PHYSICS ESSAYS 33, 4 (2020)
On a new field theory formulation and a space-time adjustment that predict the same precession of Mercury and the same bending of light as general relativity Larry M. Silverberga) and Jeffrey W. Eischen Department of Mechanical and Aerospace Engineering, North Carolina State University, Campus Box 7910, Raleigh, North Carolina 27695-7910, USA (Received 23 July 2020; accepted 8 November 2020; published online 3 December 2020) Abstract: This article introduces a new field theory formulation. The new field theory formulation recognizes vector continuity as a general principle and begins with a field that satisfies vector continuity equations. Next, independent of the new formulation, this article introduces a new space-time adjustment. Then, we solve the one-body gravitational problem by applying the space-time adjustment to the new field theory formulation. With the space-time adjustment, the new formulation predicts precisely the same precession of Mercury and the same bending of light as general relativity. The reader will find the validating calculations to be simple. The equations of motion that govern the orbital equations are in terms of Cartesian coordinates and time. An undergraduate college student, with direction, can perform the validations. VC 2020 Physics Essays Publication.[http://dx.doi.org/10.4006/0836-1398-33.4.489]
R esum e: Cet article pr esente une nouvelle formulation de la th eorie des champs. Cette formulation de la th eorie des champs reconna^ıt la continuit e vectorielle comme un principe g en eral et se base sur un champ qui satisfait les equations vectorielles de continuit e. De plus, et ind ependamment de cette nouvelle formulation, cet article introduit un nouvel ajustement de l’espace-temps. Enfin, nous r esolvons le proble`me gravitationnel a un corps, en appliquant la modification de l’espace-temps a la nouvelle formulation de la th eorie des champs. Avec cette modification de l’espace-temps, cette nouvelle formulation pr edit pr ecis ement la m^eme pr ecession de Mercure et la m^eme courbure de la lumie`re que la relativit eg en erale. Le lecteur trouvera les calculs de validation simples. Les equations du mouvement qui gouvernent les equations orbitales sont fonction de coor- donn ees du repe`re cart esien et du temps. Un etudiant de premier cycle, avec un peu d’orientation, pourra effectuer les calculs de validation soi-m^eme.
Key words: Newtonian Theory; Field Theory; General Relativity; Gravitation; Precession of Mercury; Bending of Light; Vector Continuity.
I. INTRODUCTION overlap. The formulations differ by their metrics, by their frames of reference, and by their coordinate systems. Scien- One describes mathematically any space-time field that tists have attempted to connect one theory to the other. For has flow lines that never begin, nor end, nor cross, as a four- example, as it pertains to the connection between NT and dimensional vector function that satisfies vector continuity GR, Atkinson1 examined GR in Euclidean terms and Monta- equations [see Appendix A for the development of the new nus2 developed a formulation of GR in so-called absolute field theory (FT) formulation]. The vector continuity equa- Euclidean space-time. Sideris3 asked fundamental questions tions are general equations that reduce to conservation laws, about the connections between NT and GR, and Ziefle4 mod- to wave equations, and to potential equations. Therefore, in ified NT by the introduction of “gravitons” in an attempt to retrospect, it was never a coincidence in Newtonian theory predict the precession of Mercury and the bending of light. (NT) that the gravitational potential satisfies potential equa- These and other researchers strengthened the belief that tions. It was never a coincidence in electromagnetic theory important connections exist between NT, EM, SR, and GR. (EM) that Maxwell’s equations describe fields that satisfy However, the connections continue to be confusing, and the wave equations within a given frame of reference. Vector work is not complete. continuity appears throughout NT, EM, and special relativity To address this confusion, let us now distinguish (SR). One of the novelties of the new FT formulation is that between two classes of physical theories: The class-A theory it begins with vector continuity. addresses physical behavior within a frame of reference, and NT, EM, SR, and general relativity (GR) claim different the class-B theory addresses physical behavior across two or territories in the landscape of physics, and their territories more frames of reference. This article focuses on the class-A theory, in particular, the new FT formulation and on a new a)[email protected] space-time adjustment that we apply to gravitation in the
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adjustment agree precisely with the corresponding predic- tions by GR. The comparisons with GR are based on the solution to Einstein’s field equation known as the Schwarzs- child solution—the solution that describes the gravitational field that surrounds a spherical mass, on the assumption that its electric charge, angular momentum, and universal cosmo- logical constant are all zero. The solution is a useful approxi- mation applicable to astronomical objects such as many stars and planets, including the earth and the sun. The reader will FIG. 1. The territories of analysis of FT and RT. find the new orbital equations and the steps involved in the validations to be simple. The orbital equations are quite sim- ple to program, and curvilinear coordinates are not involved. new FT formulation. We will refer to the class-B theory as An undergraduate college student, with direction, can per- relativity theory (RT). To be clear, by frame of reference, we form the validations. mean the frame in which one takes measurements and makes Appendix A contains the development of the new FT observations. This is quite different from what one means by formulation. Appendix B sets up the one-body gravitational the coordinate system. One can employ different coordinate problem and other mathematical details needed to solve the systems within a frame of reference, and one can employ the one-body gravitational problem by the new FT formulation. same coordinate system across different frames of reference Appendices C–G explain other properties of the FT formula- at the instant the frames are coincident. One employs the tion. Appendix H derives the effective radial potential in Galilean transformation between coordinate systems within a GR from the Schwarzschild solution. Finally, Appendix I frame of reference and the Lorentz transformation across reviews four-dimensional field theory. frames of reference regardless of the coordinate system. As depicted in Fig. 1, NT, EM, and GR nicely fit in FT, and SR II. INTRODUCTION TO THE NEW FT FORMULATION fits in both FT and RT. NT and EM treat time as a parameter within a three- Before proceeding to the validations, we briefly intro- dimensional frame of reference. SR was the first theory to duce the new FT formulation and compare it with the employ space-time within a four-dimensional frame of refer- present-day formulation. The present day formulation starts ence. It was also the first theory to address the differences in with a relativistic correction in inertial space of the law of behavior across different four-dimensional frames of refer- inertia in NT; the present-day field theory is a relativistic d ence. Both of these advancements were as significant to NT. It replaces the law Fr ¼ m dt vr ðr ¼ 1; 2; 3Þ, which physics as much as they were a source of confusion to those describes the interaction force vector responsible for chang- who did not appreciate the significance of the distinction. ing the state of a particle, with the relativistic law Like NT and EM, SR employed the inertial frame of refer- qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi d 2 ence. GR was the first theory to employ curved space-time Fr ¼ m dt vr= 1 ðÞv=c ðr ¼ 1; 2; 3Þ. Notice in the law and the first to make a number of predictions pertaining to of inertia, with and without the relativistic correction, gravity. GR was first to predict accurately the precession that one obtains the force vector by time differentiation. of orbiting bodies and the bending of light around massive Furthermore,qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi notice that the relativistic velocity components bodies. GR focused on a metric-changing transformation 2 between coordinate systems within a frame of reference, vr= 1 ðv=cÞ ðr ¼ 1; 2; 3Þ align with the first three commonly described as between flat and curved space-times components of a 4D unit vector that is tangent to the par- within the frame of reference. Relatively theory could ticle’s space-time path. The authors characterize the present- include parts of GR depending on how one classifies GR. day field theory as intending to accommodate the transition Ultimately, the distinction between a class-A theory and a from the particle to the field and from the spatial domain to class-B theory is important to one’s understanding of the the space-time domain. We now contrast the present-day for- relationships among NT, EM, SR, and GR, and in under- mulation with the new formulation. standing how they compare with the FT formulation devel- The new FT formulation begins differently. It replaces oped in this article. the spatial domain with the space-time domain. The The Merriam-Webster dictionary defines analysis as the kinematic quantities in the spatial domain differ from separation of a whole into its component parts. In this spirit, the corresponding quantities in the space-time domain. In the the new FT formulation begins by interpreting the whole as a spatial domain, one employs position vector components field that blankets space-time and satisfies vector continuity xr ðr ¼ 1; 2; 3Þ of particles, 3D velocity vector components equations. We refer to the whole as the 4D energy vector vr ðr ¼ 1; 2; 3Þ (time derivatives of xr), and 3D accelera- field or just as energy. We shall refer to the component parts tion vector components ar ðÞr ¼ 1; 2; 3 (time derivatives of the whole as fragments of energy. of vr). In contrast, in the space-time domain, the components The body of this article focuses on validating FT with of a field point are xr ðÞr ¼ 1; 2; 3; 4 : One employs 4D the space-time adjustment for gravitation. The validation position vector components x0r ðr ¼ 1; 2; 3; 4Þ of source consists of showing that the precession of Mercury and points, unit vector components er ðr ¼ 1; 2; 3; 4Þ (path the bending of light predicted by FT with the space-time derivatives of x0r), and 4D curvature vector components Physics Essays 33, 4 (2020) 491 kr ðr ¼ 1; 2; 3; 4Þ (path derivatives of erÞ: The path converts the two-body problem into a one-body problem derivative is a derivative with respect to the path increment (Appendix B). The following compares the treatments of the pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ds ¼ dx0rdx0r (repeated indices are summed) (Appendix I). one-body problem in FT and NT with its treatment in GR. Note that the term curvature that one uses in space-time As will be seen below, the treatments require a space-time geometry and that one uses in GR are different. With the adjustment. Table I lists the fragment of energy, energy, space-time representation of kinematic quantities in place, action force, and interaction force for the one-body gravita- the next step is to drape a 4D energy vector field over the tional problem and more generally the governing equations space-time domain. The energy vector has field lines that (change equations and laws of inertia). For the one-body never begin, nor end, nor cross, which one expresses mathe- gravitational problem, the orbital mechanics takes place in matically by 4D vector continuity equations. The vector con- the x1, x2 plane, the mass of the sun is M, and the mass of the tinuity equations reduce to wave equations (Appendix G) orbiting body is m, where the orbiting body is either Mercury and to conservation laws and potential equations or a photon. (Appendix I). Furthermore, 4D vector continuity motivates In general, GR starts with general covariance princi- the functional form of the fragment of energy, from which ples by which one determines a class of transformations one builds up the 4D energy vector Ar ðÞr ¼ 1; 2; 3; 4 between 4D Euclidean and 4D non-Euclidean (curved) (Appendix D). The form of the simple fragment of energy space-times. In the absence of electromagnetic effects, the a force vector in non-Euclidean space-time is zero; the parti- is Ar ¼ r er ðÞr ¼ 1; 2; 3; 4 , where a denotes the inten- 2 cle follows a geodesic. Note that we will not review here sity of the fragment and where r ¼ðxr x0rÞðxr x0rÞ (repeated indices are summed from 1 to 3). With this new the formidable apparatus of curved space-times. However, foundation, the next step is to define the action force vector with this apparatus, one obtains a metric for the gravita- and the interaction force vector. The components of the tional problem called the Schwarzschild metric. The action force vector by one fragment or the resultant action Schwarzschild metric leads to the so-called Schwarzschild solution. The Schwarzschild solution can be manipulated force vector by a set of fragments are Pr ðÞr ¼ 1; 2; 3; 4 to reveal an effective radial potentialb),5 and a correspond- (path derivatives of Ar), and the components of the 4D inter- action force vector by one fragment acting on another frag- ing interaction force vector in 4D Euclidean space-time, ment or the resultant interaction force vector acting on notwithstanding that the mathematical derivation lacks some reasoning6 (see Appendix H for the development of another fragment are Fr ¼ aPr ðÞr ¼ 1; 2; 3; 4 , where again a is the intensity of the fragment being acted on. Next, the effective radial potential from the Schwarzschild solu- one recognizes a unique property of path differentiation in a tion). Finally, one determines a 3D acceleration vector space-time domain; that the path derivative of any continu- from the interaction force vector. ous 4D vector field (in our case, the 4D energy vector) parti- tions mathematically into a conservative part and a A. Gravitation in the new FT formulation with no Lorentzian part. The conservative part corresponds to space-time adjustment mechanical action and the Lorentzian part to electromagnetic In the one-body gravitational problem, the field consists action (Appendix E). Finally, in the new FT formulation, we of just one fragment of energy. TABLE Ia gives its compo- employ path differentiation to set up the change equa- nents. The term A given there is the magnitude of the vector tion kr ¼ Pr ðÞr ¼ 1; 2; 3; 4 , where kr are the compo- field with components As (s ¼ 1, 2, 3, 4). Next, an action nents of the curvature vector of a fragment of energy, and force is determined. Then, one determines 4D curvature vec- where Pr are the components of the resultant action force tor components ks (s ¼ 1, 2, 3, 4) by substituting the action vector that acts on the fragment of energy (Appendix A). force vector into the change equation. With respect to the two formulations, first, note that the change equation maps to the relativistic law of inertia. One B. Gravitation in the new FT formulation with a finds that the change equation and the relativistic law of iner- space-time adjustment tia map one-to-one by comparing the time derivatives of the relativistic linear momentum components with the force Table Ib gives the components As (s ¼ 1, 2, 3, 4) of the components in the new FTformulation. The force compo- vector field of a stationary fragment using a space-time nents FEr and FNr, corresponding to the present-day formula- adjustment that we introduce for gravitation. We first apply tion andq theffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi new formulation, map one-to-one when it here to the new FT formulation and later to NT. The space-time adjustment adjusts the relationship between the F ¼ F 1 ðvÞ2: Next, note that the new formulation Er Nr c speed of the source point of the orbiting fragment upon replaces the particle and the wave with the field along which the field of a stationary fragment acts, such as with its fragment of energy, and introduces a connection Mercury or a photon, and the orbiting fragment’s corre- between mechanics and electromagnetism through path sponding angular momentum. Start by considering the circu- differentiation. lar motion of an orbiting fragment. In FT, the following expression is exact for circular motion: III. THREE FORMULATIONS
The fundamental problem of gravitation is a two-body b)Wikipedia, Two-body problem in general relativity, [Online]. Available: problem. Whether employing FT, NT, or GR, one https://en.wikipedia.org/wiki/Two-body_problem_in_general_relativity 492 Physics Essays 33, 4 (2020)
TABLE I. Gravitation in FT and NT.
