactly the integrand from part c).

9 On the y = const. wallsn ˆ = ±yˆ, E~ (E~ · yˆ) = E2yˆ and B~ · yˆ = 0, so we get 1 f~(y = {0, b}, t) = ∓ (E2 − B2)ˆy 8π · E2 πx πz = ± 0 cos2ωt sin2 sin2 8π a b µ ¶ ¸ E2c2π 1 πx πz 1 πx πz − 0 cos2 sin2 + sin2 cos2 sin2ωt yˆ 8ω2 a2 a b b2 a b the sum of the ﬁrst two integrands from part d). On the z = const. wallsn ˆ = ±zˆ,(E~ · zˆ) = 0 and B~ · zˆ = 0, so we get

B2 E2c2π πx f~(z = {0, b}, t) = ∓ zˆ = ∓ 0 zˆsin2 sin2ωt 8π 8ω2b2 a the last integrand from part d). y

F(x) s k b k k (y) F (x) (z) k (z) (z) k z s (x) F – (z) k F(x) a (y) x

Figure 3: Average total forces, surface charges and surface currents on the cavity.

10 Electromagnetism Problem 2: Waves in a Dilute Gas Solution (see Feynman Lectures on Physics, vol. II, chapter 32) a) The EM wave is travelling in the ˆx direction; it has a transverse electric ﬁeld, so assume E × ˆy = 0. Then the electron in the atom behaves classically as a damped, driven harmonic oscillator ¡ 2 ¢ −iωt me y¨ + γy˙ + ω0y = −qE0e with the solution 1 qE(t) y(t) = 2 2 . ω − ω0 + iγω me For the dipole moment per unit volume:

2 1 naq E P = na(−q)y = 2 2 ω0 − ω − iγω me Therefore the volume polarizability is, according to the deﬁnition given,

2 P 1 naq α(ω) = = 2 2 ²0E ω0 − ω − iγω ²0me

(A quantum mechanical derivation would give this same expression multiplied by the oscillator strength f for the transition.) b) With no free charges or currents, Maxwell’s equations read ∂B ∇ · D = 0; ∇ × E = − ∂t ∂D ∇ · B = 0; ∇ × H = ∂t

and B = µ0H, D = ²0E + P = ²0(1 + α)E for a single frequency ω. This gives us the following wave-equation

2 ∂ D 1 2 2 − ∇ D = 0. ∂t µ0²0(1 + α)

Now let D be that of a plane wave: D ∝ ei(kx−ωt). Then

ω2 k2 = µ ² (1 + α)ω2 = (1 + α) 0 0 c2 p ⇒ n(ω) = 1 + α(ω)

One can also get this result by using the microscopic E, B and P ﬁelds: µ ¶ 1 ∂ P ∇ · E = − ∇ · P; c2∇ × B = + E ²0 ∂t ²0

11 2 ∂ E 2 2 1 ∂P ⇒ 2 − c ∇ E = − ∂t ²0 ∂t also 2 2 ∂ P ∂P 2 naq 2 + γ + ω0P = − E. ∂t ∂t me Together these give us k2 = (1 + α)ω2/c2 for a plane wave, as before. (Note that we are neglecting dipole-dipole interactions in the dilute gas.) c) We start by noting that according to Fourier-analysis Z∞ 1 E(x, t) = dk ei(kx−ω(k)t)Eˆ(k) 2π −∞ Z∞ Z∞ 1 2 2 Eˆ(k) = dx e−ikxE(x, 0) = √ dx e−i(k−kc)x−x /(2σ ) 2πσ2 −∞ −∞ 2 2 = e−i(k−kc) σ /2

Now Taylor-expand ω(k) about k = kc: µ ¶ dω ω(k) = ω(k ) + (k − k ) + O{(k − k )2} c dk c c kc 2 ≡ kcvph + vg(k − kc) + O{(k − kc) }

where, by deﬁnition, vph and vg are the phase- and group-velocities, respectively. Now let K = k − kc. Then Z∞ 1 2 2 E(x, t) = dK eikc(x−vpht)+iK(x−vg t)−K σ /2 2π −∞

