Some Problems with the Use of the Dirac Delta Function I: What Is the Value of ∫∫∫ ∞ Δ(X)Dx?

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Some problems with the use of the Dirac delta function I: R ∞ What is the value of 0 δ(x)dx?

M. Amaku1,2 , F.A.B. Coutinho1, O. Eboli´ *3 , E. Massad4

1Universidade de SãoPaulo, Faculdade de Medicina, 05405-000, SãoPaulo, SP, Brasil. 2Universidade de SãoPaulo, Faculdade de Medicina Veterinária e Zootecnia, 05508-970, SãoPaulo, SP, Brasil. 3Universidade de SãoPaulo, Instituto de F´ısica,05508-090, SãoPaulo, SP, Brasil. 4Escola de MatemáticaAplicada, Funda¸cãoGetúlioVargas, 22250-900, Rio de Janeiro, RJ, Brasil.

Received on April 06, 2021. Revised on June 03, 2021. Accepted on June 05, 2021. This is the ﬁrst of a series of articles dealing with the casual use of distributions made by physicists and engineers. This use is very economical but sometimes leads to embarrassing conclusions that are diﬃcult to justify with rigorous theories of distributions. Keywords: Dirac delta function, distributions.

1. Motivation until the work of Schwartz [7]. The work of Dirac and Schwartz is praised in the book by Lighthill [8] as One of the authors received an e-mail from a famous follows: “To Paul Dirac who saw the it must be true; physicist about one of his articles [1]. The following Laurent Schwartz who proved it and to George Temple is a transcription of a relevant part that motivated who showed how simple it could be made.” The last this article: “As to your paper, I see it as an appeal reference is to the work George Temple [9]. It should that people stop using the now-standard notation for be mentioned the above mentioned symbolic methods the Dirac delta function, and switch to the notation of can also be justified by an almost equivalent theory by distributions. I don’t think this is practical, and I’m not J. G. Mikuzinski [10, 11]; see also [12, 13]. For further too supportive of papers that insist on this anyway.” information the history of distributions see [14, 15] and We have to respect this opinion because the physicists’ references therein. use of the Dirac delta function as if it were a function In spite of this some of the symbolic methods men- is very economical. However, in the mentioned paper, tioned above, the efforts to put them in a rigorous we were not “advocating stopping the now standard mathematical form still do not explain them clearly. notation of the Dirac function”. In fact we responded In fact, we believe that neither Schwartz [7] nor Tem- that “I disagree with you that our paper proposes a new ple [16] distribution theories can easily clarify the formal way of using the Dirac delta function. In fact we used methods commonly used by Physicists. In this series of methods that are becoming more and more common articles we try to show examples where this appears in physics and only mentioned in remarks that the to be so, comment on the possible relevance of the procedures could be done consistently by using Schwartz results, and show cases where the theory, when properly or Temple distribution theories. I am sure that you used, greatly illuminate the physics involved. The first of known examples of inconsistencies when using the now these problems presented in this article has no practical standard methods to handle blindly the Dirac function. consequences, but since it brings a lot of confusion to the Let us explain our position more clearly with a bit of literature we think it is a good idea to clarify it. More history. specifically, we focus on the value of the integral th th In the late 19 century and the beginning of the 20 Z ∞ century a number of special techniques called “improper δ(x)dx = X. (1) functions methods” or “symbolic methods” were being 0 used by engineers and mathematical physicists [2–6]. 1 Two possible choices for X are 2 and 1, that are called, The Dirac delta function became very popular because respectively, the weak and strong definitions of the Dirac of the extensive use of them in Dirac’s work. However, delta. We discuss these two options in detail in Section 2, it was introduced before by Kirchhoff and widely used while we summarize the main points of the theories of by Heaviside. These methods were difficult to justify distributions in Section 3. Once a choice for the integral in equation (1), we must * Correspondence email address: [email protected] consistently work with its chosen value when we consider