a) Gravitation in the new FTformulation with no space-time adjustment (Appendices A and B)
Fragment of energy Mc2 A ¼ 0; A ¼ 0; A ¼ 0; A ¼ 1 2 3 4 r Energy Mc2 A ¼ r Action force GM x GM x P ¼ 1 ; P ¼ 2 ; P ¼ 0; P ¼ 0 1 c2 r3 2 c2 r3 3 4
Change equations P1 ¼ k1; P2 ¼ k2; P3 ¼ k3; P4 ¼ k4
2 2 2 2 Interaction force F1 ¼ mc P1; F2 ¼ mc P2; F3 ¼ mc P3; F4 ¼ mc P4
b) Gravitation in the new FT formulation with the space-time adjustment (Appendices A and B)
Fragment of energy Mc2 1 A1 ¼ 0; A2 ¼ 0; A3 ¼ 0; A4 ¼ r v 2 1 c ! Energy 2 1=2 Mc2 h 1 v 2 A ¼ Mc2 ; h ¼ rmvb; b ¼ 1 r mc r3 c Action force GM x GM h 2 x GM x GM h 2 x P ¼ 1 þ 3 1 ; P ¼ 2 þ 3 2 P ¼ P ¼ 0 1 c2 r3 c2 mc r5 2 c2 r3 c2 mc r5 3 4
Change equations P1 ¼ k1; P2 ¼ k2; P3 ¼ k3; P4 ¼ k4
2 2 2 2 Interaction force F1 ¼ mc P1; F2 ¼ mc P2; F3 ¼ mc P3; F4 ¼ mc P4
c) Gravitation in NT with no space-time adjustmenta
Potential energy GMm V ¼ r Interaction force GMm GMm F ¼ x ; F ¼ x ; F ¼ 0 1 r3 1 2 r3 2 3
Law of inertia F1 ¼ ma1; F2 ¼ ma2; F3 ¼ ma3
d) Gravitation in NT with the space-time adjustment [or by GR,b (Appendix H)] ! Fragment of energy GMm v 2 V ¼ 0; V ¼ 0; V ¼ 0; V ¼ 1 þ 1 2 3 4 r c Potential energy GMm GMH2 V ¼ ; H ¼ rmv r c2mr3 Interaction force GMm 3GMH2 GMm 3GMH2 F ¼ þ x ; F ¼ þ x ; F ¼ 0 1 r3 c2mr5 1 2 r3 c2mr5 2 3 law of inertia F1 ¼ ma1; F2 ¼ ma2; F3 ¼ ma3 aWikipedia, “Newton’s law of universal gravitation,” [Online]. Available at https://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation bWikipedia, “Two-body problem in general relativity,” [Online]. Available at https://en.wikipedia.org/wiki/Two-body_problem_in_general_relativity
1 h 2 1 mrv fragment is constant. When one multiplies the magnitude sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 1 2 ¼ 1 þ 2 ; where h ¼ ; Mc of the stationary fragment by the right side of the v mc r v 2 r 1 1 first expression in Eq. (1), one finds that the mathematical c c expression of the stationary fragment splits into two parts, (1) with h being constant (see A in Table Ib). The second step of the space-time adjustment is to replace the left side of where h is angular momentum, m is mass, and r is the dis- Eq. (1) with the right side of Eq. (1). This eliminates the tance to the source point. To verify that this expression is dependence on v from the expression for A, leaving the exact, substitute h into the right side of the expression and expression in terms of its independent spatial coordinates then combine the two terms. The space-time adjustment alone. Using this expression, we determine the action force 2 1 consists of first multiplying the magnitude Mc r of the sta- at any point along the trajectory wherein the motion is not tionary fragment in Table Ia by the left side of the first circular expression in Eq. (1). Given that v is constant, one preserves Next, we consider that the sun and Mercury are not vector continuity (see A4 in Table Ib). In the gravitational merely simple fragments of energy but that each is a bundle problem, the velocity v of the orbiting fragment is not of fragments. The path derivative of the fragments of energy constant, whereas the angular momentum of the orbiting of each bundle has a mechanical part and an electromagnetic Physics Essays 33, 4 (2020) 493 part [see Eq. (A6)]. We assume within each bundle that the the right side of Eq. (2), which eliminates the dependence on electromagnetic part cancels. The components of the action v and expresses V in terms of its independent spatial coordi- force acting on the orbiting fragment (Mercury or a photon) nates alone. We determine the interaction force by the sun are Pr ¼ Sr ¼ @A=@xr; ðÞr ¼ 1; 2; 3; 4 . Holding h con- from Fr ¼ @V=@xr; ðÞr ¼ 1; 2; 3 ; holding H constant. stant, while taking the derivatives, we get Pr (r ¼ 1, 2, 3, 4) Again, we used the resulting expression to determine the in Table Ib. Next, we consider the interaction force on the interaction force at any point along the trajectory wherein orbiting fragment by the action of the sun. From Eq. (A10), the motion is not circular (see Fr (r ¼ 1, 2, 3, 4) in Table Id). we get Fr (r ¼ 1, 2, 3, 4) in Table Ib. Note that we no longer Again, we no longer guarantee vector continuity to be satis- guarantee vector continuity to be satisfied after making the fied after making the adjustment. adjustment. One determines the components of a 4D curva- In this article, the reader will discover, had Newton ture vector ks (s ¼ 1, 2, 3, 4) of the orbiting fragment by added the previously mentioned second term in Eq. (2) to the substituting the mechanical part of its action into the change gravitational potential, that he would have predicted the equation in Table Ib. This yields orbital equations that we same precession of Mercury as in GR. On the other hand, he will solve numerically. The reader will discover that the use would not have been inclined to pursue that because during of the second term on the right side of Eq. (1), which is the time of Newton scientists had not yet observed the pre- responsible for the adjustment, is also responsible for pre- cession of Mercury, they had not yet developed the concept dicting the same precession of Mercury and bending of light of vector continuity, nor had scientists yet developed the as in GR. apparatus of curved space-times.
C. Gravitation in NT with no space-time adjustment IV. RESULTS Referring to Table Ic, NT with no space-time adjustment begins with a potential energy function V or with interaction This section solves two one-body gravitational problems—the precession of Mercury in its orbit around the force vector components F (s ¼ 1, 2, 3). A gradient vector s sun and the bending of light grazing the surface of the sun— relates the two by F ¼ @V=@x ðÞs ¼ 1; 2; 3 : Then, one s s each by FT and by NT with and without the adjustments. determines 3D acceleration vector components as (s ¼ 1, 2, 3) by substituting the interaction force vector into the gov- Table II lists the physical constants used in both problems erning law of inertia. Note that the potential energy function and Table III lists the accelerations used to compute the in NT with no space-time adjustment is not associated with a dynamic responses. potential energy vector function. The potential energy func- A body orbits the sun in the problem of the precession of tion for gravitation will recognize that V is associated with a Mercury and in the problem of the bending of light. In the potential energy vector function. bending of light problem, the body is a massless point travel- ing at the speed of light. One calls such a point a photon. In D. Gravitation in NT with a space-time adjustment TABLE II. Physical constants. We begin with NT and now associate the potential energy a V of a body with a potential energy vector function like in the Sun new FT formulation. Next, we apply the space-time adjust- M Mass of the sun 1.989 1030 kg ment to the one-body gravitational problem in NT. Again, the rs Radius of the sun 696 000 000 m space-time adjustment considers the relationship between the b,c speed of the mass center of the orbiting body and its corre- Mercury sponding angular momentum and, again, we start by consid- m Mass of mercury 3.3 1023 kg 10 ering the circular motion of the orbiting body. In NT, the rp Perihelion radius of Mercury 4.6 10 m 10 following expression is exact for circular motion: rb Aphelion radius of Mercury 6.982 10 m A Semimajor axis of Mercury 57.91 109 m 2 v 2 H 1 e Eccentricity of the orbit 0.20566 1 þ ¼ 1 þ where H ¼ rmv; (2) 3 c mc r2 vp Perihelion velocity 58.98 10 m/s T Orbital period 87.969 Earth days ¼ 7 600 530 s where H is the angular momentum of the orbiting body. The Otherd,e space-time adjustment consists of first multiplying the gravi- -11 3 2 tational potential GM 1 of the stationary body in Table Ic G Gravitational constant 6.674 10 m /kg-s r 8 by the left side of the first expression in Eq. (2). Given that v c Speed of light 2.99 10 m/s is constant, one preserves vector continuity (see V4 in aWikipedia, Sun, [Online]. Available: https://en.wikipedia.org/wiki/Sun Table Id). Again, in the one-body gravitational problem, the bWikipedia, Mercury (planet), [Online]. Available: https://en.wikipedia.org/ speed v of the orbiting body is not constant whereas its angu- wiki/Mercury_(planet) lar momentum is constant. Again, when one multiplies the cDavid R. Williams, NASA Goddard Space Flight Center, https:// gravitational potential GM 1 by the right side of the first nssdc.gsfc.nasa.gov/planetary/factsheet/mercuryfact.html r dWikipedia, Gravitational constant, [Online]. Available: https://en.wikipe- expression in Eq. (2), one finds that the expression splits into dia.org/wiki/Gravitational_constant two parts, with H being constant (see V in Table I). Like in eWikipedia, Speed of light, [Online]. Available: https://en.wikipedia.org/ FT, the second step is to replace the left side of Eq. (2) with wiki/Speed_of_light 494 Physics Essays 33, 4 (2020)
TABLE III. Calculating the dynamic response.
a) New FT formulation without the adjustment 2 3 2 0 1 6 v1 v1 v2 7 GM x1 !6 1 7B C 2 6 c c c 7B r2 r C a1 m v 6 7B C ¼ 1 6 2 7B C a2 l c 6 v v v 7@ GM x2 A 4 1 2 1 2 5 c c c r2 r mM l ¼ m þ M b) New FT formulation with the adjustment 0 0 1 1 2 3 2 2 v v v B GM h 1 x1 C 6 1 1 2 7 @1 3 A !6 1 7B 2 þ 2 C 2 c c c B r mc r r C a m v 6 7B C 1 ¼ 1 6 7B 0 1 C a 6 2 7B 2 C 2 l c 6 v1 v2 v2 7B GM h 1 x2 C 4 1 5@ @1 þ 3 A A c c c r2 mc r2 r
h ¼ rmvb
c) NT without the adjustment a1 m GM x1 ¼ 3 a2 l r x2
d) NT with the adjustment a m GM 3GMH2 x 1 ¼ þ 1 a2 l r3 c2m2r5 x2 H ¼ rmv each problem, we determined the trajectories by numerical 6pGðM þ mÞ du ¼ ; integration of the accelerations of the orbiting body. We c2Að1 e2Þ adopted a numerical approach instead of an analytical 7 ¼ 5:047 10 rad=orbit approach, because analytical results do not yet exist for all of the cases considered and because the numerical approach ¼ 0:104093 arc sec=orbit simplifies the verification of the results. We used the ODE ¼ 43:2 arc sec=century: solver DVERK78 within MAPLESOFTTM.c) The number of Again, the numerically integrated trajectories began at digits of accuracy was set to Digits ¼ 20. DVERK78 finds a numerical solution of state equations using a seventh-eighth Mercury’s perihelion with the initial conditions x1 ¼ rp; order continuous Runge-Kutta method. x2 ¼ 0; v1 ¼ 0; v2 ¼ vp and continued to its next perihelion, Next, consider the initial conditions corresponding to all after a little more than a full revolution (see Fig. 2). In order of the cases in Table III. The initial angular momentum in to determine the precise location of the next perihelion, the qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁ time derivative of the orbit radius dr=dt ¼ðx x_ þ x x_ Þ=r 2 1 1 2 2 FT is h ¼ mrpvp= 1 vp=c : (The subscript p stands for was monitored, and the time when its value crossed zero was perihelion.) In NT, the initial angular momentum is H ¼ determined. At that instant, the predicted angle of precession rpmvp: In all cases, the parameters in all cases are calculated was calculated numerically from du ¼ x2=rp. Table IV shows the precession angles numerically determined by FT from the data in TablepffiffiffiffiffiffiffiffiffiffiffiffiffiffiII. Also a1 ¼ v_1; a2 ¼ v_2; v1 ¼ x_1; 2 2 and NT with and without the adjustments. The values v2 ¼ x_2; and v ¼ v1 þ v2 .
