Problem 1
Consider the two isolated capacitors shown in the figure. Capacitor #1 has
capacitance C1 = 1µF and capacitor #2 has capacitance C2 = 2µF . Suppose
that you have a total charge Qtot = 30µC that must be stored on the two
capacitors, with Q1 placed on C1 and Q2 placed on C2 such that Q1 +Q2 = Qtot
(since the available free charge is distributed Q1,2 ≥ 0).
(a) How much electric charge, Q1 and Q2 (in µC), would you place on each
capacitor to maximize the total amount of stored energy, Utot = U1 + U2 ,
and what is Utot (max) (in µJ)?
(b) How much electric charge, Q1 and Q2 (in µC), would you place on each
capacitor to minimize the total amount of stored energy, Utot = U1 + U2 ,
and what is Utot (min) (in µJ)?
(c)What are the potential drops, ∆V1 and ∆V2 (in volts), across each of the
two capacitors in (b)?
Solution
The charge on a capacitor and voltage across are related as
Q = CV
(1)
The energy stored is
1
1
1 Q2
QV = CV 2 =
(2)
2
2
2 C
Given that the total charge is fixed the energy spends only on Q1 and Q2 = Qtot − Q1
U=
U=
1
1
Q21 +
(Qtot − Q1 )2
2C1
2C2
(3)
(a) Eq.(3)is a quadratic function, since C1 > 0 and C2 > 0 it has a minimum.The
maximum is restricted only by other considerations.problem implies that Q1 ≥
0.Therefore maximum energy is reached when all charge is placed either on the
first or on the second capacitor.Clearly, since C1 < C2 all charge should be
placed on C1 .
1 2
Qtot = 4.5 × 10−4 = 450 µJ
(4)
Umax =
2c1
(b)The minimum
∂U
Q1
Q2
Q1
Q2
=0=
−
→
=
(5)
∂Q1
C1
C2
C1
C2
occurs when voltages are equal.This happens when capacitors are connected in
parallel.
Ctot = C1 + C2 = 3 µF
(6)
1
U=
Q2tot = 150 µJ
(7)
2Ctot
In this case
C1
C2
Q1 =
Qtot Q2 =
Qtot
(8)
C1 + C2
C1 + C2
(c)Potential drops are equal
V1 = V2 = V =
Qtot
= 10 V.
Ctot
(9)
Problem 2
Three noninteracting particles of mass m move inside a 1-dimensional infinite
square well potential of size a.
(a)Assume that the particles are distinguishable. Write down the general timeindependent wave functions, properly normalized. Label the wave functions with any appropriate quantum numbers.
(b)Using (a), find the lowest 6 energy eigenvalues in units of E0 = π 2 h̄2 /(2ma2 )
and show which wave functions they correspond to. Identify any degeneracies.
(c)Assume now that the particles are indistinguishable bosons. What is the
ground state wave function and its energy in units of E0 ?
(d)Assume now that the particles are indistinguishable fermions. What is the
ground state wave function and its energy in units of E0 ?
Solution
(a) The Schrodinger’s equation
−
h̄2 d2
ψ(x) = Eψ(x)
2m dx2
(10)
must be solved under boundary condition
π(0) = ψ(a) = 0
Thus
ψn (x) = A sin kn x where kn2 =
(11)
2mE
h̄
(12)
and
kn a = πn, n = 1, 2, 3, ...
Normalization
Z a
sin2
πnx 0
ψn =
a
r
Z
dx =
0
a
1
dx −
2
a
Z
0
1
cos
2
(13)
2πx
a
dx =
2
πnx
h̄2 πn 2
= E0 n2
sin
n =
a
a
2m a
φn1 n2 n3 = ψn1 (x1 )ψn2 (x2 )ψn3 (x3 )
E=
E0 (n21
+
n22
+
n23 )
a
2
(14)
(15)
(16)
(17)
(b) The energy is given by eq. (17), so that the lowest states have energies 12 +
12 + 12 = 3, 22 + 12 + 12 = 6, 9, 11, 12, 14, ... (in units of E0 ) and degeneracy
of each state is 3! = 6 if particles are distinct. For distinct particles states of
any permutational symmetry are allowed. If particles are distinct then there
must be something that distinguishes particles and their wave functions so that
antisymmetric products of wave functions would not cancel out. A proton and
x
neutron is a good example of a situation where particles are distinct but experience the exact same nuclear potential; in this case isospin quantum number is
introduced. Out of all permutations there is only one fully symmetric state and
one fully antisymmetric state. The question about degeneracy here is also related to Gibbs paradox which states that there cannot be a continuous transition
from non-identical to identical particles.