Copyright by Sociedade Brasileira de F´ısica.Printed in Brazil. e20210132-2 Some problems with the use of the Dirac delta function I other related problems as it is shown below. As an 2. The problem illustration, we present the relevance for physics of this choice in Section 4 of this paper where we describe and The value of the integral in equation (1) involving the deduce two routinely used formulas in electromagnetism: Dirac delta function is the cause of some perplexity in the literature. The following integral (1) The Laplacian of the Coulomb potential, namely Z ∞ 1 δ(x)dx = 1 (6) 4 = −4πδ3(~r), (2) −∞ r is well know, but the value of equation (1) is subject of where r stands for |~r|. some discussion in the literature. (2) The divergence of the Coulomb field of a unit point According to G. Barton [18] on page 33 the “strong charge, namely definition of the delta function requires X = 1,... but some books choose X = 1 ”. We quote below some books ~r 2 ∇ · = 4πδ3(~r). (3) and articles that choose this last value: r3 (1) The first reference is in page 29 of [19]. No These two formulas were not know before the theory of explanation for this is given, but we can think generalized functions. We show how the theory modifies that the reasoning is as follows. We can write our understanding of electromagnetism and clarify com- equation (1) as plex examples. A very good early paper on this is [17]. Z ∞ Z 0 Z ∞ The two formulas are well know by physics students, but 1 = δ(x)dx = δ(x)dx + δ(x)dx we show that their derivation appear to depends on the −∞ −∞ 0 value of the integral mentioned in the title. Z ∞ As mentioned above these formulas were unknown = 2 δ(x)dx (7) until the development of distribution theory and without 0 them some calculations are problematic. For instance, we since δ(x) is an even function. Here is a similar know that argument

1 1 ~r ∇ · ~r 1 Z ∞ 4 = ∇ · ∇ = −∇ · = + ~r · ∇ 1 ikx r r r3 r3 r3 δ(x) = e dk (8) 2π −∞ 3 ~r · ~r = − − 3 = 0 (4) then r3 r5 Z ∞ Z ∞ 1 ikx ~ ~ e dk dx = 1. (9) for ~r 6= 0 and that this laplacian is not defined for ~r = 0. −∞ 2π −∞ However, Since equation (8) is an even function of x we have Z Z 2 1 3 1 4πR ∇ · ∇ d ~r = ∇ · dS~ = − = −4π (5) ∞ ∞ 2 Z 1 Z 1 r≤R r S r R eikxdk dx = . (10) 0 2π −∞ 2 This was not considered a problem before the introduction of distributions. It was considered to be just (2) The second reference is given in page 791 of [20]. 1 the potential of a point charge, which, for the pioneers, The justification for using equation (1) with X = 2 didn’t exist anyway: it was an idealized case. Of course is that the n-dimensional delta function is given by if we consider equation (2) the above result becomes so natural that the above calculation is considered as a 1 δ(|~r|) δ(~r) = n−1 , (11) proof of this equation. Note, however, that since the right ωn |~r| side of this equations is the delta function, a distribution or generalized function as it is also called, the left where ωn is the surface area of the n-dimensional side must also be a generalized function. Therefore, we unit sphere. In order to the integral of δ(~r) over shall return to this in Sections 3 and 4 of this article, the whole space to be one we must use equation (1) with X = 1 . and explain how electromagnetic theory was before the 2 theory of distributions and the advantages gained after the electric field, magnetic field, electric potential, etc., Remark 1: Formula (14) is a simplified version of the are “promoted” to distributions so that both sides of argument given by Courant and Hilbert [20] described the equations are distributions. The end of Section 4 is above in equation (11). Take for instance n = 3, then especially important in this respect. ω3 = 4π.

Revista Brasileira de Ensino de F´ısica, vol. 43, e20210132, 2021 DOI: https://doi.org/10.1590/1806-9126-RBEF-2021-0132 Amaku et al. e20210132-3

