Direct derivation of Lienard Wiechert potentials, Maxwell's equations and Lorentz force from Coulomb's law
DDirect derivation of Li´enard–Wiechert potentials, Maxwell’sequations and Lorentz force from Coulomb’s law
Hrvoje Dodig ∗ University of Split, Rudera Boˇskovi´ca 37, 21000 Split, Croatia
Abstract
In 19 th century Maxwell derived famous Maxwell equations from the knowledge of threeexperimental physical laws: the electrostatic Coulomb’s law, Ampere’s force law andFaraday’s law of induction. However, theoretical basis for Ampere’s force law and Fara-day’s law remains unknown to this day. Furthermore, the Lorentz force is considered assummary of experimental phenomena, the theoretical foundation that explains genera-tion of this force is still unknown.To answer these fundamental theoretical questions, in this paper we derive relativisti-cally correct Li´enard – Wiechert potentials, Maxwell’s equations and Lorentz force fromtwo simple postulates: (a) when all charges are at rest the Coulomb’s force acts betweenthe charges, and (b) that disturbances caused by charge in motion propagate away fromthe source with finite velocity. The special relativity was not used in our derivations northe Lorentz transformation. In effect, it was shown in this paper that all the electrody-namic laws, including the Lorentz force, can be derived from Coulomb’s law and timeretardation.This was accomplished by analysis of hypothetical experiment where test charge is atrest and where previously moving source charge stops at some time in the past. Then thegeneralized Helmholtz decomposition theorem, also derived in this paper, was applied toreformulate Coulomb’s force acting at present time as the function of positions of sourcecharge at previous time when the source charge was moving. From this reformulation ofCoulomb’s law the Li´enard–Wiechert potentials and Maxwell’s equations were derivedby careful mathematical manipulation.In the second part of this paper, the energy conservation principle valid for movingcharges is derived from the knowledge of electrostatic energy conservation principle validfor stationary charges. This again was accomplished by using generalized Helmholtzdecomposition theorem. From this dynamic energy conservation principle the Lorentzforce is finally derived. Keywords:
Coulomb’s law, Li´enard–Wiechert potentials, Maxwell equations, Lorentzforce ∗ Corresponding author
Email address: [email protected] (Hrvoje Dodig) a r X i v : . [ phy s i c s . c l a ss - ph ] D ec . Introduction In his famous
Treatise [1, 2] Maxwell derived equations of electrodynamics based on theknowledge about the three experimental laws known at the time: the Coulomb’s lawdescribing the electric force between charges at rest; Ampere’s law describing the forcebetween current carrying wires, and the Faraday’s law of induction. Prior to Maxwell,magnetism and electricity were regarded to as separate phenomena. It was James ClerkMaxwell who unified these seemingly disparate phenomena into the set of equationscollectively known today as Maxwell’s equations. In modern vector notation, the fourMaxwell’s equations that govern the behavior of electromagnetic fields are written as: ∇ · D = ρ (1) ∇ · B = 0 (2) ∇ × E = − ∂ B ∂t (3) ∇ × B = µ J + 1 c ∂ E ∂t (4)where symbol D denotes electric displacement vector, E is electric field vector, B is avector called magnetic flux density, vector J is called current density and scalar ρ is thecharge density. Furthermore, there are two more important equations in electrodynamicsthat relate magnetic vector potential A and scalar potential φ to electromagnetic fields B and E : B = ∇ × A (5) E = −∇ φ − ∂ A ∂t (6)In standard electromagnetic theory, if a point charge q s is moving with velocity v s ( t )along arbitrary path r s ( t ) the scalar potential φ and vector potential A caused by mov-ing charge q s are described by well known, relativistically correct, Li´enard–Wiechertpotentials [3, 4]: φ = φ ( r , t ) = 14 π(cid:15) (cid:18) q s (1 − n s ( t r ) · β s ( t r )) | r − r s ( t r ) | (cid:19) (7) A = A ( r , t ) = µc π (cid:18) q s β s ( t r )(1 − n s ( t r ) · β s ( t r )) | r − r s ( t r ) | (cid:19) (8)where t r is retarded time, r is the position vector of observer and vectors n s ( t r ) and β s ( t r ) are: β ( t r ) = v s ( t r ) c (9) n ( t r ) = r − r s ( t r ) | r − r s ( t r ) | (10)2hese equations were almost simultaneously discovered by Li´enard and Wiechert around1900’s and they represent explicit expressions for time-varying electromagnetic fieldscaused by charge in arbitrary motion. Nevertheless, Li´enard-Wiechert potentials werederived from retarded potentials, which in turn, are derived from Maxwell equations.Maxwell’s electrodynamic equations provide the complete description of electromag-netic fields, however, these equations say nothing about mechanical forces experienced bythe charge moving in electromagnetic field. If the charge q is moving in electromagneticfield with velocity v then the force F experienced by the charge q is: F = q ( E + v × B ) (11)The force described by equation (11) is well known Lorentz force. Discovery of thiselectrodynamic force is historically credited to H.A. Lorentz [5], however, the similarexpression for electromagnetic force can be found in Maxwell’s Treatise , article 598 [2].The difference between the two is that Maxwell’s electromotive force acts on movingcircuits and Lorentz force acts on moving charges.However, it is not yet explained what causes the Lorentz force, Ampere’s force lawand Faraday’s law. Maxwell derived his expression for electromotive force along movingcircuit from the knowledge of experimental Faraday’s law. Later, Lorentz extendedMaxwell’s reasoning to discover the force acting on charges moving in electromagneticfield [5]. Nevertheless, it would be impossible for Lorentz to derive his force law withoutthe prior knowledge of Maxwell equations [6].Nowadays, the Lorentz force ( q v × B term) is commonly viewed as an effect of Ein-stein’s special relativity. For example, an observer co-moving with source charge wouldnot measure any magnetic field, while on the other hand, the stationary observer wouldmeasure the magnetic field caused by moving source charge. However, in this work, wedemonstrate that the special relativity is not needed to derive the Lorentz force andMaxwell equations. In fact, we derive Maxwell’s equations and Lorentz force from morefundamental principles: the Coulomb’s law and time retardation.There is another reason why the idea to derive Maxwell’s equations and Lorentz forcefrom Coulomb’s law may seem plausible. Because of mathematical similarity betweenCoulomb’s law and Newton’s law of gravity many researchers thought that if Maxwell’sequations and Lorentz force could be derived from Coulomb’s law that this would behelpful in understanding of gravity. These two inverse-square physical laws are written: F C = q q π(cid:15) r − r | r − r | (12) F G = − Gm m r − r | r − r | (13)The expressions for Coulomb’s force and Newton’s gravitational force are indeed similar,however, these two forces significantly differ in physical nature. The latter force is alwaysattractive while the former can be either attractive or repulsive. Nevertheless, a numberof researchers attempted to derive Maxwell’s equations from Coulomb’s law, and mostof these attempts rely on Lorentz transformation of space-time coordinates between therest frame of the moving charge and laboratory frame.3he first hint that Maxwell’s equations could be derived from Coulomb’s law andLorentz transformation can be found in Einstein’s original 1905 paper on special rela-tivity [7]. Einstein suggested that the Lorentz force term ( v × B ) is to be attributed toLorentz transformation of the electrostatic field from the rest frame of moving charge tothe laboratory frame where the charge has constant velocity. Later, in 1912, Leigh Pagederived Faraday’s law and Ampere’s law from Coulomb’s law using Lorentz transforma-tion [8]. Frisch and Willets discussed the derivation of Lorentz force from Coulomb’slaw using relativistic transformation of force [9]. Similar route to derivation of Maxwell’sequations and Lorentz force from Coulomb’s law was taken by Elliott in 1966 [10]. Kobein 1986 derives Maxwell’s equations as the generalization of Coulomb’s law using specialrelativity [11]. Lorrain and Corson derive Lorentz force from Coulomb’s law, again, byusing Lorentz transformation and special relativity[12]. Field in 2006 derives Lorentzforce and magnetic field starting from Coulomb’s law by relating the electric field toelectrostatic potential in a manner consistent with special relativity [13]. The most re-cent attempt comes from Singal [14] who attempted to derive electromagnetic fields ofaccelerated charge from Coulomb’s law and Lorentz transformation.All of the mentioned attempts have in common that they attempt to derive Maxwellequations from Coulomb’s law by exploiting Lorentz transformation or Einstein’s spe-cial theory of relativity. However, historically the Lorentz transformation was derivedfrom Maxwell’s equations [15], thus, the attempt to to derive Maxwell’s equations usingLorentz transformation seems to involve circular reasoning [16]. The strongest criticismcame from Jackson who pointed out that it should be immediately obvious that, withoutadditional assumptions, it is impossible to derive Maxwell’s equations from Coulomb’slaw using theory of special relativitiy [17]. Schwartz addresses these additional assump-tions and starting from Gauss’ law of electrostatics and by exploiting the Lorentz in-variance and properties of Lorentz transformation he derives the Maxwell’s equations[18].In addition to the criticism above, we point out that the derivations of Maxwell’sequations from Coulomb’s law using Lorentz transformation should only be consideredvalid for the special case of the charge moving along the straight line with constantvelocity. This is because the Lorentz transformation is derived under the assumptionthat electron moves with constant velocity along straight line [15]. For example, ifthe particle moves with uniform acceleration along straight line the transformation ofcoordinates between the rest frame of the particle and the laboratory frame takes thedifferent mathematical form than that of the Lorentz transformation [19]. If the particleis in uniform circular motion yet another coordinate transformation from the rest frameto laboratory frame, called Franklin transform, is valid [20]. None of the above citedpapers consider the fact that Lorentz transformation is no longer valid when the chargeis not moving along straight line with constant velocity.To circumvent problems with special relativity and Lorentz transformation we takeentirely different approach to derive Li´enard–Wiechert potentials and Maxwell’s equa-tions from Coulomb law. We start our derivation from the analysis of the followinghypothetical experiment: consider two charges at rest at present time, one called thetest charge, and the other called the source charge. The source charge was moving inthe past but it is at rest at present time. Because both charges are at rest at present theforce acting on test charge at present time is the Coulomb’s force.However, in the past when the source charge was moving, we assume that the force4 r s E c q s q s CC SOURCE CHARGE q s ISSTATIONARY AT PRESENT TIME t p AT PRESENT TIME t p ELECTRIC FIELDAT POINT ON CONTOUR r ISCOULOMB’S ELECTROSTATIC FIELD E c E c = q s πε r − r s | r − r s | ENERGY CONSERVATION LAWAT PRESENT TIME t p : H C E c · d r = SOURCE CHARGE q s STOPSMOVING AT PAST TIME t s < t p (a) present time (b) past time Figure 1: In (a) the source charge q s is at rest at present time t p . Each point on closed contour C is affected by Coulomb’s electrostatic field E c . Energy conservation principle at present time is (cid:72) C E c · d r = 0. In (b) the source charge q s is moving along arbitrary path and it stops at past time t s < t p . Dynamic energy conservation principle valid in the past is assumed to be unknown when sourcecharge was moving. acting on test charge was not the Coulomb’s force. To discover the mathematical formof this ”unknown” electrodynamic force acting in the past from the knowledge of knownelectrostatic force (Coulomb’s law) acting at present time the generalized Helmholtzdecomposition theorem was applied. This theorem, derived in Appendix A, allowed usto relate Coulomb’s force acting at present time to the positions of source charge at pasttime. From here, Li´enard–Wiechert potentials and Maxwell’s equations were derived bycareful mathematical manipulation.It should be emphasized that we did not resort to theory of special relativity nor toLorentz transformation in our derivation of Maxwell’s equations. Not less importantly,the presented derivation of Maxwell’s equations from Coulomb’s law is valid for chargesin arbitrary motion. In effect, we may say that more general physical law (Maxwell’sequations) acting at past time is derived from the knowledge of limited physical law(Coulomb’s law) acting at present time.However, from Maxwell’s equations, it is very difficult, if not entirely impossible,to derive the Lorentz force without resorting to some form of energy conservation law.As shown in Fig. 1a, at present time, the single stationary charge creates Coulomb’selectrostatic field. Known energy conservation law valid at present time states thatcontour integral of Coulomb’s field along closed contour C is equal to zero.But, this electrostatic energy conservation law valid at present is not necessarily validin the past when the source charge was moving. Thus, in the second part of this paperwe derive this ”unknown” dynamic energy conservation principle valid in the past fromthe knowledge of electrostatic energy conservation principle valid at present time. Thiswas again achieved by the careful application of generalized Helmholtz decompositiontheorem which allowed us to transform electrostatic energy conservation law valid atpresent to dynamic energy conservation law valid in the past.5his dynamic energy conservation law states that the work of non-conservative forcealong closed contour is equal to the time derivative of the flux of certain vector fieldthrough the surface bounded by this closed contour. From this dynamic energy conser-vation law the Lorentz force was finally derived.
