Classical Mechanics Problem 1: Central Potential Solution a)
Integrals of motion for a central potential V (r): Angular Momentum L = rv¡t = r2 φ˙ ¢ Energy per unit mass E = 12 r˙ 2 + vt2 + V (r) = 21 r˙ 2 + Veff (r) where vt is the tangential velocity and Veff is defined as Veff (r) = V (r) +
L2 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: dVeff =0 dr
dV L2 v2 = 3 = t = rφ˙ 2 dr r r
⇒
then L φ˙ = ωφ = 2 = r
µ
1 dV r dr
¶1/2
so for the period we get
Tφ = b)
2π = 2π ωφ
µ
1 dV r dr
¶−1/2
Write the orbit as in the statement of the problem: r(t) = r0 + ²(t)
with
dVeff (r0 ) = 0 dr
and
²2 ¿ r02 .
The energy per unit mass is now E = 12 ²˙2 + Veff (r0 + ²), and since ² is small we may Taylor-expand the potential as Veff (r0 + ²) = Veff (r0 ) +
dVeff 1 d2 Veff (r0 )² + (r0 )²2 + O(²3 ) 2 dr 2 dr | {z } =0
so then
1 2 1 d2 Veff ²˙ + (r0 )²2 + O(²3 ) = const. 2 2 dr2 In the above equation we readily recognize the equation of the simple harmonic oscillator with ¶1/2 µ 2 d Veff ωr = dr2 r=r0 E − Veff (r0 ) =
1
and its general solution is p
E − Veff (r0 ) cos [ωr (t − t0 )] ωr
²(t) = where t0 is an arbitrary constant.
Now return to writing ωr in terms of V (r) instead of Veff (r). d2 Veff d2 V 3L2 d2 V d2 V 3 dV 2 = + = + 3ω = + φ dr2 dr2 r4 dr2 dr2 r dr ¶1/2 · µ ¶¸1/2 µ 2 3 dV 1 d d V 3 dV ωr = + = r dr2 r dr r=r0 r3 dr dr r=r0
ωr2 =
And the radial period is Tr = c)
2π ωr
Stability is determined by the sign of ωr2 . For stability: ωr2 > 0, so 1 d r3 dr
µ ¶ 3 dV r >0 dr
for the Yukawa-potential V (r) = −
GM −kr e r
so the condition is µ ¶ £ ¤ £ ¤ GM 1 d 3 dV r = 3 e−kr 1 + kr − (kr)2 > 0 ⇒ 1 + kr − (kr)2 > 0 3 r dr dr r Ã√ ! Ã√ ! £ ¤ 5−1 5+1 2 1 + kr − (kr) = + kr − kr > 0 2 2 which is satisfied only if
Ã√ kr <
Therefore circular orbits are unstable for Ã√ kr > ¥
2
5+1 2
5+1 2
!
!
d)
The outermost stable circular orbit is at Ã√ r0 =
5+1 2k
!