(3) The third reference is given by Blinder [21] in his by physicists. In fact, consider the delta function in equation (7). His justification for this is that “the spherical coordinates, which is given by [18] factor 1 reflects the fact that the delta function is 2 δ(r − r )δ(θ − θ )δ(φ − φ ) located in one of the limits, so that, only half of the δ3(~r − ~r ) = 0 0 0 (18) 0 r2 sin θ delta function is within the range of integration”. or (4) The fourth reference is given by [22]. The authors 3 δ(r − r0)δ(cos θ − cos θ0)δ(φ − φ0) of this article propose a new definition of the δ (~r − ~r0) = . (19) Dirac delta function. According to them, the usual r2 definition of the delta function is given by We now show that equation (18) satisfies one of the basic properties of the delta function and comment Z b ~ δ(x)dx = 1 (12) upon it. In fact, we have for ~r0 6= 0 a Z 3 3 δ (~r − ~r0) d ~r if the interval does not contain the origin, and indeterminate if a or b equal zero, that is, if one Z ∞ Z 2π Z π 2 3 of the end points of the interval of integration = r dr dφ sin θdθ δ (~r − ~r0) coincides with zero. 0 0 0 Z ∞ Z 2π 2 δ(r − r0) They propose, assuming that a < b, that the definition = dr r 2 dφ δ(φ − φ0) 0 r 0 should be Z π δ(θ − θ )  dθ sin θ 0 Z b 1 if ab < 0 sin θ δ(x) dx = 1 if ab = 0 . (13) 0 2 = 1. (20) a 0 if ab > 0 On the other hand, the limit ~r0 → 0 is not well This definition of delta function was proposed earlier defined if we adopt the weak definition for the Dirac and independently by John von Neumann and David delta function. In order to see this fact, let us take the Hilbert [23]. This fact was pointed out by [24]. limit along a line r0 → 0 with φ0 and θ0 fixed. From One consequence of the weak definition of the delta equation (20) we obtain that function is that when you consider the three-dimensional Z delta function we should have δ3(~r) d3~r

3 δ(r) δ (~r) = (14) Z ∞ Z 2π Z π 2πr2 = δ(r)dr δ(φ − φ0)dφ δ(θ − θ0)dθ. 0 0 0 with r = |~r|. However, if you use the strong definition of (21) the delta function we have Clearly this last expression is problematic for the weak δ(r) δ3(~r) = . (15) definition since its value depends on φ0 and θ0, as well 4πr2 as it is not equal to one! Notwithstanding, this limit is In fact, consider δ3(~r) in spherical coordinates. Since the well defined if we adopt the strong definition since the delta function is at the origin it has no angular part and θ and φ are equal to zero independently of the values of is in fact δ(r) [25]. Therefore, using the weak definition θ0 and φ0. We can circumvent this dilemma as we shall of the delta function and equation (14), we can verify show in Section 4.1.2 of this paper. the consistency of the results Due to the above controversy, Gabriel Barton [18] prefers to use the so called strong definition of the Z Z ∞ δ(r) delta function, that is, the one that takes X = 1. His δ3(~r) d3~r = 4πr2 dr = 1. (16) 2 book, strongly recommended, carefully studies the two 0 2πr cases. We would like now to study this problem in the However, using equation (15) and the strong definition light of the Schwartz definition of generalized functions. also leads to a consistent result As we have already mentioned distributions is the name Z Z ∞ δ(r) that Schwartz [7] used for these mathematical objects. δ3(~r) d3~r = 4πr2 dr = 1. (17) Generalized functions is the name that G. Temple [9] 4πr2 0 prefer to call the Schwartz distributions. At this point it is already evident that the use of the weak or the strong definitions of the Dirac delta function 3. Distributions according to Schwartz implies that derived results, like equations (14) and (15) and Temple must be consistently chosen. It is difficult to obtain these results (14) and (15) In this section we present two approaches to distribu- using distribution theory in the elementary form used tions, that is the Schwartz and Temple ones. The latter

DOI: https://doi.org/10.1590/1806-9126-RBEF-2021-0132 Revista Brasileira de Ensino de F´ısica,vol. 43, e20210132, 2021 e20210132-4 Some problems with the use of the Dirac delta function I one is simpler than the Schwartz one, however, this is finite interval of the real line. Moreover, it is continuous more general. and infinitely differentiable. A classical example of a “test function” is given by

