Page 288 - Electromagnetics

Let us calculate v e for a plane wave propagating in a lossless, source-free medium where
√
ˆ
k = kω µ . By(4.216) and (4.223) we have
˜
˜
E(r,ω) = E 0 (ω)e − jβ ˆ k·r , (4.262)

ˆ
˜
k × E 0 (ω)
˜ − jβ ˆ k·r ˜ − jβ ˆ k·r
H(r,ω) = e = H 0 (ω)e . (4.263)
η
We can compute the time-average stored energydensityusing the energytheorem (4.68).
In point form we have
∂H ∂E
˜ ˜
˜ ∗
−∇ · E × + × H = 4 j w em . (4.264)
˜ ∗
∂ω ∂ω
ω= ˇω
Upon substitution of (4.262) and (4.263) we find that we need to compute the frequency
˜
˜
derivatives of E and H. Using
∂ − jβ ˆ k·r ∂ − jβ ˆ k·r dβ dβ − jβ ˆ k·r
ˆ
e = e =− jk · r e
∂ω ∂β dω dω
ˆ
and remembering that k = kβ,wehave
˜
˜
∂E(r,ω) dE 0 (ω) − jk·r dk − jk·r
˜
= e + E 0 (ω) − jr · e ,
∂ω dω dω
˜
˜
∂H(r,ω) dH 0 (ω) − jk·r dk − jk·r
˜
= e + H 0 (ω) − jr · e .
∂ω dω dω
Equation (4.264) becomes
˜ ˜
dH 0 (ω) dE 0 (ω)
˜ ∗ ˜ ∗
−∇ · E (ω) × + × H (ω)−
0
0
dω dω

˜
˜
dk
˜ ∗
˜ ∗
− jr · E (ω) × H 0 (ω) + E 0 (ω) × H (ω) = 4 j w em .
0
0
dω
ω= ˇω
The first two terms on the left-hand side have zero divergence, since these terms do not
depend on r. Bythe product rule (B.42) we have

˜ ∗ ˜ ˜ ˜ ∗ dk
E ( ˇω) × H 0 ( ˇω) + E 0 ( ˇω) × H ( ˇω) ·∇ r · = 4 w em .
0
0
dω
ω= ˇω
The gradient term is merely

dk dk x dk y dk z dk
∇ r · =∇ x + y + z = ,
dω ω= ˇω dω dω dω ω= ˇω dω ω= ˇω
hence

dk
˜ ∗ ˜ ˜ ˜ ∗
E ( ˇω) × H 0 ( ˇω) + E 0 ( ˇω) × H ( ˇω) · = 4 w em . (4.265)
0 0
dω
ω= ˇω
Finally, the left-hand side of this expression can be written in terms of the time-average
Poynting vector. By (4.260) we have
1
1
˜
ˇ
˜
˜ ∗
˜ ∗
ˇ ∗
S av = Re E × H = E 0 ( ˇω) × H ( ˇω) + E ( ˇω) × H 0 ( ˇω)
0
0
2 4
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