(c) Boson have to be in the fully symmetric state and thus all can be in the ground
state
3/2
2
πx1
πx2
(B)
φ111 (x1 x2 x3 ) =
sin
sin
E = 3E0
(18)
a
a
a
(d) Fermion wave function is antisymmetric
ψ (x )
1 1 1
p
φ123 = Âφ123 = √ ψ2 (x1 )
6 ψ (x )
3
1
ψ1 (x2 )
ψ2 (x2 )
ψ3 (x2 )
ψ1 (x3 )
ψ2 (x3 )
ψ3 (x3 )
(19)
energy of the fermion ground state is
E = E0 (32 + 22 + 12 ) = 14E0
(20)
Problem 3
The ground level of the neutral lithium atom is doubly degenerate (that is, d0
= 2). The first excited level is 6-fold degenerate (d1 = 6) and is at an energy
1.2 eV above the ground level.
(a) In the outer atmosphere of the Sun, which is at a temperature of about
6000 K, what fraction of the neutral lithium is in the excited level? Since
all the other levels of Li are at a much higher energy, it is safe to assume
that they are not significantly occupied.
(b) Find the average energy of Li atom at temperature T (again, consider only
the ground state and the first excited level).
(c) Find the contribution of these levels to the specific heat per mole, CV , and
sketch CV as a function of T .
Use 1mev = 11.6K.
Solution
(a) If the ground level energy is defined as zero and E is the energy of excited level:
X
Z=
di exp(−βεi ) = 2 + 6 exp(−βE)
i
The probability that the atom is in its excited level:
P (E) = 6 exp(−βE)/Z = 6 exp(−βE)/[2 + 6 exp(−βE)] = 3/[3 + exp(βE)]
Since E = 1.2eV , T = 6000K(∼ 0.5eV ), βE = 2.32, exp(βE) ≈ 10, we get:
P (E) = 3/(3 + 10) = 0.23.
(b) The average energy per atom is just 0.23 × 1.2 = 0.276 eV, but we can also
formally compute it using the partition function
hεi = −
ε6e−βε
3ε
1 ∂Z
=
= βε
Z ∂β
2 + 6e−βε
e +3
(c) The specific heat is:
CV =
∂hεi 3kB (βε)2 eβ
=
∂T V
(eβε + 3)2
lim CV = lim CV =
T →0
β→∞
32 −/kB T
e
kB T 2
lim CV = lim CV =
T →∞
β→0
32
16kB T 2
Problem 4
The British mathematician Samuel Earnshaw proved in 1842 that a collection
of point charges cannot be maintained in stable equilibrium solely by the action
of electrostatic interactions.
Consider four identical charges of magnitude Q > 0 located at the vertices of a
square in the x-y plane at the following four positions:
r1 = (a, 0, 0); r2 = (0, a, 0); r3 = (−a, 0, 0); r4 = (0, −a, 0).
(a) Compute the electrostatic potential energy V (r) experienced by a ‘test’
charge of magnitude q > 0 located at an arbitrary point r = (x, y, z) in
three-dimensional space. You may assume that the presence of such test
charge does not disturb the original configuration.
(b) Show that the electrostatic Coulomb force experienced by the test particle
vanishes when it is placed at the origin.
(c) Show that although the test particle is in static equilibrium at the origin,
the equilibrium is unstable. That is, there is at least one direction in threedimensional space for which the frequency of small oscillations around the
equilibrium point is purely imaginary.