A. Precession of Mercury GR predicted successfully the so-called anomalous pre- cession of Mercury, over and above the precession caused by the other planets and solar oblateness. From GR, the analyti- cally determined precession of Mercury isd),e)
c)MAPLESOFT, [Online]. Available: https://www.maplesoft.com/ d)Wikipedia, “Two-body problem in general relativity,” [Online]. Available: https://en.wikipedia.org/wiki/Two-body_problem_in_general_relativity e)Wikipedia, “Schwarzschild geodesics,” [Online]. Available: https://en. wikipedia.org/wiki/Schwarzschild_geodesics FIG. 2. (Color online) Trajectory of Mercury. Physics Essays 33, 4 (2020) 495
TABLE IV. Angle of precession for each formulation.
Precession angle
Formulation rad/orbit arc-sec/century a) FT without the adjustment 0 0 b) FT with the adjustment 5.047 10 7 43.2 c) NT without the adjustment 0 0 d) NT with the adjustment 5.047 10 7 43.2
predicted numerically by FT and NT with the adjustments agree to three decimal places with the celebrated analytical result from GR. Without any adjustment, FT and NT predict FIG. 3. (Color online) The precession angle in FT with the adjustment. a precession angle of zero. Figure 3 shows the graph of the precession angle of acceleration components of a photon under the action of a Mercury’s orbit calculated by FT with the adjustment. It is stationary source of mass M. We determined the angle of shown as a function of dr/dt in the neighborhood of the com- light bending by letting the photon speed approach the pletion of the first orbit, and it shows the zero crossing at speed of light in a limiting process. In NT, we numerically 5.047 10 7 rad. We obtained the same result by NT with integrated the photon trajectory starting at the Sun’s radius the adjustment; the two graphs overlap. Note that we also with a velocity approaching the speed of light using x ¼ directly coded the Schwarzschild geodesic equationsf) to 1 r ; x ¼ 0; v ¼ 0; v ffi c as initial conditions and continued determine the orbit using the same numerical method. s 2 1 2 forward in time. The angular momentum term h=m ¼ r v b The geodesic equations also led to a precession angle of s 2 or H=m ¼ r v was calculated at time t ¼ 0. 5.047 10 7 rad. This served as a numerical check, because s 2 Table V shows the bending angle of light numerically one derives the GR effective potential from the Schwarzs- determined by FT with the adjustment for increasing initial child geodesic equations (Appendix H). photon speeds. For an initial speed equal greater than or equal to 0.9999c, the predicted numerical value agrees to three decimal places with the celebrated analytical result. B. The bending of light Figure 5 shows the graph of the simulated bend angle The classic problem of analytically determining the versus time for FT with the adjustment for the case bending of light by GR predicts the angle of bending as7,g) v ¼ 0.99999c. From the graph, it is clear how the bend angle approaches the value of 8.534 10 6 rad asymptotically as 4GM 6 the photon travels away from the sun. dN ¼ 2 ¼ 8:534 10 rad ¼ 1:760 arc sec; c rp Table VI shows that FT without the adjustment predicts a bend angle of precisely zero. In contrast, NT without the where rp is the “distance of closest approach” (the perihe- adjustment predicts a bend angle that is precisely 50% lion) and, for the purposes of simulation, has been set equal smaller than the bend angle that GR predicts. This result is to the radius rs of the sun (see Fig. 4). In order to predict the well-known. Notably, with the space-time adjustment, FT path of a photon orbiting the sun, we considered the model predicts precisely the same bending of light that GR predicts of a photon using the equations of motion in Table III. First, in addition to precisely the same precession of Mercury that we set l ¼ mM=ðm þ MÞ to m after which the m cancelled GR predicts, per Table IV. Furthermore, it is interesting from the equations of motion. In FT with the adjustment, by that NT with no adjustment and with the adjustment, setting the photon speed to the speed of light, one would get a divide by zero error. The divide by zero arises in the term ðÞh=mc 2 ¼ ðÞr v =c 2 1 , when v is set equal to c. s 2 1 ðv=cÞ2 The expressions for the acceleration components become indeterminate (0/0), so we take the limit as v approaches c instead of evaluating v at c. One can also evaluate v at c by a canceling the two 1 ðv=cÞ2 terms to obtain 1 ¼ a2 ÀÁ 3GM v2 v1 3 v2=c r5 x1 c x2 c for the expression of the v1=c f)Wikipedia, Schwarzschild geodesics, [Online]. Available: https://en.wiki- pedia.org/wiki/Schwarzschild_geodesics g)Wikipedia, Gravitational lens, [Online]. Available: https://en.wikipe- dia.org/wiki/Gravitational_lens FIG. 4. (Color online) Bending of light. 496 Physics Essays 33, 4 (2020)
TABLE V. Predicted bending of light around the sun for FT with the TABLE VI. Bend angle predicted by each formulation. adjustment as a function of photon speed. Bend angle Photon velocity dN dN % difference at perihelion (rad) (arc-sec) from GR % difference Formulation rad arc-sec from GR 0.99c 8.620 10 6 1.778 þ1.008 0.999c 8.542 10 6 1.762 þ0.094 a) FT without the adjustment 0 0 — 6 0.9999c 8.534 10 6 1.760 0 b) FT with the adjustment 8.534 10 1.760 0 6 0.99999c 8.534 10 6 1.760 0 c) NT without the adjustment 4.266 10 0.878 50.0 d) NT with the adjustment 1.280 10 5 2.640 þ50.0
be the only way to predict accurately the precession of Mer- cury and the bending of light. This article shows with a space-time adjustment that FT predicts the precession of Mercury and the bending of light precisely (and more easily than by GR). Finally, note that the authors confirmed empiri- cally the space-time adjustment but did not rigorously derive it. The adjustment appears to be a space-time curvature term that accommodates light bending. However, without a rigor- ous derivation, for the time being, one must leave the space- time adjustment open to different interpretations. Pertaining to the development of the new FT formula- FIG. 5. (Color online) Determining the light bending angle in FT with the tion, the reader may find the vector continuity requirement in adjustment (v ¼ 0.99999c). FT of the vector field of energy and of the fragments of energy to be an appealing way to begin the formulation of a respectively, predicts precisely 50% less and then precisely field theory from both axiomatic and intuitive perspectives. 50% more bending of light than GR predicts. Axiomatically, FT parallels the tenets of analysis by begin- ning with a precise definition of energy over space-time (the V. SUMMARY whole) and of fragments of energy (the component parts) and then proceeds deductively from there. The theory pro- This article introduced a new FT formulation that begins ceeds with precise definitions of action, of interaction, from with a vector field of energy that satisfies vector continuity which the theory formulates change equations (which reduce equations and that one builds up from fragments of energy. to Newton’s law of inertia). Intuitively, there is an important The theory is explained in detail in Appendix A. The article connection between the fragment of energy and the particle also introduced a space-time adjustment for the one-body and wave conceptions. In NT and EM, the particle concep- gravitational problem. The article showed that the new FT tion corresponds to energy (mass and charge). It exists at a with the space-time adjustment predicts the same precession source point and nowhere else. The wave conception corre- of Mercury and the same bending of light that GR predicts. sponds to energy existing everywhere else while not explic- Independent of these results, this article also showed, with- itly addressing the source point. The particle and the wave out the adjustment, that FT and NT predict no precession of conceptions are existential opposites, and the fragment con- Mercury. In the bending of light problem without the adjust- ception is a union of the particle and the wave conceptions. ment, NT predicts a bend angle that is precisely 50% less Finally, in future work scientists may confirm that the new than the bending angle of light that GR predicts and, with the FT formulation is applicable to many more problems, includ- adjustment, a bend angle that is precisely 50% greater than ing those problems that are at different scales and those prob- the bending angle of light that GR predicts. The equations of lems that have complex geometries. Insightful scientists may motion are quite simple to program, and curvilinear coordi- also find that the new FT formulation unifies the representa- nates are not involved at all. Undergraduate students can tion of physical behavior. The fragment of energy is poten- employ the adjustment of the gravitational law to predict tially a universal conception at a scale of interest of an themselves the correct precession angle of Mercury and the element of energy that joins the other universal conceptions correct bending angle of light. They would simply numeri- already found in physics—that of space, of time, and of cally integrate the orbital equations given in Table III, energy as a whole. Only the future will tell. acquiring the physical constants from Table II. The comparisons presented in this article highlight the intimate connection that exists between FT, NT, and GR. GR APPENDIX A: NEW FT FORMULATION was the first theory to employ curved space-time and the first to correctly predict the precession of orbiting bodies and the Notation: Helvetica-Roman bold font for space-time bending of light around massive bodies. Both of these facts vectors (e.g., A), Times-Roman bold font for spatial vectors together led some to believe, quite naturally, that GR must (e.g., A), and italics for scalars (e.g., A) Physics Essays 33, 4 (2020) 497
Terminology (Symbols) [Universal units (in terms of length L and time T)]
Action (Pr) [1/L] Path derivative of a fragment of energy Charge ðqÞ [L] Electromagnetic intensity Conservation Satisfies conservation law (derivable from vector continuity equations) Continuity Short for scalar continuity or vector continuity
Curl (krs) [1/L] Space-time matrix for electromagnetic action
Curvature (kr) [1/L] Path derivative of space-time unit vector Electromagnetism Discipline of macroscale science describing electrical and magnetic interactions between particles of mass and charge
Electrical field (Er) [1/L] Spatial vector function from the stationary charges of particles
Energy (A or Ar) [1] Short for scalar field of energy or vector field of energy Field Spatial function or space-time function
Force (Pr or Fr) Space-time action or space-time interaction
Fragment of energy (Ar) Element of energy that satisfies vector continuity Inertial reference frame Reference frame for which fragment moves in a straight line in absence of an action upon it Intensity ðaÞ[L] Corresponds to the fragment of energy
Interaction (Fr) [1] Resultant action on a given fragment by other fragments multiplied by the negative of the intensity of the fragment
Magnetic field (Br) [1/L] Spatial vector function from the moving charges of particles Mass ðmÞ [T2/L] Mechanical intensity Matter ðaÞ [L] Intensity of a fragment of energy Mechanics Discipline of macroscale science describing mechanical interactions between particles of mass Particle Discrete point with properties of mass and/or charge Path derivative d=ds With respect to a path increment ds Potential Scalar or vector function that satisfies potential equations (derivable from vector continuity equations) Reference frame Corresponds to measurements and a perspective, works with different coordinate systems Scalar continuity Property of a function changing incrementally with incremental changes in its independent variables (also called ordinary continuity)
Source point (xar)[L] Space-time center point of a fragment or a test fragment Space-time One to three spatial coordinates and one temporal coordinate Substance Energy or matter
Unit vector (er) [1] Corresponds to a source point of a fragment or a test fragment Vector continuity Vector function whose flow lines neither begin, nor end, nor cross Wave Scalar or vector function that satisfies wave equations (derivable from vector continuity equations)
This appendix develops the new field theory formulation equations [Eq. (I18c)], and to conservations laws [Eq. (I22)]. (FT) independent of NT (Newtonian Theory), EM (Electro- Furthermore, space-time vector continuity applies to systems magnetism), SR (Special Relativity), and GR (General Rela- that undergo conformal transformations in space-time. tivity). The development employs four-dimensional field The rectangular components of the space-time vector theory. Appendix I reviews the elements of four-dimensional field and of its gradient vectors are: vector fields.
