Download 2. The surface current J = J ( x sin k y +

b sin k0 x) flows in the plane z = 0. Find the magnetic
2. The surface current J = J0 (b
x sin k0 y + y
field in the whole space.
Making use of the superposition principle, one can conveniently find separately the fields
b sin k0 y is
created by the currents directed in the x and y direction. The field of the current J0 x
described by the vector potential A = (Ax , 0, 0). The Coulomb gauge condition, ∇ · A = 0, yields
∂Ax
= 0. Then the potential obeys the equation
∂x
∂ 2 Ax ∂ 2 Ax
+
= 0.
∂y 2
∂z 2
(1)
Separation of variables at z > 0: Ax = ψ(y)f (z);
ψ 00 f 00
+
= 0;
ψ
f
(2)
f = C1 e−kz .
ψ = sin(ky + φ);
(3)
B = ∇ × A = −C1 k [ŷ sin(ky + φ) + ẑ cos(ky + φ)] e−kz ;
z>0
(4)
Analogously
B = C2 k [ŷ sin(ky + φ) − ẑ cos(ky + φ)] ekz ;
z<0
(5)
Continuity of Bz yields C1 = C2 . The boundary condition, By (z = +0) − By (z = −0) =
−(4π/c)J0 sin k0 y yields k0 = k, φ = 0 and
C1 =
2πj0
.
ck0
(6)
Now one gets
2π
B=
J0
c
− (ŷ sin k0 y + ẑ cos k0 y) e−kz ; z > 0
(ŷ sin k0 y − ẑ cos k0 y) ekz ;
z < 0.
(7)
The field of the current flowing in the y direction is found analogously. Finally one gets the
total field
2π
[x̂ sin k0 x − ŷ sin k0 y + ẑ(cos k0 x − cos k0 y)] e−kz ; z > 0
J0
Btot =
(8)
[−x̂ sin k0 x + ŷ sin k0 y + ẑ(cos k0 x − cos k0 y)] ekz ; z < 0.
c
1
3. A photon with the frequency ω propagates in the positive x direction and is scattered off an
electron moving in the negative x direction with the energy E. Find the frequency of the photon,
the energy and the direction of the electron after the scattering as a function of the scattering angle
(the angle between the photon directions after and before the scattering)
The 4-momentums of the electron and the photon before the scattering are
P =
√
1
E, − E 2 − m2 c4 , 0, 0 ;
c
Pγ =
h̄ω
(1, 1, 0, 0) ;
c
(9)
where the identity P 2 = (E/c)2 − p2 = m2 c2 was used in order to express p via E. After the
scattering,
P0 =
√
1 0 √ 02
E , E − m2 c4 cos ψ, E 02 − m2 c4 sin ψ, 0 ;
c
Pγ0 =
h̄ω 0
(1, cos θ, sin θ, 0) ;
c
(10)
where θ and ψ are angles between the x axis and the directions of motion of the scattered photon
and electron, correspondingly. We have to express ω 0 , E 0 and ψ via ω, E and θ. In order to
eliminate the parameters of the scattered
electron, let us write the conservation of 4-momentum,
0 2
0
0
P + Pγ = P + Pγ , as P + Pγ − Pγ = P 02 and take into account that P 2 = m2 c2 , Pγ2 = 0. This
yields
P · Pγ − P · Pγ0 − Pγ · Pγ0 = 0,
(11)
√
where P · Pγ = (h̄/c2 ) Eω + ω E 2 − m2 c4 etc. Now one gets an equation for ω 0 ,
√
√
Eω + ω E 2 − m2 c4 = ω 0 E + ω 0 cos θ E 2 − m2 c4 + h̄ωω 0 (1 − cos θ),
(12)
which is immediately resolved
√
E + E 2 − m2 c4
√
ω.
ω =
E + cos θ E 2 − m2 c4 + h̄ω(1 − cos θ)
0
(13)
The energy of the scattered electron is found the from the energy conservation:
E 0 = E + h̄ω − h̄ω 0 .
(14)
Conservation of the y component of the momentum vector implies
p0y = −h̄ω 0 /c.
(15)
√
Taking into account cp0 = E 02 − m2 c4 , one finds the angle between the electron direction after
the scattering and the x axis
p0y
h̄ω 0
sin ψ = 0 = − √
.
(16)
p
E 02 − m2 c4
2
4. A sphere of the radius r is filled homogeneously with the electric charge with the density
ρ. The sphere executes torsional oscillations around the diameter, ϕ = ϕ0 sin ωt. Find the timeaveraged angular distribution and the power of the radiation.
The charges move in the ϕ direction with the velocity v = r0 sin θϕ̇ = ϕ0 ωr0 sin θ cos ωt, where
0
r Rand θ are the spherical coordinates. The current j = ρv creates the magnetic moment m =
1
r0 × jdV . It follows from the symmetry, that m has only z component
2c
Z
Z
1
1
0
m=b
z
(r × j)z dV = b
z
r0 sin θjdV
2c
2c
Z
πρϕ0 ω
4π
=b
z
cos ωt r04 sin2 θd(cos θ)dr0 = b
z
ρϕ0 ω cos ωtr5 .
c
15c
The electric field of the wave emitted in the direction n is
E=
1
c2 r 0
[nm̈] =
sin θ
m̈ϕ.
b
c2 r 0
(17)
Now the radiation power in the direction n is
2πρ2 ϕ20 ω 6 r10
dP
E2
= r02 c =
sin2 θ.
dΩ
4π
225c5
(18)
The total radiation power is
2πρ2 ϕ20 ω 6 r10
P =
225c5
Z
sin2 θdΩ =
3
16π 2 ρ2 ϕ20 ω 6 r10
.
675c5
(19)