Solution
For this problem one needs to compute the electrostatic potential, as well as its first
and second derivatives. Given the validity of the ‘principle of superposition’, we do
this by assuming a single source particle of charge Q located at an arbitrary location
r0 = (x0 , y0 , z0 ). In this case the electrostatic potential between this generic source
particle and the test particle is given by
V0 (r) =
h
i−1/2
−1/2
qQ
= qQ (x−x0 )2 + (y−y0 )2 + (z −z0 )2
= qQ ξk ξk
.
|r − r0 |
where we have defined ξ1 ≡ (x − x0 ), ξ2 ≡ (y − y0 ), and ξ3 ≡ (z − z0 ) — and we have
adopted Einstein’s summation convention. Then, the first derivative of the potential
is given by the following expression:
−3/2
(r−r0 )i
∂V0
= −qQξi ξk ξk
= −qQ
.
∂xi
|r−r0 |3
In turn, the second derivatives become equal to:
2
(r−r0 )i (r−r0 )j
∂ V0
δij
= qQ 3
−
.
∂xi ∂xj
|r−r0 |5
|r−r0 |3
However, the above expressions may now be simplified as all we need are the first and
second derivatives at the origin. That is:
2
i
∂V0
qQ
∂ V0
qQ h
= 2 (r̂0 )i and
= 3 3(r̂0 )i (r̂0 )j − δij ,
∂xi 0
a
∂xi ∂xj 0
a
where (r̂0 )i is a unit vector in the direction of the source.
(a) Using the principle of superposition, the total electrostatic potential energy V (r)
is given by
1
1
1
1
V (r) = V1 (r)+V2 (r)+V3 (r)+V4 (r) = qQ
+
+
+
.
|r−r1 | |r−r2 | |r−r3 | |r−r4 |
(b) The electrostatic force experienced by the test particle when located at the origin
is given by:
qQ F = − 2 r̂1 + r̂2 + r̂3 + r̂4 = 0.
a
(c) To decide whether the equilibrium is stable or not we most analyze the potential
energy V (r) near the origin up to second order in the displacements. That is,
2
∂V
1
∂ V
+ xi xj
+ ...
V (r) = V (0) + xi
∂xi 0 2
∂xi ∂xj 0
We just concluded that the force at the origin vanishes, so the term proportional
to the first derivative is absent. Moreover, given that for the first and third
particles
2
2
∂ V1
∂ V3
qQ
=
= 3 diag(2, −1, −1),
∂xi ∂xj 0
∂xi ∂xj 0
a
whereas for the second and forth
2
2
∂ V4
∂ V2
qQ
=
= 3 diag(−1, 2, −1),
∂xi ∂xj 0
∂xi ∂xj 0
a
we obtain the following expression for the electrostatic potential energy near the
origin:
1 2 1 2
qQ
+
kx + ky − kz 2 ,
V (r) = 4
a
2
2
where we have defined the ‘spring constant’ as k ≡ 2qQ/a3 > 0. This indicates
that whereas the equilibrium is stable against small oscillations in the x-y plane,
the equilibrium is unstable against perturbations along the z-axis; visualize what
happens to the test particle if it is located at (0, 0, ). In other words, assuming a
mass m for the test
p particle, the frequency of ‘small’ oscillations along the z-axis
is given by ω3 = −2k/m, which is purely imaginary. Note that the sum of the
coefficients of the quadratic terms add up to zero. This is no coincidence as any
electrostatic potential Φ(r) = V (r)/q satisfies Laplace’s equation ∇2 Φ(r) = 0 at
any point r away from the location of the sources.
Problem 5
A cube of uniform mass density and side length 2b is placed on a fixed cylinder
of radius r in the initial position shown in the left part of the figure. The cube
may roll without slipping into the position shown in the right part of the figure.
Gravity acts in down direction.
cm
cm
b
b
r
h
r θ
Figure 1: Cube on a cylinder in the initial and in a rolled position, where cm
indicates the center of mass of the cube.
1. Calculate the height h of the center of mass (cm) of the cube as function of
θ and the parameters r, b. Hint: The condition “rolling without slipping”
relates the length of the red line, which connects the center of mass with
the horizontal line in the right figure, to an arc section of the cylinder.
Subsequently, use elementary geometry to find the height h.
