Green's functions and method of images: an interdisciplinary topic usually cast aside in physics textbooks
GGreen’s functions and method of images: aninterdisciplinary topic usually cast aside in physicstextbooks
Glauco Cohen Ferreira Pantoja ∗ Instituto de Cincias da Educao,Universidade Federal do Oeste do Par, Brazil
Walace S. Elias † Instituto Ciberespacial,Universidade Federal Rural da Amaznia, Brazil
Abstract
In the present work we discuss how to address the solution of electrostaticproblems, in professional cycle, using Green’s functions and the Poisson’s equation.By using this procedure, it was possible to verify its relation with the method ofimages as an interdisciplinary approach in didactic physics textbooks. For this, itwas considered the structural role that mathematics, specially the Green’s function,have in physical thought presented in the method of images.
Keywords:
Green’s Functions, Method of Images, Physics Textbooks
One of the usual problems on electrostatics consists in obtaining the electric field (orelectric potential) generated by a charge distribution in certain region of the space. Itis possible to solve this problem by direct integration over the charge distribution or ∗ Electronic address: [email protected] ; † Electronic address: [email protected] a r X i v : . [ phy s i c s . e d - ph ] J un ackling the Poisson’s equation, subjected to a set of boundary conditions imposed onthe field (or potential).Solving non-homogeneous differential equations using Green’s Functions is one of themost powerful forms of describing the solution for a problem of this kind. However, agreat number of classical books on electrodynamics do not explore Poisson’s Equationsolutions using this method. Instead, the common approach uses the method of images, avery interesting way to solve the problem, once it requires a deep physical interpretation.The method of images can be used when we are trying to obtain the electrostatic fieldgenerated by charge distribution near a conductive surface. This procedure takes intoaccount the symmetry of the problem adding a charge image located outside the regionof interest. From this new arrangement, it’s possible to reconstruct the same boundaryconditions of the initial problem without charge image.However, it seems at first that this procedure is barely or not related to the resolutionof Poisson’s (or Laplace) equation. The method, on the contrary, is completely compatiblewith the more general procedure of solving Poisson’s equation via Green’s functions. It isa practical and conceptually elegant mathematical tool, even though it is not general. Inaddition, it assumes the existence of virtual image charges in regions where the solution isnot valid, what seems somewhat artificial for seems introducing in teaching and learningprocesses.One of the authors in [1] developed a classification of tasks on electrostatics in whichare presented four primary classes of situations addressed to: (i) calculation of elec-trostatic fields; (ii) symbolic representation of the electrostatic field; (iii) analogicalrepresentation of the electrostatic field; (iv) description of electrostatic interactions.All these primary classes include four or five secondary classes of situations whoseclassification is based on the objects, variables and unknowns presented by the problem.According to authors in case (i) there are five classes of situations. Among them themost complicated/complex is the one related to calculation of electrostatic fields (orpotentials) due to unknown charge distributions, which include, for instance, conductorsin electrostatic equilibrium.Their arguments rely upon two main epistemological reasons related to the necessarythought operations for mastering these problems: advanced mathematical techniques forproblem-solving and conceptual deepness demanded for physical interpretation of chargeredistribution.The relation among these concepts has been barely explored and, moreover, the cal-culations by the methods of images are often restricted to the case of point charges, withexception of Reitz, Milford and Christy [2] that include the problem of linear images, casesin which it is possible to calculate the potencial due to very large, electrically charged,2ires placed in regions containing conductors.More complicated situations involving well-known and unknown electric charge distri-butions are rarely discussed using the Green’s functions method. Nevertheless, Machado[3] and Jackson [4], solve the problem of a discharged and grounded sphere in front ofa point charge through this method, a situation