Istvan Ziegler

8/10/2019 Istvan Ziegler

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and its general solution is

(t) =

E Veff(r0)

rcos[r(t t0)]

wheret0 is an arbitrary constant.

Now return to writing r in terms ofV(r) instead ofVeff(r).

2r =d2Veff

dr2 =

d2V

dr2 +

3L2

r4 =

d2V

dr2 + 32 =

d2V

dr2 +

3

r

dV

dr

r =

d2V

dr2 +

3

r

dV

dr

1/2r=r0

=

1

r3d

dr

r3

dV

dr

1/2r=r0

And the radial period is

Tr = 2

r

c) Stability is determined by the sign of 2r . For stability: 2r >0, so

1

r3d

dr

r3

dV

dr

> 0

for the Yukawa-potential

V(r) = GMr

ekr

so the condition is

1

r3d

dr

r3

dV

dr

=

GM

r3 ekr

1 + kr (kr)2> 0 1 + kr (kr)2> 0

1 + kr (kr)2=

5 1

2 + kr

5 + 1

2 kr

> 0

which is satisfied only if

kr

5 + 1

2

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d) The outermost stable circular orbit is at

r0=

5 + 12k

its energy per unit mass is

E=V(r0) +1

2(r0)

2 =V(r0) +1

2

r

dV

dr

r=r0

1

2

GM

r0ekr0(kr0 1) = GM

r0ekr0

5 1

4

> 0

Ifr0 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.

r

Veff

Figure 1: Effective potential against distance from the origin

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Classical Mechanics Problem 2: Planar Double Pendulum

Solution

l

l

a) L= T VThe moment of inertia for a uniform rod of length l and mass m is

I=1

3ml2 about one of the ends

and

Ic = 1

12ml2 about the rods center

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 ofthe center of mass of the second rod and T2,rot is its rotational energy about its centerof mass. Then

T1=1

6ml221

T2,rot = 1

24ml222

and

T2,trans=1

2m

x2c+ y2c

wherexc andyc are the coordinates of the second rods center of mass, so

xc= l sin 1+

l

2sin 2

yc= l cos 1 l2

cos 2

from which

x2c+ y2c =l

2

21+

1

422+

12(sin 1sin 2+ cos 1cos 2)

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that is7

36 2

7

6 +

3

4= 0

whose solutions are

= 3 67

,

so finally

=

3 6

7

g

l

1/2

c) To sketch the eigenmodes, find eigenvectors of the matrix in partb).

2 =g/l (low-frequency mode)

2= 3

8

3

1=

13

2

7 1

1

(2

7 1)/3> 0 and real, therefore the two pendula are in phase; 2 =+g/l (high-frequency mode)

2=

3

8

3

1=

1

3

2

7 1

1

(27 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.

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Electromagnetism Problem 1

Solutiona) Normal modes are products of harmonic standing waves in thex, y and z directions.

For their frequencies, we have

= c

k2x+ k2y+ k

2z =c

nxa

2+ny

b

2+nz

b

21/2; 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 thatny = 0, nz = 0 does not satisfy the boundary conditionE,at wall = 0). Since

we are told to pick the mode with E y, the boundary conditions require

E(r, t) =E0y sin

x

a sin

z

b cos t

The magnetic induction we can get from Faradays Law:

B

t = c

E

= xc

Eyz

zc Eyx

=

xcE0

b sin

x

a cos

z

b zcE0

acos

x

a sin

z

b

cos t

B(r, t) = cE0

x

bsin

x

a cos

z

b z

acos

x

a sin

z

b

sin t

where the frequency is (by the argument above)

= c

1

a2+

1

b2

1/2

b) At a boundary of media, the discontinuity in thenormalcomponent of the electric fieldis 4 times the surface charge density , so

Ey(x, 0, z) = 4

(x, 0, z) =E04

sinx

a sin

z

b cos t

(x,b,z) = (x, 0, z)and

(0, y , z) =(a,y,z) = (x,y, 0) = (x,y,b) 0Similarly, at the boundary of media the discontinuity of the tangential component ofthe magnetic field is given by the surface current

n B= 4c

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where nis a unit vector normal to the surface, so

(x, 0, z) = c2E04

zb

sin xa

cos zb

+ xa

cos xa

sin zb

sin t

(x,b,z) = (x, 0, z)

(0, y , z) = c2E04

y

asin

z

b sin t

(a,y,z) = (0, y , z)

