|
Lagrange multipliers, scattering |
|
Addition of angular momentum, EM waves |
|
Producing electromagnetic radiation |
|
Relativistic E&M |
|
Identical particles |
Problem 1:
Write down the electronic configuration for the Magnesium atom (Z =
12) in its ground state. Then enumerate the allowed term symbols 2S+1LJ for
the ground state from the point of view of angular momentum alone.
Solution:
- Concepts:
The energy levels of multi-electron atoms
- Reasoning:
We use the Pauli exclusion
principle and the rules for adding angular momentum to find the allowed
terms 2S+1L.
- Details of the calculation:
The electronic configuration is (1s)2, (2s)2,
(2p)6 (3s)2.
All shells are full and have L = 0, S = 0, J = 0.
There is only one allowed term, 1S0 .
Problem 2:
Consider a 3He atom (composed of a nucleus of spin 1/2 and two electros).
The electronic angular momentum is J =
L +
S, where L
is the orbital angular momentum of the electron and S is its spin. The
total angular momentum of the atom is F =
J +
I, where I
is the nuclear spin. The eigenvalues of J2 and F2 are
j(j + 1)ħ2 and f(f + 1)ħ2
respectively.
(a) What are the possible values of the quantum numbers j and f for a
3He
atom in the ground state?
(b) What are the possible values of the quantum numbers j and f for an
excited 3He atom with one electron is in the 2p and the other in the
4f state?
Solution:
- Concepts:
Addition of angular momenta
- Reasoning:
We are asked to add three
angular momenta
- Details of the calculation:
(a) For the (1s)2 ground state we have L = 0, S =
0, J =
L
+ S = 0.
F = J +
I, i = ½.
There is only possible values for f, are f = ½.
(b) For the excited 3He atom we have l1 = 1, l2
= 3.
The possible values for l are 4, 3, and 2.
The possible values for s are 0 and 1.
The possible values for j are j = 5, 4, 3, 2, 1.
F = J +
I, i = ½.
So the possible values for f are f = 11/2, 9/2, 7/2, 5/2, 3/2, ½.
Problem 3:
A bead of mass m is constrained to move without friction on a helix whose equation in
cylindrical polar coordinates is ρ = b, z = aΦ under the influence of
gravity, F = -mg k.
(a)
Use the Lagrange multiplier method and find the appropriate Lagrangian including
terms expressing the constraints.
(b) Apply the Euler-Lagrange
equations to obtain the equations of motion. Solve for the forces of
constraint in the z- and ρ-direction.
(c) If the bead starts from rest at z = 0, find its position as a function
of time.
Solution:
- Concepts:
Lagrangian Mechanics, the Lagrange multiplier method
- Reasoning:
We are instructed to use the Lagrange multiplier method to
solve the problem.
- Details of the calculation:
(a) Lagrangian not incorporating the
constraints:
L = ½m[(dρ/dt)2 + ρ2(dΦ/dt)2 +(dz/dt)2]
- mgz.
Put the equations of constraint into the form
Σk alk
dqk +
alt dt = 0, l = 1, ..., m.
We
have two equations of constraint.
(i) dρ = 0; a1ρ = 1, a1Φ
= a1z = 0.
(ii) dz - adΦ = 0 ; a2ρ = 0, a2Φ
= -a, a2z = 1.
(b) Lagrange's equation:
d/dt(∂L/∂(dqk/dt)) - ∂L/∂qk = ∑lλlalk,
Σk alk dqk + alt dt = 0.
We have:
md2ρ/dt2 - mρ(dΦ/dt)2 = λ1,
mρ2d2Φ/dt2 + 2mρ(dρ/dt)(dΦ/dt) = -aλ2,
md2z/dt2 +
mg = λ2,
and
ρ = b, dρ/dt = d2ρ/dt2 = 0,
z = aΦ, dz/dt = adΦ/dt, d2z/dt2
= ad2Φ/dt2.
We therefore have:
-mb(dΦ/dt)2 = λ1.
mb2 d2Φ/dt2 = -aλ2.
mad2Φ/dt2 +
mg = λ2.
Using
the last two equations to solve for λ2 we have
λ2 = b2mg/(a2 + b2).
Therefore the equation of motion is
d2Φ/dt2 + ag/(a2 + b2) = 0.
d2z/dt2 + a2g/(a2 + b2) = 0.
λ1 = -mb(dΦ/dt)2.
λ1 is the force of
constraint in the radial direction, λ2 is the force of constraint
in the z- direction.
(c) The bead accelerates in the z-direction with
constant acceleration az = -a2g/(a2 + b2).
z(t) = -½azt2, Φ(t) = z(t)/a.
dz/dt = azt. dΦ/dt = azt/a. λ1
= -mb(azt/a)2 = -mb(agt/(a2 + b2))2.
Problem 4:
Two equal magnitude, opposite sign charges
are located at either end of a molecule of mass M and length l. The
molecule rotates end over end (a nonrelativistic tumbling motion) with an
initial rotational period T (cT >> l). How long will it take the
molecule to lose 1/10 of its rotational energy by electromagnetic radiation?
Solution:
- Concepts:
The radiation field of an accelerating dipole, the Larmor
formula
- Reasoning:
The energy lost per unit time given by the Larmor formula
P = -dE/dt = <(d2p/dt2)2>/(6πε0c3).
(SI units)
The two point charges move in a circle and produce a rotating
dipole. Let the dipole rotate about the x-axis.
p(t) = p0cos(ωt)
k - p0sin(ωt) j.
d2p/dt2
= -ω2p, (d2p/dt2)2 = ω4p02.
- Details of the calculation:
P = -dE/dt = ω4p2/(6πε0c3)
is the radiated power.
The rotational kinetic energy is E = ½Iω2, with I = Ml2/12.
(I assume that the mass M is distributed uniformly along the length l. You
could also read the problem as implying that a mass M/2 is located at the
position of each charge and the rod is massless. Then I = Ml2/4.)
Approximate solution: -∆E = -(dE/dt)∆t =
-E/10.
∆t = Iω26πε0c3/(20ω4p2)
= Ml2πε0c3/(40ω2p2) =
MT2ε0c3/(160 πq2).
Exact
solution:
-dE/dt = ω4p2/(6πε0c3)
= 4E2p2/(6 I2πε0c3).
-dE/E2 = ω4p2/(6πε0c3)
= 4p2dt/(6 I2πε0c3).
-(6 I2πε0c3/(4p2))∫E00.9E0dE/E2
= ∫dt = ∆t.
∆t = [MT2ε0c3/(16 πq2)]
[1/0.9 - 1].
Problem 5:
A neutral conducting cube is at rest with its center at the origin in
reference frame K. A uniform electric field E = E0j is
present. Reference frame K' moves with uniform velocity v = vi
with respect to K. Find E an B inside the cube as observed in K'.
Solution: (change)
- Concepts:
Lorentz transformation of the electromagnetic fields
- Reasoning:
We are asked to transform the electromagnetic fields from the laboratory
frame K to a frame moving with uniform velocity v = vi with
respect to K.
- Details of the calculation:
In SI units the transformation of the electromagnetic fields from the
laboratory frame to a frame moving with velocity v with respect to
the laboratory frame is given by:
E'|| = E||, B'|| =
B||, E'⊥ = γ(E + v×B)⊥,
B'⊥ = γ(B - (v/c2)×E)⊥.
In K we have inside the conductor E = 0, B = 0. There is
no electric field imside the conductor
In K' we Therefore have have E' = 0, B' = 0.