## How Relativity Connects Electric and Magnetic Fields

Michael Fowler

University of Virginia

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Suppose we have an infinitely long straight wire, having a charge density of electrons of coulombs per meter, all moving at speed v to the right (recall typical speeds are centimeters per minute) and a neutralizing fixed background of positive charge, coulombs per meter. The current I in the wire has magnitude (and actually is flowing to the left). Suppose now a positive charge q outside the wire, a distance r from the axis, is also moving at speed v. (Of course, both positive and negative charges are distributed uniformly throughout the wire. We just show them separated for ease of visualization.)

What force does q feel?

The magnetic field strength is (from ), and , so

The magnetic force F on q is , pointing directly away from the wire, and of magnitude

.

Therefore, if the initial situation is as in the figure above, the charge will accelerate away from the wire.

Let us now examine the same physical system in the frame of reference in which the charge is initially at rest. In this frame, the electrons are also at rest, but the positive background charge is flowing at -v as shown.

What force does q feel in this frame? Since q is at rest, it cannot feel a magnetic force, since such forces depend linearly on speed.

Yet it looks as if it can't feel an electric force either, because of the equal densities of positive and negative charges in the wire. So, we might conclude that in this frame q feels no force at all, so will not move. But we just proved that in the other frame, q accelerates away from the wire! These statements can't both be right-if q moves away from the wire in one frame, it must in the other (recall also these frames are moving relative to each other at centimeters per minute).

The mistake in the above argument is that we neglected Fitzgerald-Lorentz contraction. If we (correctly) assume that in the frame where the wire is at rest, the positive and negative charge densities exactly cancel, they will no longer do so in the frame where we take the external charge at rest and the wire moving. In this frame, the positive charges are moving at v, so since their charge density was at rest, in this frame it must be . Since v is extremely small, we can take the extra density to be .

Similarly, the electrons are at rest in this frame, so their density is less by .

These two effects do not cancel, but add, to give in this frame a positive electrostatic charge density equal to . Using (enclosed charge), we find an electrostatic outward field of magnitude . This electrostatic field exerts a force on the charge q of magnitude

.

But , so this electrostatic force is identical in magnitude to the magnetic force we found in the other frame! So observers in the two frames will agree on the rate at which the particle accelerates away from the wire, but one will call it a magnetic force, the other an electric force. We must conclude that whether a particular force on an actual particle is magnetic or electric depends on the frame of reference-so the distinction is rather artificial.

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Copyright © 1996 Michael Fowler