A current loop is modeled as N positive point charges, at rest on a circular contour in inertial frame K, and N negative point charges moving around the contour at a constant, common speed. By computing the positions of the positive and negative charges in frame K’ at the single instant t’=0, it is shown that the electrically neutral current loop in frame K is electrically polarized in frame K’.

2.  Introduction

We begin by addressing the following question: If x_i(t), y_i(t) and z_i(t), the rectangular position coordinates of particle "i" in some set of N particles, are known in inertial frame K at any instant t, then what are the coordinates in inertial frame K’ at a given instant in K’? As usual, K’ moves in the positive x-direction of K at speed v, and the respective axes of the two frames are parallel. The clocks at the origins coincide when they mutually read zero.

Classically (i.e. before Special Relativity), the Galilean transformation provided a straightforward solution to this problem:

, (2.1a)

, (2.1b)

, (2.1c)

. (2.1d)

Eq. 2.1d stipulates that all of the clocks, distributed throughout and at rest in their respective frames, are synchronized at all times.

The Galilean transformation produces acceptable results when the relative speed of K and K’ is much less than c (the speed of light) and when the spatial domain is not very great (lab experiments, etc.). But rigorously speaking the more general Lorentz transformation must be used when the relative speed of the two frames is close to c (and/or when the distances are vast):

, (2.2a)

, (2.2b)

, (2.2c)

. (2.2d)

Eq. 2.2d implies that the clocks in K are not, in the opinion of K’, all synchronized (and vice versa). For example, if (from the perspective of K) two events occur simultaneously but at different x-coordinates, then they occur at different times according to the clocks of K’. Of course when there are multiple particles involved (and when the events are simply the coincidence of the particles with clocks at rest in K and K’ when those clocks coincide), then by the set {x_i’(t’)} K’ presumably means the x_i’-coordinates at any single instant t’ in frame K'. It will be demonstrated herein that given a knowledge of x_i(t), x_i’(t’) can always be computed if not analytically determined, with similar remarks applying to the other rectangular coordinates.

Let us focus upon the following simple system: N particles are evenly spaced around a circular contour that is centered on the origin and at rest in the xy-plane of frame K. The cases considered will be (1) the particles are at rest in K, and (2) the particles move around the contour at constant speed u. In general the particles are distinguished by integer tags (denoted by "i" subscripts above), and at t=0 they have the locations depicted in Fig. 2.1. (Only four particles are shown for the sake of clarity.)

3.  Case 1: u=0

In this case the particles are all at rest in K. That is, each particle permanently coincides with one of the clocks at rest in K. Indeed we can visualize the particles as being soldered to a rigid, circular hoop. We would like to determine where the particles are in K’, at the one instant t’=0. Or in other words, we would like to determine the hoop’s shape, viewed from K’. For openers, we note that the Lorentz transformation (Eqs. 2.2a-d) gives the spatial coordinates at many different instants in K’.

Let us first consider a particle with a positive x-coordinate. At time t=0 Eqs. 2.2a and d simplify to:

, (3.1a)

. (3.1b)

According to K’: when the K-clock at K-coordinate x reads zero, (a) the coincident K’-clock has the x’-coordinate specified by Eq. 3.1a, and (b) that K’-clock reads t’ as specified by Eq. 3.1b. In the opinion of K’ the event {particle coincides with K-clock at K-coordinate x when that clock reads t=0} occurred before t’=0. But K’ wants to know where, in K’, the particle is at time t’=0, which is to say after a time span of |t’| has elapsed.

. (3.2)

(Since the particle permanently coincides with the K-clock at x, we have dropped the "(t)" in "x(t)".) Using this new value for t (i.e. using t₁) in Eq. 2.2a produces

. (3.3)

And of course Eq. 2.2d becomes

. (3.4)

Eq. 3.3 expresses the celebrated phenomenon of length contraction. In the present case our rigid hoop, which is circular in K, will be ovoid at any instant in K’. If f is the angle which some point (x’,y’) on the oval makes with the positive x’-axis, then x’ and y’ can be expressed in terms of f and the circle’s radius in its rest frame K. (See Appendix B.)

If x(t) is less than zero, then Eq. 3.1b produces t’>0. In the opinion of K’, the event {particle coincides with K-clock at K-coordinate x when that clock reads zero} has already happened. In this case K’ wants to know which of the K’-clocks the particle coincided with when that clock read zero. Letting t₁ = -vx/c² again produces Eq. 3.3

4.  Solving for Case 1 Using a Computer

Let us again consider a positive x-coordinate, so that t’ in Eq. 3.1b is less than zero. In the previous section it might be said that we adjusted t (the reading of the coincident K-clock) to t₁ = g^-1|t’| in one fell swoop. We were able to do this because the particle permanently coincides with the K-clock at K-coordinate x. More generally, when the particle moves in K, we must "creep up" to the value t₁ of a coincident K-clock when the coincident K’ clock reads practically zero. (Although there is no reason to do this in the present case, the technique will be useful when u>0.)

An algorithm that implements this "creeping up" strategy is as follows:

For large m this algorithm will require many loops, but this presents no problem for a modern computer.

Fig. 3.1 plots 8 particles (N=8) in K and K’ at times t=0 and t’=0 respectively. (The particles are at the corners of the polygons. The data points were computed by Program "Sect4" in Appendix A.) A relative frame speed of v=.8c was used. The positions in K are plotted in blue, and the positions in K’ are plotted in red. Note the length contraction in K’. Also note that y=y’ and z=z’ for each particle.