2. Generalized Helmholtz decomposition theorem
Because generalized Helmholtz decomposition theorem is central for deriving Maxwellequations and Lorentz force from Coulomb’s law, in this section, we briefly present thisimportant theorem while the derivation itself is moved to Appendix A. There have beenseveral previous attempts in the literature to generalize classical Helmholtz decomposi-tion theorem to time dependent vector fields [21, 22, 23]. However, in none of the citedarticles the Helmholtz theorem for functions of space and time is presented in the math-ematical form usable for the mathematical developments described in this paper. Thisis probably caused by difficulties in stating such a theorem and this was clearly stated in[22]: ”There does not exist any simple generalization of this theorem for time-dependentvector fields”.However, we show that there indeed exists the simple generalization of Helmholtzdecomposition theorem for time-dependent vector fields and that it can derived fromtime-dependent inhomogeneous wave equation. To improve the clarity of this paper,the complete derivation of Helmholtz decomposition theorem for functions of space andtime is moved to Appendix A, subsection A.2. As it was shown in Appendix A, thegeneralization of Helmholtz decomposition theorem for the vector function of space andtime F ( r , t ) can be written as: F ( r , t ) = − ∇ (cid:90) R dt (cid:48) (cid:90) R (cid:18) ∇ (cid:48) · F ( r (cid:48) , t (cid:48) ) (cid:19) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) (14)+ 1 c ∂∂t (cid:90) R dt (cid:48) (cid:90) R (cid:18) ∂∂t (cid:48) F ( r (cid:48) , t (cid:48) ) (cid:19) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) + ∇ × (cid:90) R dt (cid:48) (cid:90) R (cid:18) ∇ (cid:48) × F ( r (cid:48) , t (cid:48) ) (cid:19) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) where scalar function G ( r , t ; r (cid:48) , t (cid:48) ) is the fundamental solution of time-dependent inho-mogeneous wave equation given as: ∇ G ( r , t ; r (cid:48) , t (cid:48) ) − c ∂ ∂t G ( r , t ; r (cid:48) , t (cid:48) ) = − δ ( r − r (cid:48) ) δ ( t − t (cid:48) ) (15)In the equation above, δ ( r − r (cid:48) ) = δ ( x − x (cid:48) ) δ ( y − y (cid:48) ) δ ( z − z (cid:48) ) is 3D Dirac delta function,and δ ( t − t (cid:48) ) is Dirac delta function in one dimension. Fundamental solution G ( r , t ; r (cid:48) , t (cid:48) ),sometimes called Green’s function, represents the retarded in time solution of the inho-mogeneous time dependent wave equation and it can be written as: G ( r , t ; r (cid:48) , t (cid:48) ) = δ (cid:18) t (cid:48) − t + | r − r (cid:48) | c (cid:19) π | r − r (cid:48) | (16)6here position vector r (cid:48) is the location of the source at time t (cid:48) .From equation (14) it is evident that the Helmholtz decomposition theorem for func-tions of space and time can be regarded to as a mathematical tool that allows us torewrite any vector function that is function of present time t and of present position r asvector function of previous time t (cid:48) and of previous position r (cid:48) . Furthermore, the gener-alized Helmholtz decomposition theorem (14) comes with additional limitation that it isvalid if vector function F ( r (cid:48) , t (cid:48) ) approaches zero faster than 1 / | r − r (cid:48) | as | r − r (cid:48) | → ∞ .Very similar theorem was presented in article written by Heras [24]; the differenceis that in Heras’ article the time integrals in equation (14) were a priori evaluated atretarded time t (cid:48) = t − | r − r (cid:48) | /c . As such, the generalized Helmholtz theorem presentedin [24] is not suitable for the derivation of Maxwell equations and Lorentz force from theCoulomb’s law. Reason for this, as it will become evident later in this paper, is that ifwe immediately evaluate the time integrals in equation (14) the important informationis lost from the equation.
3. Derivation of Maxwell Equations from Coulomb’s law
In this section we derive Maxwell equations from Coulomb’s law using generalizedHelmholtz decomposition theorem represented by equation (14). To begin the discussion,we consider hypothetical experiment shown in Fig. 2, where source charge q s is movingalong trajectory r s ( t ) and it stops at some past time t s . The test charge q is stationaryat all times. We assume that the disturbances caused by moving source charge propagateoutwardly from the source charge with finite velocity c . These disturbances originatingfrom the source charge at past time manifest itself as the force acting on stationary testcharge at present time. This means that there is a time delay ∆ t between the past time t s when the source charge has stopped and the present time t p when this disturbancehas propagated to the test charge:∆ t = t p − t s = | r − r s ( t s ) | c (17)At precise moment in time t p , that we call the present time, the force acting on stationarytest charge q is the Coulomb’s force because source charge and test charge are both atrest, and because the effect of source charge stopping at past time t s had enough timeto propagate to test charge. The Coulomb’s force F c experienced by the test charge q atpresent time t p can be expressed by the following equation: F c ( r , t p ) = qq s π(cid:15) r − r s ( t s ) | r − r s ( t s ) | t p = t s + ∆ t (18)Let us now consider the time t just one brief moment before the stopping time t s : t = t s − δt (19)where δt → δt is so small that we mighteven call it infinitesimally small. Then at the moment in time infinitesimally before the7 yz r q q s r s ( t s ) SOURCE CHARGE q s STOPS AT TIME t s STATIONARY TESTCHARGE q FEELS THECOULOMB FORCE ATPRESENT TIME t p > t s t p = t r + | r − r s ( t s ) | c Figure 2: Source charge q s is moving along arbitrary trajectory r s ( t ) and stops at time t s . Because q s stops moving at past time t s , at present time t p , the stationary test charge q experiences electrostaticCoulomb force. present time t p the force felt by test charge q is still the Coulomb’s force if δt →
0. Usingthese considerations, we can now rewrite equation (18) as: F c ( r , t p − δt ) = qq s π(cid:15) r − r s ( t ) | r − r s ( t ) | t = t s − δt ; δt → δt →
0. The reason whywe have written the Coulomb’s law this way is to permit slight variation of time beforestopping time t s so that we can exploit generalized Helmholtz decomposition theorem inorder to derive Maxwell’s equations from Coulomb’s law. Had we not done this then thesource charge position vector r s ( t s ) would simply be the constant vector and generalizedHelmholtz decomposition could not be used.Because the right hand side of equation (20) is now the function of time t and position r we are allowed to use the generalized Helmholtz decomposition theorem to rewrite theright hand side of equation (20). This is because generalized Helmholtz decompositiontheorem states that any vector function of time t and position r can be decomposed asdescribed by this theorem if that function meets certain criteria. Thus, using generalizedHelmholtz decomposition theorem we can rewrite the right hand side of equation (20)as: 8 c ( r , t p − δt ) = qq s π(cid:15) r − r s ( t ) | r − r s ( t ) | = (21)= −∇ (cid:90) R dt (cid:48) (cid:90) R (cid:18) ∇ (cid:48) · qq s π(cid:15) r (cid:48) − r s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | (cid:19) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) + 1 c ∂∂t (cid:90) R dt (cid:48) (cid:90) R (cid:32) ∂∂t (cid:48) qq s π(cid:15) r (cid:48) − r s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | (cid:33) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) + ∇ × (cid:90) R dt (cid:48) (cid:90) R (cid:18) ∇ (cid:48) × qq s π(cid:15) r (cid:48) − r s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | (cid:19) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) To clarify the notation in equation above note that dV (cid:48) = dx (cid:48) dy (cid:48) dz (cid:48) represents the dif-ferential volume element of an infinite volume R . As defined in Appendix A, subsectionA.1, the primed position vector r (cid:48) is written in Cartesian coordinate system as: r (cid:48) = x (cid:48) ˆx + y (cid:48) ˆy + z (cid:48) ˆz (22)where variables x (cid:48) , y (cid:48) , z (cid:48) ∈ R . Vectors ˆx , ˆy and ˆz are orthogonal Cartesian unit basisvectors. Furthermore, in Cartesian coordinates, the primed del operator ∇ (cid:48) that appearsin equation (21) is defined as: ∇ (cid:48) = ˆx ∂∂x (cid:48) + ˆy ∂∂y (cid:48) + ˆz ∂∂z (cid:48) (23)From the definition above, it follows that primed del operator ∇ (cid:48) acts only on functionsof variables x (cid:48) , y (cid:48) , z (cid:48) , and consequently, on functions of primed position vector r (cid:48) = x (cid:48) ˆx + y (cid:48) ˆy + z (cid:48) ˆz . It does not act on functions of position vector of source charge r s ( t (cid:48) ) becausethis position vector is function of variable t (cid:48) . Using these definitions we can write thefollowing simple relations: r (cid:48) − r s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | = −∇ (cid:48) | r (cid:48) − r s ( t (cid:48) ) | (24) ∇ (cid:48) · π r (cid:48) − r s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | = −∇ (cid:48) π | r (cid:48) − r s ( t (cid:48) ) | = δ ( r (cid:48) − r s ( t (cid:48) )) (25) ∇ (cid:48) × r (cid:48) − r s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | = −∇ (cid:48) × ∇ (cid:48) | r (cid:48) − r s ( t (cid:48) ) | = 0 (26)where δ ( r (cid:48) − r s ( t (cid:48) )) is 3D Dirac’s delta function. Inserting equations (25) and (26) intoequation (21), and eliminating charge q from the equation, yields the following relation: q s π(cid:15) r − r s ( t ) | r − r s ( t ) | = − ∇ (cid:90) R dt (cid:48) (cid:90) R q s (cid:15) δ ( r (cid:48) − r s ( t (cid:48) )) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) (27)+ 1 c ∂∂t (cid:90) R dt (cid:48) (cid:90) R (cid:32) ∂∂t (cid:48) q s π(cid:15) r (cid:48) − r s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | (cid:33) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48)
