(15.178) represents a rotation matrix in ℝ
2 ; see Sect. 11.2. Therefore, F(t) and F(s)
À1
are commutative. Notice, however, that we do not have a resolvent matrix as a
simple form that appears in (15.116). In fact, from (15.178) we find R s, t
ð Þ 6 ¼
F t À s
ð
Þ. It is because the matrix A given as (15.179) is not a constant but depends
on t.
Rewriting (15.90) as
x t
ð Þ ¼ x 0
ð Þℜ 0, t
ð Þ þ
Z t
0
ds a cos ωt 0
ð
Þ ℜ s, t
ð Þ
½
ð 15:181Þ
and setting BCs as x(0) ¼ (σ τ), with the individual components we finally obtain
ξ t
ð Þ ¼ σ cos e b sin ωt
À τ sin e b sin ωt
þ
a
b
sin e b sin ωt
,
ζ t
ð Þ ¼ σ sin e b sin ωt
þ τ cos e b sin ωt
À
a
b
cos e b sin ωt
þ
a
b
:
ð15:182Þ
Note that differentiating both sides of (15.182) with respect to t, we recover
original equation of (15.172) to be solved. Further setting ξ(0) ¼ σ ¼ ζ(0) ¼ τ ¼ 0 as
BCs, we have
ξ t
ð Þ ¼
a
b
sin e b sin ωt
,
ζ t
ð Þ ¼
a
b
1 À cos e b sin ωt
h
i
:
ð15:183Þ
Notice that the above BCs correspond to that the charged particle is initially
placed (at the origin) at rest (i.e., zero velocity). Deleting the argument t, we get
ξ
2
þ ζ À
a
b
2 ¼
a
b
2 :
ð15:184Þ
Figure 15.3 depicts the particle velocities ξ and ζ in the x- and z-directions,
respectively. It is interesting that although the velocity in the x-direction switches
from negative to positive, that in the z-direction is kept non-negative. This implies
that the particle is oscillating along the x-direction, but continually drifting toward
the z-direction while oscillating. In fact, if we integrate ζ(t), we have a continually
drifting term
a
b t.
Figure 15.4 shows the Lorentz force F as a function of time in the case where the
charge of the particle is positive. Again, the electric Lorentz force (represented by E)
switches from positive to negative in the x-direction, but the magnetic Lorentz force
(represented by _
x  B) holds non-negative in the z-direction all the time.
To precisely analyze the positional change of the particle, we need to integrate
(15.182) once again. This requires the detailed numerical calculation. For this
purpose the following relation is useful [7]:
15.4 Motion of a Charged Particle in Polarized Electromagnetic Wave
615
2 ; see Sect. 11.2. Therefore, F(t) and F(s)
À1
are commutative. Notice, however, that we do not have a resolvent matrix as a
simple form that appears in (15.116). In fact, from (15.178) we find R s, t
ð Þ 6 ¼
F t À s
ð
Þ. It is because the matrix A given as (15.179) is not a constant but depends
on t.
Rewriting (15.90) as
x t
ð Þ ¼ x 0
ð Þℜ 0, t
ð Þ þ
Z t
0
ds a cos ωt 0
ð
Þ ℜ s, t
ð Þ
½
ð 15:181Þ
and setting BCs as x(0) ¼ (σ τ), with the individual components we finally obtain
ξ t
ð Þ ¼ σ cos e b sin ωt
À τ sin e b sin ωt
þ
a
b
sin e b sin ωt
,
ζ t
ð Þ ¼ σ sin e b sin ωt
þ τ cos e b sin ωt
À
a
b
cos e b sin ωt
þ
a
b
:
ð15:182Þ
Note that differentiating both sides of (15.182) with respect to t, we recover
original equation of (15.172) to be solved. Further setting ξ(0) ¼ σ ¼ ζ(0) ¼ τ ¼ 0 as
BCs, we have
ξ t
ð Þ ¼
a
b
sin e b sin ωt
,
ζ t
ð Þ ¼
a
b
1 À cos e b sin ωt
h
i
:
ð15:183Þ
Notice that the above BCs correspond to that the charged particle is initially
placed (at the origin) at rest (i.e., zero velocity). Deleting the argument t, we get
ξ
2
þ ζ À
a
b
2 ¼
a
b
2 :
ð15:184Þ
Figure 15.3 depicts the particle velocities ξ and ζ in the x- and z-directions,
respectively. It is interesting that although the velocity in the x-direction switches
from negative to positive, that in the z-direction is kept non-negative. This implies
that the particle is oscillating along the x-direction, but continually drifting toward
the z-direction while oscillating. In fact, if we integrate ζ(t), we have a continually
drifting term
a
b t.
Figure 15.4 shows the Lorentz force F as a function of time in the case where the
charge of the particle is positive. Again, the electric Lorentz force (represented by E)
switches from positive to negative in the x-direction, but the magnetic Lorentz force
(represented by _
x  B) holds non-negative in the z-direction all the time.
To precisely analyze the positional change of the particle, we need to integrate
(15.182) once again. This requires the detailed numerical calculation. For this
purpose the following relation is useful [7]:
15.4 Motion of a Charged Particle in Polarized Electromagnetic Wave
615