Path:  physics insights > physics > magnetism > The Field of a Uniformly Moving Charge

There's more than one way to compute the field of a moving charge.  On this page we're going to do it by transforming the charge's field from its own rest frame, which we already know, into the frame of an observer moving relative to the charge.

Figure 1 -- Moving charge and observer:

We start with a charge, q, and orient the axes such that the charge is at the origin, at time 0, and is traveling parallel to the x axis at velocity v in the coordinate system t, x, y, z for the inertial frame S.  In the charge's rest frame, frame T, we'll use the coordinates τ, x', y', z'.  As usual we'll set c=1, and we'll be using cgs units, and metric signature (-1,1,1,1).  We'll also define

(1)    

We want to find the field seen by an observer at time 0, stationary in frame S, at location (x,y,0).

In the charge's rest frame, frame T, the field is static, so we don't need to worry about what time it is.  At time 0 in frame S, the point we're concerned with has the following coordinates in frame T:

(2)    

The E field at that point, in frame T, will be:

(3)    

In the charge's rest frame there is, of course, no magnetic field.   The Faraday tensor for frame T will then be:

(4)    

The Lorentz transform from S to T will be

(5)    

and the Faraday tensor in frame S will be

(6a)  

which, just to be painfully clear, we write out "longhand" as

(6b)  

Multiplying out, that's

(7)    

from which we can read off the E and B field components in frame S.  Plugging in the values from (3), we have:

(8)

This deserves a couple of comments.  First of all, we see that

(9)    

which means the electric field points directly toward the moving charge -- it does not "lag" the charge's position due to its motion.  (Of course, if the charge is accelerating it's a different story -- the information that the charge has changed its motion can't get to the observer until time r/c, and until that moment the field will point to where the charge would have been had it continued in a straight line at constant velocity.)

Second, the field isn't spherically symmetric.  At points on the x axis (y=0), we have

(10a)    

while at points along the y axis (x=0) we have

(10b)    

The field is "pancaked" -- it's stronger out to the sides, and weaker in front and in back of the charge.  If we view the field as a "bubble" around the charge, we can view the "bubble" as being Lorentz contracted.  However, Gauss's law still holds and the integral of the field strength over the surface of a sphere around the charge is the same in all inertial frames.  So, if the "bubble" is contracted fore-to-aft, it must bulge out to the sides, which isn't quite the same as physical Lorentz contraction. 

The Biot-Savart Law of Magnetostatics

We'll now present a brief (and highly unsatisfactory) derivation of the Biot-Savart law.

Within a wire, we have

(B.1)    

Then we can rewrite B_z from (8) as

(B.2)    

For slowly moving charges, where v << c, we can rewrite (B.2) as

(B.3)  

where θ is as shown in figure 1.  If we define

(B.4)  

then we can rewrite (B.3) as

(B.5)  

Recalling that B_x=B_y=0, and with the observation that B_z is perpendicular to the current and radius vectors, and with an application of the right-hand rule to check that the direction of the vector is correct, we can therefore write

(B.6)  

We can now integrate it over all currents, to obtain the total field due to the current in an arbitrary collection of wires; this is the Biot-Savart law:

(B.7)

Unfortunately our derivation is unsatisfactory for a couple of reasons.  First, I eliminated γ in (B.3) by assuming currents were moving much slower than c.  This is normally the case, but just the same we'd like to have some idea of whether this law applies to, say, the beam in a particle accelerator, where v approaches c.  Second, while it's true that the limit of the field due to the charge carriers in a short wire as the length approaches zero is the same as the field of a point charge, we assumed in our use of equation (2) that the motion of our point charge was uniform.  That's only true for a straight wire!  If the wire bends, the charge carriers are actually accelerating.  Again, as long as the charge carrier velocities are much smaller than c we can largely ignore the changes to the field introduced by the acceleration, but none the less the consequence is that, from this derivation, we can't tell whether the Biot-Savart law is correct for arbitrarily kinked wires and high-velocity currents, or whether it's only a low-speed approximation for use with smoothly curved wires.



Page created on 1/17/2007