3.1. The Schwartz definition 2 exp[− a ] if |x| < a ϕ(x) = a2−x2 (27) In order to define a distribution or a generalized func- 0 if |x| ≥ 0 tion, as they are also known, we recall the definition of a functional. A functional F is a mathematical object that whose support is the open set (−a, a) and it is infinitely acts on functions and produces a number. So if ϕ(x) is differentiable. a function and F is a functional we can write F [ϕ] = Physical magnitudes, that have no singularities can be Number. A simple example of a functional is given by simple “promoted” to regular distributions by replacing a function f(x) that acts on a “good function” ϕ(x) as f(x, y, z) by it in equation (23). So, consider the x follows: component of the electric field that gives rise to a regular Z ∞ distribution f(x)ϕ(x) dx = Number. (22) Z ∞ Z ∞ Z ∞ −∞ 3 GEx [ϕ] = Ex(x, y, z)ϕ(x, y, z)d V −∞ −∞ −∞ The function ϕ(x) should be a good function so that the integral is well defined. We say that the function f(x) = Number, (28) generates the functional. Equation (22) can be generalized to three dimensions where ϕ(x, y, z) is a three-dimensional test function. as follows It is common to hear that the derivative of a function that has a finite jump discontinuity has a delta Z ∞ Z ∞ Z ∞ f(x, y, z)ϕ(x, y, z) d3V = Number. multiplied by this jump. This should be clarified as −∞ −∞ −∞ follows: Suppose you have a function f(x). Promote this (23) to a distribution, that is, consider that it generates a where d3V = dxdydz and ϕ(x, y, z) is a good function distribution. The distribution so generated is infinitely 0 in three dimensions. differentiable. For example, the first derivative Gf of a A distribution or generalized function is a linear and distribution generated by f acting on a test function continuous functional acting upon the space of the ϕ(x) is by definition “good functions” ϕ(x). Distributions that are generated 0 0 by functions like in equation (22) are called regular Gf [ϕ] ≡ Gf [−ϕ ]. (29) distributions. We shall denote distributions generated by a function f(x) by We then recall the definition of the step (Heaviside) function Z Gf = f(x)ϕ(x)dx. 1 for x ≥ x1 H(x − x1) = . (30) 0 for x < x1 Distributions that are not generated by functions are called irregular. Consider now H(x − x1) as a distribution. The space The Dirac delta function in one dimension is defined of test functions ϕ(x) is the set of infinite differentiable by following functional function defined in a finite interval of the real numbers containing the point x1. So, the distribution is the DiracDelta(x − y)[ϕ(x)] ≡ ϕ(y), (24) functional Z ∞ Z ∞ where ϕ(y) is the value of ϕ at the point y, that is, GH [ϕ] = H(x − x1)ϕ(x) dx = ϕ(x) dx. a number. The Dirac delta function is an irregular −∞ x1 distribution, because there are no function that satisfies (31) equation (26). Equation (24) very often written as According to equation (29) the derivative of GH [ϕ] is δ(x − y)[ϕ] = ϕ(y). (25) Z ∞ dϕ G0 [ϕ] = −G [−ϕ0] = − dx The last equation is also written as H H x1 dx Z ∞ = −ϕ(∞) + ϕ(x ) = ϕ(x ). (32) δ(x − y)ϕ(x) dx = ϕ(y). (26) 1 1 −∞ However, this functional is the Dirac delta function, viz. The function ϕ(x) in the Schwartz theory is called a Z ∞ “test function” and is a function that have support (the δ(x − x1)[ϕ] = δ(x − x1)ϕ(x) dx = ϕ(x1). (33) interval where the function is different from zero) in a −∞

Revista Brasileira de Ensino de F´ısica, vol. 43, e20210132, 2021 DOI: https://doi.org/10.1590/1806-9126-RBEF-2021-0132 Amaku et al. e20210132-5

