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

There is an apparent paradox in the electromagnetic momentum of a moving charged parallel-plate capacitor. One expects such a system to possess, along the direction of motion, an electromagnetic momentum that can be calculated from the electromagnetic field energy within the moving capacitor. However, for the capacitor plate surfaces parallel to the direction of motion, electromagnetic momentum computed from its fields turns out to be twice that expected from the field energy of the system, while for plate surfaces perpendicular to the direction of motion, the field momentum computation yields a nil value [1, 2]. Under the assumption that the electromagnetic momentum of the moving charged parallel-plate capacitor should not depend upon the plate orientations, a suggestion has been floated [3] that the electromagnetic momentum of a system should instead be calculated using a formulation based on vector potential. However, electromagnetic momentum computed that way does not yield the total electromagnetic momentum of the system, as computed from the standard definition of the electromagnetic momentum density, there seem to some contribution to the electromagnetic momentum missing.

Elsewhere, it has recently been shown that additional contributions to momentum arise from the pressure or stress within a perfect fluid, that could be comprising an ideal gas, when the system is undergoing a bulk motion and similar is the case in electromagnetic systems [4, 5]. From that it could accordingly be construed that it is the stress, arising due to the electromagnetic forces between various constituents of the charged system, whose contribution in a moving electromagnetic system might apparently be the term that is missing in electromagnetic momentum computed from vector potential. However, doubts have been expressed [3] not only upon whether the arguments employed in the case of perfect fluids [4] are really compatible with and could thus be applied in electromagnetic systems, but even on whether the standard definition of the electromagnetic momentum density itself is proper and thus might need to be modified.

Here, we examine such doubts in detail and provide simple arguments to dispel them, clarifying the basic physics involved. For that purpose, we compute the electromagnetic momentum of a moving system from the electromagnetic stress-energy tensor, determining contributions arising from various stress terms in the Maxwell stress tensor. We show that these additional contributions could be significant even for a non-relativistic motion of the system. It should be clarified that even though we shall be dealing with non-relativistic motion of the system, all our arguments are under the aegis of the special theory of relativity, in particular the mass-energy relation, where in an electromagnetic system, the electric potential energy of the charges (or the equivalent electric field energy) contributes to the mass of the system and that there is momentum associated with the energy flux in a system. We thereby demonstrate that even without referring to the fluid description, one arrives at similar results using the stress energy tensor in electromagnetic systems. Although we do not explicitly invoke the fluid picture here, the underlying physics and the mathematics involved are essentially the same in the either description since pressure in a fluid system corresponds to stress in the electromagnetic fields, and this correspondence between the two does add to the physical perspective.

The electromagnetic energy-momentum density can be computed from the symmetric, trace-free, electromagnetic stress-energy tensor [6] (in cgs units)

Here $\eta^{\mu\nu}(=\eta_{\mu\nu})$ is the metric tensor of the special relativity [6, 7]

and $F^{\mu\nu}$ and $F_{\mu\nu}$ stand for the antisymmetric electromagnetic field tensors

We follow the convention where Greek letters $\alpha,\beta,\mu,\nu$ etc. take the values $0,1,2,3$ while Latin letters $i,j$ take the values $1,2,3$.

From the stress-energy tensor (Eq. (1)), using Eqs. (2) and (3), we get the electromagnetic energy density

and the energy flux ($c{T}^{0i}$) in $i$th direction, as well as $\it i$th component of the electromagnetic momentum density (${T}^{i0}/c$) as

while the symmetric Maxwell stress tensor, given by

denotes $\it i$th component of the momentum flux in $\it j$th direction. Kronecker delta, $\delta^{ij}=1$ for $i=j$ and zero otherwise. Even in the simplest of electromagnetic field configurations, for instance for a uniform electric field in a region, as in the case of a charged parallel plate capacitor, there is finite stress in the fields. From the Maxwell stress tensor (Eq. (6)) we find a tension or negative pressure, $-(E^{2}+B^{2})/{8\pi}$, along the field lines, however, at right angles to the field lines there is a positive pressure [6, 8].