A1 ¼ Ae1; A2 ¼ Ae2; A3 ¼ Ae3; A4 ¼ Ae4; 1. Vector continuity (A1a) Begin with a 4D vector field that satisfies vector conti- @As @As @As @As nuity in 4D space [Eqs. (I16) and (I18)]. The space-time field ; ; ; ; ðÞs ¼ 1; 2; 3; 4 ; (A1b) @x @x @x @x lines of the 4D vector field neither begin, nor end, nor cross. 1 2 3 4 Represent a general 4D vector field that satisfies vector con- where e1, e2, e3, and e4 are unit vector components. The tinuity as a linear combination of admissible 4D vector differential forms of the vector continuity equations are fields, called fragments of energy. As admissible functions, [Eqs. (I16) and (I18)] the fragments satisfy vector continuity equations but not the boundary conditions of any particular problem. As building @A1 @A2 @A3 @A4 blocks, the fragments of energy depart from the particle and 0 ¼ þ þ þ ; (A2a) @x @x @x @x the wave conceptions. The particle is a source located along 1 2 3 4 a space-time line and not elsewhere, whereas the wave is 2 2 2 2 missing the source and it exists everywhere else. In contrast, @ As @ As @ As @ As 0 ¼ 2 þ 2 þ 2 þ 2 ; ðÞs ¼ 1; 2; 3; 4 : the fragment of energy has both a source point and it exists @x1 @x2 @x3 @x4 everywhere else. (A2b) Space-time vector continuity and the fragment of energy are generalizations of other concepts. The vector continuity The integral forms of the vector continuity equations are equations reduce to potential equations [Eq. (I18a)], to wave [Eqs. (I16) and (I18)] 498 Physics Essays 33, 4 (2020)
TABLE VII. Metrics.
Space Metric Coordinates
4D flat 2 2 2 2 2 2 2 2 Rectangular ds ¼ dx1 þ dx2 þÀÁdx3 þ dx4 ðdx4 ¼ c dt Þ ds2 ¼ dr2 þ r2 dh2 þ sin2hdu2 c2dt2 Spherical 2 2 2 2 3D flat ds ¼ dx1 þ dx2 þ dx3 Rectangular 1 4D curved r 1 r Spherical (1-body) ds2 ¼ 1 s dr2 þ r2 ðdh2 þ sin2hdu2Þ c2 1 s dt2 r r 1Schwarzschild metric.
þ 4. General form of the solution 0 ¼ ðÞA1n1 þ A2n2 þ A3n3 þ A4n4 dV; (A3a) þAny linear combination of admissible vector fields is an @As @As @As @As admissible vector field 0 ¼ n1 þ n2 þ n3 þ n4 dV; @x1 @x2 @x3 @x4 As ¼ aaAas þ abAbs þ acAcs þÁÁÁ: (A4) ðÞs ¼ 1; 2; 3; 4 : (A3b) Refer to the coefficients aa; ab; etc., in the linear combina- Equations (A2a) and (A3a) correspond to Eq. (A1a), and tion as the intensities of the fragments. The vector field Eqs. (A2b) and (A3b) correspond to Eq. (A1b). By counting As ð¼ 1; 2; 3; 4Þ satisfies vector continuity. However, as field lines, the vector continuity equations ensure that field stated earlier, it does not naturally satisfy boundary condi- lines neither begin, nor end, nor cross. Equation (A3a) counts tions. A uniqueness theorem can determine the boundary the number of field lines of As ðs ¼ 1; 2; 3; 4Þ that enters conditions that produce a unique vector field solution a domain D as negative, and the number that leaves it as As ðs ¼ 1; 2; 3; 4Þ (see Appendix C). positive. In this way, the integral around the boundary of D is zero when every field line that enters D also leaves it. 5. Action and interaction When Eq. (A3a) is satisfied over an arbitrary domain contained in a given region, one says that the vector field Consider the vector field As ðs ¼ 1; 2; 3; 4Þ of a single fragment of energy. Its source point is located at As ðs ¼ 1; 2; 3; 4Þ satisfies vector continuity in that region. x0s ðÞs ¼ 1; 2; 3; 4 ; and the path of its source point is along 2. Metrics the unit vector components es ¼ dx0s=ds ðs ¼ 1; 2; 3; 4Þ [Eq. (I4)]. If the fragment is initially at rest, it will remain at Physical theories differ by the metrics they employ. rest in the absence of being acted on by another fragment Table VII shows several of the important metrics. For FT, only if the change is observed in an inertial frame of refer- we shall use the 4D flat metric and, in NT, one uses the 3D ence. Only in the inertial frame does the change in the space- flat metric. In the one-body gravitational problem, GR uses time direction of the fragment not result from itself. the given 4D curved metric. Action: The action by a fragment is the path derivative
3. Fragments of energy dA P ¼ s : (A5) s ds Table VIII lists two analytical forms of 4D vector fields As (s ¼ 1, 2, 3, 4) that satisfy space-time vector continuity Action is a 4D vector field that mathematically decomposes [Eqs. (A2) and (A3)]. As shown, the fragment of energy into a mechanical part and an electromagnetic part assumes the form of the reciprocal of spatial distance. One can show that this is not coincidental (see Appendix D). Ps ¼ Ss þ Ts; (A6)
TABLE VIII Fragment of energy and wave.
Fragment of energy and wave (admissible vector fields)
Type Ae1 e2 e3 e4
Fragment of energy1 1 0001 r 1 Wave AuðÞ e1ðÞu e2ðÞu e3ðÞu e4ðÞu 1In fundamental units
2 2 2 2 r ¼ ðÞx1 xa1 þ ðÞx2 xa2 þ ðÞx3 xa3 2 2 2 2 u ¼ u1x1 þ u2x2 þ u3x3 þ u4x4; u1 þ u2 þ u3 þ u4 ¼ 0; u4 ¼ 6i x1 ¼ u1ct; x2 ¼ u2ct; x3 ¼ u3ct ur ðr ¼ 1; 2; 3; 4Þ are constants. xa1; xa2; and xa3 are the components of source point of the fragment of energy. Physics Essays 33, 4 (2020) 499
@A disconnected from the other, was following a straight space- Ss ¼ ; (A7) @xs time line as though it belonged to a separate, unconnected world. In the disconnected state, the action forces on the @A1 @As @A2 @As fragments were set to zero. The change equations reconnect Ts ¼ e1 e2 @xs @x1 @xs @x2 the fragments to each other to reform the whole. One @A3 @As @A4 @As expresses the change in the straight-line path of the source e3 e4: (A8) @xs @x3 @xs @x4 point of a fragment by curvature vector components ks ¼ des ds ðs ¼ 1; 2; 3; 4Þ [Eq. (I8)]. The change equations are
Equation (A7) is the mechanical part (notice its conser- dAbs dAcs dAds þ þ þÁÁÁ¼kas ðÞs ¼ 1; 2; 3; 4 : (A12) vative form), and Eq. (A8) is the electromagnetic part (notice dsb dsc dsd its Lorentzian form). For the mathematical derivation of Eqs. (A6)–(A8), see Appendix E. Equation (A12) sets the resultant action force acting on Interaction: Action is a vector field that is by a frag- a fragment to the curvature vector of the fragment’s ment, also called an action force, and interaction is a vector source point. Under the nonrelativistic assumption, the that acts on a fragment, also called an interaction force. Con- change equation reduces to Newton’s law of inertia (see sider fragments a, b, c, d, etc. Define the interaction force Appendix F). Newton’s law of inertia is acting on fragment a by fragments b, c, d, etc., to be the resultant of the action forces by fragments b, c, d, etc., evalu- Fbs þ Fcs þ Fds þÁÁÁ¼ maaas ðÞs ¼ 1; 2; 3 : (A13) ated at the source point of fragment a multiplied by the nega- tive of the intensity aa of fragment a. Thus, the resultant of The following describes momentum and energy in the FT the action forces by fragments b, c, d, etc., evaluated at the formulation, each of which reduce to their classical counter- center of fragment a is parts under the nonrelativistic speed assumption [Eq. (I8c)].
Pasja ¼ ðÞjPbs þ Pcs þ Pds þÁÁÁ a: (A9) 7. Momentum and energy The interaction force acting on fragment a by fragments b, c, Based on the preceding development, we now examine d, etc., is momentum and energy in the two-body gravitational prob- 2 lem (see Appendix B). mac mechanical; Fas ¼ aaPasja; aa ¼ Linear momentum: From Eqs. (A5)–(A7), (A10), and qa electromagnetic: (A12) (A10) Fr One sees in Eq. (A10) that P j (s ¼ 1, 2, 3, 4) is a resultant ¼ mkCr ¼ makar þ mbkbr; as a c2 of other fragments. Also, note that one can drop the Lr (A14a-c) evaluation sign without causing confusion because P ðs ¼ ¼ mer ¼ maear þ mbebr; as ic 1; 2; 3; 4Þ in Eq. (A10) is never evaluated at its own source mx ¼ m x þ m x ; point. Furthermore, notice that one associates the fragment’s Cr a ar b br intensity a as with either mechanical behavior or as with a where m ¼ m þ m . Equation (A14a) defines resultant force electromagnetic behavior. The association depends on the a b components F external to the system, a resultant mass m nature of the change in Eq. (A6). r internal to the system, and curvature components kCr of the Finally, by their forms (see Table VIII), it follows that mass center of the system. Equation (A14b) defines linear mechanical interaction forces between any pair of fragments momentum components L of the system, and Eq. (A14c) are central forces. They act in equal and opposite directions r defines coordinates xCr of the system’s mass center. Under along the line between their source points. Designating two the nonrelativistic speed assumption, Eqs. (A14a)–(A14c) fragments by a and b, one expresses these properties as reduce to their classical counterparts. Angular momentum: The cross product is a 3D opera- Fas þ Fbs ¼ 0 ðÞs ¼ 1; 2; 3; 4 : tion. The more general form utilizes permutation symbols Fas ¼ CxðÞas xbs for some C 6¼ 0 [Eq. (I5c) and Fig. 6], in terms of which a resultant moment (A11) about the mass center, a resultant angular momentum about the mass center, and the relationship between them are 6. The change equation MCr ¼ erst4ðÞxasFat þ xbsFbt ; (A15a) An essential part of analysis is to understand how to divide a whole into disconnected parts and then to reconnect hCr dhCr ¼ erst4ðÞxasmaeat þ xbsmbebt ; ic ¼ MCr ¼ 0: the parts to reform the whole. The concepts of action force ic ds and interaction force are tools that facilitate analysis. In an (A15b,c) inertial space, by assuming that a fragment that is initially at rest will remain at rest, we had divided a whole into its Equation (A15a) defines resultant moment components disconnected parts. Each fragment of energy, when MCr; and Eq. (A15b) defines angular momentum 500 Physics Essays 33, 4 (2020)
that the moment about the mass center is zero, too (MC ¼ 0). The gravitational action forces are mechanical, that is, we assume that their electromagnetic parts cancelled [see Eqs. (A5)–(A8)]. From Eq. (A12)
@A Pbr ¼ ¼ kbr; (B1) @xr where A is the magnitude of the energy vector field of body a. From Eq. (A15), the angular momentum of the system is constant
FIG. 6. Right-hand rule for 4D space. hC ¼ maðÞxa1ea2 xa2ea1 þ mbðÞxb1eb2 xb2eb1 ic components hCr. Under the nonrelativistic assumption, Eqs. (A15) reduce to their classical counterparts. ¼ constant: (B2) Energy: The kinetic energy T is Next, convert this two-body problem to a one-body problem. 1 1 1 Let T ¼ mv2 ffi mv2 : (A16) 2 v 2 2 1 maear ¼ ler mbebr ¼ ler; er ¼ ebr ear c ; ; makar ¼ lkr mbkbr ¼ lkr; kr ¼ kbr kar m m Conservation of energy is determined as follows: l ¼ a b ðÞr ¼ 1; 2 : (B3) ð ma þ mb mc2 d ÀÁ 0 ¼ e2 þ e2 þ e2 þ e2 ds 2 ds 1 2 3 4 Equations (B3) satisfy Eqs. (A14). From Eqs. (A10) and ð (A12), 2 ¼ mc ðÞk1e1 þ k2e2 þ k3e3 þ k4e4 ds ð 2 2 Fbr ¼ mbc kbr ¼ lc kr: (B4) ¼ ðÞF1e1 þ F2e2 þ F3e3 þ F4e4 ds ð ð Substitute Eqs. (B3) into Eq. (B2) to get [Eq. (I8c)] 2 ¼ ðÞF1e1 þ F2e2 þ F3e3 ds mc k4e4ds ð ð hC mamb v2 v1 @V @V @V mc2 ÀÁ ¼ ðÞx1e2 x2e1 ¼ l x1 x2 ; 2 ic ma þ mb ic ic ¼ dx1 dx2 dx3 de4 ; ! @x1 @x2 @x3 2 1=2 0 1 2 2 m m v mc 1 1 l ¼ a b ; b ¼ 1 : (B5) ¼ V1 V2 B C m þ m c 2 @ v 2 v 2A a b 1 2 1 1 c c From Eq. (B5), the angular momentum is 0 1 2 2 B v2 v1 C 2 B C hC ¼ lðÞx1v2 x2v1 b: (B6) mc B c c C ¼ V1 V2 B C; 2 @ v 2 v 2A 1 2 1 1 In the gravitational potential below, ma ¼ M denotes the c c more massive body and mb ¼ m denotes the less massive body. We will replace hC above with the term where V is potential energy. It follows that: l h ¼ mxðÞ1v2 x2v1 b where hC ¼ h: (B7) E ¼ T þ V; E1 ¼ E2: (A17) m
Under the nonrelativistic assumption, Eqs. (A16) and (A17) Note that an analogous substitution arises in NT, where we reduce to their classical counterpart. replace HC ¼ lðÞx1v2 x2v1 with l H ¼ mxðÞ1v2 x2v1 where HC ¼ H: APPENDIX B: GRAVITATIONAL EQUATIONS FROM FT m
1. From the two-body problem to the one-body 2. Relationship between curvature and acceleration problem in FT Locate two bodies in an inertial frame. The system’s From Eq. (I8b), one can express the three-dimensional mass center is located at the frame’s origin, and it is station- spatial curvature vector k3 in terms of the three-dimensional ary ½xCr ¼ 0; eCr ¼ 0, and xCr ¼ 0 in Eq. (A14)]. It follows acceleration vector by Physics Essays 33, 4 (2020) 501
2 3 1 1 a v v a þ v a þ v a !1 1 2 2 !3 3 k3 ¼ 2 2 a va 2 2 v ¼ 2 6 2 7; v c v c 2 c 6 2 7 v 4 v 5 c2 1 1 c c () ! 1 1 v 2 ¼ !a 1 þ v½v a þ v a þ v a : (B8) c2 2 c 1 1 2 2 3 3 v 2 1 c
For the planar orbital motion of Mercury about the sun, 8 9 ! ! ! < 2 = 1 a1 v v1 ! k2 ¼ 2 1 þ ½v1a1 þ v2a2 þ v3a3 2 : c ; v a2 v2 c2 1 c 2 3 v 2 v v 6 1 2 1 2 7 1 6 c c c 7 !6 7 ; (B9) ¼ 2 6 2 7a v 2 4 v1 v2 v1 5 c2 1 1 c c c c where 2 3 2 3 1 2 2 6 v2 v1 v2 7 6 v1 v1 v2 7 6 1 7 6 1 7 6 c c c 7 1 6 c c c 7 6 7 ¼ 6 7 4 v v v 2 5 D 4 v v v 2 5 1 2 1 1 1 2 1 2 : c c c c c c ! ! ! v 2 v 2 v v 2 v 2 v 2 v 2 D ¼ 1 1 1 2 1 2 ¼ 1 1 þ 2 ¼ 1 c c c c c c c
It follows that: 2 3 2 ! v1 v1 v2 2 6 7 2 6 1 7 2 v 1 6 c c c 7 a ¼ c 1 6 7k2; c v 2 4 v v v 2 5 1 1 2 1 2 c c c c 2 3 v 2 v v !6 1 1 2 7 2 6 1 7 2 v 6 c c c 7 ¼ c 1 6 7k2: (B10) c 4 v v v 2 5 1 2 1 2 c c c
Substituting the action force in Table I into Eq. (B4) and substituting the result into Eq. (B10) yield the acceleration in Table III for the new FT formulation.