It is a very important result that if a function is weakly to f(x) if for any test function ϕ(x) (see above) continuous or have only finite jump discontinuities it can the limit be “promoted” to a generalized function by just using it Z ∞ Z ∞ with suitable test functions. However if a function has an lim fn(x)ϕ(x) dx = f(x)ϕ(x) dx (34) n→∞ infinite discontinuity more care is required as discussed −∞ −∞ in Section 4 with respect to the function 1/|~r|. exists in spite of that classically limn→∞ fn(x) does not It is possible and very useful to use distributions exist. Sometimes this is written as defined with test functions with support in a finite interval (a, b) of the real axis. In fact, we are going to weak lim fn(x) = f(x). n→∞ use this type of distributions in another article in this series. Here are two examples of such sequences: Distributions have limitations. For example, they can n 1 not be multiplied. Let us illustrate this point. We have Gn(x) = (35) x 1 + n2x2 seen that, to discuss the δ(x−y), it is formally convenient to write equation (26). Likewise, to discuss the meaning and R ∞ R ∞ of 0 δ(x)dx, we should consider −∞ δ(x)H(x)φ(x)dx sin(nx) W (x) = . (36) where H(x) is the step function defined in equation (30) n πx and which renders H(x)ϕ(x) discontinuous. Therefore, These examples tend to the Dirac delta function, which Z ∞ is not a function classically. But, as we shall see δ(x)H(x)ϕ(x) dx −∞ other functions can be “promoted” to distributions. Furthermore, the following sequence tends to the above does not make sense in the Schwartz distribution theory. Heaviside distribution This explains why it is difficult to solve the problem of 1 1 what is the value of X in the expression H (x) = + arctan(nx). (37) n 2 π Z ∞ δ(x) dx = X Differentiation in the Temple approach is like in the 0 Schwartz definition, in Schwartz distribution theory. d d 0 weak lim fn(x) = f(x) = f (x) (38) Also, to solve the more general problem of calculating n→∞ dx dx equation (21) with test functions defined in the interval [0, ∞] for r, [0, 2π] for φ and [0, π] for θ is of no help, so that Z ∞ Z ∞ because the test functions vanish at the end points. 0 0 So, we conclude that although f ϕ(x) dx = − fϕ (x) dx, (39) −∞ −∞ Z ∞ δ(x − x0)F (x0) dx0 = F (x) that is, the result given by equation (29) holds. −∞ Remark 2: There are other definitions and alternative when the integral is over a finite interval treatments to generalized functions. One that uses discontinuous test functions is by Kurasov [27]. This theory Z b may be used to solve some of the problems raised in [28]. δ(x − x0)F (x0) dx0 a 4. What are the values of 4 1 and ∇ · ~r ? and when the variable x0 coincides with one of the r r3 ends of the interval of integration, the above integral is In classical electromagnetism, that is, prior to the inven- undefined or requires great care. As mentioned before we tion of distributions these formulae were not known. We can circumvent this as we shall show in Subsection 4.1.2 shall discuss below several ways of obtaining them, and of this paper. the advantages they have over the approach used prior to distribution theory. 3.2. The Temple definition 1 3 The Temple [16] approach to distributions or generalized 4.1. Obtaining 4 r = −4πδ (~r) functions is simpler, but perhaps, less general than We begin discussing the case of 4 1 where r = |~r| = Schwartz approach. Two good books using this approach r px2 + y2 + z2 is the modulus of the vector ~r = x~i+ are [8, 26]. ~ ~ According to Temple generalized functions are special yj + zk and 4 is the Laplacian operator, whose expres- limits of sequences of functions. These special limits, also sion in cartesian coordinates is called weak limits, are defined as follows: A sequence of ∂2 ∂2 ∂2 4 = + + . (40) differentiable functions fn(x)(n = 1, 2,... ) converges ∂x2 ∂y2 ∂z2

DOI: https://doi.org/10.1590/1806-9126-RBEF-2021-0132 Revista Brasileira de Ensino de F´ısica,vol. 43, e20210132, 2021 e20210132-6 Some problems with the use of the Dirac delta function I

In spherical spherical coordinates the Laplacian is For the sake of continuity of the argument, let us postpone the proof of equation (46). Now, using the 1 ∂2 4 = r + ... (41) expression for δ(~r) given in equation (15) for the strong r ∂r2 definition of the delta function we can rewrite equation (45) as where we omitted contributions from the angular part of the operator. 1 4 = −4πδ3(~r). The fact that r 1 4 = −4πδ3(~r) (42) Hence, we have demonstrated equation (42). Therefore, r this result can be consistently obtained with the use is the object of much discussion in the literature. In fact, of the strong definition of the delta function provided there are countless papers about this subject. Here is a equation (46) holds true for this choice. So, let us obtain selection of the ones we find more useful [21, 29–32]. this equation. 1 In order to demonstrate equation (46) using the strong In order to calculate 4 r we have to “promote” the function 1/r to a distribution. As mentioned above and definition of the delta function we use a test function of discussed below, this is not a simple task because the the form classical function 1/r is mathematically not well defined g(r) for r > 0 for r = 0. The physical significance of the magnitude ϕ(r) = r (47) 0 for r < 0 1/r we are considering is the inverse of the length of the segment from 0 to the point whose radial coordinate is r. with g(r) infinitely differentiable and satisfying As can be anticipated, the use of the strong or weak definition of the Dirac delta function in equation (42) lim rg(r) = 0. (48) requires that the definition of the distribution associated r→0 to 1/r should be chosen carefully as showed below. Then, to demonstrate equation (46) we integrate the ∂ right-hand side of it, that is r ∂r δ(r), multiplied by the B g(r) 2 4.1.1. The definition of the distribution G 1 [ϕ] test function in a volume 4πr dr to get r r used by S. M. Blinder Z ∞ δ(r) g(r) 2 Our goal in this section is to prove equation (42) 4π r dr = 4πg(0). (49) r r using the strong definition of the Dirac delta function 0 consistently. To this end, we analyze the definition of On the other hand, the left-hand side of equation (46) the distribution corresponding to the classical function multiplied by the test function (47), and integrated in 1/r that follows was used by S. M. Blinder in the second the volume 4πr2dr gives part of his article [21]. Moreover, let us follow Temple’s Z ∞ ∂δ(r) g(r) method to define a distribution that corresponds to the 4π − r2 dr = 4πg(0), (50) classical function 1/r: 0 ∂r r when we integrate the left-hand side of last equation B 1 1 G 1 = weak lim H r − , (43) r n→∞ r n by parts. Let us do the details of this last calculation, following the page 34 of the reference [18] with modifi- 1 cations. that is, we are using a sequence of functions H(r − n ) that tends to H(r) when n → ∞. A more careful Z ∞ ∂δ(r) g(r) Z ∞ ∂δ(r) definition of 1/r, in three dimensions, will be given in − r2 dr = − [rg(r)] dr ∂r r ∂r Subsection 4.1.3 where this calculation is repeated in 0 0 more detail. We shall omit the weak lim detail Z ∞ ∂δ(r) n→∞ = − [rg(r)] dr (51) in what follows, for simplicity. Then, −∞ ∂r

