Single-Particle Electrodynamics - Assassination Science

Taking the dot-product of (6.80) with another arbitary vector b, we find
(b·∇)(n˜r·a) =
Finally, in the case when b = n˜r , we find
employing (6.72), we thus find that
(a·b) − (n˜r·a)(n˜r·b)
. (6.81)
˜r
(n˜r·∇)(n˜r·a) = (n˜r·a) − n 2˜r(n˜r·a) ;
˜r
(n˜r·∇)(n˜r·a) = (n˜r·a)
˜r
6.6.6 The monopolar inverse-square field
{
1 − r2
˜r 2 }
. (6.82)
We now consider carefully the monopolar inverse-square field, employing the
˜r-space:
n˜r
4π˜r . (6.83)
2
Our purposes in doing so are twofold. Firstly, we shall find that the ˜r-space
provides a quick yet powerful way of divining the delta-function divergence
of (6.83), without having to employ any integral theorems, or resorting to
Poisson’s equation for the scalar potential; the results found will also be
of use in the following sections. Secondly, we shall show that the process
of employing the ˜r-space for the monopolar field, and then computing the
gradient to obtain the dipolar field, is completely equivalent to performing
these two processes in the reverse order.
thus,
We first consider the spatial gradient of (6.83) in an arbitrary direction:
n˜rj
∂ i
4π˜r = 1
2 4π˜r ∂ 2 in˜rj + n˜rj
4π ∂ i
= δ ij − n˜ri n˜rj
4π˜r 3
1
˜r 2
− 2n˜rin˜rj
4π˜r 3 ;
∂ i
n˜rj
4π˜r 2 = δ ij − 3n˜ri n˜rj
4π˜r 3 . (6.84)
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