We want to determine the electromagnetic momentum of a system in the lab frame $\cal K$, where the rest frame $\cal K^{\prime}$ of the system is moving with a uniform velocity $\bf v$ along the $x$-axis. With $\beta=v/c$ and the Lorentz factor $\gamma=1/\sqrt{1-\beta^{2}}$, the matrix for Lorentz transformation between $\cal K^{\prime}$ and $\cal K$ is given by [6]

a prime on a quantity indicating it refers to the rest frame $\cal K^{\prime}$ of the charged system.

From a Lorentz transformation of the stress-energy tensor from the rest frame $\cal K^{\prime}$, where the electromagnetic momentum may be taken as zero, to the lab frame $\cal K$

Throughout, we denote the stress-energy tensor in the rest frame with a prime, $T^{\prime}$, in order to distinguish its components from those in the lab frame, for instance, $T^{11}$ represents one of the components of the Maxwell stress tensor in the lab frame while $T^{\prime 11}$ stands for the corresponding component in the rest frame. It will be shown that in the energy flux ($c{T}^{0i}$) as well as the electromagnetic momentum density (${T}^{i0}/c$) in the lab frame there are terms arising from the contribution of Maxwell stress tensor ($T^{\prime i^{\prime}j^{\prime}}$) in the rest frame. It is these additional terms which lead to the resolutions of the above-mentioned paradox in the moving charged capacitor as well as an explanation of the mysterious factor of 4/3 in the electromagnetic momentum of a moving charge, a puzzle that apparently defied a proper resolution for almost a century [9, 10, 11].

We consider a charged parallel-plate capacitor at rest in reference frame $\cal K^{\prime}$, with surface charge densities $+\sigma$ and $-\sigma$ on plates 1 and 2 respectively, to be moving in the lab frame $\cal K$ with a uniform velocity $\bf v$, which is non-relativistic ($\beta=v/c\ll 1$) so that we keep in our formulations only the lowest order terms in $\beta$. We assume the capacitor plate surfaces are perpendicular to the direction of motion. For definitiveness we assume $\bf v$ to be along the $x$-axis, with the capacitor plates in shape of circular discs, each of radius $L$, lying in the y–z plane. Then for $t=0$, $x^{\prime}=x$.

In order to calculate the electromagnetic momentum of a moving system from the vector potential, we begin by first computing the scalar potential $\phi^{\prime}$, in the rest frame $\cal K^{\prime}$, at a point between the capacitor plates that lies at a distance, say, $x^{\prime}=x$ from plate 1. For this we include contributions to $\phi^{\prime}$ from both the capacitor plates (Fig. 1), as

where $L$ is the individual plate dimension.

For $x^{\prime}\leq d\ll L$, we can write it as

Thus taking $\phi^{\prime}=0$ at $x^{\prime}=d/2$, midway between the plates, and with the restriction $|x^{\prime}|\ll L$, we get $\phi^{\prime}=2\pi\sigma d$ at plate 1 ($x^{\prime}=0)$, and $\phi^{\prime}=-2\pi\sigma d$ at plate 2 ($x^{\prime}=d$). The vector potential ${\bf A}^{\prime}=0$ everywhere, in the rest frame $\cal K^{\prime}$.

The electric field $\bf E^{\prime}$ between the capacitor plates is then obtained, in frame $\cal K^{\prime}$, as

which is constant in regions between the capacitor plates. The magnetic field is zero everywhere in frame $\cal K^{\prime}$.

Now we can calculate the potentials in frame $\cal K$ from a Lorentz transformation (Eq. (24)) of the 4-vector ($\phi,{\bf A}$) from frame $\cal K^{\prime}$ (Eq. (26)), at $t=0$ with $x^{\prime}=x$. as

The only finite component of the vector potential ${\bf A}$ is along $x$-direction, with values $A_{\rm x}=2\pi\sigma vd$ at plate 1 ($x=0)$, and $A_{\rm x}=-2\pi\sigma vd$ at plate 2 ($x=d$).

