Page 53 - Electromagnetics Handbook

y = y, (2.74)

z = z, (2.75)
v

ct = γ ct − γ x, (2.76)
c
and the chain rule gives
∂ ∂ v ∂
= γ − γ , (2.77)
2
∂x ∂x c ∂t
∂ ∂
= , (2.78)
∂y ∂y
∂ ∂
= , (2.79)
∂z ∂z
∂ ∂ ∂
=−γv + γ . (2.80)
∂t ∂x ∂t
We begin by examining Faraday’s law in the laboratory frame. In component form we
have
∂E z ∂E y ∂ B x
− =− , (2.81)
∂y ∂z ∂t
∂E x ∂E z ∂ B y
− =− , (2.82)
∂z ∂x ∂t
∂E y ∂E x ∂ B z
− =− . (2.83)
∂x ∂y ∂t
These become
∂E z ∂E y ∂ B x ∂ B x
− = γv − γ , (2.84)
∂y ∂z ∂x ∂t
∂E x ∂E z v ∂E z ∂ B y ∂ B y
− γ + γ = γv − γ , (2.85)
2
∂z ∂x c ∂t ∂x ∂t
∂E y v ∂E y ∂E x ∂ B z ∂ B z
γ − γ − = γv − γ , (2.86)
2
∂x c ∂t ∂y ∂x ∂t
after we use (2.77)–(2.80) to convert the derivatives in the laboratory frame to derivatives
with respect to the moving frame coordinates. To simplify (2.84) we consider
∂ B x ∂ B y ∂ B z
∇· B = + + = 0.
∂x ∂y ∂z
Converting the laboratory frame coordinates to the moving frame coordinates, we have
∂ B x v ∂ B x ∂ B y ∂ B z
γ − γ + + = 0
2
∂x c ∂t ∂y ∂z
or
2
∂ B x v ∂ B x ∂ B y ∂ B z
−γv =−γ + v + v .
2
∂x c ∂t ∂y ∂z
Substituting this into (2.84) and rearranging (2.85) and (2.86), we obtain
∂ ∂ ∂ B x
γ(E z + vB y ) − γ(E y − vB z ) =− ,
∂y ∂z ∂t
∂ ∂ v
∂E x
− γ(E z + vB y ) =− γ B y + E z ,
∂z ∂x ∂t c 2
∂ ∂E x ∂ v
γ(E y − vB z ) − =− γ B z − E y .
∂x ∂y ∂t c 2

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