Page 356 - Intro to Tensor Calculus

350

I 21. Show that Maxwell’s equations (2.6.75) through (2.6.78) for the magnetic field under magnetostatic
~
~
~
~
~
conditions reduce to ∇× H = J and ∇· B =0. The divergence of B being zero implies B can be derived
~
~
~
~
from a vector potential function A such that B = ∇× A.Here A is not unique, see problem 24. If we select
~
~
A such that ∇·A = 0 then show for a homogeneous, isotropic material, free of any permanent magnets, that
~
2 ~
∇ A = −µJ.
I 22. Show that under nonsteady state conditions of electrodynamics the Faraday law from Maxwell’s
~
equations (2.6.75) through (2.6.78) does not allow one to set E = −∇V. Why is this? Observe that
~
~
~
~
∇· B =0 so we can write B = ∇× A for some vector potential A. Using this vector potential show that
!
~
∂A
~
Faraday’s law can be written ∇× E + = 0. This shows that the quantity inside the parenthesis is
∂t
~
∂A
~
conservative and so we can write E + = −∇V for some scalar potential V. The representation
∂t
~
∂A
~
E = −∇V −
∂t
is a more general representation of the electric potential. Observe that for steady state conditions ∂ ~ A =0
∂t
so that this potential representation reduces to the previous one for electrostatics.
~
∂A
~
I 23. Using the potential formulation E = −∇V − derived in problem 22, show that in a vacuum
∂t
~
∂∇· A ρ
2
(a) Gauss law can be written ∇ V + = −
∂t 0
(b) Ampere’s law can be written
2 ~

∂V ∂ A
~ ~
∇× ∇× A = µ 0 J − µ 0 0 ∇ − µ 0 0 2
∂t ∂t
(c) Show the result in part (b) can also be expressed in the form
!
~
∂A ∂V
2 ~ ~ ~
∇ A − µ 0 0 −∇ ∇· A + µ 0 0 = −µ 0 J
∂t ∂t
I 24. The Maxwell equations in a vacuum have the form
~
~
∂B ∂D
~ ~ ~ ~ ~
∇× E = − ∇× H = + ρ V ∇· D = ρ ∇· B =0
∂t ∂t
~
~
~
~
2
where D = 0 E, B = µ 0 H with 0 and µ 0 constants satisfying 0 µ 0 =1/c where c is the speed of light.
~
∂A
~ ~ ~ ~
Introduce the vector potential A and scalar potential V defined by B = ∇× A and E = − −∇ V.
∂t
Note that the vector potential is not unique. For example, given ψ as a scalar potential we can write
~
~
~
B = ∇× A = ∇× (A + ∇ ψ), since the curl of a gradient is zero. Therefore, it is customary to impose some
~
~
kind of additional requirement on the potentials. These additional conditions are such that E and B are
~
~
not changed. One such condition is that A and V satisfy ∇· A + 1 ∂V =0. This relation is known as the
2
c ∂t
~
Lorentz relation or Lorentz gauge. Find the Maxwell’s equations in a vacuum in terms of A and V and show
that
2 2
1 ∂ ρ 1 ∂
2 2 ~ ~
∇ − V = − and ∇ − A = −µ 0 ρV.
2
2
c ∂t 2 0 c ∂t 2