The electric field in frame $\cal K$ is the same as ${\bf E^{\prime}}$ in frame $\cal K^{\prime}$.

which is constant in the region between the capacitor plates. The magnetic field, ${\bf B}=\nabla\times{\bf A}=0$ in $\cal K$ too.

Then integrating over the charges $\sigma S$ on plate 1 and 2, where ${S}$ is the cross-section area on each plate, we get for the electromagnetic momentum contribution from Eq. (17)

Then ${\bf P}_{\rm A}={{\cal E}{\bf v}}/{c^{2}}$, where ${\cal E}$ is the system potential energy equal to the electrostatic field energy in the system, given by [22]

Thus the electromagnetic momentum (Eq. (30)), as calculated from the vector potential, depends upon the electromagnetic energy ${\cal E}$ of the system. This momentum is due to an overall movement of the system carrying an energy ${\cal E}$ and can be understood from the Lorentz transformation where energy ${\cal E}$ in the rest frame $\cal K^{\prime}$ gives a momentum ${\cal E}{\bf v}/{c^{2}}$ in lab frame $\cal K$ for a non-relativistic motion ($\gamma\approx 1$).

However, from Eq. (9) we get in this case a conflicting value for the electromagnetic field momentum

since in this particular case the magnetic field is zero (${\bf B}=0$) throughout. Thus there seems to be a mismatch in the two approaches to calculate the electromagnetic momentum of the system.

This, of course, is quite puzzling as one would have expected the two formulations to yield consistent results for the electromagnetic field momentum. Could the discrepancy be due to a choice of gauge as Eq. (11) appears to be gauge dependent? Total electromagnetic momentum of the system, nevertheless, should not depend upon the chosen gauge. In any case, ${\bf P}_{\rm A}={\cal E}{\bf v}/{c^{2}}$, calculated from the vector potential. is giving a value, expected even on physical grounds for a system carrying an energy ${\cal E}$ and moving with velocity ${\bf v}$. Now as for Eq. (32), any gauge choice cannot turn the electromagnetic field momentum zero from an otherwise finite value, as in Eq. (30). After all a gauge for the vector potential always preserves the electric and magnetic field values. So no particular gauge choice can make the magnetic field zero so as to make the electromagnetic field momentum also zero as in Eq. (32). Therefore the discrepancy in the electromagnetic momentum values, between Eqs. (30) and (32), is due to something else that might have been overlooked and the contribution therefrom not taken into consideration.

That something is the momentum contribution, amiss from Eq. (30), that arises from the stress in the electromagnetic fields in system, given by the Maxwell stress tensor. To ascertain that missing contribution, we first calculate the stress-energy tensor in the rest frame $\cal K^{\prime}$

The negative sign in $T^{\prime 11}\;(=-E^{\prime 2}/8\pi)$ here indicates tension (negative pressure) along the field lines, which are along $x$ direction (${\bf E^{\prime}}=E^{\prime}_{\rm x}\hat{\bf x}$).

It should be noted that a Galilean transformation of the the energy-momentum tensor (23) leads to Eq. (30). On the other hand, a more appropriate Lorentz transformation to the lab frame $\cal K$ for non-relativistic velocities (Eq. (21)) yields additional, off-diagonal components for the stress-energy tensor

where $cT^{01}$ is the energy flux in $x$ direction, and $T^{10}/c$ is the $x$ component of the electromagnetic momentum density.