9n Appendix C, subsection C.1, we have shown that the time derivative that appears inthe second right hand side integral of equation (27) can be written as: ∂∂t (cid:48) q s π(cid:15) r (cid:48) − r s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | = q s π(cid:15) ∇ (cid:48) × ∇ (cid:48) × v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | − q s (cid:15) v s ( t (cid:48) ) δ ( r (cid:48) − r s ( t (cid:48) )) (28)where v s ( t (cid:48) ) is the velocity of the source charge q s at time t (cid:48) : v s ( t (cid:48) ) = ∂ r s ( t (cid:48) ) ∂t (cid:48) (29)By inserting equation (28) into equation (27) it is obtained that: q s π(cid:15) r − r s ( t ) | r − r s ( t ) | = − ∇ (cid:90) R dt (cid:48) (cid:90) R q s (cid:15) δ ( r (cid:48) − r s ( t (cid:48) )) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) (30) − c ∂∂t (cid:90) R dt (cid:48) (cid:90) R q s (cid:15) v s ( t (cid:48) ) δ ( r (cid:48) − r s ( t (cid:48) )) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) + 1 c ∂∂t (cid:90) R dt (cid:48) (cid:90) R q s π(cid:15) (cid:20) ∇ (cid:48) × ∇ (cid:48) × v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | (cid:21) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) We now make use of the following identity, also derived in Appendix Appendix C, sub-section C.2, to rewrite the last right hand side term of equation (30) as: (cid:90) R dt (cid:48) (cid:90) R q s π(cid:15) (cid:20) ∇ (cid:48) × ∇ (cid:48) × v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | (cid:21) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) = (31)= ∇ × ∇ × (cid:90) R dt (cid:48) (cid:90) R q s π(cid:15) v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) Replacing the last right hand side integral in equation (30) with equation (31) anddifferentiating the resulting equation with respect to time t yields: q s π(cid:15) ∂∂t r − r s ( t ) | r − r s ( t ) | = − ∂∂t ∇ (cid:90) R dt (cid:48) (cid:90) R q s (cid:15) δ ( r (cid:48) − r s ( t (cid:48) )) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) (32) − c ∂ ∂t (cid:90) R dt (cid:48) (cid:90) R q s (cid:15) v s ( t (cid:48) ) δ ( r (cid:48) − r s ( t (cid:48) )) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) + 1 c ∂ ∂t ∇ × ∇ × (cid:90) R dt (cid:48) (cid:90) R q s π(cid:15) v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) In the physical setting shown in Fig. 2 the coordinates of the test charge q are fixed,hence, order in which we apply operator ∇ × ∇× and second order time derivative ∂∂t can be swapped (because operator ∇ does not affect variable t ). Furthermore, becausevariables t (cid:48) , x (cid:48) , y (cid:48) and z (cid:48) are independent of time t we can move the double time derivativeunder the integral sign in the last right hand side integral of above equation:10 s π(cid:15) ∂∂t r − r s ( t ) | r − r s ( t ) | = − ∂∂t ∇ (cid:90) R dt (cid:48) (cid:90) R q s (cid:15) δ ( r (cid:48) − r s ( t (cid:48) )) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) (33) − c ∂ ∂t (cid:90) R dt (cid:48) (cid:90) R q s (cid:15) v s ( t (cid:48) ) δ ( r (cid:48) − r s ( t (cid:48) )) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) + ∇ × ∇ × (cid:90) R dt (cid:48) (cid:90) R q s π(cid:15) v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | c ∂ ∂t G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) The second order time derivative of G ( r , t ; r (cid:48) , t (cid:48) ) in the last term of equation (33) can bereplaced with equation (15) to obtain: q s π(cid:15) ∂∂t r − r s ( t ) | r − r s ( t ) | = − ∂∂t ∇ (cid:90) R dt (cid:48) (cid:90) R q s (cid:15) δ ( r (cid:48) − r s ( t (cid:48) )) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) (34) − c ∂ ∂t (cid:90) R dt (cid:48) (cid:90) R q s (cid:15) v s ( t (cid:48) ) δ ( r (cid:48) − r s ( t (cid:48) )) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) + ∇ × ∇ × (cid:90) R dt (cid:48) (cid:90) R q s π(cid:15) v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | δ ( r − r (cid:48) ) δ ( t − t (cid:48) ) dV (cid:48) + ∇ × ∇ × (cid:90) R dt (cid:48) (cid:90) R q s π(cid:15) v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | ∇ G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) Using sifting property of Dirac’s delta function allows us to rewrite the third right handside term of equation (34) as: ∇ × ∇ × (cid:90) R dt (cid:48) (cid:90) R q s π(cid:15) v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | δ ( r − r (cid:48) ) δ ( t − t (cid:48) ) dV (cid:48) = (35)= ∇ × ∇ × (cid:90) R dt (cid:48) q s π(cid:15) v s ( t (cid:48) ) | r − r s ( t (cid:48) ) | δ ( t − t (cid:48) ) == ∇ × ∇ × q s π(cid:15) v s ( t ) | r − r s ( t ) | To continue the derivation of Maxwell’s equations from Coulomb’s law we should notethat operator ∇ does not affect vector v s ( t ) because v s ( t ) is a function of variable t .Hence, the application of standard vector calculus identity ∇ × ∇ × P = ∇ ( ∇ · P ) − ∇ P yields: ∇ × ∇ × q s π(cid:15) v s ( t ) | r − r s ( t ) | = q s π(cid:15) ∇ (cid:18) ∇ · v s ( t ) | r − r s ( t ) | (cid:19) − q s π(cid:15) ∇ v s ( t ) | r − r s ( t ) | = (36)= q s π(cid:15) ∇ (cid:18) v s ( t ) · ∇ | r − r s ( t ) | (cid:19) − q s π(cid:15) v s ( t ) ∇ | r − r s ( t ) | == − q s π(cid:15) ∇ ∂∂t | r − r s ( t ) | + q s (cid:15) v s ( t ) δ ( r − r s ( t )) == q s π(cid:15) ∂∂t r − r s ( t ) | r − r s ( t ) | + q s (cid:15) v s ( t ) δ ( r − r s ( t ))11ombining equations (34), (35) and (36), after cancellation of appropriate terms, yields:0 = − ∂∂t ∇ (cid:90) R dt (cid:48) (cid:90) R q s (cid:15) δ ( r (cid:48) − r s ( t (cid:48) )) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) (37) − c ∂ ∂t (cid:90) R dt (cid:48) (cid:90) R q s (cid:15) v s ( t (cid:48) ) δ ( r (cid:48) − r s ( t (cid:48) )) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) + q s (cid:15) v s ( t ) δ ( r − r s ( t ))+ ∇ × ∇ × (cid:90) R dt (cid:48) (cid:90) R q s π(cid:15) v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | ∇ G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) In Appendix Appendix C, subsection C.3, we have derived the following mathematicalidentity: (cid:90) R dt (cid:48) (cid:90) R q s π(cid:15) v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | ∇ G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) = (38)= − (cid:90) R dt (cid:48) (cid:90) R q s (cid:15) v s ( t (cid:48) ) δ ( r (cid:48) − r s ( t (cid:48) )) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) By inserting equation (38) into equation (37) it is obtained that:0 = − ∂∂t ∇ (cid:90) R dt (cid:48) (cid:90) R q s (cid:15) δ ( r (cid:48) − r s ( t (cid:48) )) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) (39) − c ∂ ∂t (cid:90) R dt (cid:48) (cid:90) R q s (cid:15) v s ( t (cid:48) ) δ ( r (cid:48) − r s ( t (cid:48) )) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) + q s (cid:15) v s ( t ) δ ( r − r s ( t )) − ∇ × ∇ × (cid:90) R dt (cid:48) (cid:90) R q s (cid:15) v s ( t (cid:48) ) δ ( r (cid:48) − r s ( t (cid:48) )) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) If we now introduce new constant µ = c (cid:15) and divide whole equation (39) by c weobtain: 0 = − c ∂∂t ∇ (cid:90) R dt (cid:48) (cid:90) R q s (cid:15) δ ( r (cid:48) − r s ( t (cid:48) )) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) (40) − c ∂ ∂t (cid:90) R dt (cid:48) (cid:90) R q s µc v s ( t (cid:48) ) c δ ( r (cid:48) − r s ( t (cid:48) )) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) + q s µ v s ( t ) δ ( r − r s ( t )) − ∇ × ∇ × (cid:90) R dt (cid:48) (cid:90) R q s µc v s ( t (cid:48) ) c δ ( r (cid:48) − r s ( t (cid:48) )) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) Although it is perhaps not yet apparent, equation (39) is Maxwell-Ampere equation givenin introductory part of this paper as equation (4). To evaluate right hand side integrals inequation (40) we use sifting property of Dirac’s delta function (cid:82) R δ ( r (cid:48) − r s ( t (cid:48) )) f ( r (cid:48) ) dV (cid:48) =12 ( r s ( t (cid:48) )), where f ( r (cid:48) ) is function of position vector r (cid:48) , to rewrite the right hand side in-tegrals in equation (40) as: (cid:90) R