5.  Case 2: u=v.

Let us now consider the case where N particles travel at a constant speed, say u=v=.8c, around the circular contour in K. We shall suppose that the sense of rotation is counterclockwise, looking down on the xy-plane from the positive z-axis. It might be said that in frame K the particles are evenly spaced beads, sliding on a rigid, circular wire.

Before computing the x_i’(t’=0) in K’, let us briefly discuss what might be expected. Every observer must agree that all the particles are in contact with the hoop at any given moment. And the hoop must appear to be length contracted when it is moving. But since a given particle does not permanently coincide with any particular clock in K, we should not expect that y_i’ will equal y_i in this case.

Fig. 5.1 again plots x(t=0), y(t=0), and x’(t’=0), y’(t’=0) when N=8. (The data points were computed by Program "Sect5" in Appendix A.) Note in this case that we find the usual length contraction (of the hoop) in K’, but also that y’=y only for particles 2 and 6. In the case of particles off the y/y’ axis there is a shift upward (toward positive y’) in K’.

6.  The Relativity of Particle Angular Density.

Let f be an angle formed by a point on the oval (length contracted circular contour) and the positive x’-axis in the x’y’-plane of frame K’:

. (6.1)

And assume that N=36,000 particles are evenly distributed around the circular contour in Frame K. Define the particle angular density, r’(f), to be the number of particles between f and f+Df at time t’=0. (In frame K the particle density is single-valued and equal to 100/degree.)

It is clear in Figs. 4.1 and 5.1 that the angular density in frame K’ varies with f. In the case where the particles are at rest in K, we expect the particles to be more tightly packed at the top and bottom of the hoop in K’ (owing to length contraction), and Fig. 4.1 confirms with that expectation. And in the case where the particles move around the circular contour in K, Fig. 5.1 indicates that (at any given instant in K') they evidently congregate toward the top of the hoop (where their speeds relative to K’ are greatest).

Since the coordinates of any particle can be computed at time t’=0 in frame K’, the angular density can be computed and plotted as a function of f. In this case the angular densities for (a) particles at rest in K, and (b) particles moving in K are plotted at the single instant t'=0 in K'. Fig. 6.1, whose data points were computed by Program "Sect6" in Appendix A, depicts such plots. Again it was assumed that u=v=.8c in the case of particles moving in K. The moving case is plotted in red, and the resting case is plotted in blue. Note the greater density of the resting (in K) particles at f = 90^o and 270^o, and the greater density of the moving (in K) particles at the hoop’s top (f < 180^o).

7.  An Uncharged Current Loop.

Let us model an uncharged current loop as N=36,000 positively charged particles, at rest on the usual circular contour in frame K, and the same number of negatively charged particles going around the circle at constant speed u=v=.8c. At time t=0 the charges are superimposed in frame K. The positive and negative angular densities are both 100/degree in frame K.

As seen in the previous section, owing to length contraction r₊’(’(f) will vary with f. In Frame K’ the positive particles will be more densely packed at f=90^o and f=270^o. And, because the negative particles are moving in frame K, they will have their own peculiar density in K’ (see Fig. 6.1). We need only multiply each negative charge density by minus 1 and sum the two densities in order to obtain the net charge density. Fig. 7.1 plots the result. (The data points were computed by Program "Sect7" in Appendix A.)

It is clear in Fig. 7.1 that an uncharged current loop, which is unpolarized in its "rest" frame K (i.e. the rest frame of the positive charges), will be electrically polarized in Frame K’. The electric dipole moment lies in the current loop’s plane and points perpendicular to the loop’s motion in K’. It should be noted that the net charge of the loop is zero in both frames. But in K’ the loop has some of the properties of an electric dipole.

8.  A Possible Role of Polarization in the Field Transformations.

Let us consider an arbitrarily large (or infinite), uncharged sheet magnet, centered on the origin and at rest in the xy-plane of frame K. At points (x,y,z), where x, y and z are all small compared to the sheet’s x and y dimensions, B_z (the sheet’s magnetic field) is practically single-valued. And of course E=0 everywhere.

From the perspective of one of these points, we can model the magnet as a large number of microscopic current loops arrayed in the xy-plane. (R, the radius of any one of these loops, is presumably very small compared to z.)

Now according to the field transformations there is an electric field in K’:

. (8.1)

Yet presumably dB’/dt’=0 in K’ as in K! It is clear on inspection of Fig. 7.1 that the electric polarization of each microscopic current loop (in K’) may contribute to this nonzero E_y’.

9.  "Homopolar" Induction.

The fact that E may be zero in K and E’ may be nonzero in K’, despite the fact that dB/dt = dB’/dt’ = 0, is often referred to as homopolar induction. The mechanism underlying this field transformation result has been a matter of some debate.

Recent experiments in Argentina indicate that such E' fields may be measurable phenomena. In those experiments a spinning, wafer-shaped, uncharged permanent magnet was employed. Although the theoretical microscopic current loops in such cases are not at rest in any inertial frame, it is not unreasonable to expect that any one of them will be electrically polarized when the magnet spins relative to the lab frame. In such cases the electric dipole moments would point radially inward or outward, again perpendicular to the current loop's instantaneous velocity relative to the lab frame. Such electric dipoles should engender radial electric field components, even though dB/dt = 0.

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