APPENDIX C: THE UNIQUENESS THEOREM FOR 4D VECTOR FIELDS Distinguish between two solutions of the vector continuity equations (A2b) by a superscript and express the difference between the two solutions as
ðÞ1 ðÞ2 As ¼ As As ; ðÞs ¼ 1; 2; 3; 4 : (C1) 502 Physics Essays 33, 4 (2020)
Next, consider the mathematical identity [Eqs. (I12) and @A @A @A @A 0 ¼ 1 þ 2 þ 3 þ 4 ; (D1) (I13)] @x1 @x2 @x3 @x4 ð @C @C @C @C 2 2 2 2 1 2 3 4 @ As @ As @ As @ As 0 ¼ þ þ þ dx1dx2dx3dx4 0 ¼ þ þ þ ; ðÞs ¼ 1; 2; 3; 4 : @x1 @x2 @x3 @x4 2 2 2 2 þ @x1 @x2 @x3 @x4 (D2) ¼ ðÞC1n1 þ C2n2 þ C3n3 þ C4n4 dx1dx2dx3; (C2)
Consider a stationary fragment for which es ¼ ds4 ðs ¼ 1; where 2; 3; 4Þ: (drs is the Kronecker-delta function for which drs ¼ 1whenr ¼ s and drs ¼ 0 otherwise) [Eq. (I23)]. Thus, @A1 @A2 @A3 for the stationary fragment, A1 ¼ A2 ¼ A3 ¼ 0andA4 ¼ A. Cs ¼ A1 þ A2 þ A3 @xs @xs @xs From Eq. (D1), @A=@x4 ¼ 0soA isanotafunctionoftime, @A4 only of the spatial coordinates. Next, for the purposes of the þ A4 ; ðs ¼ 1; 2; 3; 4Þ @xs fragment serving as a building block, disallow giving prefer- ence to any one spatial direction, that is, disallow directional- It follows that: ity. The vector field, under this assumption, becomes a hi 1=2 X X 2 function of r x x 2 x x 2 x x 2 , @C @C @C @C s¼4 t¼4 @A ¼ ðÞ1 01 þ ðÞ2 02 þ ðÞ1 03 1 þ 2 þ 3 þ 4 ¼ s ; s 1 t 1 @x1 @x2 @x3 @x4 ¼ ¼ @xt where x0s ðs ¼ 1; 2; 3Þ are the coordinates of the fragment’s (C3) point source. Perform the following mathematical steps:
C1n1 þ C2n2 þ C3n3 þ C4n4 @A dA @r x x dA @A ¼ ¼ s 0s ðÞs ¼ 1; 2; 3 ; ¼ 0; dA1 dA2 dA3 dA4 @xs dr @xs r dr @x4 ¼ A1 þ A2 þ A3 þ A4 : (C4) dn dn dn dn 2 2 2 @ A @ xs x0s dA xs x0s d A 2 ¼ ¼ 2 @x @xs r dr r dr Substituting Eqs. (C3) and (C4) into Eq. (C2) s 2 2 r x x 2 ð ! ðÞs 0s dA @ A X X 2 þ ðÞs ¼ 1; 2; 3 ; ¼ 0: s¼4 t¼4 @A 3 2 0 ¼ s dx dx dx dx ; r dr @ x4 s 1 t 1 1 2 3 4 ¼ ¼ @xt þ dA dA dA dA ¼ A 1 þ A 2 þ A 3 þ A 4 dx dx dx : From Eq. (D2), 1 dn 2 dn 3 dn 4 dn 1 2 3 2 3 (C5) 2 2 4 x x 2 d2A r ðÞx1 x01 dA5 0 ¼ 1 01 þ First, consider in Eq. (C5) the integrand in the event integral. r dr2 r3 dr Notice that it is equal to zero only if all of the partial deriva- 2 3 2 2 tives vanish implying, for uniqueness, that As must be con- 2 2 r x2 x02 4 x2 x02 d A ðÞdA5 stant throughout the region. It follows, for uniqueness, that þ 2 þ 3 ð1Þ ðÞ2 r dr r dr the two solutions As and As can only be unique up to an 2 3 additive constant. Next, consider the integrand in the bound- 2 2 4 2 2 r ðÞx3 x03 5 ary integral. It dictates the quantities that one must specify in x3 x03 d A dA þ 2 þ 3 order to obtain a solution that is unique up to an additive r dr r dr constant. To obtain the unique solution, one must specify d2A 2 dA ¼ þ ; either As or dAs=dn for each s ¼ 1, 2, 3, 4 at each point on dr2 r dr the boundary. It is interesting to imagine the region to be the entire universe, that it contains no singularities, and that As so or dAs=dn is equal to zero at its boundary (past and present). That would imply that the universe is void of change. Fortu- dB 2 nately, the universe contains singularities—the source points 0 ¼ þ B; (D3) dr r of the fragments. in which B ¼ dA=dr. Finally, solve Eq. (D3) to get B ¼ a=r2. The form of the stationary fragment becomes APPENDIX D: THE UNIQUE PROPERTY OF THE FUNCTION A 5 1/r a A ¼ þ C; (D4) Begin with the differential form of the vector continuity r equations of As ¼ Aes ðs ¼ 1; 2; 3; 4Þ and of its gradient vec- tors, expressed in rectangular coordinates as [Eqs. (A2) and where a and C are constants and where one refers to a as the (A3)] fragment’s intensity. Physics Essays 33, 4 (2020) 503
TABLE IX. Mechanical constants and electromagnetic constants.
Mechanical constants Electromagnetic constants
Universal gravitational constant: G ¼ 6:67 10 11 m3=kg Á s2 Speed of light: c ¼ 3 109 m=s Electric constant: k ¼ 8:99 109 m3kg=C2s2 31 19 Electron mass: me ¼ 9:1 10 kg Electron charge: qe ¼ 1:6 10 C
APPENDIX E: DECOMPOSITION OF ACTION INTO A APPENDIX F: FROM THE CHANGE EQUATION TO MECHANICAL PARTAND AN ELECTROMAGNETIC PART NEWTON’S LAW OF INERTIA
Consider the 4D vector field components As ¼ Aes Below we deduce the law of inertia Fs ¼ mas; ðs ¼ (s ¼ 1, 2, 3, 4). The components of its path derivative in the 1; 2; 3Þ [Eq. (A13)] from the change equation Ps ¼ ks; direction of its unit vector are Ps ¼ dAs=ds; where dxs ¼ ðÞs ¼ 1; 2; 3; 4 [Eq. (A12)]. First, invoke the relationship 2 esds (s ¼ 1, 2, 3, 4). The following shows that one can ks ¼ as=c ; ðs ¼ 1; 2; 3Þ under the nonrelativistic speed express the components of the path derivative as: assumption [Eq. (I8c)], where as; ðs ¼ 1; 2; 3Þ are accelera- tion components. Next, designate the fragment’s intensity as Ps ¼ Ss þ Ts; ðÞs ¼ 1; 2; 3; 4 ; (E1) either mechanical or electromagnetic in which mc2 mechanical a ¼ ; (F1) @A q electromagnetic Ss ¼ ; (E2) @xs where m is mass and q is charge. Substitute these into Ts ¼ ks1e1 ks2e2 ks3e3 ks4e4: (E3) Eq. (A10), which is the expression for the interaction force, We shall refer to the path derivative as action or as action to get force. In Eq. (E1), Ssðs ¼ 1; 2; 3; 4Þ are the components of Fs ¼ mas mechanical; (F2) the mechanical part of the action force, Tsðs ¼ 1; 2; 3; 4Þ are the components of the electromagnetic part of the action q force, and k ¼ð@A =@x Þ ð@A =@x Þ,(s, t ¼ 1, 2, 3, 4) are F ¼ a electromagnetic: (F3) st t s s t s c2 s the entries of a curl matrix [Eq. (G5)]. To show Eq. (E1), adopt the summation notation for Equation (F2) is the “mechanical” law of inertia, and repeated indices and manipulate Ps ðÞs ¼ 1; 2; 3; 4 as follows: Eq. (F3) is the “electromagnetic” law of inertia. In NT and dAs @As dxt @At @At @As EM, one always adopts the mechanical law of inertia in both Ps ¼ ¼ ¼ et et ¼ Ss þ Ts; ds @xt ds @xs @xs @xt mechanics and in electromagnetism. Let us explain the rea- son for adopting the mechanical law in electromagnetism. @At @ðÞAet @et @A @A Ss ¼ et ¼ et ¼ A et þ etet ¼ ; One explains this based on how, at the atomic scale, ele- @x @x @x @x @x s s s s s ments bundle. Consider a pair of electrons. Begin by con- @At @As Ts ¼ et ¼ kstet; (E4) verting quantities expressed in terms of conventional units @xs @xt into universal units. Table IX lists the physical constants, and Table X expresses three classical laws in terms of con- since (@e /@x )e ¼ @=@x e e ¼ 0 and e e ¼ 1. This com- t s t sðÞt t t t ventional units and universal units. pletes the derivation of Eq. (E1). However, let us also put the In Table X, the subscript 0 indicates an expression in electromagnetic part of the action intoÀÁ its more familiar clas- terms of universal units. One expresses acceleration a and 1 sical Lorentzian vector form F ¼ q E þ c v B ; where distance r in terms of length L and time T, whether in univer- quantities in bold are 3D vectors. Write the electromagnetic sal units or in conventional units, so their expressions are interaction force as Fs ¼ a Ts ¼ q kstet. This yields the the same in both conventional units and universal units. classical Lorentzian vector form One finds from Table X: F=F0 ¼ m=m0; F=F0 ¼ Fe=Fe0 ¼ "# ! "# ! Fg=Fg0; and m=m0 ¼ me=me0 so the conversions are iBx E e iBx E v=ic F ¼ q ¼ q q c2 m c4 EÁ 0 e4 EÁ 0 1 e ¼ pffiffiffiffiffiffi ; e ¼ : (F4) 0 1 0 1 q m G v 1 e0 Gk e0 @ iB x þ E A @ E þ v x B A ¼ q ic ¼ q c ; From Table IX, the ratio of mechanical intensity to electro- E Á v=ic E v=ic 2 3 Á magnetic intensity and the ratio of gravitational interaction 0 B3 B2 to electromagnetic interaction are 6 7 ÀÁ B ¼ 4 B3 0 B1 5; EÁ¼ E1 E2 E3 : 2 B2 B1 0 a m c m ¼ e ¼ 5:12 107; (F5) (E5) ae qe 504 Physics Essays 33, 4 (2020)
TABLE X. Fundamental unit and conventional unit comparison.