often solved by using the method ofimages.Panofsky and Phillips [5] approach the general problem of Green’s function by dis-cussing general, mathematical and physical features, although they do not elaborate thediscussion for specific problems. The present work presents a discussion on how to ap-proach electrostatics in the professional cycle from the point of view of solving Green’sfunctions for Poisson’s equation, can be articulated to the method of images in an in-terdisciplinary approach. Our framework takes into account the structural role thatMathematics (Green’s functions) have in Physical thought (method of images).The structure of this paper is presented as follows. In Section 2 presents a briefdiscussion about mathematical structures of physical thought. In Section 3 the prob-lem of Green’s functions is presented from a historical point of view until the completemathematical formulation of the solution of the Poisson’s equation, considering three-dimensional and two-dimensional cases. Section 4 presents a set of electrostatic problemswhose solution was obtained by Green’s function to verify the relation with method ofimage. Final remarks are made in the Section 5. The relation between Mathematics and Physics is not just historical, but also epistemo-logical. The junction among Physics, Astronomy and Mathematics in the CopernicanRevolution fully stresses this fact. Expressing physical ideas in mathematical terms, onthe other hand, is much more than a predictive tool, because it envolves structuring phys-ical thought in function of mathematic enunciations. It is not necessary to defend the roleof Mathematics in Physics, because it is blatantly obvious. However, it is fundamentalto discuss which role is developed in teaching and learning these disciplines.Karam [6] and Rebello et al.[7] state that the results of studies on transference fromMathematics to Physics are strikingly clear about the hindrances faced by the studentsin this task, once using the first in the second envolves more than a simple correspon-dence relation between two distinct conceptual domains. In other words, that means thisassociation is very different from the rote use of formulas.Thus, approaching the role of Mathematics in Physics, requires differentiating itstechnical role (tool-like) and its structural role (reason-like). The first one can be assumed3hen its used in the second one. In table 2 some characteristics of the technical dimensionconcerning the role of mathematics are pointed (extracted from Karam [6]).Therefore, Karam [6] states that the technical role of Mathematics is associated withcalculations developed in a disconnected way from physical problems (e. g. plug-and-chug), while its counterpart, the structural one, is related to the use of Mathematicsto reason about the physical world, that is, to establish reference to it. Although thefirst is important for mastering the second one, the technical domain is not sufficient tolead students to the structural level [6]. The author highlights it is impossible to detachconceptual understanding and mathematical structures use, and points some importantcharacteristics of this feature, which we present in the table 2.We then seek to discuss the structural role of Green’s function in Physics by explain-ing its relation with the method of images by modeling and comprehending problemscontaining known and unknown charge distributions.TechnicalBlindy use an equation to solve quantitative problemsFocus on mechanic or algorithmic manipulationsUse arguments of authorityRote memorization of equations and rulesFragmented knowledgeIdentify superficial similarities between equationsMathematics seen as calculation toolMathematics seen as language used to represent and communicateTable 1: Technical dimension concerning the role of mathematics in physics - author:KaramStructuralDerive an equation from physical principles using logical reasoningFocus on physical interpretations or consequencesJustify the use of specific mathematical structures to model physical phenomenaStructured knowledge: connect apparently different physical assumptions through logicRecognize profound analogies and common mathematical structures behind differentphysical phenomenaMathematics seen as reasoning instrumentMathematics seen as essential to define physical concepts and structure physicalthoughtTable 2: Structural dimension concerning the role of mathematics in physics - author:Karam 4
Green’s Functions and Poisson’s equation