2. Write the potential energy U of the cube as function of θ.
3. Show that θ = 0 is an equilibrium position.
4. Is the equilibrium position at θ = 0 stable or unstable?
Solution
1. The arc section is r θ and
h = r cos θ + b cos θ + r θ sin θ .
2.
U (θ) = m g h = m g h (r cos θ + b cos θ + r θ sin θ) .
3.
dU
dθ
=
m g h (−r sin θ − b sin θ + r sin θ + r θ cos θ)
=
m g h (−b sin θ + r θ cos θ)
and
dU = 0.
dθ θ=0
So, θ = 0 is an equilibrium position.
4.
d2 U
= m g h (−b cos θ + r cos θ − r θ sin θ)
dθ2
and, therefore,
d2 U = m g h (−b + r) .
dθ2 θ=0
The equilibrium position at θ = 0 is stable for r > b and unstable for r < b.
To decide the r = b case one needs higher derivatives:
d3 U d3 U =
m
g
h
(−r
sin
θ
−
r
θ
cos
θ)
,
= 0.
dθ3 b=r
dθ3 θ=0
d4 U d4 U =
m
g
h
(−2
r
cos
θ
+
r
θ
sin
θ)
,
= −2 r m g h
dθ4 b=r
dθ4 θ=0
unstable.
Problem 6
The nucleus 113 Cd captures a thermal neutron having negligible kinetic energy,
producing 114 Cd in an excited state. The excited state of 114 Cd decays to the
ground state by emitting a photon. Find the energy of the photon. Useful
constants:
me = 0.511 MeV/c2
mn = 1.008665 u
m(113 Cd) = 112.904401 u
m(114 Cd) = 113.903359 u
1 u = 931.5 MeV/c2
1 eV = 1.6 × 10−19 J
Solution
The reaction is 113 Cd + n →114 Cd + γ.
4-vector conservation with c = 1

 
  p 2
m113
mn
m114 + p2γ
 0   0  
−pγ

 
 
 0 + 0 =
0
0
0
0
m113 + mn =
q



pγ
  pγ 
+

  0 
0
m2114 + p2γ + pγ
(recovering c)
Eγ
=
=
2m113 mn − m2114 + m2113 + m2n 2
·c
2 (mn + m113 )
9.042MeV.
pγ =
Problem 7
Two identical spin 12 fermions move in one dimension under the influence of the
infinite square well potential with V = ∞ for x < 0, x > L, and V = 0 for
0 ≤ x ≤ L.
(i) Assuming the fermions don’t interact, find the energy eigenvalue corresponding to the lowest energy state for which the particles are constrained
to be in a triplet spin state and give the corresponding eigenfunction(s).
(ii) Find the energy eigenvalue corresponding to the lowest energy state for
which the particles are constrained to be in a singlet spin state and give
the corresponding eigenfunction(s).
~ = B ẑ is applied to the system leading to the
Now suppose a magnetic field B
following additional term in the Hamiltonian,
H10 = µB(S1Z + S2Z ).
(iii) Compute the dependence of the energy levels obtained in Parts (i) and (ii)
on the field B and sketch a graph of the result. As B is increased, find the
critical value of B for which a level crossing occurs for the overall ground
state of the system. Describe this level crossing.
Now suppose that in addition to the applied field, the two particles interact
mutually via a delta function repulsion of the form
H20 = λδ(x1 − x2 )
(λ > 0).
(iv) Assuming that perturbation theory is valid discuss what happens to the
energy levels as λ is increased and find an expression (valid to first order
in λ) for the dependence on λ of the critical field for the level crossing you
found in (iii).
