A Helmholtz’ Theorem Because 1 ∇2 = −4πδ(R) (A.1) R where R = r − r with magnitude R = |R| and where δ(R)=δ(r − r)= δ(x − x)δ(y − y)δ(z − z) is the three-dimensional Dirac delta function (see Appendix B), then any sufficiently well-behaved vector function F(r)= F(x, y, z) can be represented as F(r)= F(r)δ(r − r) d3r V 1 1 = − F(r)∇2 d3r 4π R V 1 F(r) = − ∇2 d3r, (A.2) 4π V R the integration extending over any region V that contains the point r. With the identity ∇×∇×= ∇∇ · −∇2, Eq. (A.2) may be written as 1 F(r) 1 F(r) F(r)= ∇×∇× d3r − ∇∇ · d3r. (A.3) 4π V R 4π V R Consider first the divergence term appearing in this expression. Because the vector differential operator ∇ does not operate on the primed coordinates, then 1 F(r) 1 1 ∇· d3r = F(r) ·∇ d3r. (A.4) 4π V R 4π V R Moreover, the integrand appearing in this expression may be expressed as 1 1 F(r) ·∇ = −F(r) ·∇ R R F(r) 1 = −∇ · + ∇ · F(r), (A.5) R R where the superscript prime on the vector differential operator ∇ denotes differentiation with respect to the primed coordinates alone. Substitution of 422 A Helmholtz’ Theorem Eq. (A.5) into Eq. (A.4) and application of the divergence theorem to the first term then yields 1 F(r) 1 F(r) 1 ∇ · F(r) ∇· d3r = − ∇ · d3r + d3r 4π R 4π R 4π R V V V 1 1 1 ∇ · F(r) = − F(r) · nˆd2r + d3r 4π S R 4π V R = φ(r), (A.6) whichisthedesiredformofthescalarpotentialφ(r) for the vector field F(r). Here S is the surface that encloses the regular region V and contains the point r. For the curl term appearing in Eq. (A.3) one has that 1 F(r) 1 1 ∇× d3r = − F(r) ×∇ d3r 4π R 4π R V V 1 1 = F(r) ×∇ d3r. (A.7) 4π V R Moreover, the integrand appearing in the final form of the integral in Eq. (A.7) may be expressed as 1 ∇ × F(r) F(r) F(r) ×∇ = −∇ × , (A.8) R R R so that 1 F(r) 1 ∇ × F(r) 1 F(r) ∇× d3r = d3r − ∇ × d3r 4π R 4π R 4π R V V V 1 ∇ × F(r) 1 1 = d3r + F(r) × nˆd2r 4π V R 4π S R = a(r), (A.9) whichisthedesiredformofthevectorpotential. The relations given in Eqs. (A.3), (A.6), and (A.9) then show that F(r)=−∇φ(r)+∇×a(r), (A.10) where the scalar potential φ(r) is given by Eq. (A.6) and the vector potential a(r) is given by Eq. (A9). This expression may also be written as F(r)=F (r)+Ft(r), (A.11) where F (r)=−∇φ(r) ∇ · − 1 ∇ F(r ) 3 1 ∇ F(r ) · 2 = d r + nˆd r (A.12) 4π V |r − r | 4π S |r − r | References 423 is the longitudinal or irrotational part of the vector field (where ∇×F (r )= 0), and where F (r)=∇×a(r) t 1 F(r) = ∇×∇× d3r 4π |r − r| V ∇ × 1 ∇× F(r ) 3 1 ∇× F(r ) × 2 = d r + nˆd r (A.13) 4π V |r − r | 4π S |r − r | is the transverse or solenoidal part of the vector field (where ∇·F (r ) = 0). If the surface S recedes to infinity and if the vector field F(r)isregular at infinity, then the surface integrals appearing in the above expressions and Eqs. (A.12)–(A.13) become F (r)=−∇φ(r) ∇ · − 1 ∇ F(r ) 3 = d r , (A.14) 4π V |r − r | F (r)=∇×a(r) t ∇ × 1 ∇× F(r ) 3 = d r . (A.15) 4π V |r − r | Taken together, the above results constitute what is known as Helmholtz’ theorem [1]. Theorem 12. Helmholtz’ Theorem. Let F(r) be any continuous vector field with continuous first partial derivatives. Then F(r) can be uniquely ex- pressed in terms of the negative gradient of a scalar potential φ(r) and the curl of a vector potential a(r), as embodied in Eqs. (A.10) and (A.11). References 1. H. B. Phillips, Vector Analysis. New York: John Wiley & Sons, 1933. B The Dirac Delta Function B.1 The One-Dimensional Dirac Delta Function The Dirac delta function [1] in one-dimensional space may be defined by the pair of equations δ(x)=0; x =0 , (B.1) ∞ δ(x) dx =1. (B.2) −∞ It is clear from this definition that δ(x) is not a function in the ordinary mathematical sense, because if a function is zero everywhere except at a single pointandtheintegralofthisfunctionoveritsentiredomainofdefinition exists, then the value of this integral is necessarily also equal to zero. Because of