Equation Universal units Conventional units
Newton’s second law F0 ¼ m0aF¼ ma Universal law of gravitation m2 m2 F ¼ c4 e0 F ¼ G e g0 r2 g r2 Coulomb law of electrostatics q2 q2 F ¼ e0 F ¼ k e e0 r2 e r2
F G m 2 consistent with the experimentally verified motion of a g ¼ e ¼ 2:34 10 43: (F6) Fe k qe charged particle. Thus, one employs Eq. (F2) and not Eq. (F3) in both NT and EM (at the usual scales). Equations (F5) and (F6) show that, for a pair of electrons, mass dominates over charge while electromag- APPENDIX G: FROM VECTOR CONTINUITY TO netic interaction force dominates over gravitational interac- MAXWELL’S EQUATIONS tion force. One expresses these relationships as the two properties Consider a domain that contains a fragment of energy A0s (s ¼ 1, 2, 3, 4) that excites a vector field of energy with components A (s ¼ 1, 2, 3, 4) in a given domain. Ps ¼ Sms þ Tes ffi Tes ðÞSes ffi 0 ; s (F7) Other so-called excitations could exist outside of the a ¼ am þ ae ffi am ðÞaq ffi 0 : domain. One would treat their influences on the compo- nents As (s ¼ 1, 2, 3, 4) of the vector field of energy in the Per Eq. (F7), in the electron’s bundle, its mass dominates domain through boundary conditions. Next, assume that over its charge, and its electromagnetic action force domi- one is observing continuity in an inertial frame of refer- nates over its gravitational action force, both for reasons that ence. All of the considerations are in a single frame of ref- are associated with the nature of the bundle at the atomic erence. One has no need to invoke RT. Finally, assume scale, which is beyond the scope of this article. The elec- that As (s ¼ 1, 2, 3, 4) and the components of its gradient tron’s aggregate curvature is ks ¼ 1=aaðÞmkms þ aqkqs ¼ vectors satisfy space-time continuity except at the source 2 ðae=amÞTs ¼ðq=mc ÞTs. Finally, under the nonrelativistic point of the fragment of energy A0s (s ¼ 1, 2, 3, 4) 1 speed assumption, ma ¼ qT ¼ qðE þ c ðÞv B , which is [Eqs. (I12), (I13), and (I29)]
vector continuity integral form differential form þ A 0 ¼ A Á ndV ٗ Á A ¼ 0 s : (G1a,b) ð þ
As ðÞs ¼ 1; 2; 3; 4 4pas d3dE ¼ ٗAs Á ndV 4pasd3 ¼ ٗ Á ٗAsٗ
In Eq. (G1), we switched to a vector notation consis- In addition to vector continuity of the curl vectors, we will tent with the customary notation of Maxwell’s equa- call upon the curl identities. Their integral and differential tions. To derive Maxwell’s equations, particular interest forms are [Eqs. (I10) and (I11)] will lie in vector continuity of the curl vectors þ Ks ¼ð@As=@xÞ ð@A=@xsÞ,(s ¼ 1, 2, 3, 4). From Eq. integral form 0 ¼ krsdxrdxs (G1), the space-time vector continuity equations of the ðÞr;s;t curl vectors are @krs @kst @ktr differential form 0 ¼ xrst ¼ þ þ @xt @xr @xs ð þ 8 > <> ðÞ2; 4; 3 for u ¼ 1 integral form 4pas d3dE ¼ Ks Á ndV; ðÞ3; 4; 1 for u ¼ 2 ðÞr; s; t ¼ : (G2a,b) > ðÞ4; 2; 1 for u ¼ 3 X : t¼4 @k ðÞ1; 2; 3 for u ¼ 4 differential form 4pa d ¼ ts : s 3 t 1 ¼ @xt (G3a,b) Physics Essays 33, 4 (2020) 505
Show that Eqs. (G2) and (G3) generate Maxwell’s equations
ð integralþ formð differential form 4p 1 @ 4p 1 @E j Á ndS ¼ B Á dl E Á ndS jd ¼r B c þ c @t c c @t 4pq ¼ E Á ndS 4pqd ¼rÁE þ ð : (G4a-d) 1 @ 1 @B 0 ¼ E Á dl þ B Á ndS 0 ¼ þr E þ c @t c @t 0 ¼ B Á ndS 0 ¼rÁB
Begin to show Eqs. (G4) by defining the electric field vector First, express the two integrals on the right side of Eq. (G6) E and the magnetic field vector B in terms of the elements of as
the curl matrix K and rewrite the needed relationships in the þ Þ convenient forms V i B Á dldxr ðÞr ¼ 1; 2; 3 r Á ndS ¼ Þ ; E Á ndSðÞ r ¼ 4 2 3 2 3 ð Ð (G7a,b) 0 k k 0 iB E E ndSdx r 1; 2; 3 6 k12 13 14 7 6 iB3 2 1 7 Á r ðÞ¼ 6 7 6 7 kr4dV ¼ : 6 7 6 7 0 ðÞr ¼ 4 6 k21 0 k23 k24 7 6 iB3 0 iB1 E2 7 6 7 6 7 K ¼ 6 7 ¼ 6 7; 6 k k32 0 k 7 6 iB iB1 0 E 7 4 31 34 5 4 2 3 5 Equation (G7a) follows from the mathematical manipula- tions given below: k41 k42 k43 0 E1 E2 E3 0 0 1 0 1 0 1 0 1 0 iB iB
B C B 3 C B 2 C þ þ 0
B C B C B C V B C V B iB C B 0 C B iB C @ A B 3 C B C B 1 C 1 Á ndS ¼ iB3 Á ndS 1 ¼ B C K2 ¼ B C K3 ¼ B C @ iB2 A @ iB1 A @ 0 A iB2 hiÀÁÀÁ E1 E2 E3 þ þ ¼ i B3 B3 dx3 þ B2 B2 dx2 dx1 0 1 þ E1 0 1 0 1
B C i B dl dx ;
B C k14 k23 ¼ Á 1 V B C B C B E2 C 0 about 1
¼ B C E ¼ B k C iB ¼ B k C:
4 B C @ 24 A @ 31 A þ þ iB3 @ E3 A V B C k34 k12 Á ndS ¼ @ 0 A Á ndS 0 2 iB1 (G5) hiÀÁÀÁ þ þ ¼ iB3 B3 dx3 B1 B1 dx1 dx2 Below, we first derive the integral form of Maxwell’s equa- þ tions, Eq. (G5a), from the integral form of the continuity ¼ i B Á dl dx2; equations of the curl vectors, Eq. (G2a), and from the inte- 0 about 21
gral form of the curl identities, Eq. (G3a). Then, we derive þ þ iB2 the differential form of Maxwell’s equations, Eq. (G11b), V B C 3 Á ndS ¼ @ iB1 A Á ndS from the differential form of the continuity equations of the 0 curl vectors, Eq. (G2b), and the differential form of the curl ÂÃÀÁÀÁ ¼ i Bþ B dx þ Bþ B dx dx identities, Eq. (G3b). þ 2 2 2 1 1 1 3
¼ i B Á dl dx3; 0 about1 3 1. Integral form
þ þ E1 þ V B C Write the integral that expresses the space-time vector 4 Á ndS ¼ @ E2 A Á ndS ¼ E Á ndS: continuity of the curl vectors in Eq. (G2a) over a time slice E3 dx4 [Eq. (I22)]as
þ þ ð
V V @ ndS k dV Equation (G7b) follows from the next set of mathematical r Á ndV ¼ r Á þ r4 dx4: (G6) @x4 manipulations: 506 Physics Essays 33, 4 (2020) ð ð !ð ð !
k14dV ¼ E1dS dx1 k24dV ¼ E2dS dx2; 2 3 plane 3 1 plane ð ð !ð
k34dV ¼ E3dS dx3 k44dV ¼ 0: 1 2 plane
Next, let us consider the curl identities. Express Eq. (G3a) as 8 "#ð þ > þ > @ < i B Á ndS þ E Á dl dx4 for u ¼ 1; 2; 3 @x4 ðÞr;s;t about u krsdxrdxs ¼ > þ : (G8) ðÞr;s;t > : i B Á ndS for u ¼ 4
Equation (G8) follows from the next set of mathematical manipulations: þ ÀÁ ÀÁ ÀÁ þ þ þ krsdxrdxs ¼ E2 E2 dx2 dx4 E3 E3 dx4 dx3 iB1 B1 dx3 dx2 ðÞ2;4;3 "#ð þ @ ¼ i B Á ndS þ E Á dl dx4; @x4 ðÞ2;3 about 1 þ ÀÁ ÀÁ ÀÁ þ þ þ krsdxrdxs ¼ E3 E3 dx3 dx4 E1 E1 dx4 dx1 iB2 B2 dx1 dx3 ðÞ3;4;1 "#ð þ @ ¼ i B Á ndS þ E Á dl dx4; @x4 ðÞ3;1 about 2 þ ÀÁ ÀÁ ÀÁ þ þ þ krsdxrdxs ¼ E2 E2 dx4 dx2 iB3 B3 dx2 dx1 þ E1 E1 dx1 dx4 ðÞ4;2;1 "#ð þ @ ¼ i B Á ndS þ E Á dl dx4; @x4 ðÞ1;2 about 3 þ ÀÁ ÀÁ ÀÁ þ þ þ krsdxrdxs ¼ iB3 B3 dx1 dx2 þ iB1 B1 dx2 dx3 þ iB2 B2 dx3 dx1 ðÞ1;2;3 þ ¼ i B Á ndS:
Write the left side and the right side of Eq. (G2a) for (r ¼ 1, 2, 3) as ð ð ð ð 4p 4p 4pa d dE ¼ 4p ae d dVdx ¼ j d dVdx ¼ i j Á ndS dx dx r 3 r 3 4 ic r 3 4 c r 4
þ þ ð þ ð ; V @ 1 @ r Á ndV ¼ i B Á dldxr E Á ndSdxr dx4 ¼ i B Á dl E Á ndS dxr dx4 @x4 c @t where ar ¼ aer ¼ qvr=ic ¼ jr=ic for ðÞr ¼ 1; 2; 3 and a4 ¼ q. Equating the left side and the right side yields Maxwell equation (G4a). Next, write the left side and the right side of Eq. (G2a) for (r ¼ 4) as
ð ð þ þ V 4pa4 d3dE ¼ 4pa d3dVdx4 ¼ 4pqdx4 r Á ndV ¼ E Á ndS dx4:
Equating the left side and the right side yields Maxwell equation (G4b). Finally, notice that Eq. (G8) for (u ¼ 1, 2, 3) is Max- well equation (G4c) and that Eq. (G8) for (u ¼ 4) is Maxwell equation (G4d).