2 1 B 1 ∂ where we used that g(r) vanishes for r < 0 in the 4 = 4G 1 = H(r) (44) r r r ∂r2 last step. Now, equation (47) allow us to write the last or integral as Z ∞ 1 1 ∂ 1 d 4 = r δ(r) = − δ(r), (45) δ(r) [rg(r)] dr r r2 ∂r r2 −∞ dr Z ∞ dg(r) where the last equality was obtained using = δ(r) r + g(r) dr = g(0). (52) −∞ dr ∂ − r δ(r) = δ(r). (46) ∂r This demonstrates the result.

Revista Brasileira de Ensino de F´ısica, vol. 43, e20210132, 2021 DOI: https://doi.org/10.1590/1806-9126-RBEF-2021-0132 Amaku et al. e20210132-7

KH SW 4.1.2. The definition of the distribution G 1 [ϕ] 4.1.3. The definition of the distribution G1/r [ϕ] r used by Ben Kuang-Yu Hu used by Ray Skinner and John A. Weil Let us now see that using a different definition of the Another and more clear way to define the distribution distribution that corresponds to the classical function 1/r is given by Ray Skinner and John A. Weil [17]. 1/r, we can demonstrate equation (42) using the weak We now present this approach to show that equation (42) definition of the delta function. We follow here the article is true, making clear that this framework is more by Ben Yu-Kuang Hu [33], but with some modifications. illuminating than the above demonstrations. We follow First we define new spherical coordinates with r closely the presentation of [17]. ranging from −∞ to ∞ and the polar angle θ ranging The approach is to consider that both sides of the from 0 to π/2. Then, we can define the distribution equation (42) must be interpreted as distributions. That is, the classical operator and the classical function KH 1 1 1 for r > 0 G 1 = sign(r) = . (53) 1/r must be “promoted” to a generalized operator r r r −1 for r < 0 acting on generalized functions and generalized function Once again, the magnitude r is physically the distance respectively. from the origin to a point with coordinate r. Let us begin by “promoting” 1/r to a generalized Now we calculate function. The classical function 1/r and its derivatives 2 3 2 −1/r , 2/r , etc. “blow up” at r = 0. Following KH 1 ∂ 2 ∂ 4 G 1 = sign(r) = δ(r) Skinner and Weil we can define the generalized function r r ∂r2 r ∂r corresponding to 1/r as δ(r) = −2 = −4πδ3(~r), (54) Z Z r2 SW G 1 [ϕ] = ϕ(~r) r dr sin θdθdφ = ϕ(~r) r drdΩ where now we have used the results of equations (46) r S S (57) and (14). Hence, we have demonstrated equation (42) where ϕ is any test function and dΩ = sin θdθdφ is the but using in the last step the weak definition of the delta solid angle subtended from the origin to an element of function, that is equation (14). volume containing by ~r. Similarly, the classical function Once again we used the result given in equation (46), 1/r2 corresponds to the generalized function so let us derive it for the weak definition of the delta function. To do so, we use a test function ϕ(r) whose Z SW support contains the origin and integrate its product G 1 [ϕ] = ϕ(~r) drdΩ. (58) r2 with the left-hand side equation (46) S Z ∞ ∂δ(r) Z ∞ ∂(rϕ(r)) Equations (57) and (58), however, do not define univo- − ϕ(r)r dr = δ(r) dr (55) ∂r ∂r cally, from the mathematical point of view, a generalized −∞ −∞ function corresponding to 1/r and 1/r2 respectively, as Z ∞ ∂ϕ(r) we shall see at the end of this section. Nevertheless, = r + ϕ(r) δ(r) dr −∞ ∂r Physics chooses the definitions (57) and (58) uniquely! We can now calculate formula (42) which requires Z ∞ = δ(r)ϕ(r) dr (56) that we use for the generalized ∇ operator following a −∞ definition in agreement with equation (29). We start by where the last