Istvan Ziegler
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Transcript of 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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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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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)
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