Single-Particle Electrodynamics - Assassination Science

+ η 0 (2ε) 2 ( 4
3 πε3 ) −1 ∫
V d
d 3 r d
{ 1
5 − 3 8
( ) rd
+ 1 ( ) rd 3 ( ) }
3 rd 5
− δ ij . (6.31)
2ε 4 2ε 40 2ε
Now, as will be shown shortly, the angular integration of two factors of n d
yields
thus, (6.31) yields
η 0
( 4
3 πε3 ) −2 ∫
∫
d 3 r
r≤ε
∫V d
d 3 r d n i dn j d f(r d) = 1 3 δij ∫
d 3 r ′ r i sr j s
r ′ ≤ε
V d
d 3 r d f(r d );
( ) 4 −1 ∫ { 2ε 1
= η 0 (2ε) 2 3 πε3 4πrd 2 dr d
0 5 − 1 ( ) rd
+ 1 ( ) rd 2 ( ) }
1 rd 5
−
2 2ε 3 2ε 30 2ε
= 2ε2
5 η 0
= 1 3 η −2. (6.32)
That this result, η −2 /3, is indeed correct follows from the fact that we could
have alternatively computed it as the integral of r i dr j d ; the normals ni dn j d contribute
the factor of 1/3, and the magnitudes r 2 d contribute the integral η −2 .
We now employ clairvoyance once again, and predict that the only integrals
involving r i sr j s
that we shall have need to perform will be of the form
η 0
( 4
3 πε3 ) −2 ∫
∫
d 3 r
r≤ε
d 3 r ′ rd −m rs 2 n i sns j f(n d ), (6.33)
r ′ ≤ε
where m = 2 or 3 only. While the integral over n d cannot be performed
without the explicit f(n d ) being supplied, we can nevertheless perform the
complete r s integral, as well as the integral over r d , following the same method
as above. When this is done for m = 2, one finds
η 0
( 4
3 πε3 ) −2 ∫
∫
d 3 r
r≤ε
d 3 r ′ rd −2 r2 s n i sns j f(n d )
∫ { 3
= η 0 d 2 n d
2 δij − 1 }
2 ni dn j d f(n d ) (6.34)
r ′ ≤ε
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