Page 223 - Electromagnetics

We can also write Ampere’s law as
c ˜
˜
˜ i
∇× H = J + jω˜ E
c
where ˜ is the complex permittivity
2 2
˜ σ(ω) ω 0 ω ν
c p p
˜ (ω) = ˜ (ω) + = 0 1 − − j . (4.76)
2
2
2
jω ω + ν 2 ω(ω + ν )
˜
If we wish to describe the plasma in terms of a polarization vector, we merely use D =
˜
˜
˜
˜
˜
˜
0 E + P = ˜ E to obtain the polarization vector P = (˜ − 0 )E = 0 ˜χ e E, where ˜χ e is the
electric susceptibility
ω 2
p
˜ χ e (ω) =− .
2
ω + ν 2
˜
˜
We note that P is directed opposite the applied field E, resulting in ˜ < 0 .
The plasma is dispersive since both its permittivity and conductivity depend on ω.
2
2
c
As ω → 0 we have ˜ → 0 r where r = 1 − ω /ν , and also ˜ c ∼ 1/ω, as remarked
p
3
2
in (4.28). As ω →∞ we have ˜ − 0 ∼ 1/ω and ˜ c ∼ 1/ω , as mentioned in (4.29).
c
When a transient plane wave propagates through a dispersive medium, the frequency
dependence of the constitutive parameters tends to cause spreading of the waveshape.
We see that the plasma conductivity (4.75) is proportional to the collision frequency ν,
and that, since ˜ c < 0 by the arguments of § 4.5, the plasma must be lossy. Loss arises
from the transfer of electromagnetic energy into heat through electron collisions. If there
are no collisions (ν = 0), there is no mechanism for the transfer of energy into heat, and
the conductivity of a lossless (or “collisionless”) plasma reduces to zero as expected.
In a lowloss plasma (ν → 0) we may determine the time-average stored electromagnetic
energy for sinusoidal excitation at frequency ˇω. We must be careful to use (4.59), which
holds for materials with dispersion. If we apply the simpler formula (4.64), we find that
for ν → 0
1 1 ω 2 p
ˇ 2
ˇ 2
w e = 0 |E| − 0 |E| .
4 4 ˇ ω 2
For those excitation frequencies obeying ˇω< ω p we have w e < 0, implying that the
material is active. Since there is no mechanism for the plasma to produce energy, this is
obviously not valid. But an application of (4.59) gives
2 2
1 ∂ ω p 1 1 ω p
ˇ 2
ˇ 2
ˇ 2
w e = |E| 0 ω 1 − = 0 |E| + 0 |E| , (4.77)
4 ∂ω ω 2 4 4 ˇ ω 2
ω= ˇω
which is always positive. In this expression the first term represents the time-average
energy stored in the vacuum, while the second term represents the energy stored in the
kinetic energy of the electrons. For harmonic excitation, the time-average electron kinetic
energy density is
1
∗
w q = Nm e ˇ v · ˇ v .
4
Substituting ˇ v from (4.72) with ν = 0 we see that
2
1 1 Nq e ˇ 2 1 ω 2 p
ˇ 2
∗
Nm e ˇ v · ˇ v = |E| = 0 |E| ,
4 4 m e ˇω 2 4 ˇ ω 2
which matches the second term of (4.77).
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