Retarded potentials and special relativity?

Question

I'm curious. . . is there a substantial difference between a classical retarded potential, such as for the electric/magnetic fields in EM, and special relativistic formulations of EM? What i'm meaning is that it seems by merely assuming a finite speed of propagation you could get something that looks like the transformations or behavior of EM fields given they were relativistic. There is a difference, however, between a classical spacetime with time retarded potentials and being in a Minkowksi spacetime where by the nature of spacetime structure you get the same set of equations. How would I know the difference? Is there something unique about a relativistic retarded potential that differs from one which is set in a classical spacetime?

Answer 1

Retarded solutions are a general feature of the wave equation. The response of the electromagnetic field to a current can be given by a retarded solution, but that's also true of the response of water waves to a splash, or sound waves to a clap. There's nothing inherently relativistic about having such solutions.

Light is different because if you boost to a new reference frame, its speed stays the same. By contrast, the speed of a sound wave or water wave would change. That's permitted by the wave equation, because it only applies in the rest frame of the air or water.

However, in prerelativistic physics, there would be nothing apparently special about light. You would by default assume the same logic holds for light waves: under a boost their speed becomes $c+v \neq c$, and that's because Maxwell's equations are only valid in the rest frame of the ether.

In other words, what changes between pre-relativity and relativity is knowing how to relate electromagnetic fields in reference frames with relative motion. Just knowing that retarded solutions exist doesn't tell us this.

Answer 2

The Lienard-Wiechert formulation for EM waves is special. It gives the correct retarded results that fit special relativity.

You have a charge at a source time and location, and the charge may have a velocity and acceleration. You can predict the force at another location at the time that the force arrives. So the assumption is that the target location is stationary. We do not use the frame of the source.

A charge at the target location-and-time can have a velocity, and that can have an effect depending on the direction of the velocity.

Obviously the force the target experiences will be the same independent of the frame we chose. And the L-W formula gets the right result.

Take a simple example. We use the electric and magnetic forces predicted by L-W, simplified for our example. No acceleration. Velocity $\beta$ as fraction of lightspeed is parallel to the direction vector $n$ from source to target.

$F = K \frac{1}{(1-\beta)^3} \frac{n-\beta}{R^2}$

Where $K$ represents a collection of boring constants,

$n$ is the unit direction vector from source to target,

$\beta$ is the source velocity vector divided by $c$ lightspeed,

$R$ is the distance from source to target.

For source charge and target charge both stationary, distance R apart, $F=\frac{Kn}{R^2}$ .

Now change the frame. Source is moving toward target at $\beta=\frac{c}{2}$. Target is moving the same velocity. Nothing has changed except the frame.

Force that leaves the target travels $2R$ distance before it reaches the target.

$F=K\frac{1}{(1-0.5)^3} \frac{(1-0.5)n}{(2R)^2} = K\frac{n}{(0.5)^2(2R)^2} = \frac{Kn}{R^2} $

We get the same force, just as we ought to. But the force travels twice as far and it takes twice as long to arrive.

The formula gets the right result independent of frame, except that the times and distances come out wrong. Just fudge the arrival time and distance the right amount, and you have special relativity.