2. Differential form Next, we independently derive the differential form of Maxwell’s equations, Eq. (G4), from Eqs. (G2b) and (G3b). From Eqs. (G5) and (G2b) Physics Essays 33, 4 (2020) 507
j1 @ðÞ0 @ðÞ iB3 @ðÞiB2 @ðÞ E1 Integrating Eq. (H2) by parts, recognizing that r, t, and 4p d3 ¼ þ þ þ ; ic @x1 @x2 @x3 @x4 u are independent ð j2 @ðÞiB3 @ðÞ0 @ðÞ iB1 @ðÞ E2 4p d3 ¼ þ þ þ ; 0 ¼ df ic @x1 @x2 @x3 @x4 ð j3 @ðÞ iB2 @ðÞiB1 @ðÞ0 @ðÞ E3 @f @f @f @f 4p d3 ¼ þ þ þ ; ¼ dr þ dr_ þ dt_þ du_ ; ic @x1 @x2 @x3 @x4 _ "#@r @r_ @t @u_ @ðÞE @ðÞE @ðÞE @ðÞ0 ð 4pqd ¼ 1 þ 2 þ 3 þ : @f d @f d @f d @f 3 @x @x @x @x ¼ dr dt du ; 1 2 3 4 @r ds @r_ ds @t_ ds @u_ In vector notation, j @E from which 4p d ¼ ir B 4pqd ¼rÁE : (G9a,b) ic 3 @x 3 4 @f d @f d @f d @f 0 ¼ ; 0 ¼ ; 0 ¼ : Next, consider the differential form of the curl identities, @r ds @r_ ds @t_ ds @u_ Eq. (G3b). From Eq. (G5) (H3) @ðÞiB3 @ðÞiB1 @ðÞiB2 0 ¼ þ þ From Eqs. (H2) and (H3), @x @x @x 3 1 2 @ðÞE2 @ðÞ E3 @ðÞ iB1 @f r 2E @f 2H 0 ¼ þ þ ; ¼ 21 s t_ ¼ ¼ 2r2u_ ¼ C @x3 @x2 @x4 @t_ r mc2 @u_ l @ðÞE3 @ðÞ E1 @ðÞ iB2 mM 0 ¼ þ þ l ¼ ; @x1 @x3 @x4 m þ M @ðÞ E2 @ðÞ iB3 @ðÞE1 0 ¼ þ þ : and so @x1 @x4 @x2 In vector notation, m E2 mc2 H 2 m GMm r_2 ¼ C þ 2 2mc2 2 l 2r2 r @B 0 ¼ irÁB 0 ¼ r E i : (G10a,b) H 2 GMm @x4 þ C : (H4) cl r3 Equation (G9a) yields the first Maxwell’s equation, Eq. (G4a). Equation (G9b) yields the second Maxwell’s Equation (H4) expresses conservation of energy equation, Eq. (G4b). Equation (G10a) yields the third Max- well’s equation, Eq. (G4c). Equation (G10b) yields the C ¼ ðÞTr þ Tu þ ðÞV þ VGR ; (H5) fourth Maxwell’s equation, Eq. (G4d). 2 2 2 in which Tr ¼ðm=2Þr_ and Tu ¼ ðÞHC=l m=2r are the radial and tangential components of kinetic energy and APPENDIX H: GRAVITATIONAL EQUATIONS FROM GR 2 3 V ¼ GMm=r and VGR ¼ ðÞHC=cl GMm=r are the Below we obtain an exact solution to the one-body grav- Newtonian potential and the effective radial potential. The itational problem beginning from the Schwarzschild metric. effective radial potential VGR is the term in the potential Note that a perturbation solution is also available.8 Begin energy that GR introduces. From Eq. (H4), h) with the Schwarzschild metric, written as GMm H 2 GMm 2 2 2 V þ V ¼ C : (H6) ds ¼ c ds GR r cl r3 ÀÁ 2 rs 2 1 2 2 2 2 ¼ c 1 dt r dr dh þ sin h du : r 1 s Equation (H6) is the gravitational potential given in r Table Ib, in which H/m ¼ H /l. (H1) C
In Eq. (H1), s is proper time and r ¼ 2GM=c2 is the s APPENDIX I. ELEMENTS OF 4D VECTOR FIELDS Schwarzschild radius. For the one-body gravitational prob- lem, let h ¼ p=2 and d=ds ¼ ðÞ_ : From Eq. (H1), 1. Space r 1 The geometry of space: One deduces the geometry of c2 ¼ frðÞ; r_; t_; u_ ¼ c2 1 s t_2 r_2 r2u_ 2: rs an dimensional space from the algebraic relations of its r 1 n- r associated n-dimensional cubes. In 3D space, the 3-cube (H2) (ordinary cube) has three pairs of positive and negative faces, each representing a 2-cube (square). In 4D space, the h)David R. Williams, NASA Goddard Space Flight Center, https:// 4-cube has four pairs of positive and negative faces, each nssdc.gsfc.nasa.gov/planetary/factsheet/mercuryfact.html representing a 3-cube. 508 Physics Essays 33, 4 (2020)
The right-hand rule: In 3D space, one customarily sets Notice that the curvature vector and the unit vector are per- d d up the coordinates in a right-handed order to ensure consis- pendicular, because 0 ¼ ds ð1Þ¼ds ðe Á eÞ¼2k Á e: tency in subsequently defined operations. In 4D space, the The perpendicular vector: Next, let us define the per- right-handedness of the four possible triads is not obvious. pendicular vectors for 2D, 3D, and 4D spaces. In a 2D space, For example, is (1, 4, 3) or (1, 3, 4) right-handed when 4 is let the 2D vector w be perpendicular to x. In a 3D space, let associated with geometric time? One determines the correct the 3D vector w be perpendicular to x and y, and in a 4D “right-handedness” from the assembly of the 3-cubes into a space let the 4D vector w be perpendicular to x, y, and z. 4-cube. Figure 6 shows an easy way to remember the result- One expresses the formulas for the perpendicular vectors in ing right-hand rules. As shown, one starts with a tetrahedron. terms of the permutation symbols as follows: The vertices of its base are labeled 1, 2, and 3 in the right- X2 handed order, and its top vertex is labeled as 4. Next, one For 2D spaces: w ¼ e x ; opens the tetrahedron like the pedals of a flower and lays it r rs s s¼1 flat. The right-handed orders of the triads express the right- X3 X3 hand rules. For 3D spaces: wr ¼ erstxsyt; (I5a-c) As shown, the triads (1, 2, 3), (2, 4, 3), (3, 4, 1), and s¼1 t¼1 X4 X4 X4 (4, 2, 1) are each right-handed and so too are the quads (1, 2, For 4D spaces: w ¼ e x y z : 3, 4), (2, 4, 3, 1), (3, 4, 1, 2), and (4, 2, 1, 3). Switching any r rstu s t u s¼1 t¼1 u¼1 two adjacent indices converts a right-handed space into a left-handed space and vice-versa. Mathematically, one can Equation (I5) expresses the components of the perpendicular express the right-hand rules for a 4D space by the permuta- vectors in an index notation. In the 2D case, one more com- tion symbol erstu. One defines it such that erstu ¼ 0 when any monly expresses Eq. (I5a) as w ¼ ix, in which w and x are two indices repeat, such that erstu ¼ 1 when the indices are in complex numbers and where i is the imaginary number, rec- the right-handed order, and such that erstu ¼ 1 when the ognizing that a complex number is equivalent to a 2D vector indices are in the left-handed order. and that the imaginary number i represents a 90 counter- clockwise rotation of the 2D vector. In the 3D case, one 2. Vectors more commonly expresses Eq. (I5b) as the cross The vector rules: Let x, y, and z denote n-dimensional product w ¼ x y. vectors and let a and b denote numbers (scalars). Write each The 4D space-time vector: Let us now partition the 4D T position vector x into a 3D spatial position vector x and a vector as a column of n numbers x ¼½x1 x2 x3 x4 and define vector addition and multiplication by a scalar as fol- temporal position vector x4 ¼ c0t. The partitioned position lows: Add two vectors by adding together the numbers in vector, unit vector, and curvature vector are their corresponding entries. Multiply the vector by a scalar ! ! by multiplying each entry by the scalar. Let n-dimensional x v=c0 x ¼ ; e ¼ b0 ; vectors satisfy the following two commutative rules, two c0t 1 "# ! ! (I6a-c) associative rules, and distributive rule: 2 2 b0 a b0 v=c0 k ¼ 2 va ; x þ y ¼ y þ x ax ¼ xa c0 0 c0 1 ðx þ yÞþz ¼ x þðy þ zÞðxaÞb ¼ xðabÞ: (I1a-e) where aðx þ yÞ¼ax þ by 0 1 x1 Next, define the dot product of the two vectors x and y by @ A sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 x ¼ x2 and b0 ¼ : v 2 x3 1 þ x Á y ¼ x1y1 þ x1y1 þÁÁÁþxnyn: (I2) 2 c0
Using the dot product operation, the magnitude of a vector x Under the nonrelativistic speed assumption, for which and the component x of a vector x in the direction of the y v c0, the unit vector and the curvature vector are approxi- vector y are mated by (see Fig. 7) pffiffiffiffiffiffiffiffiffiffiffi x 0 1 jxj¼ x Á x; xy ¼ x Á : (I3a,b) a jxj 1 v 1 @ a v A e ffi ; k ffi 2 Á : (I7a,b) c0 c0 c0 The path: Next, let x represent a position vector over a c0 path. The path increment is ds. The unit vector tangent to the path is the path derivative of the position vector, and the cur- vature vector is the path derivative of the unit vector tangent The speed of light: In space-time physics, one produces to the path, written geometric time x4 by rotating the radially directed propaga- tion length ct of an electromagnetic wave to a geometric dx de time axis. This is expressed mathematically by the equation e ¼ k ¼ : (I4a,b) ds ds x4 ¼ ict where, again, one recognizes that i is a 90 rotation Physics Essays 33, 4 (2020) 509
around closed volumes into integrals in open events. The transverse theorems are similar. When performing the manipulations, we will denote a differential line element ð1Þ ð2Þ ð2Þ by dx1 , a differential surface element by dx1 dx2 , a differ- ð3Þ ð3Þ ð3Þ ential volume element bydx1 dx2 dx3 , and a differential ð4Þ ð4Þ ð4Þ ð4Þ event by dx1 dx2 dx3 dx4 . Longitudinal theorems: Denote differential circulation ð1Þ ð1Þ (1) by dC ¼ A1 dx1 , where A1 is the component of A in (1) FIG. 7. e and k under the nonrelativistic speed assumption. the direction of the differential line element dx1 . Also, ð2Þ ð2Þ ð2Þ ð2Þ denote differential curl by dK ¼ k12 dx1 dx2 , where k12 ¼ ð2Þ ð2Þ ð2Þ ð2Þ ð@A2 =@x1 Þ ð@A1 =@x2 Þ denotes curl density and ð3Þ ð3Þ ð3Þ ð3Þ denote differential coil by dX ¼ x dx1 dx2 dx3 where of a 2D vector (in radial-time). Comparing x4 ¼ c0t above ð3Þ ð3Þ ð3Þ ð3Þ ð3Þ ð3Þ ð3Þ x ¼ð@k12 =@x3 Þþð@k23 =@x1 Þþð@k31 =@x2 Þ denotes Eq. (I6) and x4 ¼ ict above, one finds that c0 ¼ ic. From Eq. (I6), coil density. First, consider an open finite surface divided into differ- "# ential surface elements. Differential circulations along v=ic b2 a b2 v=ic e ¼ b ; k ¼ va ; differential line elements from adjacent differential surfaces 1 c2 0 ic 1 cancel. The only differential circulations that are left are (I8a,b) along the face of the finite surface. It follows that the circula- qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi tion around the boundary of the finite surface is equal to the curl inside the surface. It also follows that both are equal to where b ¼ 1= 1 ðvÞ2. From Eq. (I8), under the nonrela- c zero if the surface is closed. Next, consider a finite volume tivistic speed approximation divided into differential volume elements. Differential curls 0 1 along differential surface elements from adjacent differential a 1 v 1 @ A volumes cancel. The only differential curls that are left are e ¼ ; k ¼ ia Á v : (I8c) ic ic c2 along the face of the finite volume. It follows that the curl c around the face of the finite volume is equal to the coil inside the volume. It also follows that both are equal to zero if the Summary: In 3D-vector analysis, space-time paths are volume is closed and that the coil density inside the closed kinematic, expressed in terms of a position vector, a velocity volume is identically equal to zero. The following lists the vector, and an acceleration vector. In 4D-vector analysis, in longitudinal theorems: particular, in space-time analysis, space-time paths are geo- þ metric, expressed in terms of a position vector, a unit vector ð1Þ ð1Þ circulation C ¼ A1 dx1 ; tangent to the path, and a curvature vector of the path þ 0 1 Cclosed ¼ Kopen ð3Þ ð2Þ ð2Þ x curl K ¼ k12 dx1 dx2 ; B 1 C Kclosed ¼ X ¼ 0 B C dx dv þ 3D analysis: x ¼ @ x2 A v ¼ a ¼ ; ð3Þ ð3Þ ð3Þ ð3Þ dt dt coil X ¼ x dx1 dx2 dx3 : x3 ! (I10) x dx de 4D analysis: x ¼ e ¼ k ¼ : One can also show by simple substitution that x4 ds ds (I9a,b) @kð3Þ @kð3Þ @kð3Þ 0 xð3Þ 12 23 31 : (I11) ¼ ¼ ð3Þ þ ð3Þ þ ð3Þ @x3 @x1 @x2