dt (cid:48) (cid:90) R q s (cid:15) δ ( r (cid:48) − r s ( t (cid:48) )) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) = (cid:90) R q s (cid:15) G ( r , t ; r s ( t (cid:48) ) , t (cid:48) ) dt (cid:48) (41) (cid:90) R dt (cid:48) (cid:90) R q s µc v s ( t (cid:48) ) c δ ( r (cid:48) − r s ( t (cid:48) )) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) = (cid:90) R q s µc v s ( t (cid:48) ) c G ( r , t ; r s ( t (cid:48) ) , t (cid:48) ) dt (cid:48) (42)To evaluate right hand side integrals in equations above we now replace Green’s function G ( r , t ; r s ( t (cid:48) ) , t (cid:48) ) in these equations with equation (16) to obtain: (cid:90) R dt (cid:48) (cid:90) R q s (cid:15) δ ( r (cid:48) − r s ( t (cid:48) )) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) = (43)= (cid:90) R q s (cid:15) δ (cid:18) t (cid:48) − t + | r − r s ( t (cid:48) ) | c (cid:19) π | r − r s ( t (cid:48) ) | dt (cid:48) (cid:90) R dt (cid:48) (cid:90) R q s µc v s ( t (cid:48) ) c δ ( r (cid:48) − r s ( t (cid:48) )) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) = (44)= (cid:90) R q s µc v s ( t (cid:48) ) c δ (cid:18) t (cid:48) − t + | r − r s ( t (cid:48) ) | c (cid:19) π | r − r s ( t (cid:48) ) | dt (cid:48) The right hand side integrals in equations (43) and (44) can be evaluated by making useof the following standard mathematical identity involving Dirac’s delta function: δ ( f ( u )) = δ ( u − u ) (cid:12)(cid:12) ∂∂u f ( u ) (cid:12)(cid:12) u = u (45)where f ( u ) is real function of real argument u , and u is the solution of equation f ( u ) =0. Using identity (45), the Dirac’s delta function in equations (43) and (44) can bewritten as as: δ (cid:18) t (cid:48) − t + | r − r s ( t (cid:48) ) | c (cid:19) = δ ( t (cid:48) − t r ) (cid:12)(cid:12)(cid:12) − c v s ( t (cid:48) ) · ( r − r s ( t (cid:48) )) | r − r s ( t (cid:48) ) | (cid:12)(cid:12)(cid:12) = δ ( t (cid:48) − t r )1 − c v s ( t (cid:48) ) · ( r − r s ( t (cid:48) )) | r − r s ( t (cid:48) ) | (46)= δ ( t (cid:48) − t r )1 − β ( t r ) · n ( t r )where β ( t r ) and n ( t r ) are given by equations (9) and (10), respectively. From equation(45) it follows that the time t r is the solution to the following equation: t − t r − | r − r s ( t r ) | c = 0 (47)Evidently, the time t r is the time when the disturbance created by moving source chargeat position in space r s ( t r ) was created. This disturbance moves through the space with13nite velocity c and reaches the position r of the test charge at time t . In the electro-magnetic literature this time t r is commonly known as retarded time.To proceed with derivation of Maxwell equations, we now insert equation (46) intoequations (43) and (44), and evaluate the integrals over t (cid:48) using the sifting property ofDirac’s delta function to obtain: (cid:90) R dt (cid:48) (cid:90) R q s (cid:15) δ ( r (cid:48) − r s ( t (cid:48) )) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) = 14 π(cid:15) q s | r − r s ( t r ) | (1 − β ( t r ) · n ( t r ))(48) (cid:90) R dt (cid:48) (cid:90) R q s µc v s ( t (cid:48) ) c δ ( r (cid:48) − r s ( t (cid:48) )) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) = µc π q s β s ( t r ) | r − r s ( t r ) | (1 − β ( t r ) · n ( t r ))(49)By inserting equations (48) and (49) into equation (40), and rearranging, it is obtained: ∇ × ∇ × µc π q s β s ( t r ) | r − r s ( t r ) | (1 − β ( t r ) · n ( t r )) = q s µ v s ( t ) δ ( r − r s ( t )) (50) − c ∂∂t ∇ π(cid:15) q s | r − r s ( t r ) | (1 − β ( t r ) · n ( t r )) − c ∂ ∂t µc π q s β s ( t r ) | r − r s ( t r ) | (1 − β ( t r ) · n ( t r ))The first right hand side term of the equation above can be identified as the current J of the point charge distribution moving with velocity v s ( t ) multiplied by constant µ : µ J = µq s v s ( t ) δ ( r − r s ( t )) (51)We now define scalar function θ ( r , t ) and vector function Q ( r , t ) as: θ ( r , t ) = 14 π(cid:15) q s | r − r s ( t r ) | (1 − β ( t r ) · n ( t r )) (52) Q ( r , t ) = µc π q s β s ( t r ) | r − r s ( t r ) | (1 − β ( t r ) · n ( t r )) (53)With the aid of scalar function θ ( r , t ), vector function Q ( r , t ), and expression µ J givenby equation (51) the equation (50) can be written as: ∇ × ∇ × Q ( r , t ) = µ J − c ∂∂t (cid:18) −∇ θ ( r , t ) − ∂∂t Q ( r , t ) (cid:19) (54)Furthermore, we now define two vector functions M and N as: M = ∇ × Q ( r , t ) (55) N = −∇ θ ( r , t ) − ∂∂t Q ( r , t ) (56)14sing definitions of vector functions M and N given by equations (55) and (56) we canrewrite equation (54) as: ∇ × M = J + 1 c ∂ N ∂t (57)We shall now investigate the mathematical properties of vector fields M and N . Notethat because for any differentiable vector field P we can write ∇ · ∇ × P = 0, fromequation (55) it follows that: ∇ · M = 0 (58)The curl of the gradient of any differentiable scalar function ψ is equal to zero, i.e. ∇ × ∇ ψ = 0. Thus, taking the curl of equation (56) yields: ∇ × N = −∇ × ∂∂t Q ( r , t ) = − ∂∂t ∇ × Q ( r , t ) = − ∂ M ∂t (59)Finally, in Appendix Appendix C, subsection C.4, we have shown that the divergence ofvector field N is: ∇ · N = q s (cid:15) δ ( r − r s ( t )) = ρ ( r , t ) (cid:15) (60)which completes the derivation of electrodynamic equations from Coulomb’s law.To compare these equations to Maxwell’s equations, in Table 1 we have summa-rized governing equations for scalar potential θ ( r , t ), vector potential Q ( r , t ), vector field M and vector field N which are all derived from Coulomb’s law. By comparison withLi´enard–Wiechert potentials given in Table 2, we see that scalar potential θ ( r , t ) is iden-tical to Li´enard–Wiechert scalar potential φ ( r , t ) and vector potential Q ( r , t ) is identicalto Li´enard–Wiechert magnetic vector potential A ( r , t ). Furthermore, by comparing Ta-ble 1 and Table 2 we find that vector field M is identical to magnetic flux density B andthat vector field N is identical to electric field E .In Table 3 we have compared Maxwell’s equations governing fields B and E withdifferential equations governing vector fields M and N . Clearly, left hand side of Table3 is identical in the mathematical form to the right hand side of the same table, hence,differential equations governing vector fields M and N are identical to those governingvector fields B and E . This is expected, because we already know that vector field N = E and vector field M = B .Thus, it should be evident by now that we have derived Maxwell equations andLi´enard–Wiechert potentials directly from Coulomb’s law. This was achieved by math-ematically relating known electrostatic Coulomb’s law acting on test charge at presenttime to ”unknown” electrodynamic fields acting at past. The mathematical link betweenthe static case in the present and dynamic case in the past was provided by general-ized Helmholtz theorem. The derived equations are valid for arbitrarily moving sourcecharge and these equations are not confined to motions along straight line. Furthermore,it should be noted that we have derived the Maxwell equations and Li´enard–Wiechertpotentials directly from Coulomb’s law without resorting to special relativity or Lorentztransformation. 15 able 1: Potentials and vector fields derived from Coulomb’s law. symbol equation description θ ( r , t ) 14 π(cid:15) q s | r − r s ( t r ) | (1 − β ( t r ) · n ( t r )) scalar potential derived fromCoulomb’s law Q ( r , t ) µc π q s β s ( t r ) | r − r s ( t r ) | (1 − β ( t r ) · n ( t r )) vector potential derived fromCoulomb’s law M ∇ × Q ( r , t ) vector field M derived fromCoulomb’s law N −∇ φ ( r , t ) − ∂∂t Q ( r , t ) vector field N derived fromCoulomb’s law Table 2: Standard electromagnetic theory expressions for Li´enard–Wiechert scalar potential φ ( r , t ),Li´enard–Wiechert vector potential A ( r , t ), magnetic flux density B and electric field E . symbol equation description φ ( r , t ) 14 π(cid:15) q s | r − r s ( t r ) | (1 − β ( t r ) · n ( t r )) Li´enard–Wiechert scalarpotential A ( r , t ) µc π q s β s ( t r ) | r − r s ( t r ) | (1 − β ( t r ) · n ( t r )) Li´enard–Wiechert vectorpotential B ∇ × A ( r , t ) magnetic flux density E −∇ φ ( r , t ) − ∂∂t A ( r , t ) electric field Table 3: Maxwell equations for electromagnetic fields E and B compared with differential equations forvector fields M and N derived from Coulomb’s law. Maxwell equation description equations derived fromCoulomb’s law ∇ × B = J + c ∂ E ∂t Maxwell-Ampere equation ∇ × M = J + c ∂ N ∂t ∇ × E = − ∂ B ∂t Faraday’s law ∇ × N = − ∂ M ∂t ∇ · E = ρ(cid:15) Gauss’ law for electric field ∇ · N = ρ(cid:15) ∇ · B = 0 Gauss’ law for magneticfield ∇ · M = 016 yz r q s r s ( t s ) C MOVING SOURCE CHARGE q s STOPS AT TIME t s CLOSED CONTOUR C R = c ( t p − t s ) SPHERE OFRADIUS R Figure 3: Source charge q s is moving along arbitrary trajectory r s ( t ) and it stops at past time t s .Closed contour C is at rest at all times. All the points r on contour C are inside the sphere of radius R = c ( t p − t s ). At present time t p > t s all the points on contour C are affected only by Coulomb’selectrostatic field.
