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Three-vector and scalar field identities and uniqueness theorems in ...

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Theorem U1. The divergence <strong>and</strong> curl of a twice cont<strong>in</strong>uouslydifferentiable (static) three-<strong>vector</strong> <strong>field</strong>, whichvanishes sufficiently rapidly at <strong>in</strong>f<strong>in</strong>ity, uniquely determ<strong>in</strong>esthe three-<strong>vector</strong> <strong>field</strong> over an unbounded volumeV of R 3 .In other words, we must specify∇ × F(x, y, z) = j(x, y, z),∇ · F(x, y, z) = ρ(x, y, z),(1a)(1b)over the volume V. In an electromagnetic context j isa circulation current density <strong>and</strong> ρ is a source chargedensity. A proof of a version of this theorem over a f<strong>in</strong>itevolume of R 3 is given <strong>in</strong> Ref. 11.The second <strong>uniqueness</strong> theorem, which is typicallycalled a Helmholtz theorem, can be stated as follows: 6Theorem H1. A general cont<strong>in</strong>uous three-<strong>vector</strong> <strong>field</strong>def<strong>in</strong>ed everywhere <strong>in</strong> R 3 that along with its first derivativesvanishes sufficiently rapidly at <strong>in</strong>f<strong>in</strong>ity may beuniquely represented as a sum of an irrotational <strong>and</strong> asolenoidal part, up to a possible additive constant <strong>vector</strong>.A proof of theorem H1 is given <strong>in</strong> Ref. 8. To prove thistheorem it is sufficient to prove only that the three-<strong>vector</strong>F can be written as 11F(x, y, z) = −∇Φ(x, y, z) + ∇ × A(x, y, z). (2)Then the three-<strong>vector</strong> identity∇ × ∇Φ = 0, (3)<strong>and</strong> the irrotational <strong>field</strong> assumption ∇ × F I = 0 impliesthat F I = −∇Φ is irrotational. And the three-<strong>vector</strong>identity∇ · ∇ × A = 0, (4)<strong>and</strong> the solenoidal <strong>field</strong> assumption ∇ · F S = 0 impliesthat F S = ∇ × A is solenoidal. Thus Eq. (2) is a sum ofan irrotational <strong>and</strong> a solenoidal part.The most straightforward proof of Eq. (2) (see Refs. 9<strong>and</strong> 10) <strong>and</strong> therefore of theorem H1 is to obta<strong>in</strong> a Helmholtzidentity 8 that is of the same form as Eq. (2). Aproof of this identity is presented here so that our analysisis self conta<strong>in</strong>ed. This proof is based on the assumptionthat there exists a solution for a three-<strong>vector</strong>F of an <strong>in</strong>homogeneous <strong>vector</strong> Poisson equation <strong>in</strong> Cartesiancoord<strong>in</strong>ates, <strong>and</strong> that each separate Cartesian componentof F is a solution of a <strong>scalar</strong> Poisson equation.The <strong>scalar</strong> Poisson equation can then be solved <strong>in</strong> termsof a two-po<strong>in</strong>t <strong>scalar</strong> Green’s function G(r, r ′ ) that connectsits unit delta function source located at the sourcepo<strong>in</strong>t r ′ = (x ′ , y ′ , z ′ ) to a measurement at the <strong>field</strong> po<strong>in</strong>tr = (x, y, z). Because∇ 2 14πr = −δ3 (r − r ′ ), (5)for all r <strong>and</strong> r ′ with r ≡ |r − r ′ |, the Poisson equationdef<strong>in</strong><strong>in</strong>g the Green’s function,∇ 2 G(r, r ′ ) = −δ 3 (r − r ′ ), (6)leads to G(r, r ′ ) = 1/4πr. The delta function property ofthe <strong>vector</strong> identity (5) can be applied to any well-behavedfunction F(x, y, z) as∫F(x, y, z) = F(x ′ , y ′ , z ′ )δ 3 (r − r ′ )dV ′ (7a)V∫′ ( ) −1= F(x ′ , y ′ , z ′ )∇ 2 dV ′ . (7b)V 4πr′Because the Laplacian operator acts only on the <strong>field</strong>po<strong>in</strong>t coord<strong>in</strong>ates, it can be brought outside of the <strong>in</strong>tegrationover the source po<strong>in</strong>t coord<strong>in</strong>ates. We can usethe well known three-space identity∇ 2 A = ∇(∇ · A) − ∇ × (∇ × A) (8)to rewrite Eq. (7b) as [typo corrected from journal article]∫F(x, y, z) = − ∇ ∇ · F(x′ , y ′ , z ′ )dV ′V 4πr′∫+ ∇ × ∇ × F(x′ , y ′ , z ′ )dV ′ , (9)V 4πr′where one of the ∇ operators has been brought back<strong>in</strong>to each of the <strong>in</strong>tegrals, which is allowed because theyoperate only on the <strong>field</strong> coord<strong>in</strong>ates. Although Eq. (9)is <strong>in</strong> a form like Eq. (2) as required to prove theorem H1,it is necessary to obta<strong>in</strong> an identity that is suitable toprove the three-space theorem U1 as well. If we use the<strong>vector</strong> <strong>identities</strong> 10∇ · (φA) = A · ∇φ + φ∇ · A,∇ × (φA) = −A × ∇φ + φ∇ × A,(10a)(10b)(note that the divergence <strong>and</strong> curl equal zero with respectto the <strong>field</strong> coord<strong>in</strong>ates of a <strong>vector</strong> A which is a functiononly of the source coord<strong>in</strong>ates) <strong>and</strong> use the identity∇(1/r) = −∇ ′ (1/r), we f<strong>in</strong>d∫F(x, y, z) = ∇ F(x ′ , y ′ , z ′ ) · ∇ ′ 1V 4πr dV ′′∫+∇ ×V ′ F(x ′ , y ′ , z ′ ) × ∇ ′ 14πr dV ′ . (11)The <strong>vector</strong> <strong>identities</strong> (10) are now applied aga<strong>in</strong>, this timeon the <strong>in</strong>tegr<strong>and</strong>s of Eq. (11) where the ∇ operators acton the source coord<strong>in</strong>ates. We obta<strong>in</strong> four terms; the onesconta<strong>in</strong><strong>in</strong>g ∇ · (φA) <strong>and</strong> ∇ × (φA), respectively, becomesurface <strong>in</strong>tegrals via the follow<strong>in</strong>g <strong>identities</strong>, 10 (that is,lett<strong>in</strong>g T = φA),∫∮ ∮∇ · T dV = T · dS = T · n dS, (12a)VSS∫∮∮∇ × T dV = n × T dS = − T × n dS, (12b)VSwhere n is the unit surface normal of S bound<strong>in</strong>g V .These <strong>in</strong>tegrals vanish as r → ∞ for the <strong>field</strong> F, whichis assumed to fall off sufficiently rapidly at <strong>in</strong>f<strong>in</strong>ity. TheS2

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