its energy per unit mass is 1 1 E = V (r0 ) + (r0 ωφ )2 = V (r0 ) + 2 2 GM −kr0 1 GM −kr0 e (kr0 − 1) = e 2 r0 r0
µ ¶ dV r dr r=r0
Ã√
5−1 4
! >0
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
r
Figure 1: Effective potential against distance from the origin
3
Classical Mechanics Problem 2: Planar Double Pendulum Solution
q1
l
q2
a)
l
L=T −V The moment of inertia for a uniform rod of length l and mass m is I=
1 2 ml 3
about one of the ends
and
1 ml2 about the rod’s center 12 The kinetic energy term we can decompose into three parts: Ic =
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 T1 = ml2 θ˙12 6 1 T2,rot = ml2 θ˙22 24 and ¢ 1 ¡ T2,trans = m x˙ 2c + y˙ c2 2 where xc and yc are the coordinates of the second rod’s center of mass, so l sin θ2 2 l yc = −l cos θ1 − cos θ2 2
xc = l sin θ1 +
from which
¸ · 1 x˙ 2c + y˙ c2 = l2 θ˙12 + θ˙22 + θ˙1 θ˙2 (sin θ1 sin θ2 + cos θ1 cos θ2 ) 4 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 V1 = −mg cos θ1 ; V2 = −mg l cos θ1 + cos θ2 2 2 The full Lagrangian is then L = T1 + T2,rot + T2,trans − V1 − V2 ¸ · · ¸ 1 3 2 1 = ml2 θ˙12 + θ˙22 + θ˙1 θ˙2 cos (θ1 − θ2 ) + mgl cos θ1 + cos θ2 3 6 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 Since we are going to look for normal modes with θj = θˆj exp (iωt), where the θˆj ¿ 1, we immediately see that in the term θ˙1 θ˙2 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˙12 + θ˙22 + θ˙1 θ˙2 − mgl θ12 + θ22 + const. 3 6 4 4 The Euler-Lagrange equations are ∂L d ∂L = ˙ dt ∂ θj ∂θj so in the specific case: 4¨ θ1 + 3 1¨ θ1 + 2
1¨ θ2 + 2 1¨ θ2 + 3
µ ¶ g 3 θ1 = 0 l 2 µ ¶ g 1 θ2 = 0 l 2
if we now look for normal modes, as mentioned, the above set of equations takes the form #" # " ¡ ¢ 4 2 3g 1 2 θˆ1 3ω − 2 l 2ω ¡1 2 1 g¢ =0 1 2 θˆ2 2ω 3ω − 2 l Non-trivial solutions exist if the determinant of the matrix on the left is zero. Denoting ω 2 = λg/l, we can write this condition as ¶µ ¶ µ 3 1 1 λ2 4 λ− λ− − = 0, 3 2 3 2 4 5
that is whose solutions are
7 2 7 3 λ − λ+ =0 36 6 4 6 λ± = 3 ± √ , 7
so finally
·µ ω± =
c)
6 3± √ 7
¶ ¸1/2 g l
To sketch the eigenmodes, find eigenvectors of the matrix in part b). • ω 2 = λ− g/l (low-frequency mode) µ ¶ ´ 3 8 ˆ 1³ √ ˆ θ2 = − θ1 = 2 7 − 1 θˆ1 λ 3 3 √ (2 7 − 1)/3 > 0 and real, therefore the two pendula are in phase; • ω 2 = λ+ g/l (high-frequency mode) µ ¶ ´ 3 8 ˆ 1³ √ θˆ2 = − θ1 = −2 7 − 1 θˆ1 λ 3 3 √ (−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 ¸1/2 ·³ q πnx ´2 ³ πny ´2 ³ πnz ´2 ω = c kx2 + ky2 + kz2 = c + + ; a b b