4. Derivation of Electrodynamic Energy Conservation Law and Lorentz Force
To derive the electrodynamic energy conservation law from Coulomb’s law we first con-sider hypothetical physical setting shown in Fig. 3 where the source charge q s is movingalong arbitrary trajectory r s ( t ). Then the source charge q s stops at some time in thepast t s . In this physical setting, closed contour C is at rest at all times. At presenttime t p > t s all the points inside the sphere of radius R = c ( t p − t s ) are affected onlyby electrostatic Coulomb’s field. The known energy conservation law valid at presentdictates that contour integral of electrostatic field along any closed contour immersedinside the sphere of radius R equals to zero: (cid:73) C E c ( r , t p ) · d r = (cid:73) C q s π(cid:15) r − r s ( t s ) | r − r s ( t s ) | · d r = 0 (61)where E c is Coulomb’s electrostatic field, r s ( t s ) is the position vector of source chargewhen it stopped moving, and vector r is the position vector of the point on contour C . This electrostatic energy conservation law, valid at present time t p , states that nonet work is done in transporting the unit charge along any closed contour immersed inelectrostatic field.To proceed, we assume that in the past, when the source charge was moving, theenergy conservation law is unknown. However, generalized Helmholtz decomposition17heorem allows us to derive this ”unknown” electrodynamic energy conservation lawvalid in the past from the knowledge of electrostatic energy conservation law valid atpresent. To derive this unknown electrodynamic electrodynamic conservation law weconsider the contour integral (61) at the moment t infinitesimally before the time whenthe source charge stopped: t = t s − δt (62)where δt is infinitesimally small time interval. If time interval δt approaches zero ( δt → t : (cid:73) C E c ( r , t p − δt ) · d r = (cid:73) C q s π(cid:15) r − r s ( t ) | r − r s ( t ) | · d r = 0 (63)Because the integrand on the right hand side of equation (63) is the function of varyingtime t and position vector r the generalized Helmholtz decomposition theorem can beapplied to rewrite this integrand as the function of past positions and velocities of thesource charge. In fact, such expression is already derived in previous section as equation(33), repeated here for clarity: q s π(cid:15) r − r s ( t ) | r − r s ( t ) | = − ∇ (cid:90) R dt (cid:48) (cid:90) R q s (cid:15) δ ( r (cid:48) − r s ( t (cid:48) )) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) (64) − c ∂∂t (cid:90) R dt (cid:48) (cid:90) R q s (cid:15) v s ( t (cid:48) ) δ ( r (cid:48) − r s ( t (cid:48) )) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) + 1 c ∂∂t (cid:90) R dt (cid:48) (cid:90) R q s π(cid:15) (cid:20) ∇ (cid:48) × ∇ (cid:48) × v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | (cid:21) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) Substituting the first two right hand side terms of equation (64) with equations (48) and(49) and combining the result with equations (52) and (53), and using c = 1 /µ(cid:15) yields: q s π(cid:15) r − r s ( t ) | r − r s ( t ) | = −∇ θ ( r , t ) − ∂∂t Q ( r , t ) + K ( r , t ) (65)where vector function K ( r , t ) is equal to the last right hand side term of equation (64): K ( r , t ) = 1 c ∂∂t (cid:90) R dt (cid:48) (cid:90) R q s π(cid:15) (cid:20) ∇ (cid:48) × ∇ (cid:48) × v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | (cid:21) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) (66)Replacing the first two terms on the right hand side of equation (65) with equation (56)yields: q s π(cid:15) r − r s ( t ) | r − r s ( t ) | = N ( r , t ) + K ( r , t ) (67)Then, by inserting equation (67) into right hand side of equation (63) it is obtained that:0 = (cid:73) C E c ( r , t p − δt ) · d r = (cid:73) C ( N ( r , t ) + K ( r , t )) · d r (68)18he space-time integral on the right hand side of equation (66) is very difficult to evaluate.However, we can eliminate vector field K ( r , t ) from the right hand side of equation (68)by the application of Stokes’ theorem:0 = (cid:73) C E c ( r , t p − δt ) · d r = (cid:73) C N ( r , t ) · d r + (cid:90) S ∇ × K ( r , t ) · d S (69)From here, we take the curl of both sides of equation (67) and by combining with equation(59) it is obtained that: ∇ × K ( r , t ) = −∇ × N ( r , t ) = ∂∂t M ( r , t ) (70)Because surface S and contour C are stationary we can write that ∂∂t M ( r , t ) = ddt M ( r , t ).Inserting equation (70) into equation (69) and taking into account that surface S andcontour C are not moving yields:0 = (cid:73) C E c ( r , t p − δt ) · d r = (cid:73) C N ( r , t ) · d r + ddt (cid:90) S M ( r , t ) · d S (71)The right hand side of equation (71) is unknown energy conservation principle valid forvarying in time dynamic fields N ( r , t ) and M ( r , t ) and it is derived from electrostaticenergy conservation principle valid at present time. If N is replaced by E and if M is replaced by B it can be seen that we have just obtained the physical law known inelectrodynamics as Faraday’s law.From equation (71) the conclusion can be drawn about the nature of Faraday’s law. Itrepresents the energy conservation principle valid for non-conservative dynamic fields andit is dynamic equivalent of electrostatic energy conservation principle valid for Coulomb’selectrostatic field.However, even the Faraday’s law itself can be considered as consequence of somethingelse. To see this, consider simply connected volume V bounded by surface ∂V as shownin Fig. 4. The surface ∂V is union of two surfaces S and S bounded by respectivecontours C and C . Contours C and C consist of exactly the same spatial points,however, the Stokes’ orientation of these contours is opposite C = − C . Then, using ∇ × N ( r , t ) = − ∂∂t M ( r , t ) the first right hand side contour integral of equation (71) canbe written as: (cid:73) C N ( r , t ) · d r = − (cid:73) C N ( r , t ) · d r = − (cid:90) S ∇ × N ( r , t ) · d S = ddt (cid:90) S M ( r , t ) · d S (72)Replacing the first right hand side term of equation (71) with equation (72) yields dif-ferent form of dynamic energy conservation law:0 = (cid:73) C E c ( r , t p − δt ) · d r = ddt (cid:73) ∂V M ( r , t ) · d S (73)If we replace M ( r , t ) with B we see that right hand side of equation (73) is time derivativeof Gauss’ law for magnetic fields. The standard interpretation of Gauss’ law for mag-netic fields is that magnetic monopoles do not exist. However, from equation (73) weconclude that alternative interpretation of this law is that its time derivative represents19he dynamic energy conservation law. From the derivations presented, we might evensay that Faraday’s law is the consequence of Gauss’ law for magnetic fields. It should benoted that these energy-conservation equations were all derived from simple electrostaticCoulomb’s law.From dynamic energy conservation law the derivation of Lorentz force is straight-forward: we now assume that all the points on surface ∂V shown in Fig. 4 have somedefinite velocity v such that | v | << c . Then the surface ∂V is the function of time,hence, C = C ( t ) and S = S ( t ). Hence, we can rewrite equation (73) as the sum of twosurface integrals: ddt (cid:73) ∂V M ( r , t ) · d S = ddt (cid:73) S ( t ) M ( r , t ) · d S + ddt (cid:73) S ( t ) M ( r , t ) · d S = 0 (74)where ∂V = S ( t ) ∪ S ( t ). The Leibniz identity [25] for moving surfaces states that forany differentiable vector field P we can write: ddt (cid:90) S ( t ) P · d S = (cid:90) S ( t ) (cid:20) ∂∂t P + ( ∇ · P ) v (cid:21) · d S − (cid:73) C ( t ) v × P · d r (75)Applying the Leibniz identity to the surface integral over surface S ( t ) in equation (74)and using ∇ · M ( r , t ) = 0 yields: ddt (cid:73) S ( t ) M ( r , t ) · d S + (cid:73) S ( t ) ∂∂t M ( r , t ) · d S − (cid:73) C ( t ) v × M ( r , t ) · d r = 0 (76)Using the result from previous section, i.e. ∇ × N ( r , t ) = − ∂∂t M ( r , t ), and applying theStokes’ theorem yields: ddt (cid:73) S ( t ) M ( r , t ) · d S − (cid:73) C ( t ) N ( r , t ) · d r − (cid:73) C ( t ) v × M ( r , t ) · d r = 0 (77)Because curves C ( t ) and C ( t ) comprise of same points, however, Stokes’ orientation ofcurves C ( t ) and C ( t ) is opposite, i.e. C ( t ) = − C ( t ), we can rewrite equation (77) as: (cid:73) C ( t ) [ N ( r , t ) + v × M ( r , t )] · d r + ddt (cid:90) S ( t ) M ( r , t ) · d S = 0 (78) = U CS C S ∂ V = S ∪ S C Figure 4: Closed surface ∂V that bounds volume V is union of two surfaces S and S . Contour C bounds surface S and contour C bounds surface S . Contours C and C are identical, however theyhave different Stokes’ orientation. C ( t ). Because the first term in equation (78)is contour integral of vector field we can conclude that this term represents the non-zero work done by non-conservative electrodynamic force in transporting the unit chargealong contour C ( t ).Hence, just as the left hand side of equation (73) represents the work done by con-servative electrostatic force in transporting the unit charge along contour C , the contourintegral on the left of equation (78) represents the work done by non-conservative elec-trodynamic force in transporting the unit charge along the same contour. The purposeof surface integral on the left hand side of equation (78) is to balance non-zero work ofnon-conservative electrodynamic force along contour C . Thus, it can be concluded thatthe electrodynamic force F D on charge q moving with velocity v along contour C is: F D = q N ( r , t ) + q v × M ( r , t ) (79)Finally, in previous section we have shown that N = E and that M = B . Thus, byreplacing N with E and M with B it is obtained that: F D = q ( E + v × B ) (80)which is expression for well known Lorentz force. It was derived theoretically from theknowledge of electrostatic energy conservation law which, in turn, can be derived fromCoulomb’s law. Thus, we may say that we have just derived the Lorentz force fromsimple electrostatic Coulomb’s law.
5. Conclusion
In this paper we have presented the theoretical framework that explains Maxwellequations and the Lorentz force on more fundamental level than it was previously done.Maxwell derived Maxwell equations from experimental Ampere’s force law and experi-mental Faraday’s law, and Lorentz continued work on Maxwell’s theory to discover theLorentz force. In last 150 years, no successful theory was presented that would explainMaxwell’s equations and Lorentz force on more fundamental level.To accomplish this, relativistically correct Li´enard–Wiechert potentials, Maxwellequations and the Lorentz force were derived directly from electrostatic Coulomb’s law.In contrast to frequently criticized previous attempts to derive Maxwell’s equations fromCoulomb’s law using special relativity and Lorentz transformation, the Lorentz transfor-mation was not used in our derivations nor the theory of special relativity. In fact, inthis work, dynamic Li´enard–Wiechert potentials, Maxwell equations and Lorentz forcewere derived from Coulomb’s law using the following two simple postulates:21a) when charges are at rest the Coulomb’s law describes the force acting betweencharges(b) disturbances caused by moving charges propagate outwardly from moving chargewith finite velocityThe derivation of these dynamic physical laws from electrostatic Coulomb’s law wouldnot be possible without generalized Helmholtz decomposition theorem also derived inthis paper. This theorem allows the vector function of present position and presenttime to be written as space-time integral of positions and velocities at previous time. Incontrast, standard Helmholtz decomposition theorem is valid for functions of space onlyand it ignores time.To derive the Lorentz force from Coulomb’s law, in section 4, the ”unknown” dynamicenergy conservation law valid in the past was derived from the knowledge of electrostaticenergy conservation law valid at present. The link between the present and the pastwas again provided by generalized Helmholtz decomposition theorem. This ”unknown”dynamic energy conservation principle turned out to be Faraday’s law of induction. Ad-ditionally, it was shown that Faraday’s law of induction can be considered equivalentto time derivative of Gauss’ law for magnetic field. From these energy conservationconsiderations the Lorentz force was derived.From the presented analysis one important question naturally arises: are Maxwell’sequations and Lorentz force the consequence of electrostatic Coulomb’s law? They aremost probably not. It is rather the opposite, Coulomb’s law is the limiting case of Lorentzforce when the source charge becomes stationary. However, as it was shown in this paper,it is entirely possible to deduce dynamic Maxwell equations and Lorentz force from theknowledge of simple electrostatic Coulomb’s law.Finally, this paper attempts to answer another important question: how can wededuce more general dynamic physical laws from the limited knowledge provided bystatic physical law? The significance of answering this question is that in the future itwill perhaps become possible that similar reasoning could deepen the understanding ofphysical laws other than Maxwell equations and Lorentz force.
Appendix A. Derivation of generalized Helmholtz decomposition theorem
In this appendix, we derive the generalized Helmholtz decomposition theorem for vectorfunctions of space and time. However, in effort to enhance the readability of this work,we first start by considering some basic identities given in section A.1 of this appendix.