In this section the problem of Green’s function is presented from a historical point of viewand it is discussed the apparent contradiction between the fact that differential operatorsapplied in Green’s Functions are expressed in terms of the Dirac Delta ”function” , ini-tially elaborated by Paul Dirac and, then, formalized by Laurent Schwartz in XX century[8]. How could Green alive between XVIII and XIX centuries, write his formulation in XXcentury notation? The reason is: he did not do that. In the following section, we discusshow was Green’s approach and how it is different from the version used in this paper. George Green was born on July 14th of 1793, Nottingham, England and died on May 31stof 1841, in the same town. It was one of the biggest exponents in Mathematical-Physicsof the region, being the first to introduce the concept of Potential and the method ofGreen’s functions, larged used until the present days in many fields on Physics. However,it seems he has been forgotten for a while, what would imply posthumous recognition forhis work, due to his popularization in works of William Thomson, known as Lord Kelvin[8]. Nonetheless, if his work was so important both for Mathematics and Physics, whyit remained obscure in history? Cannell [8] enumerates factores like: his prematuredeath, at the age 47; the fact of going to Cambridge to study lately and then returningto Nottingham, without establishing personally in the former city; his graduation inmath in a relatively advanced age; the development of abstract works for the period helived, without drawing attention of the scientific community, more worried with practicalquestions at that time; the advanced nature of his work, barely understood for muchscientists of the poque .Electromagnetism was not a commonplace subject at Green’s time, it became sosolely after Kelvin and Faraday. Green knew, however, the works of Laplace, Legendreand Lacroix and had access to a translation to english of the
Mcanique Celeste due toPierre Laplace, made by John Toplin, his tutor in
Nottingham Free Grammar School ,a Leibnizian (what explains his preference for ”d-ism” instead of the Newtonian ”dot-ism”). Green also deeply knew the work of Poisson in Magnetism, probably accessed byattending to the
Nottingham Subscription Library . The mathematician was interested,in his essay on electricity and magnetism (1828), in inverse-type problem related to the Actually, the Dirac Delta function is a limit of a sequence, that is, a distribution for which there ismathematical foundation and formulation. ∇ ϕ = − ρ(cid:15) (1) ϕ = 14 π(cid:15) (cid:90) V ρ ( (cid:126)r (cid:48) ) | (cid:126)r − (cid:126)r (cid:48) | dV (cid:48) . (2)In other words, Green was interested in determining the charge distribution fromoperations on the potential function, whose negative gradient would result in the forceon an unit charge exerted on this conductor (nowadays, we interpret it as the electricfield). For this, Green developed a work in mathematical analysis and constructed whatwe know by Green’s theorem. It is derived nowadays using the divergence theorem, due toGauss-Ostrogradsky, by integrating by parts. After using the theorem, Green investigatedwhat happens in the neighborhood of a point charge located in (cid:126)r = (cid:126)r (cid:48) (modern notation),evaluating the limit of the solution for (cid:126)r → (cid:126)r (cid:48) , that is, when the calculation of thepotential is made near the point which the charge is placed. He then carried out to thefollowing function (modern notation) : G ( (cid:126)r, (cid:126)r (cid:48) ) = 14 π | (cid:126)r − (cid:126)r (cid:48) | . (3)Green also applied his solution and succeeded in finding a formula relating the un-known surface charge density in a condutor with the known potential in its surface; hissolution is likewise discontinuous, what is physically feasible, once electric charges (orfluid, at that time), were known to stay concentrated in the conductor’s surface. Themathematician checked if the function satisfies Laplace’s equation outside the source andconsidered the Green function as a response to an unitary impulse [8], exactly as is donenowadays.In 1930, 102 years later, Paul Dirac introduced his famous “delta functions” withoutproper mathematical rigor, although with a significant practical value. In modern nota-tion the differential equation satisfied by Green’s functions are presented in function ofthese “improper functions”, as Dirac called them. Nevertheless, the formalization of suchmathematical elements just turned possible after the work of Laurent Schwartz, in the At that time, the Poisson’s equation was written like ” ∇ ϕ = − πρ ”, what changes the solution to ϕ = (cid:82) V ρ ( (cid:126)r (cid:48) ) | (cid:126)r − (cid:126)r (cid:48) | dV (cid:48) . Once we changed the original Poisson’s equation for the international system, the original Greenfunction was G ( (cid:126)r, (cid:126)r (cid:48) ) = | (cid:126)r − (cid:126)r (cid:48) | id est , a distribution. This is the reason why, in our calculus, we use modernnotation to find Greens’ functions. We can clearly see how Green was ahead of his time. Electric charges are held stationary by other forces than the ones of electric origin, suchas molecular binding forces. Since charges are stationary, no electric currents and, thus,no magnetic fields are presented ( (cid:126)B = 0). For a stationary electric charge distribution,described by ρ ( (cid:126)r ), the associated electrostatic field satisfies the following set of differentialequations, ∇ · (cid:126)E = ρ(cid:15) , (4) ∇ × (cid:126)E = 0 . (5)Accordingly to the Helmholtz’s theorem [11], once both divergence and curl of a suffi-ciently smooth, rapid decaying, vector field are known, the problem can solved. For theelectrostatic field, the solution for (cid:126)E can be written as the gradient of a scalar funcion ϕ ( (cid:126)r ), since it is irrotational: (cid:126)E = −∇ ϕ, (6)where ϕ ( r ) is well-known as the electrostatic potential. Replacing (6) in the equation (4)leads to the Poisson’s equation: ∇ ϕ = − ρ(cid:15) , (7)and when regions without electric charge distribution are considered, ρ = 0, becomes ∇ ϕ = 0 , (8)equation known as Laplace’s equation.In electrostatics problems the solution will be unique if the boundary conditions areimposed on the potential ϕ ( (cid:126)r ) (on the electrostatic field (cid:126)E ( (cid:126)r )) in some point of the space,accordingly to the Uniqueness theorem [4]. If we impose the boundary conditions on ϕ ( (cid:126)r ), these are known as Dirichlet boundary conditions. However, when the boundaryconditions are applied to (cid:126)E ( (cid:126)r ), they are denoted as Neumann boundary conditions [4].Another possibility is to apply mixed boundary conditions, both on ϕ ( (cid:126)r ) and (cid:126)E ( (cid:126)r ). Inthis case, we call Robbin’s boundary conditions.7 .3 Green’s functions In general, solving the scalar differential equation (7) for ϕ is easier than solving vectordifferential equations for (cid:126)E , (4) and (5) equations. We can apply Green’s function in (7),and from this follows the n-dimensional equation below, ∇ G ( (cid:126)r, (cid:126)r (cid:48) ) = − δ ( n ) ( (cid:126)r − (cid:126)r (cid:48) ) . (9)Considering the Green’s identities [9, 10], it is possible to obtain an expression forElectrostatic Potential ∇ · ( ϕ ∇ G ) = ϕ ∇ · ( ∇ G ) + ∇ ϕ · ∇ G, (10) ∇ · ( G ∇ ϕ ) = G ∇ · ( ∇ ϕ ) + ∇ ϕ · ∇ G, (11)what leads to (cid:90) (cid:0) ϕ ∇ G − G ∇ ϕ (cid:1) dV = (cid:73) ( ϕ ∇ G − G ∇ ϕ ) · ˆ ndS. (12)Then, using (7) and Green’s Identity (10) − (11) in equation (12), one can obtain ϕ ( (cid:126)r ) = 1 (cid:15) (cid:90) GρdV (cid:48) + (cid:73) G ∂ϕ∂n dS (cid:48) − (cid:73) ϕ ∂G∂n dS (cid:48) , (13)which is the general solution for an electrostatic potential and, consequently, for the elec-tric field. The second and third terms in equation (13) are associated with the choice ofthe boundary conditions to which the electric charge density is subject. Once the bound-ary conditions are defined, equation (13) will have a unique and well-defined solutionacoording to the uniquess theorem.In a great number of physical problems that includes conductors in electrostatic equi-librium and zero potential, it is adequate to apply both on the Potential and on theGreen’s Function,the Dirichlet’s boundary conditions. This implies, ϕ ( (cid:126)r ) = 1 (cid:15) (cid:90) GρdV (cid:48) . (14)For this kind of problem is always possible to add into the Green’s function a solutionto Laplace’s equation, denoted by ( G L ( (cid:126)r, (cid:126)r (cid:48) )), which satisfies physical and mathematicalboundary conditions. Therefore, the full Green’s function will be written as G ( (cid:126)r, (cid:126)r (cid:48) ) = G D ( (cid:126)r, (cid:126)r (cid:48) ) + G L ( (cid:126)r, (cid:126)r (cid:48) ) , (15)where G D ( (cid:126)r, (cid:126)r (cid:48) ) depends exclusively the dimensions of Laplacian operator, whereas8 L ( (cid:126)r, (cid:126)r (cid:48) ) depends on the boundary conditions. In the next section, we shall determinethe expression for G D in 3-dimensional and 2-dimensional cases for Laplacian operator. The analytical expression for the Green’s function in three dimensions will be determined.It is necessary to apply a Fourier Transform, leading Green’s function to