2 2 e(x−vg t) /(2σ ) ikc(x−vpht) ikc(x−vpht) E(x, t) = e √ = e N(x − vgt, σ) 2πσ2 d) From part c) µ ¶ µ ¶ dω dk −1 c d log n −1 v = = = 1 + . g dk dω n d log ω √ For the dilute gas, n = 1 + α ≈ 1 + α/2, which we will write as n = nr + ini (for α is complex)

2 2 2 naq ω0 − ω nr ≈ 1 + 2² m 2 2 2 2 2 0 e (ω0 − ω ) + γ ω 2 naq γω ni ≈ 2² m 2 2 2 2 2 0 e (ω0 − ω ) + γ ω

12 Here the real part nr of the index of refraction determines the dispersion, and the imaginary part ni determines the absorption/gain coeﬃcient. At ω = ω0:

2 d log nr naq nr = 1 and = − 2 d log ω 2²0meγ

µ 2 ¶−1 µ 2 ¶ naq naq vg = 1 − 2 c ≈ 1 + 2 c 2²0meγ 2²0meγ

Note that vg > c at ω = ω0. This is called anomalous dispersion. It does not vio- late causality because signals (information) cannot travel faster than the minimum of (vph, vg), and now vph = c (since nr = 1). Also, the waves are damped by the electronic resonance maximally at ω = ω0.

13 Quantum Mechanics Problem 1 Solution

1. The ground state will have no nodes, so we can pick the even part of the general solution of the free Schr¨odingerequation inside the well. Outside the well, square-integrability demands the solutions to vanish at inﬁnity. The wave-function for the ground state is then

|x| < w |x| > w ψ(x) = cos kx ψ(x) = Ae−α|x|

Both the wave-function and its derivative has to be continuous at the boundaries of the well:

ψ : cos kw = Ae−αw dψ : −k sin kw = −αAe−αw dx ⇒ k tan kw = α

Directly from Schr¨odinger’sequation:

|x| < w |x| > w ~2k2 ~2α2 E = E = − + V 2m 2m 0

~2k2 ~2α2 ⇒ = − + V 2m 2m 0 From which we get the transcendental equation: · ¸ 2mV 1/2 k tan kw = 0 − k2 ~2

Let k∗ denote the positive root of the equation above, and introduce the following notation: √ 2mV π k = 0 and k = . c ~ max 2w

Clearly, the LHS of the equation diverges at kmax; and the RHS describes a circle with radius kc, as shown in Fig. 4. For the energy we have ~2k∗2 E = 2m

14 k tankw kc k* * tan k* w* H 2- 2L 1/2 (–)kc k

k * * k kc kmax

Figure 4: Graphical representation of the solution of the transcendental equation

2. Write the result of part 1 in the non-dimensional form: · ¸ 2mw2V 1/2 kw tan kw = 0 − (kw)2 ~2 According to the condition given in the statement of the problem, the radius of the circle on the RHS (that in Fig. 4) goes to inﬁnity, therefore

k → kmax and ~2k2 ~2π2 E → max = 2m 8mw2 3. The potential barrier on the low-potential side of the well, denoted a in the figure, will be finite (for any E), so the particle will eventually escape by the tunnel-effect. 4. ∆E = 0, because the perturbation is odd (and therefore its integral with the square of the ground-state wave-function vanishes). 5. Za 1 p F = dx 2m(V (x) − B) ~ w

V (x) = V0 − eEx V − B B = V − eEa ⇒ a = 0 0 eE

15 V() x

V 0

a x

–w0 w

√ Za 2m p F = dx (V − B) − eEx ~ 0 w √ µ ¶ ¯ ¯a 2m 2 1 3/2¯ = − (V0 − B − eEx) ¯ ~ 3 eE w √ 2m 2 1 = (V − B − eEw)3/2 ~ 3 eE 0

6. Write the energy of the particle as 1 B = mv2. 2 Then 2B v2 = . m The time it takes for the particle to bounce back and forth once is 4w T = , v so it hits the right wall with frequency v ν = 4w