(x,y, 0) = c2E04

y

bsin

x

a sin t

(x,y,b) = (x,y, 0)

c) Since there is no charge on theb b sides, the force there is purely magnetic and isgiven by

F(t) = 1

2c

bb

B

d2x

F(x= 0, t) = E20 c

2

82a2x

b0

dy

b0

dz sin2z

b sin2t

= x

c

4

b

aE0sin t

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 inthe equation above).

d) Start with the sides where y = const. The magnetic component of the force can bewritten as above

Fmag(y= 0, t) = E20 c

2

82 y

a0

dx

b0

dz

1

a2cos2

x

a sin2

z

b +

1

b2sin2

x

a cos2

z

b

sin2t

= y 12

c2

E0sin t2 1

4

b

a+

a

b

To simplify this result further, use from part a)

2 = (c)2

1

a2+

1

b2

1=ab(c)2

b

a+

a

b

1Then

Fmag(y = 0, t) = y

E04

sin t

2ab

23

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The electric component of the force can be written as

Fel(y= 0, t) =

1

2 E d2x= y

E208

cos2t

a0

dx

b0

dz sin2x

a sin2

z

b

= y1

2

E04

cos t

2ab

3

Ftot(y= 0, t) = Fel(y= 0, t) + Fmag(y= 0, t)

= y

E04

2ab

23

cos2t sin2t

= yE04 2

ab

23cos2t

andFtot(y= b, t) = Ftot(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

F(z = 0, t) = E20 c

2

82b2 z

b

0

dy

a

0

dx sin2 x

a sin2

t

= z ab

c4

E0sin t2

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

f= 1

4E( E n) E

2

8n +

1

4B( B n) B

2

8n

On thex= const. walls n= x, E= 0 and B x= 0, so

f(x= {0, a}, t) = B2

8x= E

20 c

2

82a2x sin2

z

b sin2t

which is exactly the integrand from part c).

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On they= const. walls n= y, E( E y) =E2y and B y= 0, so we get

f(y = {0, b}, t) = 18

(E2 B2)y

=

E208

cos2t sin2x

a sin2

z

b

E20 c

2

82

1

a2cos2

x

a sin2

z

b +

1

b2sin2

x

a cos2

z

b

sin2t

y

the sum of the first two integrands from part d).

On thez= const. walls n= z, ( E z) = 0 and B z= 0, so we get

f(z=

{0, b

}, t) =

B2

8

z =

E20 c

2

82

b2

z sin2x

a

sin2t

the last integrand from part d).

F(z)

(y)

(z)

(x)

x

b

b

a

z

y

(y)

(z)

(x)

F(x)

F(x)F(z)

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

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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 thexdirection; it has a transverse electric field, so assumeE y = 0. Then the electron in the atom behaves classically as a damped, drivenharmonic oscillator

me

y+ y+ 20y

= qE0eitwith the solution

y(t) = 1

2 20+ iqE(t)

me.

For the dipole moment per unit volume:

P =na(q)y = 1

20 2 inaq2E

me

Therefore the volume polarizability is, according to the definition given,

() = P

0E =

1

20 2 inaq2

0me

(A quantum mechanical derivation would give this same expression multiplied by theoscillator strength f for the transition.)

b) With no free charges or currents, Maxwells equations read

D= 0; E= B

t

B= 0; H= Dt

and B = 0H, D = 0E + P= 0(1 + )E for a single frequency . This gives us thefollowing wave-equation

2D

t2 1

00(1 + )2D= 0.

Now let D be that of a plane wave: D ei(kxt). Then

k2 =00(1 + )2 = (1 + )

2

c2

n() =

1 + ()

One can also get this result by using the microscopicE, B and Pfields:

E= 10

P; c2 B= t

P

0+ E

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2E

t2

c2

2E=

1

0

P

talso

2P

t2 +

P

t + 20P=

naq2

meE.

Together these give us k2 = (1 +)2/c2 for a plane wave, as before. (Note that weare neglecting dipole-dipole interactions in the dilute gas.)

c) We start by noting that according to Fourier-analysis

E(x, t) = 1

2

dk ei(kx(k)t)E(k)

E(k) =

dx eikxE(x, 0) = 122

dx ei(kkc)xx2/(22)

=ei(kkc)22/2

Now Taylor-expand (k) aboutk= kc:

(k) =(kc) +

d

dk

kc

(k kc) + O{(k kc)2}

kcvph+ vg(k kc) + O{(k kc)2}where, by definition,vph andvg are the phase- and group-velocities, respectively. Now

letK= k kc. Then

E(x, t) = 1

2

dK eikc(xvpht)+iK(xvgt)K22/2

E(x, t) = eikc(xvpht)e(xvgt)

2/(22)

22

=eikc(xvpht)N(x vgt, )

d) From part c)

vg =d

dk =

dk

d

1=

c

n

1 +

d log n

d log

1.