A.1. Preliminary considerations
To clarify notation used throughout this paper we first define position vectors r and r (cid:48) as: r = x ˆx + y ˆy + z ˆz (A.1.1) r (cid:48) = x (cid:48) ˆx + y (cid:48) ˆy + z (cid:48) ˆz (A.1.2)22here ˆx , ˆy and ˆz are Cartesian, mutually orthogonal, unit basis vectors. Variables x, y, z ∈ R and x (cid:48) , y (cid:48) , z (cid:48) ∈ R are linearly independent variables. Furthermore, throughoutthis paper we use position vector r s ( t (cid:48) ) to indicate the position of the source charge. Thisposition vector r s ( t (cid:48) ) is defined as: r s ( t (cid:48) ) = x s ( t (cid:48) ) ˆx + y s ( t (cid:48) ) ˆy + z s ( t (cid:48) ) ˆz (A.1.3)where x s ( t (cid:48) ), y s ( t (cid:48) ) and z s ( t (cid:48) ) are all functions of real variable t (cid:48) ∈ R which is independentof variables x, y, z ∈ R and x (cid:48) , y (cid:48) , z (cid:48) ∈ R . The time derivative of position vector r s ( t (cid:48) ) isvelocity v s ( t (cid:48) ) of the source charge: v s ( t (cid:48) ) = ∂ r s ( t (cid:48) ) ∂t (cid:48) (A.1.4)On many occasions in this paper we have used differential operators ∇ and ∇ (cid:48) definedas: ∇ = ˆx ∂∂x + ˆy ∂∂x + ˆz ∂∂x (A.1.5) ∇ (cid:48) = ˆx ∂∂x (cid:48) + ˆy ∂∂y (cid:48) + ˆz ∂∂z (cid:48) (A.1.6)Operator ∇ acts only on functions of variables x, y, z , hence, on functions of positionvector r . On the other hand, operator ∇ (cid:48) acts only on functions of variables x (cid:48) , y (cid:48) , z (cid:48) ,thus, it acts on functions of position vector r (cid:48) . For example, if function f is the functionof position vector r , that is f = f ( r ) we can generally write: ∇ f ( r ) (cid:54) = 0 ∇ (cid:48) f ( r ) = 0 (A.1.7)On the other hand, if function f is the function of position vector r (cid:48) , that is if f = f ( r (cid:48) )we can write: ∇ f ( r (cid:48) ) = 0 ∇ (cid:48) f ( r (cid:48) ) (cid:54) = 0 (A.1.8)Furthermore, because variable t (cid:48) is independent of variables x, y, z and x (cid:48) , y (cid:48) , z (cid:48) neitheroperator ∇ nor ∇ (cid:48) acts on position vector r s ( t (cid:48) ) and velocity vector v s ( t (cid:48) ). Using theseconsiderations we see that the following equations are correct: ∇ · r s ( t (cid:48) ) = 0 ∇ · v s ( t (cid:48) ) = 0 ∇ (cid:48) · r s ( t (cid:48) ) = 0 ∇ (cid:48) · v s ( t (cid:48) ) = 0 (A.1.9)However, both operators ∇ and ∇ (cid:48) act on Green’s function G ( r , t ; r (cid:48) , t (cid:48) ) given by equation(16). In fact, one can easily verify that the following equations hold: ∇ G ( r , t ; r (cid:48) , t (cid:48) ) = −∇ (cid:48) G ( r , t ; r (cid:48) , t (cid:48) ) ∇ G ( r , t ; r (cid:48) , t (cid:48) ) = ∇ (cid:48) G ( r , t ; r (cid:48) , t (cid:48) ) ∂∂t G ( r , t ; r (cid:48) , t (cid:48) ) = − ∂∂t (cid:48) G ( r , t ; r (cid:48) , t (cid:48) ) ∂ ∂t G ( r , t ; r (cid:48) , t (cid:48) ) = ∂ ∂t (cid:48) G ( r , t ; r (cid:48) , t (cid:48) ) (A.1.10)23 .2. Generalized Helmholtz decomposition theorem To start deriving generalized Helmholtz decomposition theorem for vector functions ofspace and time we first consider inhomogeneous transient wave equation: ∇ G ( r , t ; r (cid:48) , t (cid:48) ) − c ∂ ∂t G ( r , t ; r (cid:48) , t (cid:48) ) = − δ ( r − r (cid:48) ) δ ( t − t (cid:48) ) (A.2.11)where G ( r , t ; r (cid:48) , t (cid:48) ) is the function called fundamental solution or Green’s function and δ is Dirac’s delta function. The Green’s function for inhomogeneous wave equation is wellknown and it represents an outgoing diverging spherical wave: G ( r , t ; r (cid:48) , t (cid:48) ) = δ (cid:18) t − t (cid:48) − | r − r (cid:48) | c (cid:19) π | r − r (cid:48) | (A.2.12)Let us now suppose that vector field F is the function of both space r and time t , i.e. F = F ( r , t ). Using sifting property of Dirac delta function we can write vector function F ( r , t ) as the volume integral over infinite volume R and over all the time R as: F ( r , t ) = (cid:90) R dt (cid:48) (cid:90) R F ( r (cid:48) , t (cid:48) ) δ ( r − r (cid:48) ) δ ( t − t (cid:48) ) dV (cid:48) (A.2.13)where differential volume element dV (cid:48) is dV (cid:48) = dx (cid:48) dy (cid:48) dz (cid:48) . We now replace δ ( r − r (cid:48) ) δ ( t − t (cid:48) )in equation above with left hand side of equation (A.2.11) to obtain: F ( r , t ) = − (cid:90) R dt (cid:48) (cid:90) R F ( r (cid:48) , t (cid:48) ) (cid:20) ∇ G ( r , t ; r (cid:48) , t (cid:48) ) − c ∂ ∂t G ( r , t ; r (cid:48) , t (cid:48) ) (cid:21) dV (cid:48) (A.2.14)From the discussion presented in section A.1 of this appendix, we know that D’Alambertoperator ∇ − c ∂ ∂t does not act on variables x (cid:48) , y (cid:48) , z (cid:48) and t (cid:48) nor does it act on vectorfunction F ( r (cid:48) , t (cid:48) ). Hence, we can write the D’Alambert operator ∇ − c ∂ ∂t in front ofthe integral: F ( r , t ) = − (cid:18) ∇ − c ∂ ∂t (cid:19) (cid:90) R dt (cid:48) (cid:90) R F ( r (cid:48) , t (cid:48) ) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) (A.2.15)Using standard vector calculus identity ∇ × ∇ × P = ∇ ( ∇ · P ) − ∇ P we can rewriteequation (A.2.15) as: F ( r , t ) = ∇ × ∇ × (cid:90) R dt (cid:48) (cid:90) R F ( r (cid:48) , t (cid:48) ) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) (A.2.16) − ∇ (cid:18) ∇ · (cid:90) R dt (cid:48) (cid:90) R F ( r (cid:48) , t (cid:48) ) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) (cid:19) + 1 c ∂ ∂t (cid:90) R dt (cid:48) (cid:90) R F ( r (cid:48) , t (cid:48) ) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) Because operators ∇ and ∂∂t do not act on variables x (cid:48) , y (cid:48) , z (cid:48) and t (cid:48) we can moveoperator ∇ and partial derivative ∂∂t under right hand side integrals in equation (A.2.16).24hen using standard vector calculus identities ∇ × ( ψ P ) = ∇ ψ × P + ψ ∇ × P and ∇ · ( ψ P ) = ∇ ψ · P + ψ ∇ · P , and noting that ∇ × F ( r (cid:48) , t (cid:48) ) = 0 and ∇ · F ( r (cid:48) , t (cid:48) ) = 0 wecan rewrite equation (A.2.16) as: F ( r , t ) = ∇ × (cid:90) R dt (cid:48) (cid:90) R ∇ G ( r , t ; r (cid:48) , t (cid:48) ) × F ( r (cid:48) , t (cid:48) ) dV (cid:48) (A.2.17) − ∇ (cid:90) R dt (cid:48) (cid:90) R F ( r (cid:48) , t (cid:48) ) · ∇ G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) + 1 c ∂∂t (cid:90) R dt (cid:48) (cid:90) R F ( r (cid:48) , t (cid:48) ) ∂∂t G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) We now use identities ∇ G ( r , t ; r (cid:48) , t (cid:48) ) = −∇ (cid:48) G ( r , t ; r (cid:48) , t (cid:48) ) and ∂∂t G ( r , t ; r (cid:48) , t (cid:48) ) = − ∂∂t (cid:48) G ( r , t ; r (cid:48) , t (cid:48) )to rewrite the right hand side integrals in equation (A.2.17) as: F ( r , t ) = − ∇ × (cid:90) R dt (cid:48) (cid:90) R ∇ (cid:48) G ( r , t ; r (cid:48) , t (cid:48) ) × F ( r (cid:48) , t (cid:48) ) dV (cid:48) (A.2.18)+ ∇ (cid:90) R dt (cid:48) (cid:90) R F ( r (cid:48) , t (cid:48) ) · ∇ (cid:48) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) − c ∂∂t (cid:90) R dt (cid:48) (cid:90) R F ( r (cid:48) , t (cid:48) ) ∂∂t (cid:48) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) Using vector calculus identity ∇× ( ψ P ) = ∇ ψ × P + ψ ∇× P and the the form of divergencetheorem (cid:82) V ∇ × P dV = (cid:72) ∂V P × d S we rewrite the first right hand side integral over R as: (cid:90) R ∇ (cid:48) G ( r , t ; r (cid:48) , t (cid:48) ) × F ( r (cid:48) , t (cid:48) ) dV (cid:48) = (cid:73) ∂ R G ( r , t ; r (cid:48) , t (cid:48) ) F ( r (cid:48) , t (cid:48) ) × d S (cid:48) (A.2.19) − (cid:90) R G ( r , t ; r (cid:48) , t (cid:48) ) ∇ (cid:48) × F ( r (cid:48) , t (cid:48) ) dV (cid:48) Note that the surface ∂ R is an infinite surface that bounds an infinite volume R .Furthermore, for the surface integral in the equation above, position vector r (cid:48) is locatedon infinite surface ∂ R , i.e. r (cid:48) ∈ ∂ R . Hence, if vector function F ( r (cid:48) , t (cid:48) ) decreases fasterthan 1 / | r − r (cid:48) | as | r − r (cid:48) | → ∞ the surface integral in equation (A.2.19) vanishes. Inthat case, we can write: (cid:90) R ∇ (cid:48) G ( r , t ; r (cid:48) , t (cid:48) ) × F ( r (cid:48) , t (cid:48) ) dV (cid:48) = − (cid:90) R G ( r , t ; r (cid:48) , t (cid:48) ) ∇ (cid:48) × F ( r (cid:48) , t (cid:48) ) dV (cid:48) (A.2.20)Using similar considerations, vector calculus identity ∇ · ( ψ P ) = ∇ ψ · P + ψ ∇ · P andstandard divergence theorem (cid:82) V ∇ · P dV = (cid:72) ∂V P · d S it is obtained that: (cid:90) R F ( r (cid:48) , t (cid:48) ) · ∇ (cid:48) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) = − (cid:90) R G ( r , t ; r , t (cid:48) ) ∇ (cid:48) · F ( r (cid:48) , t (cid:48) ) dV (cid:48) (A.2.21)25o treat the last integral on the right hand side of equation (A.2.18) we use the followingidentity: F ( r (cid:48) , t (cid:48) ) ∂∂t (cid:48) G ( r , t ; r , t (cid:48) ) = ∂∂t (cid:48) ( F ( r (cid:48) , t (cid:48) ) G ( r , t ; r , t (cid:48) )) − G ( r , t ; r , t (cid:48) ) ∂∂t (cid:48) F ( r (cid:48) , t (cid:48) ) (A.2.22)Using the identity above and noting that t (cid:48) is independent of x (cid:48) , y (cid:48) and z (cid:48) we can rewritethe last right hand side integral of equation (A.2.18) as: (cid:90) R dt (cid:48) (cid:90) R F ( r (cid:48) , t (cid:48) ) ∂∂t (cid:48) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) = (cid:90) R dt (cid:48) ∂∂t (cid:48) (cid:90) R F ( r (cid:48) , t (cid:48) ) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) (A.2.23) − (cid:90) R dt (cid:48) (cid:90) R G ( r , t ; r (cid:48) , t (cid:48) ) ∂∂t (cid:48) F ( r (cid:48) , t (cid:48) ) dV (cid:48) By integrating over t (cid:48) , it can be shown that the first right hand side integral in