Exercises with Magnetic Monopoles - Kurt Nalty

∫ ρ=∞
ρ=0
ρ 2
∫ ρ=∞
(
ρ2 + (z − R/2) 2) 3/2 (
ρ2 + (z + R/2) 2) ρdρ = ρ 2
3/2
ρ=0 (ρ 2 ) 3/2 ( ρ 2 + (R) 2) ρdρ 3/2
=
∫ ρ=∞
ρ=0
dρ
(ρ 2 + R 2 ) 3/2
Now
∫ ρ=∞
ρ=0
dρ
(ρ 2 + R 2 ) 3/2 =
[
] ρ=∞
ρ
R 2√ ρ 2 + R 2
ρ=0
= 1 R 2
In a similar fashion, at the other pole, we get the same value. This is
satisfying, as our values are continuous across the poles. We are now able to
finish evaluating our integral for total field angular momentum.
⃗L = −⃗a z
µq e q m R
8π
∫
z
[ ∫ ρ=∞
ρ=0
]
ρ 2
(
ρ2 + (z + R/2) 2) 3/2 (
ρ2 + (z − R/2) 2) 3/2 d(ρ2 ) dz
⃗L = −⃗a z
µq e q m R
8π
= −⃗a z
µq e q m R
8π
= −⃗a z
µq e q m R
8π
[ ∫ ∞ ∫
dz R/2
R/2 4z + 2 −R/2
[ [
− 1 ] ∞ [ z
] R/2
+ +
4z
R/2
R 2 −R/2
[(
0 + 2 ) ( ) R
+ +
4R R 2
∫
dz −R/2
R + 2 −∞
[
− 1
4z
]
dz
4z 2
] −R/2
−∞
( 2
4R + 0 )]
]
⃗L = −⃗a z
µq e q m
4π
This results agrees with Goldhaber and Jackson. We have the same magnitude,
irregardless of distance, and the momentum points from the positive
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