Energy flow and particle motion in a standing wave
Consider a plane, monochromatic, elliptically polarized standing wave (which may be generated by
reflection of a propagating wave) from a mirror), whose electric field can be written as
E0
E0
−iωt
E=
,
(1)
cos(kx) (ŷ cos(ωt) + δẑ sin(ωt)) = Re
(ŷ + iδẑ) cos(kx)e
(1 + δ 2 )1/2
(1 + δ 2 )1/2
where 0 < δ < 1.
a) Write down the expression of the magnetic field B corresponding to the electric field (1) and
determine for which values of δ the fields E are B either perpendicular or parallel to each other.
Find the expression of the flux of EM energy, both instantaneous and cycle-averaged.
b) Solve the equation of motion (in steady conditions) for a point particle of charge q and mass m
taking only the effect of the electric field (1) into account. Discuss how the oscillation amplitude and
the emitted radiation power vary with the particle position along x.
c) Now evaluate the cycle-averaged force along x,
Fx = hqv × Bix ,
(2)
by using for the particle velocity v the result of point b). Discuss the motion along x under the
action of Fx , characterizing the equilibrium points and calculating the frequency of small amplitude
oscillations along such points.
1
Solution
a) In complex notation we write the fields as E = Re(Ẽe−iωt ), etc. By using ∂t B = −∇ × E we
obtain
E0
(−ẑ + iδŷ) sin(kx) ,
(3)
−iω B̃ = −k
(1 + δ 2 )1/2
thus
B̃ =
E0
(δŷ + iẑ) sin(kx) .
c(1 + δ 2 )1/2
(4)
Si ha quindi
E0
E0
(δŷ + iẑ) sin(kx) =
(ŷδ cos(ωt) + ẑ sin(ωt)) sin(kx) .
B = Re
2
1/2
c(1 + δ )
c(1 + δ 2 )1/2
(5)
The scalar and vector products between E and B are proportional to
E · B ∝ δ cos2 (ωt) + δ sin2 (ωt) = δ ,
E × B ∝ x̂ cos(ωt) sin(ωt) − δ 2 sin(ωt) cos(ωt)x̂ = x̂(1 − δ 2 ) sin(ωt) cos(ωt) ,
(6)
(7)
then E · B = 0 if δ = 0 and E × B = 0 if δ = 1: the fields are either perpendicular or parallel to each
other for linear and circular polarization, respectively.
The energy flux S = ε0 c2 E × B is zero for every t if δ = 1, while it is an oscillating function with
zero average for δ 6= 1.
b) The equation dv/dt = qE/m has the steady solution v = Re(ṽe−iωt ) with ṽ = (iq/mω)Ẽ. Thus
the oscillation amplitude varies with the position as cos(kx). The radiated power P ∝ |dv/dt|2 ∝
cos2 (kx). Both v and P are zero for kx = π/2 + nπ, with n = 0, ±1, ±2, . . ..
c) We have
2
1
q2
iq
E02
∗
hqv × Bi =
Ẽ × B̃ =
Re
sin(kx) cos(kx)Re (i(ŷ + iδẑ) × (δŷ − iẑ))
2
mω
2mcω 1 + δ 2
q2
= x̂
E02 sin(2kx) ,
(8)
4mcω
since (ŷ + iδẑ) × (δŷ − iẑ) = −iŷ × ẑ + iδ 2 ẑ × ŷ = −i(1 + δ 2 )x̂. The cycle-averaged force does not
depend on the polarization of the wave.
The equilibrium points (Fx = 0) are where sin(2kx) = 0. These points are stable when they
correspond to minima of the “potential energy’ U such that Fx = −dU/dx, i.e.
q2
E02 cos(2kx) .
(9)
8mcωk
Thus the equilibrium is stable at x = xm with 2kxm = (2n + 1)π, n = 0, ±1, ±2, . . .. Around an
equilibroum point, since cos(2kx) ≃ −1 + (2k(x − xm ))2 /2, we have
2
qE0
m
√
(x − x0 )2 ,
(10)
U ≃ Umin +
2
2mc
√
from which we obtain the frequency of small amplitude oscillations as Ω = qE0 /( 2mc).
U=
2