In Eq. (34), the term $T^{\prime 00}\beta$ is due to the movement of the capacitor having stored energy density $T^{\prime 00}=E^{2}/8\pi$, while the other term, $T^{\prime 11}\beta$, represents the contribution of stress to the energy flux and the momentum density. The volume integral of $T^{\prime 00}{\bf v}/c^{2}$, is the same as the momentum computed from the vector potential. The other term gives an additional contribution to momentum from stress $T^{\prime 11}$ in fields, which as we show below, is due to the forces of attraction between the capacitor plates. Since $T^{\prime 11}=-T^{\prime 00}$ (Eq. (33)), the net electromagnetic momentum, obtained from $T^{10}/c$ in Eq. (34) is zero, the same as in Eq. (32).

From a physical perspective, the genesis of the additional terms in the energy flux and momentum density, can be traced to the work done on each moving plate by the electric field caused by the opposite plate. Owing to the motion of the system, there is thus a subtle additional contribution ${\bf P}_{\rm add}$ to the electromagnetic momentum, arising from the mutual electric forces of attraction on the moving charged plates, a contribution that might not be so obvious.

The capacitor plates moving with velocity ${\bf v}$ form two current sheets, each of thickness, say $\delta$, carrying current densities, ${\bf j}_{1}=\sigma{\bf v}/\delta$ on plate 1, and ${\bf j}_{2}=-\sigma{\bf v}/\delta$ on plate 2. The electric field at the location of each plate, due to the other plate, is ${\bf E}=2\pi\sigma\hat{\bf x}$ [22]. As a result, on plate 1 the electric field does work at a rate ${\bf j}_{1}\cdot{\bf E}$ per unit volume, while on plate 2 it is ${\bf j}_{2}\cdot{\bf E}$. Then total work being done per unit time on plate 1 is ${\bf j}_{1}\cdot{\bf E}S\delta=2\pi\sigma^{2}Sv$, while it is ${\bf j}_{2}\cdot{\bf E}S\delta=-2\pi\sigma^{2}Sv$ on plate 2. Equivalently, due to mutual force of attraction, ${\bf F}_{1}=2\pi\sigma^{2}S\,\hat{\bf x}$ on plate 1 and ${\bf F}_{2}=-2\pi\sigma^{2}S\,\hat{\bf x}$ on plate 2 (Fig. 2) [22], work is being done at a rate ${\bf F}_{1}\cdot\bf{v}$ on plate 1 and ${\bf F}_{2}\cdot\bf{v}$ on plate 2. Even though there is no net change in the system energy, plate 1 gains energy at the cost of plate 2, at a temporal rate $2\pi\sigma^{2}Sv$. This implies a continuous flow of energy $2\pi\sigma^{2}Sv$ per unit time, taking place across the distance of plate 1 from plate 2, due to the electric self-fields of the moving capacitor. This transport of energy across a distance $d$ per unit time forms an additional electromagnetic momentum contribution

to the system.

Such a transport of energy taking place between the two capacitor plates would be quite obvious if the two plates were not constrained to stay together (by some non-electromagnetic forces!) and were thus free to move under the influence of the electric field. In that case, plate 1 would have gained a velocity increment above ${\bf v}$ while the velocity of plate 2 will have an equal reduction below ${\bf v}$.

The momentum term ${\bf P}_{\rm add}$ is in addition to ${\bf P}_{\rm A}$, calculated in Eq. (30) using the vector potential. The total electromagnetic momentum of the system, therefore, is

which is consistent with the conclusion reached from Eq. (34), and is in agreement with electromagnetic momentum value being nil, as arrived at in Eq. (32). Thus the momentum in fields actually represents the total electromagnetic momentum (${\bf P}_{\rm em}\equiv{\bf P}_{\rm f}$) of the system.

The electromagnetic momentum calculated in Eq. (30) accounts for only the first term, viz. $2\pi\sigma^{2}\tau{\bf v}/{c^{2}}$, in Eq. (36). The electromagnetic momentum computed from the vector potential in Eq. (30) does not seem to account for momentum in the system due to the energy transport in the moving system owing to the mutual electric force between the system constituents. Thus we also see that the value of momentum in the electromagnetic fields of the capacitor, as calculated using Eq. (36), which is the same as in Eq. (32), though derived here without any reference to the fluid picture, yet turns out to be identical to the value derived from the fluid description [4].

