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Contribution of electric self-forces to electromagnetic momentum in a moving system

In moving electromagnetic systems, a certain formulation comprising the vector potential can be employed to calculate the electromagnetic momentum, which as we show, depends upon the potential energy of the system. However that does not account for the total electromagnetic momentum of the system. There is a subtle contribution from the electric forces between various constituents of the moving system that contribute to the electromagnetic momentum of the system, which can be true even for a non-relativistic motion. This, in fact, leads to resolution of some long-standing enigmas in the electromagnetic momentum of such systems, for instance in the apparent paradox of a {\em nil} electromagnetic momentum in a charged parallel-plate capacitor, with finite electromagnetic energy, moving along the plate separation. It also provides a natural solution to the century-old, apparent paradox of the famous factor of 4/3 in the electromagnetic momentum of a moving charge.

In moving electromagnetic systems, a certain formulation comprising the vector potential can be employed to calculate the electromagnetic momentum, which as we show, depends upon the potential energy of the system. However that does not account for the total electromagnetic momentum of the system. There is a subtle contribution from the electric forces between various constituents of the moving system that contribute to the electromagnetic momentum of the system, which can be true even for a non-relativistic motion. This, in fact, leads to resolution of some long-standing enigmas in the electromagnetic momentum of such systems, for instance in the apparent paradox of a nil electromagnetic momentum in a charged parallel-plate capacitor, with finite electromagnetic energy, moving along the plate separation. It also provides a natural solution to the century-old, apparent paradox of the famous factor of 4/3 in the electromagnetic momentum of a moving charge.

I. INTRODUCTION
Recently it has been shown how the energy-momentum densities of a perfect fluid with a bulk motion, comprising an ideal gas, can be derived from first principles. 1 Starting from the energy density and pressure in the rest frame of the gas, where the gas molecules may be moving randomly but with no bulk motion, and then using a Lorentz transformation of the velocity vectors of molecules to another inertial frame moving with respect to the rest frame of the fluid, the energy-momentum density of the system could be computed. Elsewhere, it has been demonstrated, 2 from a physical point of view, how a couple of pressure-dependent terms in a perfect fluid, represented in its rest frame by a diagonal stress-energy tensor, contribute to the energy-momentum density of the fluid, when undergoing a bulk motion.
A very similar situation, from a physical perspective, exists in the case of a moving electrically charged system, where the electromagnetic energy-momentum density is calculated from the symmetric, trace-free, electromagnetic stress-energy tensor 3 Here F µν and F µν stand for the antisymmetric electromagnetic field tensor and η µν (= η µν ) is the metric tensor of the special relativity 3 From the stress-energy tensor (Eq. (1)), using Eqs. (2) and (3), we get which gives the electromagnetic energy density, while T i0 /c yields i th component of the electromagnetic momentum density and is the symmetric Maxwell stress tensor, with T ij denoting i th component of the momentum flux in j th direction. It was shown explicitly 2 how like the contribution of pressure to the energy-momentum density in the case of a perfect fluid with bulk motion, non-zero terms in the Maxwell stress tensor (Eq. (6)) contribute to the energymomentum densities (Eqs. (4) and (5)) in a moving electrically charged system. However, some doubts have been expressed on the standard definition of the electromagnetic momentum density and whether the arguments based on the perfect fluid, as in the article, 2 can really be applied to the electromagnetic systems. 4 Here we provide simple arguments to dispel these doubts, clarifying the basic physics involved. That way, we show how one is led to the resolutions of some long-standing riddles in the electromagnetic momentum of such systems, for instance, in the apparent paradox of a nil electromagnetic momentum in a charged parallel-plate capacitor, with finite electromagnetic energy, moving along the plate separation or a natural solution to the famous, century-old but still intriguing factor of 4/3 in the electromagnetic momentum of a moving electric charge 5,6 , which shows up even in a non-relativistic motion 7 . We thereby show that even without invoking the fluid picture, one arrives at the same results though the former does add to the physical perspective. We use the cgs system of units throughout. Electromagnetic momentum in the fields of a charge parallel-plate capacitor, computed from the volume integral of Eq. (5) gives a finite value for the capacitor moving parallel to the plate orientations, however, it yields a nil value (with B = 0) for the same capacitor with a motion normal to the plate orientations. 2 It has, however, been argued 4 that the value of the electromagnetic momentum cannot depend upon the plate orientations of the system and that the fault may lie with Eq. (5). Taking a cue from the generalized force on a point charge q i due to velocity-dependent electromagnetic potentials 8,9 d dt where m i v i is the mechanical momentum of the point charge moving with an instantaneous non-relativistic velocity v i , and φ i and A i are scalar and vector potentials at the location of the charge q i , an assertion has been made 4 that instead of Eq. (5), the expression to be used for the contribution of electromagnetic field momentum from a point charge q i is q i A i /c, associated with electromagnetic momentum by Maxwell 10 . This form of electromagnetic momentum has been discussed at some length in the literature [11][12][13][14] . Accordingly, in case of n total charges, vector potential A i at the location of the charge q i is computed from the summation where |x i − x j | is the distance of charge q j , moving with velocity v j , from the location of the charge q i . Here j = i, of course. The contribution of q i to electromagnetic momentum then is The total electromagnetic momentum of the system is calculated from the sum Applying Eqs. (10) and (11) to a moving charged capacitor, it has been purportedly claimed 4 that the electromagnetic momentum of the system does not depend upon the plate orientations with respect to the direction of motion and a conclusion has been drawn therefrom that the fluid concept used in literature 2 cannot be applied to the analysis of electrodynamical systems. Our contention here is that Eq. (7) actually is a much more general expression, valid even when there are no electric charges in the system, for instance it can be employed to calculate the momentum carried by an electromagnetic wave. Further, as we will show, the electromagnetic momentum computed using vector potential A, is already being taken into account while calculating the electromagnetic field momentum of systems like a moving charged capacitor from Eq. (7).
In order to comprehend how q i A i /c may be contributing to the electromagnetic field momentum, we can rewrite Eq.(10) as where m ij is the mass equivalent of the potential energy of pair-wise charges q i and q j When v i and v j are different, there may be ambiguity whether the contribution to the electromagnetic field momentum from charge pair q i and q j should be m ij v j or m ij v i , though Eqs. (7) should yield an unambiguous answer. However, if all charges in the electromagnetic system are moving with the same velocity vector, v (the case we are going to deal with here), then one can write the total electromagnetic momentum of the system as Factor of 2 in the denominator is because in the summation each charge pair is getting counted twice. For a continuous charge distribution, described by a charge density ρ, an element of the electromagnetic momentum can be written as Here A is determined from the volume integral with |x − x | as the distance of the charge element ρ(x ) dτ , moving with velocity v(x ), from the field point x where A is to be evaluated. Then Eq. (14) becomes where is the electromagnetic mass equivalent of the potential energy of the system.

