Single-Particle Electrodynamics - Assassination Science

The case of (6.33) with m = 3 is more subtle, again due to the fact that
the radial r d integral is, due to its inverse-cube nature, infinite. The term
proportional to n i dn j d in (6.33) avoids this fate (due to the fact that it contains
an extra power of r d /2ε in all terms), but the term proportional to δ ij does
not. However, it will be found that, in fact, terms involving δ ij will all cancel
in our final expressions, before they need to be integrated; let us therefore
simply write its integral, unperformed, as I ij
Performing, then, the integral of the n i dn j d part explicitly, we find
η 0
( 4
3 πε3 ) −2 ∫
∫
d 3 r
r≤ε
r ′ ≤ε
3 .
∫
d 3 r ′ r −3 r2 s n i sns j f(n d ) = −η 1
d
d 2 n d n i dn j d f(n d) + I ij
3 ,
(6.38)
Again, the result (6.38) may be equivalently incorporated into our calculations
by making the simple substitution
rd
−3 r2 s n i sn j s −→ I ij
3 − rd −1 ni dn j d , (6.39)
and then performing the r d integration as if there had been no r s -dependence
at all.
Unfortunately, it is not so simple to devise such a straightforward crosscheck
of (6.39) as was performed earlier for (6.35), due to the presence of
the divergent integral I ij
3 . Instead, the author has thoroughly verified this
result by directly comparing the term-by-term integrations of rd
−3 with those
of rd
0 (which latter have been verified above), which are of course in the
ratio (n + 1)/8ε 3 (n − 2) if the overall integrand (including 4πrd
2 factor) has
dependence rd n . (This can be seen quickly by integrating the polynomials from
first principles, and noting that the extra factor of r 3 d contributes (2ε) 3 .)
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