nx , ny , nz ∈ 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 ~ k yˆ, the boundary conditions require we are told to pick the mode with E ~ t) = E0 yˆ sin πx sin πz cos ωt E(r, a b The magnetic induction we can get from Faraday’s Law: ³ ´ ~ ∂B ∂Ey ∂Ey ~ =x = −c ∇ × E ˆc − zˆc ∂t ∂z ∂x ³ π πx πz π πx πz ´ = x ˆcE0 sin cos − zˆcE0 cos sin cos ωt b a b a a b µ ¶ ˆ πx πz zˆ πx πz ~ t) = − πcE0 x B(r, sin cos − cos sin sin ωt ω b a b a a b where the frequency ω is (by the argument above) µ ω = πc b)
1 1 + 2 a2 b
¶1/2
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 E0 πx πz sin sin cos ωt 4π a b σ(x, b, z) = −σ(x, 0, 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 ~ = n ˆ×B 7
4π ~κ c
where n ˆ is a unit vector normal to the surface, so µ ¶ c2 E0 zˆ πx πz x ˆ πx πz ~κ(x, 0, z) = sin cos + cos sin sin ωt 4ω b a b a a b ~κ(x, b, z) = −~κ(x, 0, z) πz c2 E0 yˆ sin sin ωt 4ω a b ~κ(a, y, z) = −~κ(0, y, z) ~κ(0, y, z) = −
c2 E0 yˆ πx sin sin ωt 4ω b a ~κ(x, y, b) = −~κ(x, y, 0)
~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 ³ ´ 1 ~ ~ d2 x F (t) = ~κ × B 2c b×b
E 2 c2 π F~ (x = 0, t) = − 0 2 2 x ˆ 8ω a µ x = −ˆ
Zb
Zb
dz sin2
dy 0
0
c b E0 sin ωt 4ω a
πz sin2 ωt b
¶2
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 E 2 c2 π F~mag (y = 0, t) = − 0 2 yˆ 8ω
Za dx 0
µ
Zb dz 0
1 πx πz 1 πx πz cos2 sin2 + 2 sin2 cos2 2 a a b b a b
¶ ´2 1 µ b 1³ c a E0 sin ωt + = −ˆ y 2 2ω 4π a b To simplify this result further, use ω from part a) µ ¶−1 µ ¶−1 1 1 b a −2 ω −2 = (πc)−2 + = ab(πc) + a2 b2 a b Then
¶2 µ ab E0 ~ sin ωt Fmag (y = 0, t) = −ˆ y 4 2π 3 8
¶ sin2 ωt
The electric component of the force can be written as Z F~el (y = 0, t) =
= yˆ
1 ~ 2 E2 σ E d x = yˆ 0 cos2 ωt 2 8π 1 2
µ
E0 cos ωt 4
¶2
Za
Zb dz sin2
dx 0
0
πx πz sin2 a b
ab π3
F~tot (y = 0, t) = F~el (y = 0, t) + F~mag (y = 0, t) µ ¶2 ¢ E0 ab ¡ 2 cos ωt − sin2 ωt = yˆ 4 2π 3 µ ¶2 E0 ab = yˆ cos 2ωt 4 2π 3 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 outward 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 E 2 c2 π F~ (z = 0, t) = − 0 2 2 zˆ 8ω b
Zb
Za dx sin2
dy 0
0
πx sin2 ωt a
´2 a³ c E0 sin ωt = −ˆ z b 4ω 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 B2 1 ~ ~ f~ = E(E · n ˆ) − n ˆ+ B(B · n ˆ) − n ˆ 4π 8π 4π 8π ~ = 0 and B ~ ·x On the x = const. walls n ˆ = ±ˆ x, E ˆ = 0, so E 2 c2 π πz B2 ˆ = ∓ 02 2 x ˆ sin2 sin2 ωt f~(x = {0, a}, t) = ∓ x 8π 8ω a b which is exactly the integrand from part c).
9
~ E ~ · yˆ) = E 2 yˆ and B ~ · yˆ = 0, so we get On the y = const. walls n ˆ = ±ˆ y , E( 1 f~(y = {0, b}, t) = ∓ (E 2 − B 2 )ˆ y 8π · 2 E0 πx πz =± cos2 ωt sin2 sin2 8π a b µ ¶ ¸ πz 1 E02 c2 π 1 πx 2 2 πx 2 2 πz 2 sin + 2 sin cos − cos sin ωt yˆ 8ω 2 a2 a b b a b the sum of the first two integrands from part d). ~ · zˆ) = 0 and B ~ · zˆ = 0, so we get On the z = const. walls n ˆ = ±ˆ z , (E B2 E 2 c2 π πx f~(z = {0, b}, t) = ∓ zˆ = ∓ 0 2 2 zˆ sin2 sin2 ωt 8π 8ω b a the last integrand from part d).