equality follows from the fact the deriva- calculating the generalized gradient of the generalized function corresponding to 1/r given by equation (57). tive ϕ is continuous and then r∂ϕ/∂r|r=0 = 0. There- fore, equation (46) also holds for the weak definition of We have the Dirac delta function. Z Z SW 1 3 In brief, we have obtained the well know result ∇G 1 [ϕ] ≡ − [∇ϕ(~r)]rdrdΩ = − [∇ϕ(~r)] d V r S S r 1 (59) 4 = −4πδ3(~r) r Next we evaluate the Laplacian of the generalized SW function G 1 corresponding to 1/r. We have, once again using both the weak and the strong definitions of the r delta function. using equation (29) This should be no surprise. The two results are the SW SW 4G 1 [ϕ] = ∇ · ∇G 1 [ϕ] same but they result from two different definitions of r r the distribution that correspond to the classical function Z 1/r. Another more careful definition of this distribution = ∇ · [∇ϕ(~r)]rdrdΩ is going to be presented, from the point of view of S Z ~r Schwartz, in Section 4.1.3 below. This definition is = ∇ϕ(~r) · drdΩ (60) parts r much more comprehensive than the two we have just S Z ∂ϕ presented. The same distribution, from the point of view = (~r) drdΩ of Temple is going to be presented in the new subsection. S ∂r

DOI: https://doi.org/10.1590/1806-9126-RBEF-2021-0132 Revista Brasileira de Ensino de F´ısica,vol. 43, e20210132, 2021 e20210132-8 Some problems with the use of the Dirac delta function I

h i = 4π lim ϕ(~r) − lim ϕ(~r) field as a limit of a set of vector functions that tend to r→∞ r→0 it as = −4πϕ(~0), 1 1 ~r 1 weak lim EH~ r − = weak lim H r − . where we “integrated by parts” in the second line. n→∞ n n→∞ r2 r n The integral of the angular part in the above last line (63) is not straightforward. It can be evaluated expanding ϕ(~r) in spherical harmonics, and we leave for the careful The action of the electric field E~ as a distribution in reader to perform it; see reference [18] page 33 for further a test function ϕ(~r) is, using spherical coordinates, information. Hence, we obtain equation (42), that is Z G [ϕ] = E~ ϕ(~r) d3V 1 E~ 4 = −4πδ3(~r). r Z ∞ Z 2π Z π ~er 2 A very important point is to show that the definitions = lim 2 ϕ(~r) r dr sin θdθdφ, n→∞ 1 0 0 r of 1/r and 1/r2, as distributions, are determined by n (64) Physics. In fact the definitions (57) and (58) of the above quantities are unique for r > 0. In fact, take where ~er = ~r/r is the radial unit vector. for example the definition given by equation (57) of Now let us evaluate the divergence of the electric field the distribution corresponding to the classical function in spherical coordinates 1/r. We could add to it any linear combination of δ3(~r) and its derivatives and we would still represent 1/r 1 ∂ 2 1 ∂ ∇ · E~ = (r Er) + (Eθ sin θ) for r > 0. However, the electrical potential defined by r2 ∂r r sin θ ∂θ equation (57) is the one that satisfies Maxwell equations 1 ∂E + φ (65) in generalized form as shown in [17]. So, the generalized r sin θ ∂φ form of the Maxwell equations defines the distributions we need to write them. which vanishes for r > 0. Analogously to equation (29), The conclusion of this section is that we can work the divergence of this vector distribution is with either value of X in equation (1), provided we ∇ · G [ϕ] = G [ϕ] do it consistently. Notwithstanding, we have to worry E~ ∇·E~ about what are the definition as distributions of the Z 1 = weak lim −∇ϕ(~r) · EH~ r − d3V. magnitudes we are dealing. We summarize a few features n→∞ n of the different approaches presented above in Table 1. (66)