k − space. Thenthe inverse Fourier transform must be used to find the solution in the coordinates space.The Fourier transform and its inverse for Green’s function G D are presented below,respectively G Dk ≡ G D ( (cid:126)k, (cid:126)r (cid:48) ) = (cid:90) ∞−∞ G D ( (cid:126)r, (cid:126)r (cid:48) ) e i(cid:126)k · (cid:126)r d r, (16) G D ( (cid:126)r, (cid:126)r (cid:48) ) = 1(2 π ) (cid:90) ∞−∞ G Dk e − i(cid:126)k · (cid:126)r d k. (17)Applying the Fourier transform on equation (9) with n = 3 and integrating by partswe obtain the following expression G Dk = e i(cid:126)k · (cid:126)r (cid:48) (cid:0) k x + k y + k z (cid:1) , (18)which represents the Green’s function in k − space. Applying inverse Fourier transform in(18), we will recover the expression of G D in coordinates space, G D ( (cid:126)r, (cid:126)r (cid:48) ) = 1(2 π ) (cid:90) ∞−∞ e − i(cid:126)k · (cid:126)R (cid:0) k x + k y + k z (cid:1) d k, (19)with (cid:126)R = (cid:126)r − (cid:126)r (cid:48) . The integral in equation (19) becomes simpler by an adequate variablechange. Once the integrand does not depend on variable φ , integration results in numer-ical factor equals to 2 π . Rewriting the exponent in equation (19) as (cid:126)k (cid:126)R = kRcosθ , weintegrate in the variable θ to find the following result G D ( (cid:126)r, (cid:126)r (cid:48) ) = 12 π R (cid:90) ∞ sin( kR ) k dk. (20)The integral in (20) can be taken to the complex plane, with part of it being anintegral along the real axis and the other one along a contour Γ extending to infinity. ByJordan’s Lemma [10] the second integral mentioned vanishes in infinity, once this functionis obviously analytic. 9t is also possible to write the sine function in exponential form, what implies twointegrals (cid:73) sin( z ) z dz = (cid:90) ∞−∞ sin( z ) z dz = 12 i (cid:73) Γ e iz − e − iz z dz, (21)whose integration on path Γ is, by convention, positive (counterclockwise).We chose the path Γ for the first integral so that it circles (and excludes) the pole z = 0 coming from the left ( −∞ → + ∞ ) along the real axis, oriented counterclockwise(positive). We chose path Γ for the second integral in a way that the pole is included andsingularity removed, also coming from the left along the real axis, but oriented clockwise(negative). Then, we find the following result (cid:90) ∞−∞ sin( z ) z dz = 2 πi i e = π. (22)Now, we can find the the Green’s function for the Laplace’s equation for the three-dimensional case, G D ( (cid:126)r, (cid:126)r (cid:48) ) = 14 π | (cid:126)r − (cid:126)r (cid:48) | . (23)We will discuss the two-dimensional Laplacian operator case in sequence. Two-dimensional case
Evoking the the Green Function for the two-dimensional Poisson’s equation (9) with n = 2, ∇ G D ( (cid:126)r, (cid:126)r (cid:48) ) = − δ ( (cid:126)r − (cid:126)r (cid:48) ) , (24)enunciating both Fourier direct and inverse transforms G Dk = G D ( (cid:126)k, (cid:126)r (cid:48) ) = (cid:90) ∞−∞ G D ( (cid:126)r, (cid:126)r (cid:48) ) e i(cid:126)k · (cid:126)r d r, (25) G D ( (cid:126)r, (cid:126)r (cid:48) ) = 1(2 π ) (cid:90) ∞−∞ G Dk e − i(cid:126)k · (cid:126)r d k, (26)we can follow the similar procedure in three-dimensional case and obtain G Dk = e i(cid:126)k · (cid:126)r (cid:48) k x + k y , (27) G D ( (cid:126)r, (cid:126)r (cid:48) ) = 1(2 π ) (cid:90) ∞−∞ e i(cid:126)k · (cid:126)r (cid:48) e − i(cid:126)k · (cid:126)r k x + k y d k. (28)To solve the integration in equation (28), we can change variables to polar coordinates10nd apply the scalar product G ( D (cid:126)r, (cid:126)r (cid:48) ) = 1(2 π ) (cid:90) ∞ (cid:90) π e ikR cos θ k dkdθ, (29)recognizing the integral in θ as 2 πJ ( kR ), where J ( kR ) is the zero-order Bessel function G D ( (cid:126)r, (cid:126)r (cid:48) ) = 1(2 π ) (cid:90) ∞ J ( kR ) k dkdθ. (30)The integration on (30) can be done if we derivate the Green function with respectto variable R , dG D ( (cid:126)r, (cid:126)r (cid:48) ) dR = 1(2 π ) (cid:90) ∞ ∂∂R (cid:20) J ( kR ) k (cid:21) dkdθ = 1(2 π ) (cid:20) J ( kR ) R (cid:21)(cid:12)(cid:12)(cid:12)(cid:12) ∞ = − πR ) , (31)and, carrying out the integration with respect to variable R , we have G D ( (cid:126)r − (cid:126)r (cid:48) ) = − π ln | (cid:126)r − (cid:126)r (cid:48) | , (32)where the equation (32) represents the Green’s function for two-dimensional Laplace’scase. In previous section we argued about the expressions of Green’s functions in three andtwo-dimensional cases. Now, we will direct our attention to obtain the solution a setof problems using the Green’s function and verifying the relation with the method ofimages.