III. A CHARGED PARALLEL-PLATE CAPACITOR MOVING NORMAL TO PLATE ORIENTATIONS
We consider a charged parallel-plate capacitor moving normal to the plate orientations. For definitiveness we assume the capacitor to be moving with velocity v, along x-axis, with the capacitor plates lying in the y-z planes.
Let us calculate the electromagnetic momentum of a moving system from the vector potential as has been suggested. 4 We begin by first computing the scalar potential φ, at a point between the capacitor plates that lies at a distance, say, x from plate 1. For this we include contributions to φ from both the capacitor plates ( Fig. 1), as where L is the individual plate dimension. For x ≤ d L, we can write it as Thus taking φ = 0 at x = d/2, midway between the plates, we get φ = 2πσd at x = 0 and φ = −2πσd at x = d.

The vector potential is
The only finite component of the vector potential A is along x-direction.
The electric field E between the capacitor plates is then obtained as which is constant in the region between the capacitor plates, i.e., for 0 < x < d. The magnetic field, B = ∇ × A = 0 is zero everywhere. Then summing over all charges Q = σΣ, we get for the electromagnetic momentum contribution from charges on plate 1 and 2 as where τ = Σd is the capacitor volume between the plates. This is the electromagnetic momentum (Eq. (24)), calculated from the vector potential, which depends upon FIG. 1. Calculation of the scalar potential, φ at a point x between the charged capacitor plates, due to circular rings of radii r, centered on O1 and O2, from the charged capacitor plates 1 and 2, lying in the y-z planes. The surface charge densities are +σ and −σ, on plates 1 and 2 respectively. the potential energy, E = 2πσ 2 τ , of the system and is, of course, independent of the plate orientations. However that is not the total electromagnetic momentum of the system. There is a subtle contribution to the electromagnetic momentum due to the mutual force of attraction between the capacitor plates in the moving system, a contribution that might not be so obvious. This is because the capacitor system is moving and there is force of attraction on plate 1 F 1 = 2πσ 2 Σx, while on plate 2 it is F 2 = −2πσ 2 Σx (Fig. 2). As a result, on plate 1, moving with a velocity v, the electric attractive force F 1 does work at a rate F 1 · v. On the other hand, the force F 2 on plate 2 is in a direction opposite to v, therefore work being done on plate 2 (at a rate F 2 · v), is equal and opposite to that being done on plate 1. Even though there is no net change in the system energy as a result of these electric interactions in the moving capacitor system, there is a rate of continuous transport of energy taking place from plate 2 to plate 1 across the plate separation d, through their mutual force of electric attraction. This transport of energy per unit time forms a part of the total electromagnetic momentum of the system. If the capacitor plates were not clamped together and were thus free to move under the force of attraction, then plate 1 will have gained a velocity increment above v while the velocity of plate 2 would have an equal reduction below v. There is thus a transport of energy per unit time at a rate 2πσ 2 (Σ d) v from plate 2 to plate 1, through electric attraction, across separation d of the capacitor plates, each of cross-section area Σ. The resulting momentum is in a direction opposite to v.