Here we take the rest frame $\cal K^{\prime}$ of the charged capacitor to be moving in the lab frame $\cal K$ with a uniform non-relativistic velocity $\bf v$ ($v\ll c$), along the $x$-axis but with capacitor plates lying in the x–z planes and the plate separation along the $y$-axis,

The scalar potential $\phi^{\prime}$, in frame $\cal K^{\prime}$, in this case a distance $y^{\prime}$ from plate 1 (Fig. 3), is calculated as

which, for $y^{\prime}\leq d\ll L$, yields

The vector potential ${\bf A}^{\prime}=0$ in rest frame $\cal K^{\prime}$, everywhere.

The electric field $\bf E^{\prime}$ between the capacitor plates, in frame $\cal K^{\prime}$, is then obtained as

while the magnetic field is zero everywhere, in frame $\cal K^{\prime}$.

We can compute the potentials in frame $\cal K$ from a Lorentz transformation of the 4-vector ($\phi,{\bf A}$) from frame $\cal K^{\prime}$ (Eq. (40)) to $\cal K$, using (Eq. (24)). As $y^{\prime}$ in frame $\cal K^{\prime}$ corresponds to $y$ in $\cal K$, we get

Here too the only finite component of vector potential ${\bf A}$ is along $x$-direction. Potentials $\phi$ and $A_{\rm x}$ depend only upon $y$ and are independent of $x$.

The electric field in frame $\cal K$, for $\gamma\approx 1$, is given by

The electric field, ${\bf E}$, in frame $\cal K$ is thus along $y$-axis, and is constant between the capacitor plates.

The magnetic field, ${\bf B}=\nabla\times{\bf A}$, is then given by

which is consistent with ${\bf B}=\mbox{\boldmath$\beta$}\times{\bf E}$ [22].

We can get the electromagnetic momentum, ${\bf p}_{\rm A}$, as

The field momentum on the other hand is

Here we see that the electromagnetic momentum (Eq. (45)), calculated from the vector potential, is only half of that calculated from fields (Eq. (46)).

We can also obtain the electromagnetic momentum from the stress-energy tensor, which in the rest frame, $\cal K^{\prime}$ is given by

Here $T^{\prime 11}=E^{\prime 2}/8\pi$ show a positive pressure along the $x$-axis, which is perpendicular to the field lines.

While a Galilean transformation of the the energy-momentum tensor (23) leads to Eq. (45), a more appropriate Lorentz transformation for non-relativistic velocities (Eq. (21)) gives two additional, off-diagonal components for the stress-energy tensor in the lab frame $\cal K$ as

where in the lab frame $cT^{01}$ is the energy flux in $x$ direction and $T^{10}/c$ is the $x$ component of the electromagnetic momentum density.

In Eq. (48), the term proportional to $T^{\prime 00}\beta$, corresponds to ${\bf P}_{\rm A}$ (Eq. (45)), the momentum calculated from the vector potential. The other term, proportional to $T^{\prime 11}\beta$, corresponds to an additional energy flux as well as an additional momentum density. In consistency with Eq. (46), the net electromagnetic momentum, obtained from volume integral of $T^{10}/c$ in Eq. (48) is ${E^{2}\tau{\bf v}}/{4\pi c^{2}}$ since $T^{\prime 11}=T^{\prime 00}=E^{2}/{8\pi}$ here (Eq. (47)).