Probability to escape v ⇒ = e−2F unit time 4w 4w ⇒ Lifetime ∼ e2F v

16 Quantum Mechanics Problem 2 Solution 1. Drop the t-label for simplicity. Then we have · ¸ cos θ sin θe−iωt H = B , sin θeiωt − cos θ and for the eigenvectors solve · ¸ · ¸ · ¸ cos θ sin θe−iωt x x = ± sin θeiωt − cos θ y y with the normalization condition |x|2 + |y|2 = 1. From the vector-equation sin θ y = eiωt x cos θ ± 1 and with the normalization, we end up with · ¸ · ¸ cos (θ/2) sin (θ/2) |+i = |−i = sin (θ/2)eiωt − cos (θ/2)eiωt

2. Decompose the state-vector as

|ψi = c+|+i + c−|−i and write Schr¨odinger’sequation in terms of these vectors: d i~ |ψi = H|ψi · dt ¸ d d i~ c˙ |+i + c |+i +c ˙ |−i + c |−i = B [c |+i − c |−i] + + dt − − dt + − or in the (|+i, |−i) basis " # " #" # d d d c+ B − i~h+| |+i −i~h+| |−i c+ i~ = dt dt dt d d c− −i~h−| dt |+i −B − i~h−| dt |−i c− which, with the given concrete form of the vectors, is " # " #" # 2 d c+ B + ~ω sin (θ/2) −~ω cos (θ/2) sin (θ/2) c+ i~ = . 2 dt c− −~ω sin (θ/2) cos (θ/2) −B + ~ω cos (θ/2) c− Now use the identities 1 sin2(θ/2) = (1 − cos θ) 2 1 cos2(θ/2) = (1 + cos θ) 2 1 sin (θ/2) cos (θ/2) = sin θ 2

17 to get " # " #" # d c+ 1 2B − ~ω cos θ −~ω sin θ c+ i~ = . dt c− 2 −~ω sin θ −2B + ~ω cos θ c− Note that in the last equation we dropped the part of the Hamiltonian that was pro- portional to the identity, since that gives only a time dependent phase that is identical for the coeﬃcients c−, c+. This we can rewrite in the form: " # " #" # d c+ Dz Dx c+ i~ = . dt c− Dx −Dz c−

with the solution " # ( " #) " # c+ −i Dz Dx 1 = exp c− ~ Dx −Dz 0 Ã ! Ã ! |D~ |t |D~ |t = cos − iDˆ · ~σ sin . ~ ~

And so Ã ! Ã ! |D~ |t Dz |D~ |t c+ = cos − i sin ~ |D~ | ~ Ã ! Ã ! |D~ |t D2 |D~ |t |c |2 = cos2 − i z sin2 + ~ D2 ~

3. For B À ~ω, Dz → D, so 2 |c+| → 1

(Adiabatic theorem)

18 Statistical Mechanics and Thermodynamics Problem 1 Thermodynamics of a Non-Interacting Bose Gas Solution a)

1 p2 np = Ep = eβ(Ep−µ) − 1 2m

At and below TBEC µ = 0. At exactly TBEC, there are no atoms in the condensate and Z µ ¶ Z∞ V d3p 2m 3/2 x2dx N = = (2π~)−3 V 4π 2π~3 eβp2/(2m) − 1 β ex2 − 1 |0 {z } =I1 µ ¶ 1 2mkT 3/2 n = I 2π2 ~2 1 µ ¶ 2π2n 2/3 ~2 kTBEC = I1 2m

(2 points) b) The above integral with µ = 0 also applies below TBEC, but it then gives the number of non-condensed atoms. So on an isotherm below Vcritical

• Nnon-condensed is constant • T is constant ⇒ p is constant (think of the kinetic origin of pressure)

p ClassicalGaspµ 1/ V BoseGas Phase TransitionLine –5/3 pBEC µ V

(2 points)

19 c) Z µ ¶ Z∞ V p2 d3p 2m 5/2 4π x4dx U = = (2π~)−3 V 2π~3 2m eβp2/(2m) − 1 β 2m ex2 − 1 |0 {z } =I µ ¶ 2 I2 2m 1 I2 5/2 = Nc = Nc kT ∝ T I1 β 2m I1