Table 1: Comparison between the weak and strong deﬁnitions To proceed we integrate by parts, remembering that the of the Dirac delta function. surface term vanishes to the boundary condition of ϕ, and use the identity Quantity Weak value Strong value ~ ~ ~ R ∞ ∇ · (fE) = ∇f · E + f∇ · E (67) −∞ δ(x)dx 1 1 to obtain that R ∞ δ(x)dx 1 1 0 2 Z 1 ~er 3 δ(r) δ(r) G∇·E~ [ϕ] = lim ∇H r − · 2 δ (r) 2πr2 4πr2 n→∞ n r 1 3 3 4 −4πδ (~r) −4πδ (~r) 1 ~er r + H r − ∇ · ϕ(~r) d3V n r2 Z ∞ Z 2π Z π 1 ~r = lim δ r − ~er (68) 3 n→∞ 4.2. Obtaining ∇ · r3 = 4πδ (~r) 0 0 0 n

The potential of a unit charge placed at the origin is ~er 2 1 · 2 ϕ(~r) r dr sin θdθdφ given by ψ(~r) = r . Its gradient is r Z 1 ∂ 1 ~r ~r = 4πδ3(~r)ϕ(~r) d3V = 4πϕ(~0), ∇ = + ··· = − . (61) r ∂r r r r3 where we employed equation (15) to go from the second Therefore, the associated Coulomb field to this charge is to the third above lines. Remember that equation (15) ~r is valid when we work with the strong definition of the E~ = −∇ψ = . (62) r3 Dirac delta function. Therefore, we have proved that Following Temple’s approach as in Section 4.1.1, we ~r ∇ · = 4πδ3(~r). (69) can define a distribution corresponding to this electric r3

Revista Brasileira de Ensino de F´ısica, vol. 43, e20210132, 2021 DOI: https://doi.org/10.1590/1806-9126-RBEF-2021-0132 Amaku et al. e20210132-9

In a close analogy with was presented in Section 4.1.2 [23] J. Von Neumann Collected Works (Pergamon, Oxford, we can also show that this result is valid for the weak 1994) v. 1 p. 111. definition of the delta function. [24] F.A. Muller, Am. J. Phys. 62, 11 (1994). Once again, this section shows clearly that using [25] R.N. Bracewell, The Fourier Transform and Its Appli- careful definitions of physical magnitudes we can obtain cations (McGraw-Hill, Boston, 2000). well known expressions using a sound mathematical [26] T. Schucker Distributions, Fourier Transforms, And formulation. Some of their Applications to Physics (World Scientific, Singapore, 1991). Acknowledgments [27] P. Kurasov, J. Math. Anal. 201, 297 (1996). [28] F.A.B. Coutinho, Y. Nogami and F.M. Toyama, Rev. Bras. Ens. F´ıs. 31, 4302 (2009). The authors thank prof. Luiz Nunes de Oliveira for a [29] A. Gsponer, Eur. J. Phys. 28, 267 (2007). critical reading of the manuscript. M.A., F.A.B.C. and [30] V. Hnizdo, Eur. J. Phys. 32, 287 (2011). O.J.P.E. are supported in part by Conselho Nacional [31] C.P. Frahm, Am. J. Phys. 51, 826 (1983). de Desenvolvimento Cient´ıfico e Tecnológico(CNPq). [32] J. Franklin, Am J Phys. 78, 1225 (2010). O.J.P.E. is also supported by Funda¸cãode Amparo à [33] B. Yu-Kuang Hu, Am. J. Phys. 72, 409 (2004). Pesquisa do Estado de SãoPaulo (FAPESP).

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DOI: https://doi.org/10.1590/1806-9126-RBEF-2021-0132 Revista Brasileira de Ensino de F´ısica,vol. 43, e20210132, 2021