Lets consider a point electric charge, q , placed a distance d along the z axis of an infinitethin grounded plate along the xy plane.What is electrical potential produced in a region z > z = 0. Meanwhile, the Green’s Function reduces the problem of acontinuous, and in this case unknown, distribution to the one of a point charge, exactlywhat the method of images proposes.Considering the equation (23), the full Green’s function to this problem will be G ( (cid:126)r, (cid:126)r (cid:48) ) = 14 π (cid:112) ( x − x (cid:48) ) + ( y − y (cid:48) ) + ( z − z (cid:48) ) + G L ( (cid:126)r, (cid:126)r (cid:48) ) , (33)On the boundary (cid:126)S = ( x, y, z = 0), for every point located on the plate we must havethe Green’s function equals zero G ( S, (cid:126)r (cid:48) ) = 14 π (cid:112) ( x − x (cid:48) ) + ( y − y (cid:48) ) + z (cid:48) + G L ( S, (cid:126)r (cid:48) ) = 0 . (34)Thus, it is possible to see, by inspection, that the function G L must have the followingexpression G L ( (cid:126)r, (cid:126)r (cid:48) ) = − π (cid:112) ( x − x (cid:48) ) + ( y − y (cid:48) ) + ( z + z (cid:48) ) , (35)to ensure the condition given by equation (34) will be valid. Therefore, insofar the onlyknown electric charge is a point one, it can be modeled by a Dirac Delta function chargedensity, whose infinity point is located at point (0 , , d ).In other words, ρ ( x, y, z ) = q δ ( z − d ) δ ( y − δ ( x −
0) in such a way that integratingover the volume in equation (14) leads to: ϕ ( x, y, z ) = q π(cid:15) (cid:112) ( x ) + ( y ) + ( z − d ) − q π(cid:15) (cid:112) ( x ) + ( y ) + ( z + d ) . (36)We can verify that this expression is a solution for Laplace’s Equation for z > z = d where it diverges. Besides, the solution satisfy the imposed boundaryconditions. The result is the same obtained by the method of images as seen in the figure1. It is important to highlight the symmetry of Green’s Function in the inversion betweenpoint and source locations. Structuring Physical thought (withdraw the plane by a pointcharge) in such a way to make the Electrical Potential to vanish in that surface is a matterrelated to the mathematical point of view. 12igure 1: Contour lines for the electrostatic potential (36). We can verify where we mustplace an image charge (blue) in a way that maintains a null potential over the groundedplane conductor (black thick line) Electric Field
After determining the electric potential, it is possible to find the associated electric field.Therefore, using the equation (6), we find (cid:126)E = q π(cid:15) (cid:16) E x ˆ i + E y ˆ j + E z ˆ k (cid:17) , (37)where the components of electric field are given by E x = (cid:34) x (( d − z ) + r ) / − x (( d + z ) + r ) / (cid:35) ,E y = (cid:34) y (( d − z ) + r ) / − y (( d + z ) + r ) / (cid:35) ,E z = (cid:34) d − z (( d − z ) + r ) / − d + z (( d + z ) + r ) / (cid:35) , (38)with r = x + y . 13 .1.1 Infinite charged wire placed near a Grounded Infinite Plane Conductor Consider an infinite charged wire placed at distance d along the x axis, near a groundedinfinite plane conductor. It is possible to use the Green function in two-dimensional case(32), to adjust a solution to Laplace’s equation G L . The potential is zero on the chargedplane, what leads to the full Green’s function G ( x, x (cid:48) , y, y (cid:48) ) = − π ln (cid:115) ( x − x (cid:48) ) + ( y − y (cid:48) ) ( x + x (cid:48) ) + ( y − y (cid:48) ) , (39)where we considered the Dirichlet for the determination of the Green’s function G L . Theequation (39) represents the same result obtained by the method of images.To obtain the electrostatic potential for this case, we must integrate the Green functionover the volume in equation (14). Considering the charge density function given by ρ = λδ ( x − d ) δ ( y − ϕ ( x, y ) = − λ π(cid:15) ln (cid:115) ( x − d ) + ( y ) ( x + d ) + ( y ) . (40)From equation (40) it is possible to study the equipotential surfaces, if the argumentof the logarithm function is a constant( x − d ) + y ( x + d ) + y = m. (41)Thus represents a circumference with equation (cid:20) x − (cid:18) d (1 + m )(1 − m ) (cid:19)(cid:21) + y = (cid:18) md − m (cid:19) . (42)For the case m = 1 in equation (41), the radius of the circumference will be infinite,which represents a plane. As long as the solution fits ϕ ( S ) = 0 and x = ∞ , theequipotentials are on the plane and at infinity. Considering the cases with m <
1, theequipotentials surfaces represent circles of radii r = md − m centered at point x = d (1+ m )(1 − m ) ,as presented in figure 2. 14igure 2: Circular equipotential surfaces for the electric potential due to a charged wirenext to an infinite plane Electric Field