y b F(x) z
s b F(z)
k(x)
k(y)
k(z) a
k(z) –s
F(z) k(x)
k(y)
F(x) 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 x ˆ direction; it has a transverse electric field, so assume E×y ˆ = 0. Then the electron in the atom behaves classically as a damped, driven harmonic oscillator ¡ ¢ me y¨ + γ y˙ + ω02 y = −qE0 e−iωt with the solution y(t) =
ω2
1 qE(t) . 2 − ω0 + iγω me
For the dipole moment per unit volume: P = na (−q)y =
ω02
1 na q 2 E − ω 2 − iγω me
Therefore the volume polarizability is, according to the definition given, α(ω) =
P 1 na q 2 = 2 ²0 E ω0 − ω 2 − iγω ²0 me
(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 ∂t ∂D ∇×H= ∂t
∇ · D = 0;
∇×E=−
∇ · B = 0;
and B = µ0 H, D = ²0 E + P = ²0 (1 + α)E for a single frequency ω. This gives us the following wave-equation ∂2D 1 − ∇2 D = 0. 2 ∂t µ0 ²0 (1 + α) Now let D be that of a plane wave: D ∝ ei(kx−ωt) . Then k 2 = µ0 ²0 (1 + α)ω 2 = (1 + α) ⇒ n(ω) =
p 1 + α(ω)
ω2 c2
One can also get this result by using the microscopic E, B and P fields: µ ¶ 1 ∂ P 2 ∇ · E = − ∇ · P; c ∇ × B = +E ²0 ∂t ²0 11
∂2E 1 ∂P − c2 ∇2 E = − ∂t2 ²0 ∂t
⇒ also
∂2P ∂P na q 2 2 E. + γ + ω P = − 0 ∂t2 ∂t me
Together these give us k 2 = (1 + α)ω 2 /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 1 E(x, t) = 2π
Z∞ ˆ dk ei(kx−ω(k)t) E(k) −∞
Z∞ ˆ E(k) =
dx e
−ikx
E(x, 0) = √
−∞
=e
Z∞
1
2
2πσ 2
dx e−i(k−kc )x−x
/(2σ 2 )
−∞
−i(k−kc )2 σ 2 /2
Now Taylor-expand ω(k) about k = kc : µ ¶ dω ω(k) = ω(kc ) + (k − kc ) + O{(k − kc )2 } dk kc ≡ kc vph + vg (k − kc ) + O{(k − kc )2 } where, by definition, vph and vg are the phase- and group-velocities, respectively. Now let K = k − kc . Then 1 E(x, t) = 2π E(x, t) = e d)
ikc (x−vph t) e
For the dilute gas, n = is complex)
√
dK eikc (x−vph t)+iK(x−vg t)−K
dω = dk
√
µ
2
σ 2 /2
−∞
(x−vg t)2 /(2σ 2 )
From part c) vg =
Z∞
2πσ 2
dk dω
¶−1 =
= eikc (x−vph t) N (x − vg t, σ)
c n
µ ¶−1 d log n 1+ . d log ω
1 + α ≈ 1 + α/2, which we will write as n = nr + ini (for α
nr ≈ 1 + ni ≈
ω02 − ω 2 na q 2 2²0 me (ω02 − ω 2 )2 + γ 2 ω 2
na q 2 γω 2 2²0 me (ω0 − ω 2 )2 + γ 2 ω 2 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 : nr = 1
vg =
µ 1−
d log nr na q 2 =− d log ω 2²0 me γ 2
and
na q 2 2²0 me γ 2
¶−1 c≈
µ 1+
na q 2 2²0 me γ 2
¶ c
Note that vg > c at ω = ω0 . This is called anomalous dispersion. It does not violate 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¨ odinger equation 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) = Ae−α|x|
ψ(x) = cos kx
Both the wave-function and its derivative has to be continuous at the boundaries of the well: cos kw = Ae−αw
ψ: dψ : dx
−k sin kw = −αAe−αw ⇒
k tan kw = α