Solution
The one particle wave functions and energy eigenvalues for the infinite square well are
r
nπx
h̄2 π 2 2
2
sin
,
En =
n = n2 K, n = 1, 2, 3, · · ·
ψn (x) =
L
a
2mL2
where
K=
h̄2 π 2
2mL2
Part (i) The lowest energy spin-triplet states are
0
ψTriplet
1
= √ [ψ1 (x1 )ψ2 (x2 ) − ψ2 (x1 )ψ1 (x2 )] ×
2
(
↑1 ↑2
√1 [↑1 ↓2 +
2
↓1 ↓2
)
↓1 ↑2 ]
and the energy of these states is
0
ETriplet
= E1 + E2 = (1 + 4)K = 5K
Part (ii) The lowest energy spin-singlet state is
1
0
ψSinglet
= ψ1 (x1 )ψ1 (x2 ) × √ [↑1 ↓2 − ↓1 ↑2 ]
2
and the energy of this state is
0
ESinglet
= E1 + E1 = 2K
Part (iii) The term
H10 = µB(S1z + S2z ) = µBSz
(21)
leaves the singlet energy unchanged and splits the triplet degeneracy.
1
ESinglet
(B) = 2K

 5K + µB,
1
5K,
ETriplet
(B) =

5K − µB,
When B = Bc =
3K
µ
Sz = +1
Sz = 0
Sz = −1
(22)
the triplet state with Sz = −1 becomes the lowest energy state.
E
Sz = 1
Sz = 0
5K
2K
Sz = -1
0
B
Bc
Part (iv) The first-order shift of the singlet ground state energy is simply the
expectation value of the perturbation H20 ,
∆ESinglet
=
hH20 iSinglet
Z
L
=
4λ
L2
dx2 ψ1∗ (x1 )ψ1∗ (x2 )λδ(x1 − x2 )ψ1 (x1 )ψ1 (x2 )
0
0
=
L
Z
dx1
L
Z
dx1 sin4
0
πx1
3λ
=
L
2L
and, for the triplet states,
∆ETriplet
=
hH20 iTriplet
=
α
L
Z
0
dx1 |ψTriplet
(x1 , x1 )|2
0
=
0
0
because ψTriplet
(x, x) = 0. Thus, within first order perturbation theory
ESinglet ' 2K +
3λ
2L
and
ETriplet (Sz ) = 5K + µBSz
The effect of adding the repulsive term λ is to reduce the applied field, Bc , needed
to produce a level crossing,
3λ
Bc = 3K −
/µ
(23)
2L
Problem 8
Two non-identical spin-1/2 particles (say an electron and a muon) with “frozen”
spatial degrees of freedom interact via the following spin-spin Hamiltonian:
Ĥ =
ε0
Ŝ1 · Ŝ2 ,
h̄2
where ε0 > 0 is a positive constant that sets the energy scale for the problem.
(a) Obtain the eigenvalues and eigenvectors of the Hamiltonian.
(b) Assume that at time t = 0 the two-particle system is prepared in a state
such that the electron has its spin up and the muon its spin down relative
to the z-axis. That is,
1 1E
≡ | + −i.
|ψ(0)i = , −
2 2
Find the probability that after a time t > 0 both particles would have
flipped their spin. That is, what is the probability of measuring the state
| − +i after a time t.
Solution
(a) Defining the total angular momentum operator as Ŝ = Ŝ1 + Ŝ2 , we obtain
Ŝ 2 = Ŝ12 + Ŝ22 + 2Ŝ1 · Ŝ2 .
This suggests that the Hamiltonian may be written as
3 2
ε0
2
h̄
,
Ŝ
−
Ĥ =
2
2h̄2
where we have used the fact that for spin-1/2 particles Ŝ12 = Ŝ22 = 3h̄2 /4. Hence,
the eigenstates of the Hamiltonian are exactly the same as the eigenstates of Ŝ 2 ,
namely,
1
Ĥ|s = 1, mi = ε0 |s = 1, mi
4
and
3
Ĥ|s = 0, m = 0i = − ε0 |s = 0, m = 0i.
4
That is, the triplet state is 3-fold degenerate and has an energy that is ε0 times
larger than the singlet state.
(b) We are told that at t = 0 the two-particle system is prepared in the state | + −i.
However, since such an initial state is not an eigenstate of the Hamiltonian, it
is convenient to express it in terms of states of total angular momentum |s, mi.