Now here due to motion of the system, which is along the $x$-axis, no work is done by the forces of mutual attraction between the plates, or by the mechanical forces keeping the plates separated, as they are along the $y$-axis. However, there are electromagnetic forces of repulsion along the plate surfaces, on similar charges within each plate, forces that might not be so apparent [1]. There is a continuous energy flow taking place along the direction of motion, from the work done, by the forces of repulsion within each capacitor plate, which contributes to the momentum of the system moving along the $x$-axis. Its evaluation, computationally rather involved [1], yields momentum contribution, ${\bf P}_{\rm add}=2\pi\sigma^{2}\tau{\bf v}/{c^{2}}$, which turns out to be the same as calculated from the stress component $T^{\prime 11}$ ($=E^{2}/8\pi$) in the electromagnetic stress-energy tensor (Eq. (47)), implying a momentum density $E^{2}{\bf v}/(8\pi c^{2})$ (Eq. (48)). This ${\bf P}_{\rm add}$ when added to ${\bf P}_{\rm A}$ in Eq. (45), gives the total electromagnetic momentum of the system

which agrees with the electromagnetic momentum computed from fields (Eq. (46)).

There are, of course, stabilizing forces that keep the charges fixed in spite of the forces of repulsion between the charges on each plate and these contribute a mechanical momentum, ${\bf P}_{\rm m}=-{\bf P}_{\rm add}=-E^{2}\tau{\bf v}/(8\pi c^{2})$, to the system. The total momentum of the system then is

Thus, for a system of rest-frame energy ${\cal E}$, moving with velocity ${\bf v}$, it is the total momentum (electromagnetic + mechanical) that transforms as the component of a 4-vector, to yield ${\cal E}{\bf v}/{c^{2}}$.

An argument has been put forward in the literature [3] that a proper explanation of the famous Trouton-Noble experiment [24] lies in the fact that the electromagnetic momentum, ${\bf P}_{\rm A}$, of the system, calculated using the vector potential (Eqs. (30) and (45), is independent of the plates orientations with respect to the direction of motion. However, this argument is not correct as we have shown here that ${\bf P}_{\rm A}$ accounts for the electromagnetic momentum only partially, and that the total electromagnetic momentum differs for different plate orientations. The real explanation of the Trouton-Noble experiment, and which has been shown explicitly [2, 25], is that once the system is in equilibrium in the rest-frame under the electromagnetic forces and the stabilization forces, it also remains in equilibrium in all frames of reference, since both types of forces will transform relativistically in exactly the same manner from one frame to another.

We consider a spherical shell of radius $\epsilon$ and a surface area $S=4\pi\epsilon^{2}$, with a uniform surface charge density $\sigma$, and thus of a total charge $e=4\pi\epsilon^{2}\sigma$. The electric field of the charge in rest frame ${\cal K}^{\prime}$ is ${\bf E}^{\prime}=(e/{r^{2}})\hat{\bf r}$ for $r>\epsilon$, but zero for $r<\epsilon$. From ${\bf r}=x\hat{\bf x}+y\hat{\bf y}+z\hat{\bf z}$, we can express ${\bf E}^{\prime}=e{\bf r}/r^{3}=e(x\hat{\bf x}+y\hat{\bf y}+z\hat{\bf z})/{r^{3}}$ for $r>\epsilon$. The magnetic field of course is zero in ${\cal K}^{\prime}$. Writing $\mbox{\boldmath$\beta$}={\bf v}/c$ with $\mbox{\boldmath$\beta$}=\beta\hat{\bf x}$, electromagnetic fields in lab frame ${\cal K}$ for a non-relativistic motion of the sphere, to a first order in $\beta$, with $\gamma\approx 1$, are given by ${\bf E}={\bf E}^{\prime}$ and ${\bf B}=\mbox{\boldmath$\beta$}\times{\bf E}$ [22].

The electromagnetic energy, ${\cal E}^{\prime}$, of the system is the self-potential energy, evaluated from a double surface integral over the charged sphere, equal to a volume integral of the field energy density $E^{\prime 2}/8\pi$ [22],

where $\tau=4\pi\epsilon^{3}/3$ is the total volume of the sphere.

With the rest frame field energy ${\cal E}^{\prime}$, one can associate an electromagnetic mass, ${\cal E}^{\prime}/c^{2}$, and thereby, for a motion with velocity v in lab frame $\cal K$, an electromagnetic momentum