Directly from Schr¨odinger’s equation: |x| < w E=
|x| > w
~2 k 2 2m
E =−
~2 α2 + V0 2m
~2 k 2 ~2 α2 =− + V0 2m 2m From which we get the transcendental equation: ⇒
·
2mV0 k tan kw = − k2 ~2
¸1/2
Let k ∗ denote the positive root of the equation above, and introduce the following notation: √ 2mV0 π kc = and kmax = . ~ 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 E=
14
~2 k ∗ 2 2m
k tankw kc
H L
* k* * tan k* w
(kc2- –k2)1/2
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: · ¸1/2 2mw2 V0 2 kw tan kw = − (kw) ~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 E→
2 ~2 kmax ~2 π 2 = 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. 1 F = ~
Za dx
p 2m(V (x) − B)
w
V (x) = V0 − eEx B = V0 − eEa
15
⇒
a=
V0 − B eE
V(x)
V0 a
0
–w
√ F = √
2m ~
Za dx
x
w
p
(V0 − B) − eEx
w
¯a µ ¶ 2m 2 1 3/2 ¯¯ = − (V0 − B − eEx) ¯ ~ 3 eE w √ 2m 2 1 3/2 (V0 − B − eEw) = ~ 3 eE 6. Write the energy of the particle as B=
1 mv 2 . 2
Then
2B . m The time it takes for the particle to bounce back and forth once is v2 =
T =
4w , v
so it hits the right wall with frequency ν=
⇒
v 4w
v −2F Probability to escape = e unit time 4w
⇒
Lifetime ∼
16
4w 2F e v
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 θeiωt
sin θe−iωt − cos θ
¸·
x y
¸
· =±
x y
¸
with the normalization condition |x|2 + |y|2 = 1. From the vector-equation y = eiωt
sin θ x cos θ ± 1
and with the normalization, we end up with ¸ · cos (θ/2) |+i = sin (θ/2)eiωt
· |−i =
sin (θ/2) − cos (θ/2)eiωt
¸
2. Decompose the state-vector as |ψi = c+ |+i + c− |−i and write Schr¨odinger’s equation in terms of these vectors: d |ψi = H|ψi dt · ¸ d d i~ c˙+ |+i + c+ |+i + c˙− |−i + c− |−i = B [c+ |+i − c− |−i] dt dt i~
or in the (|+i, |−i) basis " # " #" # d d c+ |+i −i~h+| dt |−i c+ B − i~h+| dt d i~ = d d dt c− c− −i~h−| dt |+i −B − i~h−| dt |−i which, with the given concrete form of the vectors, is " # " #" # c+ c+ B + ~ω sin2(θ/2) −~ω cos (θ/2) sin (θ/2) d i~ = . dt c− c− −~ω sin (θ/2) cos (θ/2) −B + ~ω cos2(θ/2) Now use the identities 1 (1 − cos θ) 2 1 cos2 (θ/2) = (1 + cos θ) 2 1 sin (θ/2) cos (θ/2) = sin θ 2 sin2 (θ/2) =
17
to get d i~ dt
"
c+ c−
#
1 = 2
"
#"
2B − ~ω cos θ
−~ω sin θ
−~ω sin θ
−2B + ~ω cos θ
c+ c−
# .
Note that in the last equation we dropped the part of the Hamiltonian that was proportional 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: " # " #" # c+ Dz Dx c+ d i~ = . dt c− Dx −Dz c− with the solution "
c+
#
( = exp
c−
à = cos
−i ~
"
~ |D|t ~
Dz
Dx
Dx
−Dz
#) "
!
#
1 0
à ˆ · ~σ sin − iD
~ |D|t ~
! .
And so à c+ = cos à 2
|c+ | = cos
3. For B À ~ω, Dz → D, so
2
~ |D|t ~
~ |D|t ~
!
!
Dz −i sin ~ |D|
18
~ |D|t ~
!