µ ¶3/2 µ ¶ 5 I2 V 2mkT 5 3/2 cv = Nck = 2 2 k I2 ∝ T 2 I1 2π ~ 2

(2 points) d) From the reversibility of the Carnot-cycle:

dS1 = −dS2 for 1 cycle

∆S1 = −∆S2 for the entire process ¯ ¯ ¯ ¯ ∂S ¯ ∂U ¯ 3/2 dU = T dS − pdV ⇒ T ¯ = ¯ = cv = |aT{z } |{z} ∂T V ∂T V =0 from c) dS c ⇒ = v dT T Therefore the entropy transfer in the entire process is

ZT0 ZT0 c 2 ³ ´ ∆S = v dT = a T 1/2dT = a T 3/2 − T 3/2 i T 3 0 i Ti Ti 1 ³ ´ ∆S + ∆S = 0 ⇒ T 3/2 = T 3/2 + T 3/2 1 2 0 2 1 2

ZT0 2 ³ ´ Heat transferred to F : Q = T dS = a T 5/2 − T 5/2 . 1 1 5 0 1 T1 ZT2 2 ³ ´ Heat transferred from F : Q = T dS = a T 5/2 − T 5/2 . 2 2 5 2 0 T0 Therefore the total work done by the Carnot-machine is 2 ³ ´ W = Q − Q = a T 5/2 + T 5/2 − 2T 5/2 2 1 5 1 2 0

(4 points)

20 Statistical Mechanics and Thermodynamics Problem 2 Phase Transition in a Superconductor Solution a) ¯ ¯ ¯ ¯ ∂Q¯ ∂S ¯ cH ≡ ¯ = T ¯ ∂T H ∂T H ¯ ¯ ¯ ¯ ∂S ¯ ∂S ¯ dS = ¯ dT + ¯ dM ∂T M ∂M T ¯ ¯ ¯ ¯ ∂S ¯ ∂S ¯ ∂S ¯ ∂M ¯ ¯ = ¯ + ¯ ¯ ∂T ¯ ∂T ¯ ∂M ¯ ∂T ¯ H M T | {z H} =0 where the last term is zero because M is independent of T . Then ¯ ¯ ¯ ¯ ∂S ¯ ∂Q¯ cH = T ¯ = ¯ ≡ cM ∂T M ∂T M

(2 points) b) The transition takes place at constant T and H. The thermodynamic function whose variables are T and H is the Gibbs-potential:

dG = −SdT − MdH

Gsuper = Gnormal at every point on HC(T ), so dGS = dGN which we then write as

−S dT − M dH = −S dT − M dH S S N |{z}N =0 ¯ ¯ dH ¯ dHC SN − SS 4π ¯ = = = − (SN − SS) dT trans. dT MS VHC(T ) line

(3 points) c) By the third law S → 0 as T → 0. But the ﬁgure shows HC(T = 0) is ﬁnite. Therefore

dH C → 0 as T → 0. dT

The transition is second order where SN − SS = 0, that is, the latent heat equals zero.

V dH S − S = − H (T ) C N S 4π C dT

21 • At T = 0 the transition is second order because both entropies go to zero.

• At T = TC(H = 0) the transition is second order since HC(T ) = 0 and dHC/dT is ﬁnite.

• At all other temperatures the transition is ﬁrst order since both HC(T ) and dHC/dT are ﬁnite.

(2 points) d) Use H and T as variables ¯ ¯ ¯ ¯ ∂S ¯ ∂S ¯ dS(H,T ) = ¯ dH + ¯ dT ∂H T ∂T H

¯ ¯ ¯ ¯ ¯ ¯ ∂S ¯ cH ∂S ¯ ∂M ¯ ¯ = ¯ = − ¯ = 0 ∂H T T ∂T H ↑ ∂T H Maxwell relation

Z c a S = H dT = T 3V T < T T 3 C b = T 3V + γT V T > T 3 C µ ¶ b − a S − S = T 3V + γT V T = T (H = 0) N S 3 C

µ ¶ b − a γ = T 2 3 C µ ¶ 3γ 1/2 T (H = 0) = C b − a

(3 points)