Since the z-component of the total angular momentum is equal to zero, only
two states are required. That is,
1 | + −i = √ |s = 1, m = 0i + |s = 0, m = 0i
2
1 | − +i = √ |s = 1, m = 0i − |s = 0, m = 0i
2
Then,
1 1 |ψ(0)i = | + −i = √ |s = 1, m = 0i + |s = 0, m = 0i ≡ √ |s = 1i + |s = 0i .
2
2
Having written the initial state in terms of the eigenstates of the Hamiltonian,
the wave function at an arbitrary time t > 0 is obtained by multiplying each
eigenstate by the appropriate phase—which is proportional to the energy of the
eigenstate. That is,
1 |ψ(t)i = √ e−iE1 t/h̄ |s = 1i + e−iE0 t/h̄ |s = 0i
2
1 −iΩt −iωt
= √ e
e
|s = 1i + e+iωt |s = 0i
2
h
E
Ei
−iΩt
=e
cos(ωt) +− − i sin(ωt) −+ ,
where we have defined Ω ≡ −ε0 /4h̄ and ω ≡ ε0 /2h̄. In this way the probability
that both particles have flipped their spin is given by
2
ε0 t
Pflip (t) = h−+ |ψ(t)i = sin2 (ωt) = sin2
.
2h̄
Problem 9
(a) What is the heat capacity at constant volume of a (classical) ideal di-atomic
gas?
(b) Consider a reversible adiabatic process. If the pressure (P ) of such an ideal
di-atomic gas decreases by a factor of λ, by what factor does the temperature (T ) change, assuming the volume is held constant?
(c) Considering a column of such a gas in the gravitational field of Earth close
to Earth’s surface. By mechanically balancing the effects of gravity and
pressure changes, find the dependence of dP/dz on P and T , where z is
the altitude (height). Express your answer in terms of the Earth’s gravitational acceleration g, molar mass of the gas M , and the gas constant R.
(d) Assuming the air changes its pressure and temperature adiabatically, find
the rate of change of temperature with the altitude, dT /dz. Using R =
J
, g = 9.8m/s2 , and the molar mass of air M = 28g/mol, express
8.314 molK
the change of temperature in Kelvin per kilometer.
Solution
(a) If we have N molecules of a diatomic gas, then by equipartition theorem, we have
the internal energy U = N kB ( 32 + 1)T and so CV = 52 nR, where n is the number
of mols of the diatomic gas.
(b) For an adiabatic process, dU = −P dV or CV dT = −P dV . Using P V = nRT ,
eliminate dV to get
CV dT
dP
= 1+
.
P
nR T
Therefore,
CV
ln P = 1 +
ln T + const.
nR
and
P
P
=
= const.0
CV
1+ nR
T 7/2
T
So, if P → λP , then T → λ2/7 T .
(c)
Adzρg = −AdP ; ⇒
dP
nM
MP
= −gρ = −g
= −g
.
dz
V
RT
(d) Because
we have
dP
gM
=−
dz
P
RT
CV dT
gM
1+
=−
dz.
nR T
RT
or
dT
gM
1
2 gM
=−
=−
.
V
dz
R 1+ C
7 R
nR
Substituting, we get
K
dT
≈ −9.4
.
dz
km
Problem 10
(a) Consider a thin cylindrical shell of mass m and radius R, rolling down an
incline plane without slipping, starting at rest from the height h. Determine the speed of the center of mass at the point when the cylinder travels
the vertical distance h.
(b) Same as (a) but the cylinder is filled and of uniform mass density. Compare
the velocity with the one found in (a).
Solution
To prevent slipping, the angular velocity ω satisfies ω = v/R. The kinetic energy
change is
1
1
1 I 2
1
mv 2 + Iω 2 = mv 2 +
v .
2
2
2
2 R2
and the potential energy change is mgh. The have to equal to each other. Therefore,
s
s
2mgh
2gh
v=
=
.
I
m + RI2
1 + mR
2
(a) I = mR2 which gives v =
√
gh.
(b) I = 12 mR2 which gives v =
q
4
gh.
3
It is faster than in (a).
Problem 11
A non-uniform, non-conducting disk of mass M , radius R, and total charge Q
has a surface charge density σ = σ0 r/R, where r is the distance from the center
of the disk, and a mass per unit area σm = M σ/Q. The disk rotates with
angular velocity ω about its axis.