From the electrostatic potential (40), the resultant electric field can be found using theequation (6), given by the following expression (cid:126)E ( (cid:126)r ) = E x ˆ i + E y ˆ j, (43)where E x = λ dπ(cid:15) (cid:26) d − x + y [( d − x ) + y ][( d + x ) + y ] (cid:27) , (44)and E y = λ dπ(cid:15) (cid:26) x y [( d − x ) + y ][( d + x ) + y ] (cid:27) , (45)represents the two cartesian coordinates of electric field. We shall solve the classical problem of finding the potential inside a grounded sphere ofradius R , centered at the origin, due to a point charge inside the sphere at position (cid:126)r (cid:48) , asshowed at figure 3. The full Green Function for this problem is given by15igure 3: Diagram illustrating the Laplace’s equation for a sphere of radius R, with apoint charge located at (cid:126)r (cid:48) . G ( r, θ, r (cid:48) , θ (cid:48) ) = 14 π (cid:112) r + r (cid:48) − rr (cid:48) cos ( θ − θ (cid:48) ) + G L ( r, θ, r (cid:48) , θ (cid:48) ) . (46)where we already considered the Green’s function (23) and the spherical symmetry of theproblem to write the distance | (cid:126)r − (cid:126)r (cid:48) | .In the similar way, it is necessary to add G L into the full Green’s function (46). Overthe surface of the sphere, for any polar angle, θ , the electrical potential always will benull. This is equivalent to make the Green function (46) vanish for r = R , what leads toan expression for G L G L ( R, θ, r (cid:48) , θ (cid:48) ) = − π (cid:112) R + r (cid:48) − Rr (cid:48) cos ( θ − θ (cid:48) ) , (47)and, by inspection, we verify that in the point r the G L has the following form, G L ( r, θ, r (cid:48) , θ (cid:48) ) = − π (cid:113) r (cid:48) r R + R − rr (cid:48) cos( θ − θ (cid:48) ) . (48)which corresponds to the Green’s function inside the sphere, for a point image charge q (cid:48) outside it at point r (cid:48) = R r (cid:48) . The equation (48) is the only one that leads to a vanishingGreen’s function over the surface of the sphere.Now, we must find the associated electrostatic potential by integrating over the volumein equation (14), assuming a charge distribution like ρ ( (cid:126)r ) = qr (cid:48) δ ( r (cid:48) − d ) δ ( θ (cid:48) − δ ( φ (cid:48) − ϕ ( r, θ ) = 14 π(cid:15) q √ r + d − rd cos θ − ( qR/d ) (cid:113)(cid:0) r + R d − r R d cos θ (cid:1) . (49)16he result obtained in equation (49) can be derived by using the method of images.Considering a negative image charge placed a distance r (cid:48) = R d , from the centre of thespherical shell and a charge q (cid:48) = − ( q R/d ) produces the same results showed in (49) asrepresented in the figure 4. Electric Field
From the electrostatic potential (49), it is possible to find the electric field using theequation (7), which is expressed in spherical coordinates as (cid:126)E ( r, θ ) = E r ˆ r + E θ ˆ θ, (50)where E r = q π(cid:15) ( r − d cos θ )[ r − rd cos θ + d ] / − (cid:18) Rd (cid:19) (cid:16) r − R d cos θ (cid:17)(cid:2) r − r R d cos θ + R d (cid:3) / (51)and E θ = q π(cid:15) (cid:40) d sin θ ( d − dr cos θ + r ) / − Rd R d sin θ (cid:0) R d − rR cos θd + r (cid:1) / (cid:41) (52)and the lines of force are represented in the figure 4.Considering that the inner charge lies on the z-axis, the induced charge density atsurface of the sphere will be described by a function of the polar angle θσ ( θ ) = (cid:15) ∂V∂r (cid:12)(cid:12)(cid:12)(cid:12) r = R = − q π ( R − d ) R ( R + d − dR cos θ ) / , (53)and the total charge on the surface of sphere can be found by integrating over all angles, Q t = (cid:90) π (cid:90) π σ ( θ ) d Ω = − q. (54)What would happen if the charge q was outside of the grounded sphere? In this case,this problem can be solved using this procedure in similar way. Assuming the charge q is located at position (cid:126)r (cid:48) = d outside of a grounded sphere of radius R , the electrostaticpotential outside is given by the sum of the potentials of the charge and its image charge q (cid:48) inside the sphere. 17igure 4: Lines of force due to the electrostatic field (cid:126)E ( (cid:126)r ) and the equipotential surfacesfor a positive point charge (red) inside the spherical shell of radius R . The blue chargerepresents the image charge q (cid:48) , and it guarantees the electric field is null over the surfaceof the spherical shell. It has been showed in this paper that it is possible to establish comparison betweenGreen’s function and the method of images in electrostatic problems. The method ofimages relies upon a strong sense of physical interpretation, while the technique of Green’sfunction is a powerful form of solving problems involving differential equations.The solution attachedto the image charge appears as a solution for Laplace’s equation,satisfying the boundary conditions associated. On the other hand, Green’s functionmethod is more general technique than the one due to calculation by image charges.However, in a physics problem, without the interpretation, connecting these two in-stances, the mathematical knowledge relates in an non-substantive way and may be an-chored to non relevant prior knowledge. Therefore, it leads to non elaborated ideas as,for example, “problems involving conductors are solved