Page 406 - Electromagnetics

and

2 2 ˜ ˜
∂ 2 1 ∂ E r ∂ H r
˜
+ k (r E φ ) = + jω ˜µ , (5.158)
∂r 2 sin θ ∂φ∂r ∂θ

2 2 ˜ ˜
∂ 2 ∂ E r 1 ∂ H r
˜
+ k (r E θ ) = + jω ˜µ . (5.159)
∂r 2 ∂θ∂r sin θ ∂φ
Hence we can represent the electromagnetic field in a source-free region in terms of
˜
˜
the two scalar quantities E r and H r . Superposition allows us to solve the TE case with
˜
˜
E r = 0 and the TM case with H r = 0, and combine the results for the general expansion
of the field.
TE–TM decomposition in terms of potential functions. If we allow the vector
potential (or Hertzian potential) to have only an r-component, then the resulting fields
are TE or TM to the r-direction. Unfortunately, this scalar component does not satisfy
the Helmholtz equation. If we wish to use a potential component that satisfies the
Helmholtz equation then we must discard the Lorentz condition and choose a different
relationship between the vector and scalar potentials.

1. TM fields. To generate fields TM to r we recall that the electromagnetic fields
may be written in terms of electric vector and scalar potentials as
˜
˜
E =− jωA e −∇φ e , (5.160)
˜
˜
B =∇ × A e . (5.161)
In a source-free region we have by Ampere’s law
1 1
˜ ˜ ˜
E = ∇× B = ∇× (∇× A e ).
jω ˜µ˜ c jω ˜µ˜ c
˜
˜
Here φ e and A e must satisfy a differential equation that may be derived by examining
˜
˜
c ˜
2 ˜
∇× (∇× E) =− jω∇× B =− jω( jω ˜µ˜ E) = k E,
2
c
2
where k = ω ˜µ˜ . Substitution from (5.160) gives
˜
˜
˜
˜
2
∇× ∇× [− jωA e −∇φ e ] = k [− jωA e −∇φ e ]
or
k 2
˜ 2 ˜ ˜
∇× (∇× A e ) − k A e = ∇φ e . (5.162)
jω
˜
We are still free to specify ∇· A e .
At this point let us examine the effect of choosing a vector potential with only an
˜
˜
r-component: A e = ˆ rA e . Since
˜
ˆ
˜
θ ˆ ∂ A e φ ∂ A e
˜
∇× (ˆ rA e ) = − (5.163)
r sin θ ∂φ r ∂θ
˜
we see that B =∇ × A e has no r-component. Since
ˆ
ˆ
2 ˜
2 ˜
2 ˜
˜
ˆ r 1 ∂
∂ A e 1 ∂ A e θ ∂ A e φ ∂ A e
˜
∇× (∇× A e ) =− sin θ + + +
r sin θ r ∂θ ∂θ r sin θ ∂φ 2 r ∂r∂θ r sin θ ∂r∂φ
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