2
$\begingroup$

According to Maxwell's Equations, the electromagnetic waves in vacuum travel at the speed of light $c$. While solving Maxwell's equations using Lorenz gauge conditions (or basically evaluating scalar and vector potentials via D'Alembert operator) we find retarded potential.

The retarded potential is basically evaluated at the time when the field began to propagate from the point where it was emitted to an observer, also known as retarded time.

Calculation of Lienard-Wiechert potential shows that the magnitude of electric field at a point due to a moving charge is from the position of the particle at retarded time, however the direction of the field at the same point is in the outward direction from the present position of the positive charge as shown in Fig. 1. A commonly attributed reason is that the signal takes a finite time to propagate from a point in the charge or current distribution (the point of cause) to another point in space (where the effect is measured).

However, if any information cannot travel faster than $c$, how can the direction at the observing point $A$ be outwards from the present position? If the information containing direction can reach the observer instantaneously, why not the magnitude?

Fig. 1: Magnitude of the electric field at A appears to be as if the charge is at K, however, the direction at A is as if it is at L.

Image courtesy: From an answer by Frobenius

$\endgroup$
  • $\begingroup$ Does this help? math.ucr.edu/home/baez/physics/Relativity/GR/grav_speed.html IIRC, there are existing questions on this site about this effect, both in terms of the electric field, and gravity. Here's one in terms of gravity: physics.stackexchange.com/questions/27845/… $\endgroup$ – PM 2Ring Apr 22 at 14:58
  • 1
    $\begingroup$ "...the direction of the field at the same point is in the outward direction from the present position of the positive charge." This is true for a charge moving with constant velocity. What would happen if the charge turned a corner? $\endgroup$ – Chiral Anomaly Apr 22 at 23:55
  • $\begingroup$ @PM2Ring Thanks for that article by Baez. It is an interesting read. What is it that he talks about the cancellation of retarded effects in GR? I had a course on ED and STR, but not GR, so I don't know. But that still does not answer the question. I thought there could be a definitive answer to this seeming paradox. $\endgroup$ – exp ikx Apr 23 at 7:21
  • $\begingroup$ @ChiralAnomaly So are you asking about an accelerated charged particle? We get one more term in L-W potentials to account for acceleration, apart from accounting for velocity. I don't know how exactly to find the direction since it involves vector triple product, but through calculations, I found out the case for only constant velocity. Moreover, the expression for potential will still be evaluated at retarded time. $\endgroup$ – exp ikx Apr 23 at 7:43
  • 2
    $\begingroup$ It must be the case that a test charge is accelerated toward the present position, because the laws of electrodynamics are invariant under Lorentz transformation and a test charge is accelerated to the (present) position of the field source in the inertial frame in which the field source is at rest. $\endgroup$ – John Dvorak Apr 23 at 10:20
1
$\begingroup$

Information of a particle's current position can not reach the observer instantly. However, if the movement is uniform, this information of current position can be calculated from the particle's past.

The field is "outward" from the current position, but this is not caused by the particle at the current position, but caused by the particle's previous contribution at "retarded time". The field looks like "outward" from the current position only when the movement is uniform.

$\endgroup$
  • $\begingroup$ Are you saying that the field's direction is pre-calculated from the expected position at time $t$, from the information of velocity $v$ available at $t' = t - \frac{R}{c}$? Here $R$ is the separation between source and observer. $\endgroup$ – exp ikx Apr 25 at 16:07
  • $\begingroup$ There is no aberration of forces. If a charge is in uniform motion, the force vector will always be directed to its current position without any delay so you needn't calculate its actual position. You can know its instant position "right now". Just change frames and think that charge is at rest and an observer moves in the field of this "stationary" charge, that clears everything. That doesn't mean that forces propagate infinitely fast. mathpages.com/home/kmath562/kmath562.htm $\endgroup$ – Albert Apr 25 at 16:25
  • $\begingroup$ @exp-ikx Yes, the field's direction is pre-calculated, at observer at time t, from the information of velocity v available at t′=t−R/c. But no, the field is not caused by the expected source position at time t; it just happens that the field at the observer looks like from the expected source position at time t.(This sounds like a coincidence, but it has special relativity reasons.)At a time just before t, if the source is suddenly pushed away from its expected route, the observer will still see the field's direction as if it is from the originally expected position of the source at time t. $\endgroup$ – verdelite Apr 26 at 14:14
  • $\begingroup$ @Albert Agreed. Your argument is the same as the one in my last comment, "This sounds like a coincidence, but it has special relativity reasons". $\endgroup$ – verdelite Apr 26 at 14:21
  • 1
    $\begingroup$ @exp-ikx Your question itself has the mathematical argument. Basically you calculated the retarded potentials (scalar and vector) at the observer. You differentiated them and got the current field. This field has nothing to do with the current source position (and be coincident with it only when the movement is uniform) . It is all caused by the retarded potential made by the source from the past. The detailed calculation of this procedure is in Griffith (3rd) section 10.3.2, and in relativity example 12.13 for uniform movement. Thus we have both explanations and the constraint of c is true. $\endgroup$ – verdelite Apr 28 at 2:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.