(a) Calculate the magnetic moment µ of the disk.
(b) Show that the magnetic moment µ and the angular momentum L are
related by µ = QL/2M , where L = Iω and I is the moment of inertia.
Solution
(a) Break the disk up into a series of concentric rings at radial distance r from the
rotation axis, and with thickness dr (see figure). The magnetic moment of a
current loop is defined to be the current flowing around the loop multiplied by
the area of the loop. Firstly, we can find the current due to our ring of charge
by noting that the charge of the ring is
dQ = σdA = σ2πr dr = σ0
r
2πr dr.
R
Now all of this charge passes a fixed point in the plane of the disk and at radius
r (say on the x axis) in a time T = 2π/ω, so we have that the current from this
loop is
r
dQ
ω
r2
dI =
= σ0 2πr dr ·
= σ0 ω dr.
T
R
2π
R
This current loop has an area πr2 so we have an element of the magnetic
moment of
dµ = dI · πr2 ,
and so the magnetic moment is
Z
σ0 πω R 4
σ0 πω R5
π
µ=
r dr =
= σ0 ωR4 .
R
R
5
5
0
Although it is not necessary we can also find the total charge and so find the
magnetic moment in terms of the total charge,
Z R
σ0 r
2π
Q=
2πr dr =
σ0 R2 ,
R
3
0
so that σ0 = 3Q/2πR2 and
µ=
3
QR2 ω ẑ.
10
(b) To find the angular momentum we need the moment of inertia, which we find
using the mass density and by doing the integral in the same way as before.
Note that since the mass and charge densities are proportional, we have the
mass density in terms of the total mass
σm =
3M r
2πR2 R
so that
Z
Z
Z R
Z
3M
r 2
3M R 4
3
I = dM r2 = σm dA r2 =
2πr
dr·
r
=
r dr = M R2 ,
2πR2 0
R
R3 0
5
and we have that
L = Iω = Iω ẑ =
and we see that µ = QL/2M .
3
M R2 ω ẑ,
5
Problem 12
A neutron (mass 940 MeV/c2 = 1.6749 × 10−27 kg) is bouncing off a flat horizontal surface in the Earth’s gravitational field (g = 9.8 m/s2 ).
(a) Estimate the ground state energy of the neutron and how high does it
bounce. 1
(b) Is this motion relativistic and at what values of g relativity becomes important?
(c) Is the finite lifetime of the neutron (about 15 minutes) important, and at
what g it becomes important?
Solution
(a) The neutron is moving in a wedge-like potential: an infinite surface V (z) = ∞
for z < 0 and V (z) = mgz for z > 0. The motion is confined in the region
between z = 0 and z0 , where z0 is determined by the energy mgz0 = E. Using
uncertainty principle z0 ∆p ∼ h̄ where E ∼ ∆p2 /m we have
mgh̄
mgh̄
∼ √
∆p
mE
q
3
E ∼ mg 2 h̄2 .
E∼
(24)
(25)
The same answer follows from the dimensional analysis. For numerical estimates
in natural units one can use g/c2 = 1.11 × 10−16 m−1 = 1.11 × 10−31 fm−1 .
r
g 2
3
= 7.6 × 10−19 MeV = 1.21394 × 10−31 J
(26)
E ∼ (mc2 )(h̄c)2 2
c
The height of bouncing is z0 = E/(mg) = 7.4µm
(b) The problem is non-relativistic since E mc2 . For situation to be relativistic
mc2 must be comparable with E. Therefore E should be about 21 order of
magnitude bigger, which requires g to be about 30 orders of magnitude bigger.
(c) Energy uncertainty due to finite lifetime Γ ∼ h̄/τ ≈ 7 × 10−25 MeV is six orders
of magnitude smaller than the energy in Eq. (26). Therefore the finite lifetime
is not important. If the strength of gravity is reduced 9 orders of magnitude
then the finite lifetime becomes important.
1 h̄
= 1.05457173 × 10−34 m2 kg/s; h̄c = 197.32697 MeV fm