Single-Particle Electrodynamics - Assassination Science

6.7.2 Inverse-cube integral with two normals
We now turn to the integral
η 0
( 4
3 πε3 ) −2 ∫
∫
d 3 r
r≤ε
d 3 r ′ ˜r d −3 n˜r d in˜rd j. (6.110)
r ′ ≤ε
As we found earlier, we must be careful to replace n˜rd
expression in terms of n d :
n˜rd ≡ r d
˜r d
n d .
by its equivalent
The integral over n d then proceeds as usual, leaving the result δ ij /3. The
integral over r d is now the same as it was for (6.102), except that the integrand
is multiplied by the factor r 2 d /˜r 2 d . This does not, of course, have any effect on
the integrals of the second and third terms (in the limit w d → 0): the result
is the same as before:
( ) 4 −1 ∫ {
2ε
η 0
3 πε3 4πrd 2 dr d − 3
0
2
( ) rd
+ 1 ( ) }
rd
3
2ε 2 2ε
rd
2
(rd 2 + = −4η 0
w2 d )5/2 ε + O(w d).
3
On the other hand, the integral of the first term is modified by the appearance
of the extra factor r 2 d /˜r 2 d : we have an extra factor of u 2 over that present in
(6.104):
r 2 d
( ) 4 −1 ∫ 2ε
η 0
3 πε3 4πrd 2 dr d
0 (rd 2 + = 3η ∫ 2ε/(4ε 2 +wd 2)1/2
0
du
w2 d )5/2 ε 3 1 − u . 2
We now note the convenient identity
0
u 4
1 − u 2 ≡ u2
1 − u 2 − u2 ; (6.111)
in other words, the multiplication by u 2 is equivalent to simply an extra
added term −u 2 ; we thus immediately find
η 0
( 4
3 πε3 ) −2 ∫
∫
d 3 r
r≤ε
r ′ ≤ε
u 4
d 3 r ′ rd 2 ˜r d −5 = −8η 3 ′ + 3η 3. ′′ (6.112)
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