equation(A.2.23) vanishes : (cid:90) R dt (cid:48) ∂∂t (cid:48) (cid:90) R F ( r (cid:48) , t (cid:48) ) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) = (cid:20)(cid:90) R F ( r (cid:48) , t (cid:48) ) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) (cid:21) t (cid:48) →∞ t (cid:48) →−∞ (A.2.24)If t (cid:48) → ±∞ , and if t is finite, then from equation (A.2.12) follows that G ( r , t ; r (cid:48) , t (cid:48) ) = 0,thus, the right hand side of equation (A.2.24) is equal to zero. Inserting this result intoequation (A.2.23) yields: (cid:90) R dt (cid:48) (cid:90) R F ( r (cid:48) , t (cid:48) ) ∂∂t (cid:48) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) = − (cid:90) R dt (cid:48) (cid:90) R G ( r , t ; r (cid:48) , t (cid:48) ) ∂∂t (cid:48) F ( r (cid:48) , t (cid:48) ) dV (cid:48) (A.2.25)By inserting equations (A.2.20), (A.2.21) and (A.2.25) into equation (A.2.18) we obtainthe generalized Helmholtz theorem for vector functions of space and time: F ( r , t ) = − ∇ (cid:90) R dt (cid:48) (cid:90) R (cid:18) ∇ (cid:48) · F ( r (cid:48) , t (cid:48) ) (cid:19) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) (A.2.26)+ 1 c ∂∂t (cid:90) R dt (cid:48) (cid:90) R (cid:18) ∂∂t (cid:48) F ( r (cid:48) , t (cid:48) ) (cid:19) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) + ∇ × (cid:90) R dt (cid:48) (cid:90) R (cid:18) ∇ (cid:48) × F ( r (cid:48) , t (cid:48) ) (cid:19) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) The theorem is valid for functions F ( r (cid:48) , t (cid:48) ) that decrease faster than 1 / | r − r (cid:48) | as | r − r (cid:48) | →∞ . Appendix B. Novel vector calculus identities
In this appendix we prove two novel vector calculus identities, without which, itwould be very difficult, perhaps even not possible, to derive Maxwell’s equations from26oulomb’s law. These two novel vector calculus identities are given by the following twoequations: (cid:90) V ψ ∇ P dV = (cid:73) ∂V ψ ( d S · ∇ ) P − (cid:90) V ( ∇ ψ · ∇ ) P dV (B.1) (cid:90) V P ∇ ψdV = (cid:73) ∂V P ( ∇ ψ · d S ) − (cid:90) V ( ∇ ψ · ∇ ) P dV (B.2)where P is differentiable vector field, ψ is differentiable scalar function, volume V ⊂ R issimply connected volume, ∂V is the bounding surface of volume V and d S is differentialsurface element of ∂V such that d S = n dS . Vector n is an outward unit normal to thesurface ∂V . In Cartesian coordinate system the product ψ ∇ P can be written in termsof Cartesian components as: ψ ∇ P = ˆ x ψ ∇ P x + ˆ y ψ ∇ P y + ˆ z ψ ∇ P z (B.3)where P x , P y and P z are Cartesian components of vector P and vectors ˆx , ˆy and ˆz are Cartesian unit basis vectors. Using standard vector calculus identity ∇ · f T = ∇ f · T + f ∇ · T , valid for some scalar function f and for some vector function T , we canrewrite equation (B.3) as: ψ ∇ P = ˆ x ∇ · ( ψ ∇ P x ) − ˆ x ∇ ψ · ∇ P x (B.4)+ ˆ y ∇ · ( ψ ∇ P y ) − ˆ y ∇ ψ · ∇ P y + ˆ z ∇ · ( ψ ∇ P z ) − ˆ z ∇ ψ · ∇ P z To proceed, we now expand the identity ( ∇ ψ · ∇ ) P in terms of its Cartesian componentsas: ( ∇ ψ · ∇ ) P = (cid:18) ∂ψ∂x ∂∂x + ∂ψ∂y ∂∂y + ∂ψ∂z ∂∂z (cid:19) ( ˆx P x + ˆy P y + ˆz P z ) (B.5)= ˆx ∇ ψ · ∇ P x + ˆy ∇ ψ · ∇ P y + ˆz ∇ ψ · ∇ P z Inserting equation (B.5) into (B.4) it is obtained that: ψ ∇ P = ˆ x ∇ · ( ψ ∇ P x ) + ˆ y ∇ · ( ψ ∇ P y ) + ˆ z ∇ · ( ψ ∇ P z ) − ( ∇ ψ · ∇ ) P (B.6)We now integrate equation (B.6) over volume V and apply the divergence theorem (cid:82) V ∇ · T dV = (cid:72) ∂V T · d S to obtain: (cid:90) V ψ ∇ P dV =ˆ x (cid:73) ∂V ψ ∇ P x · d S + ˆ y (cid:73) ∂V ψ ∇ P y · d S + ˆ z (cid:73) ∂V ψ ∇ P z · d S (B.7) − (cid:90) V ( ∇ ψ · ∇ ) P dV x (cid:73) ∂V ψ ∇ P x · d S + ˆ y (cid:73) ∂V ψ ∇ P y · d S + ˆ z (cid:73) ∂V ψ ∇ P z · d S = (cid:73) ∂V ψ ( d S · ∇ ) P (B.8)Inserting equation (B.8) into (B.7) yields: (cid:90) V ψ ∇ P dV = (cid:73) ∂V ψ ( d S · ∇ ) P − (cid:90) V ( ∇ ψ · ∇ ) P dV (B.9)which we intended to prove. To prove equation (B.2) we rewrite P ∇ ψ in terms ofCartesian components as: P ∇ ψ = ˆx P x ∇ ψ + ˆy P y ∇ ψ + ˆz P z ∇ ψ (B.10)By using standard differential calculus identity f ∇ h = ∇ · ( f ∇ h ) − ∇ f · ∇ h , where f and h are differentiable scalar functions, equation (B.10) can be written as: P ∇ ψ = ˆx ∇ · ( P x ∇ ψ ) − ˆx ∇ P x · ∇ ψ + (B.11) ˆy ∇ · ( P y ∇ ψ ) − ˆy ∇ P y · ∇ ψ + ˆz ∇ · ( P z ∇ ψ ) − ˆz ∇ P z · ∇ ψ Inserting equation (B.5) into equation (B.11) yields: P ∇ ψ = ˆx ∇ · ( P x ∇ ψ ) + ˆy ∇ · ( P y ∇ ψ ) + ˆz ∇ · ( P z ∇ ψ ) − ( ∇ ψ · ∇ ) P (B.12)Integrating equation (B.12) over volume V and applying divergence theorem (cid:82) V ∇ · T dV = (cid:72) ∂V T · d S it is obtained that: (cid:90) V P ∇ ψdV = (cid:73) ∂V P ( ∇ ψ · d S ) − (cid:90) V ( ∇ ψ · ∇ ) P dV (B.13)which we intended to prove. Appendix C. Derivation of auxiliary mathematical identities
In this appendix we derive auxiliary mathematical identities that we find useful forthe derivation of Maxwell equations from Coulomb’s law.
C.1. Derivation of equation (28)
Equation (24) allows us to rewrite the time derivative in the second right hand sideintegral in equation (27) as: ∂∂t (cid:48) q s π(cid:15) r (cid:48) − r s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | = − q s π(cid:15) ∂∂t (cid:48) ∇ (cid:48) | r (cid:48) − r s ( t (cid:48) ) | (C.1.1)28ecause coordinates x (cid:48) , y (cid:48) and z (cid:48) are independent of time t (cid:48) we can swap operator ∇ (cid:48) and time derivative with respect to time t (cid:48) as: ∂∂t (cid:48) q s π(cid:15) r (cid:48) − r s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | = − q s π(cid:15) ∇ (cid:48) ∂∂t (cid:48) | r (cid:48) − r s ( t (cid:48) ) | (C.1.2)Furthermore, because coordinates x (cid:48) , y (cid:48) and z (cid:48) are independent of time t (cid:48) , the timederivative of r (cid:48) is equal to zero ∂ r (cid:48) ∂t (cid:48) = 0. The inner time derivative in the equation(C.1.2) can now be written as: ∂∂t (cid:48) | r (cid:48) − r s ( t (cid:48) ) | = v s ( t (cid:48) ) · ( r (cid:48) − r s ( t (cid:48) )) | r (cid:48) − r s ( t (cid:48) ) | = −∇ (cid:48) · v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | (C.1.3)where v s ( t (cid:48) ) is the velocity of the source charge q s at time t (cid:48) given by equation (A.1.4).Inserting equation (C.1.3) into equation (C.1.2) yields: ∂∂t (cid:48) q s π(cid:15) r (cid:48) − r s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | = q s π(cid:15) ∇ (cid:48) (cid:18) ∇ (cid:48) · v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | (cid:19) (C.1.4)To proceed with derivation, we now make use of standard vector calculus identity ∇ ×∇ × P = ∇ ( ∇ · P ) − ∇ P , valid for any differentiable vector function P . This identityallows us to rewrite the equation (C.1.4) as: ∂∂t (cid:48) q s π(cid:15) r (cid:48) − r s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | = q s π(cid:15) ∇ (cid:48) × ∇ (cid:48) × v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | + q s π(cid:15) ∇ (cid:48) v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | (C.1.5)Since Laplacian operator ∇ (cid:48) does not have effect on velocity vector v s ( t (cid:48) ) the last righthand side term in equation (C.1.5) can be written using 3D Dirac’s delta function as: q s π(cid:15) ∇ (cid:48) v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | = q s (cid:15) v s ( t (cid:48) ) ∇ (cid:48) π | r (cid:48) − r s ( t (cid:48) ) | = − q s (cid:15) v s ( t (cid:48) ) δ ( r (cid:48) − r s ( t (cid:48) )) (C.1.6)Hence, replacing the last right hand side term of equation (C.1.5) with equation (C.1.6)yields: ∂∂t (cid:48) q s π(cid:15) r (cid:48) − r s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | = q s π(cid:15) ∇ (cid:48) × ∇ (cid:48) × v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | − q s (cid:15) v s ( t (cid:48) ) δ ( r (cid:48) − r s ( t (cid:48) )) (C.1.7)which proves equation (28). C.2. Derivation of equation (31)
To derive equation (31) we make use of standard vector calculus identity ∇ × ( ψ P ) = ∇ ψ × P + ψ ∇ × P , where ψ is a scalar function and P is a vector function, to rewritethe integrand in the last right hand side term of equation (30) as:29 ∇ (cid:48) × ∇ (cid:48) × v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | (cid:21) G ( r , t ; r (cid:48) , t (cid:48) ) = (C.2.8)= ∇ (cid:48) × (cid:20)(cid:18) ∇ (cid:48) × v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | (cid:19) G ( r , t ; r (cid:48) , t (cid:48) ) (cid:21) − ∇ (cid:48) G ( r , t ; r (cid:48) , t (cid:48) ) × (cid:20) ∇ (cid:48) × v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | (cid:21) Integrating equation (C.2.8) with respect to variables x (cid:48) , y (cid:48) , z (cid:48) and t (cid:48) , and making use ofa standard form of divergence theorem (cid:82) V ∇ × P dV = (cid:72) ∂V d S × P it is obtained that: (cid:90) R dt (cid:48) (cid:90) R (cid:20) ∇ (cid:48) × ∇ (cid:48) × v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | (cid:21) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) = (C.2.9)= (cid:90) R dt (cid:48) (cid:73) ∂ R d S (cid:48) × (cid:18) ∇ (cid:48) × v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | (cid:19) G ( r , t ; r (cid:48) , t (cid:48) ) − (cid:90) R dt (cid:48) (cid:90) R ∇ (cid:48) G ( r , t ; r (cid:48) , t (cid:48) ) × (cid:20) ∇ (cid:48) × v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | (cid:21) dV (cid:48) where dV = dx (cid:48) dy (cid:48) dz (cid:48) , ∂ R is an infinite surface that bounds R , and d S is differentialsurface element of the surface ∂ R . Because ∂ R is an infinite surface, the first righthand side integral vanishes. To see this, we can use standard vector identity ∇ × ( ψ P ) = ∇ ψ × P + ψ ∇ × P and using ∇ (cid:48) × v s ( t (cid:48) ) = 0 to rewrite the first term in the first righthand side integrand as: ∇ (cid:48) × v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | = − r (cid:48) − r s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | × v s ( t (cid:48) ) (C.2.10)Because r (cid:48) ∈ ∂ R this means that | r (cid:48) | → ∞ . Provided that charge q s is moving withfinite velocity v s ( t (cid:48) ) it is clear that right hand side term of equation (C.2.10) vanishes as | r (cid:48) | → ∞ . Thus, we can now write equation (C.2.9) as: (cid:90) R dt (cid:48) (cid:90) R (cid:20) ∇ (cid:48) × ∇ (cid:48) × v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | (cid:21) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) = (C.2.11)= − (cid:90) R dt (cid:48) (cid:90) R ∇ (cid:48) G ( r , t ; r (cid:48) , t (cid:48) ) × (cid:20) ∇ (cid:48) × v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | (cid:21) dV (cid:48) There is another useful property of Green’s function G ( r , t ; r (cid:48) , t (cid:48) ) which enables us toproceed with the derivation of equation (31). This property can be written as follows: ∇ (cid:48) G ( r , t ; r (cid:48) , t (cid:48) ) = −∇ G ( r , t ; r (cid:48) , t (cid:48) ) (C.2.12)Using this property and standard vector calculus identity ∇ × ( ψ P ) = ∇ ψ × P + ψ ∇ × P allows us to rewrite the integrand in equation (C.2.9) as:30 (cid:48) G ( r , t ; r (cid:48) , t (cid:48) ) × (cid:20) ∇ (cid:48) × v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | (cid:21) = (C.2.13)= −∇ G ( r , t ; r (cid:48) , t (cid:48) ) × (cid:20) ∇ (cid:48) × v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | (cid:21) == −∇ × (cid:20) G ( r , t ; r (cid:48) , t (cid:48) ) ∇ (cid:48) × v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | (cid:21) Equation above is valid because operator ∇ does not act on velocity vector v s ( t (cid:48) ), nordoes it act on position vectors r (cid:48) and r s ( t (cid:48) ). It only acts on Green’s function G ( r , t ; r (cid:48) , t (cid:48) )because it is a function of position vector r . Inserting equation (C.2.13) into equation(C.2.11) yields: (cid:90) R dt (cid:48) (cid:90) R (cid:20) ∇ (cid:48) × ∇ (cid:48) × v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | (cid:21) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) = (C.2.14)= (cid:90) R dt (cid:48) (cid:90) R ∇ × (cid:20) G ( r , t ; r (cid:48) , t (cid:48) ) ∇ (cid:48) × v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | (cid:21) dV (cid:48) Because differential volume element is dV (cid:48) = dx (cid:48) dy (cid:48) dz (cid:48) and because operator ∇ does notact on variables x (cid:48) , y (cid:48) , z (cid:48) and t (cid:48) we can write operator ∇ in front of the integral: (cid:90) R dt (cid:48) (cid:90) R (cid:20) ∇ (cid:48) × ∇ (cid:48) × v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | (cid:21) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) = (C.2.15)= ∇ × (cid:90) R dt (cid:48) (cid:90) R G ( r , t ; r (cid:48) , t (cid:48) ) ∇ (cid:48) × v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | dV (cid:48) Using the same trick again, that is, by using standard vector calculus identity ∇× ( ψ P ) = ∇ ψ × P + ψ ∇ × P , using ∇ (cid:48) G ( r , t ; r (cid:48) , t (cid:48) ) = −∇ G ( r , t ; r (cid:48) , t (cid:48) ) and noting that operator ∇ does not act on variables x (cid:48) , y (cid:48) and z (cid:48) we can rewrite the integrand in equation (C.2.15)as: G ( r , t ; r (cid:48) , t (cid:48) ) ∇ (cid:48) × v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | = (C.2.16)= ∇ (cid:48) × (cid:20) G ( r , t ; r (cid:48) , t (cid:48) ) v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | (cid:21) + ∇ × (cid:20) G ( r , t ; r (cid:48) , t (cid:48) ) v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | (cid:21) Then, by inserting equation (C.2.16) into equation (C.2.15) and using a form of standarddivergence theorem (cid:82) V ∇ × P dV = (cid:72) ∂V d S × P we obtain that: (cid:90) R dt (cid:48) (cid:90) R (cid:20) ∇ (cid:48) × ∇ (cid:48) × v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | (cid:21) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) = (C.2.17)= ∇ × (cid:90) R dt (cid:48) (cid:73) ∂ R d S (cid:48) × v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | + ∇ × (cid:90) R dt (cid:48) (cid:90) ∂ R ∇ × (cid:20) G ( r , t ; r (cid:48) , t (cid:48) ) v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | (cid:21) dV (cid:48) ∂ R is an infinite surface the magnitude of position vector r (cid:48) is infinite,hence the surface integral in the first right hand side term of equation (C.2.17) vanishes.Furthermore, because operator ∇ does not act on variables x (cid:48) , y (cid:48) , z (cid:48) and t (cid:48) we can writeoperator ∇ in front of the second right hand side space-time integral. Hence, equation(C.2.17) can be written as: (cid:90) R dt (cid:48) (cid:90) R (cid:20) ∇ (cid:48) × ∇ (cid:48) × v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | (cid:21) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) = (C.2.18)= ∇ × ∇ × (cid:90) R dt (cid:48) (cid:90) ∂ R G ( r , t ; r (cid:48) , t (cid:48) ) v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | dV (cid:48) thus, proving the equation (31). C.3. Derivation of equation (38)
Using the mathematical identity ∇ G ( r , t ; r (cid:48) , t ) = ∇ (cid:48) G ( r , t ; r (cid:48) , t ), valid for Green’sfunction G ( r , t ; r (cid:48) , t ), one can rewrite the left hand side integral in equation (38) as: (cid:90) R dt (cid:48) (cid:90) R q s π(cid:15) v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | ∇ G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) = (C.3.19)= (cid:90) R dt (cid:48) (cid:90) R q s π(cid:15) v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | ∇ (cid:48) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) In Appendix Appendix B, we have derived two novel vector calculus identities. Subtract-ing vector identity (B.1) from vector identity (B.2) yields: (cid:90) V P ∇ ψdV = (cid:90) V ψ ∇ P dV + (cid:73) ∂V P ( ∇ ψ · d S ) − (cid:73) ∂V ψ ( d S · ∇ ) P (C.3.20)Using vector identity (C.3.20) we can rewrite equation (C.3.19) as: (cid:90) R dt (cid:48) (cid:90) R q s π(cid:15) v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | ∇ G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) = (C.3.21)= (cid:90) R dt (cid:48) (cid:90) R q s π(cid:15) G ( r , t ; r (cid:48) , t (cid:48) ) ∇ (cid:48) v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | dV (cid:48) + (cid:90) R dt (cid:48) (cid:73) ∂ R q s π(cid:15) v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | (cid:0) ∇ (cid:48) G ( r , t ; r (cid:48) , t (cid:48) ) · d S (cid:48) (cid:1) − (cid:90) R dt (cid:48) (cid:73) ∂ R G ( r , t ; r (cid:48) , t (cid:48) ) (cid:0) d S (cid:48) · ∇ (cid:48) (cid:1) q s π(cid:15) v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | Because surface ∂ R is an infinite surface the magnitude of position vector r (cid:48) ∈ ∂ R hasan infinite magnitude, | r (cid:48) | → ∞ . In that case, both right hand side surface integrals oversurface ∂ R vanish in equation (C.3.21). Hence, equation (C.3.21) becomes:32 R dt (cid:48) (cid:90) R q s π(cid:15) v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | ∇ G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) = (C.3.22)= (cid:90) R dt (cid:48) (cid:90) R q s (cid:15) G ( r , t ; r (cid:48) , t (cid:48) ) ∇ (cid:48) v s ( t (cid:48) )4 π | r (cid:48) − r s ( t (cid:48) ) | dV (cid:48) The operator ∇ (cid:48) does not affect vector v s ( t (cid:48) ) which is a function of variable t (cid:48) . Thus, wecan rewrite the Laplacian in the equation above as: ∇ (cid:48) v s ( t (cid:48) )4 π | r (cid:48) − r s ( t (cid:48) ) | = v s ( t (cid:48) ) ∇ (cid:48) π | r (cid:48) − r s ( t (cid:48) ) | = − v s ( t (cid:48) ) δ ( r (cid:48) − r s ( t (cid:48) )) (C.3.23)Inserting equation (C.3.23) into right hand side of equation (C.3.22) yields: (cid:90) R dt (cid:48) (cid:90) R q s π(cid:15) v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | ∇ G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) = (C.3.24) − (cid:90) R dt (cid:48) (cid:90) R q s (cid:15) v s ( t (cid:48) ) δ ( r (cid:48) − r s ( t (cid:48) )) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) which proves equation (38). C.4. Derivation of equation (60)
Equation (60) can be derived from equation (67) by taking the divergence of bothsides of this equation to obtain: ∇ · N ( r , t ) + ∇ · K ( r , t ) = ∇ · q s π(cid:15) r − r s ( t ) | r − r s ( t ) | = q s (cid:15) δ ( r − r s ( t )) (C.4.25)To find ∇ · K ( r , t ) we can write operator ∇ under the right hand side integral of equation(66) and then apply identity ∇ G ( r , t ; r (cid:48) , t ) = −∇ (cid:48) G ( r , t ; r (cid:48) , t ) to obtain: ∇ · K ( r , t ) = − c ∂∂t (cid:90) R dt (cid:48) (cid:90) R q s π(cid:15) (cid:20) ∇ (cid:48) × ∇ (cid:48) × v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | (cid:21) · ∇ (cid:48) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) (C.4.26)From here, using standard vector identity ∇ · ( ψ P ) = ∇ ψ · P + ψ ∇ · P and divergencetheorem it is obtained that: ∇ · K ( r , t ) = (C.4.27)= − c ∂∂t (cid:90) R dt (cid:48) (cid:73) ∂ R q s π(cid:15) (cid:20) ∇ (cid:48) × ∇ (cid:48) × v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | (cid:21) G ( r , t ; r (cid:48) , t (cid:48) ) · d S (cid:48) + 1 c ∂∂t (cid:90) R dt (cid:48) (cid:90) R q s π(cid:15) (cid:20) ∇ (cid:48) · ∇ (cid:48) × ∇ (cid:48) × v s ( t (cid:48) ) | r (cid:48) − r s ( t (cid:48) ) | (cid:21) G ( r , t ; r (cid:48) , t (cid:48) ) dV (cid:48) | r (cid:48) | →
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