Exercises with Magnetic Monopoles - Kurt Nalty

We now write the field angular momentum as
∫∫∫ ( )
⃗L = ɛ ⃗r × ⃗E × B ⃗ dxdydz
∫∫∫ ( (
= ɛ ⃗E ⃗r · ⃗B
)
− B ⃗ (
⃗r · ⃗E
))
dxdydz
= q ∫∫∫
e 1
( (
⃗r ⃗r ·
4π r ⃗B
)
− ⃗ )
B (⃗r · ⃗r) dxdydz
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= q ∫∫∫
e 1
( (⃗a r ⃗a r ·
4π r
⃗B
)
− B ⃗ )
dxdydz
= − q ∫∫∫ [( ]
e ⃗B · ∇ ⃗
)⃗a r dxdydz
4π
I am good to this point. The step, I haven’t verified to my satisfaction. The
claim is to integrate by parts to achieve
⃗L = − q ∫∫ ( )
e
⃗a r ⃗B · dS ⃗ + q ∫∫∫ ( )
e
⃗a r ⃗∇ · B ⃗ dxdydz
4π
4π
Given this step, the next step is to assert that the surface integral goes to
zero at infinity. We then identify
⃗B = − µq ( )
m
∇
4π ⃗ 1
ρ m
⃗∇ · ⃗B = − µq m
4π (−4πδ (⃗r − R⃗a z)) = µq m δ (⃗r − R⃗a z )
⃗L = q ∫∫∫ ( )
e
⃗a r ⃗∇ · B ⃗ dxdydz
4π
= q ∫∫∫
e
⃗a r (µq m δ (⃗r − R⃗a z )) dxdydz
4π
= µq eq m
4π
⃗a z
We have postive electric charge, positive magnetic charge, and ⃗ L points
from electric to magnetic charge.
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