By Hammar G.W.
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Additional info for A Preliminary Report on the Magnetic Susceptibilities of Some Cases
A -n = (V'V'w) . a (34) OXmOXn and V' X (aw) = (I em m In view of identities o~J X aw = (I em :~) X a = V'w X a m t35) . V' (37) Since the quantity eiklr-r'l == 41J"r I G(r,r') - 'I r (38) is known as the free-space scalar Green's function, it is appropriate refer to the quantity == ( r(r,r') U + b VV) 4:1:lr=r'~'1 == ( U + b VV) G(r,r') to (39) as the free-space dyadic Green's function. fr(r,r') H(r) = fVG(r,r') • J(r')dV' X J(r')dV' - fVG(r,r') X Jm(r')dV' (40) + iWEfr(r,r') . 2 Free-space Dyadic Green's Function In the previous section we derived the free-space dyadic Green's function using the potentials and antipotentials as an intermediary.
2) we get f p(r')(V2 + lc )G(r,r')dV' 2 = - (7) p(r) where the Laplacian operator operates with respect to the unprimed coordinates only. Then with the aid of the Dirac 0 functionl which permits p to be represented as the volume integral p(r) = fp(r')o(r (r in V) - r')dV' (8) we see that Eq. (7) can be written as f p(r')[(V2 + k2)G(r,r') + o(r - r')]dV' (9)" = 0 From this it follows that G must satisfy the scalar Helmholtz V2G(r,r') + k2G(r,r') = - equation o(r - r') (10) Since G satisfies Eq.
2) and condition (5) when G is given by expression (14), the desired solution of Eq. (2) can be written as the Helmholtz integral 1 q,(r) = E J p(r') eiklr-r'l 4 I 7I"r - r '1 dV' (15) Now the related problem of finding A can be easily handled. Clearly, the appropriate solution of Eq. (3) must be the Helmholtz integral A(r) = p. J J(r') eiklr-r'l (16) 471"Ir_ r/I dV' because it has the proper behavior on the sphere at infinity and it satisfies Eq. (3). To show that it satisfies Eq. (3), one only has to operate on Eq.