# How can direction of electric field due to a moving charge be from the present position of charge?

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

• 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/… – PM 2Ring Apr 22 at 14:58
• "...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? – Chiral Anomaly Apr 22 at 23:55
• @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. – exp ikx Apr 23 at 7:21
• @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. – exp ikx Apr 23 at 7:43
• 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. – John Dvorak Apr 23 at 10:20

• 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. – exp ikx Apr 25 at 16:07