For the dilute gas, n= 1 + 1 +/2, which we will write as n= nr+ ini (for is complex)

nr 1 + naq2

20me

20 2(20 2)2 + 22

ni naq2

20me

(20 2)2 + 22

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Here the real part nr of the index of refraction determines the dispersion, and the

imaginary partni determines the absorption/gain coefficient. At = 0:

nr = 1 and d log nr

d log = naq

2

20me2

vg =

1 naq

2

20me2

1c

1 +

naq2

20me2

c

Note that vg > c at = 0. This is called anomalous dispersion. It does not vio-late causality because signals (information) cannot travel faster than the minimum of(vph, vg), and nowvph = c (sincenr = 1). Also, the waves are damped by the electronicresonance maximally at = 0.

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Quantum Mechanics Problem 1

Solution

1. The ground state will have no nodes, so we can pick the even part of the general solutionof the free Schrodinger equation inside the well. Outside the well, square-integrabilitydemands the solutions to vanish at infinity. The wave-function for the ground state isthen

|x| < w |x| > w(x) = cos kx (x) =Ae|x|

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

: cos kw = Aew

d

dx : k sin kw = Aew

k tan kw =

Directly from Schrodingers equation:

|x| < w |x| > w

E=2k2

2m E=

22

2m + V0

2k2

2m =

22

2m + V0

From which we get the transcendental equation:

k tan kw =

2mV02

k21/2

Let k denote the positive root of the equation above, and introduce the followingnotation:

kc =

2mV0

and kmax=

2w.

Clearly, the LHS of the equation diverges at kmax; and the RHS describes a circle withradiuskc, as shown in Fig. 4.

For the energy we have

E=

2k2

2m

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kc kmax

k

k*

k

*

k*

tan w

kc

ktankw

kc2 k

2

1/2( )

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

2. Write the result of part 1 in the non-dimensional form:

kw tan kw =

2mw2V0

2 (kw)2

1/2According to the condition given in the statement of the problem, the radius of thecircle on the RHS (that in Fig. 4) goes to infinity, therefore

k kmaxand

E 2k2max2m

=

22

8mw2

3. The potential barrier on the low-potential side of the well, denoteda in the figure, willbe 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 ofthe ground-state wave-function vanishes).

5.

F = 1

aw

dx2m(V(x) B)V(x) =V0 eEx

B = V0 eEa a= V0 BeE

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x

V x( )

V0

w w0

a

F =2m

aw

dx

(V0 B) eEx

=

2m

2

3

1

eE

(V0 B eEx)3/2aw

=

2m

2

3

1

eE (V0 B eEw)3/2

6. Write the energy of the particle as

B=1

2mv2.

Then

v2 =2B

m.

The time it takes for the particle to bounce back and forth once is

T = 4w

v ,

so it hits the right wall with frequency

= v

4w

Probability to escapeunit time

= v

4we2F

Lifetime 4wv

e2F

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Quantum Mechanics Problem 2

Solution1. Drop the t-label for simplicity. Then we have

H=B

cos sin eit

sin eit cos

,

and for the eigenvectors solve cos sin eit

sin eit cos

x

y

=

x

y

with the normalization condition|x|2 + |y|2 = 1. From the vector-equation

y= e

it sin

cos 1 xand with the normalization, we end up with

|+ =

cos(/2)

sin(/2)eit

| =

sin(/2)

cos(/2)eit

2. Decompose the state-vector as

| =c+|+ + c|and write Schrodingers equation in terms of these vectors:

id

dt| =H|

i

c+|+ + c+ ddt

|+ + c| + c ddt

|

= B [c+|+ c|]

or in the (|+, |) basis

id

dt

c+

c

=

B i+| ddt |+ i+| ddt |

i| ddt |+ B i| ddt |

c+

c

which, with the given concrete form of the vectors, is

id

dt

c+

c

=

B+ sin2(/2) cos(/2)sin(/2) sin(/2)cos(/2) B+ cos2(/2)

c+

c

.