FIG. 2.
In a charged parallel-plate capacitor moving with velocity v, the plate 1 is experiencing an attractive force F1 along x, while plate 2 is attracted with force F2 in the opposite direction. As both plates 1 and 2 are moving with velocity v, work is being done on plate 1 at a rate F1 · v, while the work done on plate 2, at a rate F2 · v, is equal and opposite to that being done on plate 1. This implies a transport of energy from plate 2 to plate 1. This transport of energy across the plate separation forms a part of the total electromagnetic momentum of the capacitor system. This momentum is in addition to Ev/c 2 , the momentum due to an overall movement of the system carrying an energy E, and the total electromagnetic momentum of the system, therefore, is Field momentum calculated from the volume integral in Eq. (7) is since B = 0 everywhere. Thus we see that the value of momentum in the electromagnetic fields of the capacitor, as calculated using Eq. (26), is the same as in Eq. (25), the latter derived here although without any reference to the fluid picture, yet turns out to be identical to the value derived from the fluid description. 2   FIG. 3. A spherical shell of radius with a uniform surface charge density σ, moving with velocity v along the x-axis. Two symmetrically placed circular rings, each of radius sin θ and of angular width dθ, on the two opposite sides of the spherical shell along the x-axis, separated by a distance l = 2 cos θ, are shown.

IV. ELECTROMAGNETIC MOMENTUM OF A MOVING CHARGED SPHERE
We consider a spherical shell of radius with a uniform surface charge density σ, and thus of a total charge e = 4πσ 2 , moving with a non-relativistic velocity v. The electromagnetic energy, E, of the system is the selfpotential energy of the charged sphere where τ = 4π 3 /3 is the total volume of the sphere. With this field energy E, one can associate an electromagnetic mass, E/c 2 , and an electromagnetic momentum Ev/c 2 , using the vector potential (Eq. (17)). The total momentum of the system includes, in addition, a term due to an energy flow associated with the self-repulsive forces within the constituents of the charge sphere and has been calculated for a relativistically moving system. 15 We show here that this can be appreciable even for a non-relativistic motion of the system. Consider two infinitesimal surface elements in the shape of circular rings, each of radius sin θ and of angular width dθ, separated by a distance l = 2 cos θ on two opposite sides of the spherical shell along the x-axis (Fig 3). Now, each ring of surface area dΣ = 2π 2 sin θ dθ, has a radially outward force 2πσ 2 dΣ, 16 which has a component 2πσ 2 cos θ dΣ along the x-axis. As the whole system is moving with a velocity v along the x-axis, the work being done per unit time by the electric self-repulsive force on the left ring is −2πσ 2 cos θ dΣ v and an equal amount of work per unit time, 2πσ 2 cos θ dΣ v by the electric selfrepulsive force is being done on the right ring, which lies at a distance l = 2 cos θ. In other words there is an energy flow at a rate 2πσ 2 cos θ dΣ v l = 4πσ 2 v cos 2 θ dΣ, through a cross-section dΣ cos θ of the system. Integration over the total cross-section of the system yields a net momentum term due to this energy-flow This momentum results from the existence of the forces of self-repulsion between the left-half and the right-half of the moving charged spherical system, and is present even for non-relativistic velocities.
The total electromagnetic momentum of the system therefore is This explains the famous factor of 4/3 in the electromagnetic momentum of a moving charged particle, even for non-relativistic velocities. [5][6][7] Onoochin's argument 4 for the explanation of the Trouton-Noble experiment, 17 that the electromagnetic momentum is independent of orientation of the charged plates with respect to the direction of motion of the capacitor, is not appropriate. Actually, in such a system in equilibrium, but having some net electromagnetic forces, like the mutual forces of attraction between the capacitor plates or the forces of self-repulsion between the constituents of a charged sphere, there must exist some stabilization forces (Poincaré stresses 18 ), e.g., the 'mechanical' forces of a rod keeping the two capacitor plates apart, or the (yet unknown!) forces keeping the charged sphere intact, and these stabilization forces must be equal and opposite to the electric forces to keep the system in equilibrium. Once the system is in equilibrium under the electromagnetic forces and the stabilization forces, 19,20 both types of forces will make equal and opposite contributions to the energy-momentum of the total system, which does explain successfully 20 the null results of the famous Trouton-Noble experiment. 17 However, the contribution of these stabilizing forces would not show up in the energy-momentum formulation, e.g. Eqs. (4) and (5), which can account for only the electromagnetic interactions present in the system.

V. DISCLOSURES AND DECLARATIONS
The author has no conflicts of interest/competing interests to declare that are relevant to the content of this article. No funds, grants, or other support of any kind was received from anywhere for this research.