! Ã ~ Dz2 2 |D|t − i 2 sin D ~
|c+ |2 → 1
(Adiabatic theorem)
Ã
Statistical Mechanics and Thermodynamics Problem 1 Thermodynamics of a Non-Interacting Bose Gas Solution a) np =
1
Ep =
eβ(Ep −µ) − 1
p2 2m
At and below TBEC µ = 0. At exactly TBEC , there are no atoms in the condensate and V N= 2π~3
Z
µ
d3 p eβp2 /(2m)
−3
−1
= (2π~)
V
2m β
¶3/2
Z∞ 4π 0
|
x2 dx −1 {z }
ex 2
=I1
n=
1 2π 2
µ
µ kTBEC =
2mkT ~2 2π 2 n I1
¶3/2
¶2/3
I1 ~2 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 Classical Gas p µ 1/V Bose Gas Phase Transition Line –5/3 pBEC µ V
V (2 points)
19
c) V U= 2π~3
I2 = Nc I1 cv =
Z
µ
p2 d3 p −3 = (2π~) V 2 2m eβp /(2m) − 1
2m β
¶
µ
2m β
¶5/2
4π 2m
Z∞
x4 dx ex2 − 1 0 | {z } =I2
1 I2 = Nc kT ∝ T 5/2 2m I1
5 I2 V Nc k = 2 I1 2π 2
µ
2mkT ~2
¶3/2 µ ¶ 5 k I2 ∝ T 3/2 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
from c)
=0
cv dS = dT T Therefore the entropy transfer in the entire process is ⇒
ZT0 ∆Si =
cv dT = a T
Ti
∆S1 + ∆S2 = 0
ZT0 T 1/2 dT =
´ 2 ³ 3/2 3/2 a T0 − Ti 3
Ti
⇒
3/2
T0
=
´ 1 ³ 3/2 3/2 T1 + T2 2
ZT0 Heat transferred to F1 :
Q1 =
T dS =
´ 2 ³ 5/2 5/2 a T0 − T1 . 5
T dS =
´ 2 ³ 5/2 5/2 a T2 − T0 . 5
T1
ZT2 Heat transferred from F2 :
Q2 = T0
Therefore the total work done by the Carnot-machine is W = Q2 − Q1 =
´ 2 ³ 5/2 5/2 5/2 a T1 + T2 − 2T0 5 (4 points)
20
Statistical Mechanics and Thermodynamics Problem 2 Phase Transition in a Superconductor Solution a) ¯ ¯ ∂S ¯¯ ∂Q ¯¯ = T ∂T ¯H ∂T ¯H ¯ ¯ ∂S ¯¯ ∂S ¯¯ dS = dT + dM ∂T ¯M ∂M ¯T ¯ ¯ ¯ ¯ ∂S ¯¯ ∂S ¯¯ ∂M ¯¯ ∂S ¯¯ = + ∂T ¯H ∂T ¯M ∂M ¯T ∂T ¯H | {z }
cH ≡
=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 − M dH Gsuper = Gnormal at every point on HC (T ), so dGS = dGN which we then write as −SS dT − MS dH = −SN dT − MN dH |{z} =0
¯ 4π SN − SS dH ¯¯ dHC = = − (SN − SS ) = ¯ dT trans. dT MS V HC (T ) line
(3 points)
c)
By the third law S → 0 as T → 0. But the figure shows HC (T = 0) is finite. Therefore dHC → 0 as dT
T → 0.
The transition is second order where SN − SS = 0, that is, the latent heat equals zero. SN − SS = −
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V dHC HC (T ) 4π dT
• 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 dS(H, T ) =
¯ ¯ ∂S ¯¯ ∂S ¯¯ dH + dT ∂H ¯T ∂T ¯H
¯ ∂S ¯¯ cH = ¯ ∂H T T
¯ ∂S ¯¯ ∂T ¯H
= ↑
Maxwell relation
¯ ∂M ¯¯ − =0 ∂T ¯H
Z S=
cH a dT = T 3 V T 3 b = T 3 V + γT V 3 µ ¶ b−a SN − SS = T 3 V + γT V 3
T < TC T > TC T = TC (H = 0)
µ
¶ b−a TC2 3 µ ¶1/2 3γ TC (H = 0) = b−a γ=
(3 points)
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