by Green’s function” or “prob-lems involving conductors are solved by the method of images”, what places this kind ofrelation closer to the rote learning pole and further from the meaningful learning pole[12].Neverthless, in parallel, the methods may be meaningful both in Physics and Math-18matics, once it is possible to learn about conductors in electrostatic equilibrium whileconceptually and operationally tackling only using Green’s functions.The authors defend, as does Karam[6], that mathematical knowledge structures phys-ical thought and that gives meaning to mathematical knowledge through situations thatmake the concept of Green’s function useful and meaningful in the field of physics [13],permitting transference to the domain of Mathematics[7].The value of this article underlies in showing a deep relation between physical thoughtand mathematical structure in a case of electromagnetism (professional cycle). Offering awider view on the role of Mathematics in Physics than the common views of Mathematicsas tool (operationalistic function) or as a merely language (restricted communicativefunction).Another intricate point in the discussion is the fact that this knowledge is necessarilytied to epistemological features that can not be cast aside. Green himself obviously didnot knew the Dirac Delta function, neither Dirac himself had a formal proof of its validity,which was developed by Schwartz, but this did not stopped them from doing elaboratedMathematics.Similar epistemological difference can be found among the works of Newton (or Leib-niz) and the ones by Weierstrass [14]. For as much the notion of function due to thelatter mathematician approaches the concept of number (static view), the one due to theformer in closer to the concept of variable (dynamic view) [14].Related to this, is the unmentionable wide failure in Calculus teaching in the firstyear of any course of Exact Sciences [14], whose cause is, partially, associated with thedisregarding of this feature into teaching-learning processes: students often study text-books approaching the concept in a Weierstrassian perspective, which is much further(and much more formal) from students’ prior knowledge than it should be. It is reason-able to be expected for reproduction all over Brazil. In spite of the existence of greatteachers and students in theses courses, this epistemological features is beyond their willpower or applied didactical methodology in the teaching processes.Returning to the discussion of Green’s function, we advise that its interpretationshould be approached to the notion of point source, as Green himself did, because this canprovide conditions for comprehension of more modern concepts as, for example, the DiracDelta Function. Without this epistemological ingredient, the process of interdisciplinaryinteraction between Mathamatics and Physics in classroom can blatantly fail in reachingits objective of coming up with conditions for meaningful learning [12]The authors expect to contribute, by means of discussion of these simple examples, todemonstrate the feasibility of discussing in an integrated manner the method of images(with high degree of physical interpretation) and the technique of Green’s function (with19igh degree of mathematical power) in classroom.The authors also look forward to discuss principles related to providing conditionnot just for comprehension of the secondary class of situations Γ E pointed in [1], butseeking for interdisciplinary integration between Physics and Mathematics in a mannerof promoting reasoning founded in the thesis that Mathematics structure Physical thought[6] and that Physics may give sense to concepts of Mathematics [13]. References [1] G. C. Pantoja and M. A. Moreira, Latin – American Journal of Physics Education, , 4, 4304-1 (2019).[2] J. Reitz, F. Milford, R. Christy, Fundamentos da Teoria Eletromagntica , (Elsevier,Rio de Janeiro, 1982), p. 516.[3] K. D. Machado,
Teoria do eletromagnetismo , (UEPG, Ponta Grossa, 2000), v. 1, p.929.[4] J. D. Jackson,
Classical Electrodynamics , (John Wiley and Sons, New York, 1962),3th ed, p. 808.[5] W. Panofsky and M. Phillips,
Classical Electricity and Magnetism , (Dover Publica-tions, Mineola, 2005), 2nd ed, p. 494.[6] R. Karam, Physical Review Special Topics - Physics Education Research , , 010119(2014).[7] N. S. Rebello, L. Cui, A. G. Bennett, D. A. Zollmann, and D.J. Ozimek, Learn toSolve Complex Scientific Problems , edited by D. Jonassen, (Elsevier, Rio de Janeiro,2007), 6th ed. op. cit., pag. 736.[8] D. Cannell, The Mathematical Gazette, , 26 (1993).[9] J. M. F. Bassalo and M. S. Cattani, Elementos de Fsica Matemtica , (Editora Livrariada Fsica, So Paulo, 2011), p. 158.[10] G. Arfken and H. Weber,
Mathematical Methods for Physicists , (Elsevier, New York,2005), 6th ed, p. 1182.[11] D. J. Griffiths,
Introduction to Electrodynamics , (Prentice Hall, New Jersey, 1999),3th ed, p. 576. 2012] D. P. Ausubel,