Single-Particle Electrodynamics - Assassination Science

the remaining pieces glued together—is the volume of integration for r s ; we
shall call this intersection volume V s . Taking into account the Jacobian (of
value 1/2 3 = 1/8), we thus have
∫
∫
d 3 r
r≤ε
d 3 r ′ = 1 8
r ′ ≤ε
6.4.2 Symmetry arguments
∫V d
d 3 r d
∫
V s
d 3 r s . (6.17)
Clearly, integrals in r d –r s space are not straightforward with such a strange
region of integration as that described in the previous section: at the least,
we have lost spherical symmetry for r s .
However, there is one consideration that allows us to simplify matters
considerably. It will be noted that, since the two defining three-spheres in
r s -space are offset by ±r d , then a parity operation r d → −r d in r d -space
leaves the r s integration region (and hence the value of the inner r s integral)
invariant. Since the r d integral is itself over a three-sphere, we thus find that
integrals odd in r d vanish, regardless of their dependence on r s .
Likewise, we could have alternatively presented the above with the r s
integral outermost (since, fundamentally, the definitions (6.11) of r d and r s
are symmetrical), and thus conclude that integrals odd in r s vanish, again
regardless of their dependence on r d .
6.4.3 Explicit inner integrals
We now consider the explicit expressions involved in performing integrals
over the hypervolume specified by V d and V s .
We consider the r d integral to be performed “outermost”, and the r s integral
“innermost”. The process may be visualised by analogy with a computer
program with six nested loops: we are here specifying that the three components
of r d are to be incremented in the three outermost loops; the three
components of r s are incremented in the three innermost loops. The starting
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