Radiation from orbiting charges
Two identical point charges are kept in rotation with constant angolar velocity ω along the circular
orbit x2 + y 2 = R2 , z = 0.
a) Write the most general trajectory for the charges both in polar coordinates ri = ri (t), θi = θi (t)
and in cartesian coordinates xi = xi (t), yi = yi (t) (where i = 1, 2 is the charge index) and calculate
the electric dipole moment of the system.
b) Characterize the dipole radiation from the two charges system, by discussing how the power
depends on the initial conditions and finding the polarization of the radiation emitted along the x̂,
ŷ and ẑ directions.
c) Answer questions a) and b) again for the case in which the charges are orbiting with opposite
angular velocity.
d) Now consider a system of three identical charges on the circular orbit with the same angular
velocity. Find the initial conditions for which the radiation power is either zero or has its maximum.
e) Determine whether the magnetic dipole moment gives some contribution to the radiation, for each
of the above specified cases.
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Solution
a) Let assume the two charges to be separated by an azimuthal angle φ, which is a constant since
the charges rotate at constant velocity. The trajectories may be written in a symmetrical form
r1 = r2 = R ,
θ1 (t) = ωt − φ/2,
θ2 (t) = ωt + φ/2 .
(1)
In Cartesian coordinates
x1 (t) = R cos(ωt − φ/2),
x2 (t) = R cos(ωt + φ/2),
y1 (t) = R sin(ωt − φ/2) ,
y2 (t) = R sin(ωt + φ/2) .
(2)
(3)
The dipole moment is p = q(x1 + x2 ) with components
px = qR[cos(ωt − φ/2) + cos(ωt + φ/2)] = 2qR cos(φ/2) cos ωt ,
py = qR[sin(ωt − φ/2) + sin(ωt + φ/2)] = 2qR cos(φ/2) sin ωt ,
(4)
(5)
i.e. p has a constant module p = 2qR cos(φ/2) and rotates in the z = 0 plane with frequency ω.
b) In the dipole approximation the polarization of the radiation emitted along the n̂ direction is
parallel to
(p × n̂) × n̂ = p⊥ .
(6)
Since
(p × x̂) × x̂ ∼ ŷp sin ωt ,
(p × ŷ) × ŷ ∼ x̂p cos ωt ,
(7)
the polarization in the x̂ (ŷ) direction is linear and along ŷ (x̂). In the ẑ direction
(p × ẑ) × ẑ ∼ p(x̂ cos ωt + ŷ sin ωt) ,
(8)
i.e. the polarization is circular.
The total radiated power is proportional to (omitting uninteresting constant factors)
P ∼ |p̈|2 = 4q 2 R2 ω 4 cos2 (φ/2) ,
(9)
which trivially vanishes when p = 0, i.e for φ = π/2 (charges on opposite sides of a diameter) and
has a maximum for φ = 0 (charges in the same position).
c) In this case we notice that the charges meet each other along their orbit, and we assume the
encounter occurs at θ = 0 for t = 0. Thus the trajectories can be written simply as
r1 = r2 = R ,
θ1 (t) = ωt ,
θ2 (t) = −ωt ,
(10)
and
x1 (t) = R cos ωt ,
y1 (t) = R sin ωt ,
x2 (t) = R cos ωt ,
2
y2 (t) = −R sin ωt .
(11)
The total dipole p = (2qR cos ωt)x̂. There is no radiation in the direction parallel to p, while the
radiation emitted in the perpendicular plane is linearly polarized. The radiated power is proportional
to
P ∼ 4q 2 R2 ω 4 cos2 ωt .
(12)
d) The trajectories for the three charges can be written as
θ1 (t) = ωt; ,
θ2 (t) = ωt + φ21 ,
xi = cos θi ,
yi = sin θi ,
θ3 (t) = ωt + φ31 ,
(i = 1, 2, 3) .
(13)
(14)
The electric dipole moment vanishes for the symmetrical configuration in which the three charges
are on the vertices of an equilateral triangle (φ21 = −φ31 = π/3), while the maximum value is when
the charges overlap (φ21 = φ31 = 0).
e) The magnetic dipole moment
Z
m ∼ J × rd3 x ,
(15)
is constant and equal to qR2 ω for a charge orbiting at constant velocity (notice that m is proportional
to the angular momentum of the orbiting charge). Thus the magnetic dipole never contributes to
radiation, because the fields are proportional to m̈.
This problem may help to understand which a circular coil with constant current does not radiate,
although we may consider the current as produced by charges moving on circular orbits and thus
subject to acceleration.
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