3. Vector fields Transverse theorems: First, for 2D spaces, denote dif- ð1Þ ð1Þ (1) Let us now examine continuous vector fields in 2D, 3D, ferential stream by dD ¼ A2 dx1 ; where A2 is the compo- (1) and 4D spaces. In particular, we present the integral theo- nent of A perpendicular to the differential line element dx1 rems and the differential theorems that they obey. We will ð2Þ ð2Þ and denote differential leak by dX ¼ ndx1 dx2 , where n is do this systematically, by distinguishing between integrals of leak density. For 3D spaces, denote differential flux by vector components along lines of integration and integrals of ð2Þ ð2Þ ð2Þ (2) dU ¼ A3 dx1 dx2 , where A3 is the component of A vector components that are perpendicular to lines of integra- (2) (2) perpendicular to the differential surface element dx1 dx2 tion, called longitudinal theorems and transverse theorems, ð3Þ ð3Þ ð3Þ respectively. The longitudinal theorems turn integrals around and denote differential divergence by dY ¼ wdx1 dx2 dx3 , closed lines into integrals in open surfaces, integrals around where w is divergence density. For 4D spaces, denote differ- ð3Þ ð3Þ ð3Þ ð3Þ (3) closed surfaces into integrals in open volumes, and integrals ential strength by dW ¼ A4 dx1 dx2 dx3 , where A4 is the 510 Physics Essays 33, 4 (2020) component of A perpendicular to the differential volume where (3) (3) (3) element dx1 dx2 dx3 and denote differential dilatation ð2Þ ð2Þ ð3Þ ð3Þ ð3Þ by dZ ¼ fdxð4Þdxð4Þdxð4Þdxð4Þ where f is dilatation density. @A @A @A @A @A 1 2 3 4 n ¼ 1 þ 2 w ¼ 1 þ 2 þ 3 For 2D spaces, consider an open finite surface divided ð2Þ ð2Þ ð3Þ ð3Þ ð3Þ @x1 @x2 @x1 @x2 @x3 into differential surface elements. Differential streams along differential line elements from adjacent differential surfaces @Að4Þ @Að4Þ @Að4Þ @Að4Þ f ¼ 1 þ 2 þ 3 þ 4 : (I13) cancel. The only differential streams that are left are along @xð4Þ @xð4Þ @xð4Þ @xð4Þ the face of the finite surface. It follows that the stream 1 2 3 4 around the boundary of the finite surface is equal to the leak inside the surface. For 3D spaces, consider an open finite 4. Vector continuity volume divided into differential volume elements. Differen- tial fluxes on differential surface elements from adjacent dif- Appendix Section I 3 enforced continuity of the individ- ferential volumes cancel. The only differential fluxes that are ual components of a vector field, that is, scalar continuity. left are along the face of the finite volume. It follows that the This section considers, in addition to scalar continuity, conti- flux around the boundary of the finite volume is equal to the nuity of the vector field as a whole. Toward this end, one divergence inside the volume. For 4D spaces, consider an defines the concept of a field line. open finite event divided into differential event elements. The field line: The number of field lines that passes Differential strengths on differential volume elements from through a differential face of an n-dimensional body is (see adjacent differential events cancel. The only differential Fig. 8) strengths that are left are along the face of the finite event. It dN ¼ AdF ¼ A dF ¼ A Á ndF; (I14) follows that the strength around the boundary of the finite ? ? event is equal to the dilatation inside the event. The follow- where dF is a differential face, n is the outward normal to ing summarizes these results: the differential face, and where ðÞ? means perpendicular þ component. ð1Þ ð1Þ stream D ¼ A2 dx1 Vector continuity requires the number of field lines that D ¼ X ð ; leaves an n-dimensional domain (through its faces) to be ð2Þ ð2Þ leak X ¼ ndx1 dx2 equal to zero, written þ þ þ 0 ¼ dN ¼ A Á n dF: (I15) ð2Þ ð2Þ ð2Þ flux U ¼ A3 dx1 dx2 U ¼ Y ð ; (I12) Note from Eq. (I14) that a field line that enters a domain is ð3Þ ð3Þ ð3Þ equal to the negative of a field line that leaves it. divergence Y ¼ wdx1 dx2 dx3 Vector continuity of A: By comparing the transverse integral theorems developed in Appendix Section I 3 with þ Eq. (I15), it follows: for 2D spaces that the stream and leak ð3Þ ð3Þ ð3Þ ð3Þ strength W ¼ A4 dx1 dx2 dx3 density are zero, for 3D spaces that the flux and divergence W ¼ Z ð ; density are zero, and for 4D spaces that the strength and dila- ð4Þ ð4Þ ð4Þ ð4Þ tation density are zero dilatation Z ¼ fdx1 dx2 dx3 dx4
differential forms integral forms @A @A 0 ¼ 1 þ 2 For 2D spaces: þ @x1 @x2 leak 0 ¼ A Á n dl ðÞor 0 ¼ Ú Á A ; þ For 3D spaces: @A1 @A2 @A3 0 ¼ A Á n dS 0 ¼ þ þ (I16a-c) divergence @x1 @x2 @x3 ðÞor 0 ¼rÁA ; For 4D spaces: þ dilatation @A1 @A2 @A3 @A4 0 ¼ A Á n dV 0 ¼ þ þ þ @x1 @x2 @x3 @x4 : ðÞor 0 ¼ ٗ Á A Physics Essays 33, 4 (2020) 511
Vector continuity of the gradient vectors: Next, write the gradient vectors of the components of the vector fields as @A @A T C i i For 2D spaces: i ¼ Ci ¼ ÚAi; @x1 @x2 T @Ai @Ai @Ai For 3D spaces: Ci ¼ Ci ¼rAi ; (I17a-c) @x1 @x2 @x3 T :For 4D spaces: @Ai @Ai @Ai @Ai Ci ¼ ٗ Ai Ci ¼ @x1 @x2 @x3 @x4
Write the enforcement of vector continuity of the gradient vectors in Eqs. (I17a–c)as þ 2 2 @ Ai @ Ai 0 ¼ ÚAi Á n dl 0 ¼ 2 þ 2 @x1 @x2 For 2D spaces: ðÞi ¼ 1; 2 ; ðÞor 0 ¼ Ú Á ÚAi þ @2A @2A @2A 0 ¼ i þ i þ i For 3D spaces: 0 ¼ rAi Á n ds @x2 @x2 @x2 1 2 3 ; (I18a-c) ðÞi ¼ 1; 2; 3 ðÞor 0 ¼rÁrAi
For 4D spaces: þ 2 2 2 2 @ Ai @ Ai @ Ai @ Ai þ 2 þ 2 þ 2 2 ¼ 0 ٗ 0 ¼ Ai Á n dV @x @x @x @x 1 2 3 4 : ðÞi ¼ 1; 2; 3; 4 ðÞor 0 ¼ ٗ Á ٗAi
Equation (I18a) are also called potential equations and, by dV1 ¼ dx3dx2dx4 ¼ dS23dx4 dV2 ¼ dx1dx3dx4 ¼ dS31dx4; letting x4 ¼ ict, Eq. (I18c) is identically the wave equation dV ¼ dx dx dx ¼ dS dx dV ¼ dx dx dx ¼ dV: 2 2 2 2 3 2 1 4 12 4 4 1 2 3 @ Ai @ Ai @ Ai 1 @ Ai 0 ¼ 2 þ 2 þ 2 2 @ 2 : The enforcement of the vector @x1 @x2 @x3 c t (I19a-d) continuity equations, along with the appropriate boundary conditions, yields a solution that is unique up to an additive Partition the vector field as follows: constant (Appendix C). ! ! A v 1 Vector continuity over a time increment: In the fol- A ¼ ¼ Ae ¼ A pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 lowing, we consider the integral form of vector continuity in A4 c0 v þ c 0 (I20a,b) a 4D space, Eq. (I16c), over a time increment. The domain is pffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 1 now finite in the spatial dimensions and infinitesimal in the A Á n ¼ Av Á n ¼ A4v Á n: v2 þ c2 c0 temporal dimension. It is an infinitesimally thin slice in time 0 of a finite spatial domain. Over this time slice, we express Substitute Eqs. (I19) and (I20) into Eq. (I16c) to get Eq. (I16c) as the sum of two terms, one associated with flow þ þ ð lines leaving a spatial volume, and the other associated with @ A Á n dV ¼ A Á n dS þ A4dV dx4: (I21) a time rate of generation of field lines in the spatial volume. @x4 This will yield the classical 3D form of conservation. The faces of the time slice consist of four pairs of posi- In Eq. (I21), the flow out of the eight 3D cubes that make up tive and negative 3D cubes, each pair having the differential Þthe faces of a 4DÐ cube now flows out of the two integrals volume elements AÞÁ n dS and A4dV. It flows out of the sixÐ spatial faces of A Á n dS and the two temporal faces of A4dV. Letting @=@x4 ¼ 1=c0 @=@t, from Eq. (I21), one expresses the satis- faction of vector continuity over a time slice as þ ð @ 0 ¼ A Á n dS þ A4dV; @x4 or (I22a,b) þ ð @ 0 ¼ A v Á n dS þ A dV: 4 @t 4
One conventionally refers to Eqs. (I22) as a conservation FIG. 8. Field lines. law. 512 Physics Essays 33, 4 (2020) þ ð 5. The point singularity @A Á n dS ¼ 4p⍜ d dV: (I25) @ 3 Assume that an n-dimensional region contains a point x singularity. To assist with the treatment of the singularity, we introduce the n-dimensional Kronecker-delta function Next, consider the differential form of Eq. (I25), given in and the Dirac-delta function, defined as follows: Eq. (I18b) at any point except at the point singularity. Using the 3D Dirac-delta function, we can make Eq. (I18b) apply to the point singularity, too, by writing it as 1 r ¼ s d ¼ ðr; s ¼ 1; 2; …; nÞ; (I23) rs 0 r 6¼ s @2A @2A @2A 4pHd3 ¼ 2 þ 2 þ 2 : (I26) ð @x1 @x2 @x3 1 Dn 62 0 dnðxÞ¼0forx 6¼ 0; dnðxÞdDn ¼ : D 0 Dn 2 0 The left side of Eq. (I26) is equal to zero for any point (I24) that is not at the singularity. The notation implies over the volumetric neighborhood of the singularity that one sets the volume integral on the right side to the surface integral given The Dirac-delta function is a unit impulse at the origin. The in Eq. (I25) to yield a flux 4pH. value of the function approaches infinity as the size of the Finally, let us consider Eq. (I21) for @A=@x. We now infinitesimal region surrounding it approaches zero in recip- regard the 3D vector field @A=@x as the spatial part of the rocal proportion, such that the integral of the function over 4D vector field @A=@x. Using the 3D Dirac-delta function, any domain containing 0 is equal to one. The Kronecker- we can write delta function and the Dirac-delta function make it possible to, respectively, evaluate an element of a vector, and to eval- þ ð ð @A ⍜ ⍜ uate an integrandP of a function at a point. For example, the Á n dV ¼ 4p d3dV dx4 ¼ 4p d3dE: n @x sum fr ¼ s¼1 drsfs selectsÐ the rth component of the vector f (I27) and the integral f ð0Þ¼ Df ðxÞdnðxÞdDn evaluates the inte- grand at x ¼ 0. Now consider a scalar field A that is continuous in a 3D To accommodate the singularity, we write the differen- region except at its singularity. First, examine the flux that tial form given in Eq. (I18c) as the 3D vector field @A=@x produces. We say that the flux 2 2 2 2 that leaves the surface of a volume that contains the singular- @ A @ A @ A @ A 4pHd3 ¼ 2 þ 2 þ 2 þ 2 : (I28) ity is equal to 4pH. The flux that leaves the surface of any @x1 @x2 @x3 @x4 volume that does not contain the singularity is equal to zero. Using the 3D Dirac-delta function, we can write both of Doing the same for the components @Ar=@x ðr ¼ 1; 2; 3; 4Þ, these facts together as accounting for the singularity, we write
integral form differential form ð þ 2 2 2 2 ⍜ @Ar ⍜ @ Ar @ Ar @ Ar @ Ar 4p r d3dE ¼ Á n dV 4p rd3 ¼ 2 þ 2 þ 2 þ 2 : (I29) @x @x1 @x2 @x3 @x4
ðÞr ¼ 1; 2; 3; 4 ðÞor 4p⍜rd3 ¼ ٗ Á ٗAr
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