Now use the identities

sin2(/2) = 1

2(1 cos )

cos2(/2) = 1

2(1 + cos )

sin(/2)cos(/2) = 1

2sin

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18/22

to get

iddt c+

c

= 12 2B cos sin sin 2B+ cos c+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 identicalfor the coefficientsc, c+. This we can rewrite in the form:

id

dt

c+

c

=

Dz Dx

Dx Dz

c+

c

.

with the solution

c+c = expi Dz Dx

Dx Dz 1

0 = cos

| D|t

iD sin

| D|t

.

And so

c+= cos

| D|t

i Dz| D| sin

| D|t

|c+|2 = cos2| D|t i

D2zD2

sin2| D|t

3. ForB , Dz D, so|c+|2 1

(Adiabatic theorem)

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Statistical Mechanics and Thermodynamics Problem 1

Thermodynamics of a Non-Interacting Bose GasSolution

a)

np = 1

e(Ep) 1 Ep = p2

2m

At and belowTBEC= 0. At exactlyTBEC, there are no atoms in the condensate and

N= V

23

d3p

ep2/(2m) 1 = (2)3 V

2m

3/24

0

x2dx

ex2 1

=I1

n= 1

22

2mkT

2

3/2I1

kTBEC=

22n

I1

2/3

2

2m

(2 points)

b) The above integral with = 0 also applies below TBEC, but it then gives the numberofnon-condensedatoms. So on an isotherm below Vcritical

Nnon-condensed is constant T is constant

p is constant(think of the kinetic origin of pressure)

V

p

Classical Gas 1/p V

Phase Transition Line

pBEC V

5/3

Bose Gas

(2 points)

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20/22

c)

U= V

23

p2

2m

d3p

ep2/(2m) 1= (2)3 V

2m

5/2 42m

0

x4dx

ex2 1 =I2

=NcI2I1

2m

1

2m =Nc

I2I1

kT T5/2

cv = 5

2

I2I1

Nck= V

22

2mkT

2

3/2k

5

2I2

T3/2

(2 points)

d) From the reversibility of the Carnot-cycle:

dS1= dS2 for 1 cycleS1= S2 for the entire process

dU=T dSpdV=0

T ST

V

= U

T

V

=cv =aT3/2

from c)

dSdT

=cvT

Therefore the entropy transfer in the entire process is

Si =

T0Ti

cvTdT =a

T0Ti

T1/2dT =2

3 aT3/20 T3/2i

S1+ S2= 0 T3/20 =1

2

T3

/21 + T

3/22

Heat transferred to F1: Q1=

T0T1

T dS= 2

5a

T5/2

0 T5/21

.

Heat transferred from F2: Q2=

T2

T0T dS=

2

5a

T5

/22 T5/20

.

Therefore the total work done by the Carnot-machine is

W =Q2 Q1= 25

a

T5/21 + T

5/22 2T5/20

(4 points)

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Statistical Mechanics and Thermodynamics Problem 2

Phase Transition in a SuperconductorSolution

a)

cH QT

H

= T S

T

H

dS= S

T

M

dT+ S

M

T

dM

S

T

H

= S

T

M

+ S

M

T

M

T

H =0

where the last term is zero because Mis independent ofT. Then

cH= TS

T

M

= Q

T

M

cM

(2 points)

b) The transition takes place at constantT and H. The thermodynamic function whosevariables are T andH is the Gibbs-potential:

dG= SdT M dH

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

SSdT MSdH= SNdT MN=0

dH

dH

dT

trans.

line

=dHC

dT =

SN SSMS

= 4V HC(T)

(SN SS)

(3 points)

c) By the third lawS 0 as T 0. But the figure showsHC(T= 0) is finite. Therefore

dHC

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

SN SS= V4

HC(T)dHCdT

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AtT= 0 the transition is second order because both entropies go to zero. AtT =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) anddHC/dT are finite.

(2 points)

d) UseH and Tas variables

dS(H, T) = S

H

T

dH+ S

T

H

dT

S

H

T

= cH

T

S

T

H

=Maxwellrelation

MT

H

= 0

S=

cH

T dT =

a

3T3V T < T C

= b

3T3V + T V T > T C

SN SS=

b a3

T3V + T V T =TC(H= 0)

=

b a

3

T2C

TC(H= 0) =

3

b a1/2

(3 points)

22

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