Page 374 - Electromagnetics

Under the Lorentz gauge condition

∂φ h
∇· A h =−µ
∂t
this reduces to
2 ρ i
2
∇ φ h − µ ∂ φ h =− m .
∂t 2 µ
Expanding the curl-curl operation in (5.45) we have

2 ∂
∂ A h
i
2
∇(∇· A h ) −∇ A h = J − µ − µ ∇φ h ,
m
∂t 2 ∂t
which, upon substitution of the Lorentz gauge condition gives
2
2 ∂ A h i
∇ A h − µ =− J . (5.46)
m
∂t 2
We can also derive a Hertzian potential for the case of magnetic current. Letting
∂Π h
A h = µ (5.47)
∂t
and employing the Lorentz condition we have

∂Π h
D =−µ ∇× ,
∂t
2
∂ Π h
H =∇(∇· Π h ) − µ .
∂t 2
The wave equation for Π h is found by substituting (5.47) into (5.46) to give

2
∂ 2 ∂ Π h 1 i
∇ Π h − µ =− J . (5.48)
m
∂t ∂t 2 µ
i
Defining M through
∂M i
i
J = µ ,
m
∂t
we write the wave equation as
2
i
2
∇ Π h − µ ∂ Π h =−M .
∂t 2
i
i
We can think of M as a convenient way of representing J , or we can conceive of an
m
impressed magnetization current that is independent of H and defined through B =
i
µ 0 (H + M + M ). With the help of (5.48) we can also write the fields as
H =∇ × (∇× Π h ) − M i ,
∂Π h
D =−µ ∇× .
∂t

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