this, it is more appropriate to regard δ(x) as a functional quantity with a certain well-defined symbolic meaning. For example, one can consider a sequence of functions δ(x, ε) that, with increasing values of the parameter ε, differ appreciably from zero only over a decreasing x-interval about the origin and which are such that ∞ δ(x, ε) dx = 1 (B.3) −∞ for all values of ε. Although it may be tempting to try to interpret the Dirac delta function as the limit of such a sequence of well-defined functions δ(x, ε) as ε →∞, it must be recognized that this limit need not exist for all values of the independent variable x. However, the limit ∞ lim δ(x, ε) dx = 1 (B.4) ε→∞ −∞ must exist. As a consequence, one may interpret any operation that involves the delta function δ(x) as implying that this operation is to be performed with a function δ(x, ε) of a suitable sequence and that the limit as ε →∞is to be taken at the conclusion of the calculation. The particular choice of the sequence of functions δ(x, ε) is immaterial, provided that their oscillations (if any) near the origin x = 0 are not too violent [2]. Each of the following functions forms a sequence with respect to the parameter ε that satisfies the required properties. 426 B The Dirac Delta Function ε 2 2 δ(x, ε)=√ e−ε x , π δ(x, ε) = rect1/ε(x), ε δ(x, ε)= sinc(εx), π δ(r, ε)=circ1/ε(r), ε δ(r, ε)= J1(2πεr), r where rect1/ε(x) ≡ ε/2when|x| < 1/ε and is zero otherwise, circ1/ε(r) ≡ ε2/π when r<1/ε and is zero otherwise, and sinc(x) ≡ sin (x)/x when x =0 and is equal to its limiting value of unity when x = 0, where the last two of the above set of functions are appropriate for polar coordinates. Let f(x) be a continuous and sufficiently well-behaved function of x ∈ (−∞, ∞) and consider the value of the definite integral ∞ ∞ f(x)δ(x − a)dx = lim f(x)δ(x − a, ε)dx. −∞ ε→∞ −∞ When the parameter ε is large, the value of the integral appearing on the right-hand side of this equation depends essentially on the behavior of f(x) in the immediate neighborhood of the point x = a alone, and the error that results from the replacement of f(x)byf(a) may be made as small as desired by taking ε sufficiently large. Hence ∞ ∞ lim f(x)δ(x − a, ε)dx = f(a) lim δ(x − a, ε)dx, ε→∞ −∞ ε→∞ −∞ so that ∞ f(x)δ(x − a)dx = f(a). (B.5) −∞ This result is referred to as the sifting property of the delta function. That is, the process of multiplying a continuous function by δ(x − a) and integrating over all values of the variable x is equivalent to the process of evaluating the function at the point x = a. Notice that, for this result to hold, the domain of integration need not be extended over all x ∈ (−∞, ∞); it is only necessary that the domain of integration contain the point x = a in its interior, so that a+∆2 f(x)δ(x − a)dx = f(a), (B.6) a−∆1 where ∆1 > 0, ∆2 > 0. It is then seen that f(x) need only be continuous at the point x = a. The above results may be written symbolically as f(x)δ(x − a)=f(a)δ(x − a), (B.7) B.1 The One-Dimensional Dirac Delta Function 427 the meaning of such a statement being that the two sides yield the same result when integrated over any domain containing the point x = a. For the special case when f(x)=xk with k>0anda = 0, Eq. (B.7) yields xkδ(x)=0, ∀k>0. (B.8) Theorem 13. Similarity Relationship (Scaling Law). For all a =0 1 δ(ax)= δ(x). (B.9) |a| Proof. In order to prove this relationship one need only compare the integrals of f(x)δ(ax)andf(x)δ(x)/|a| for any sufficiently well-behaved continuous function f(x). For the first integral one has (for any a =0) ∞ 1 ∞ 1 f(x)δ(ax)dx = ± f(y/a)δ(y)dy = f(0), −∞ a −∞ |a| where the upper or lower sign choice is taken accordingly as a>0ora<0, respectively, and for the second integral one obtains ∞ 1 1 f(x) δ(x)dx = f(0). −∞ |a| |a| Comparison of these two results then shows that δ(ax)=δ(x)/|a|,aswasto be proved. ! For the special case a = −1, Eq.