# Do electrostatic fields really obey “action at a distance”?

In an electromagnetic theory class, my professor introduced the concept of "action at a distance in physics". He said that:

If two charges are at some very large distance, and if any one of the charge moves, then the force associated with the charges changes instantaneously. But according to Einstein, no information can travel faster than the speed of light. So photons (the information carriers in electromagnetic force) cannot instantaneously deliver information. So that we associate a field with the two charges and if any charge moves, there is a deformation in that field and this deformation travels with the speed of light and conveys the information.

If the field deformation information cannot travel more than the speed of light, how does the force instantaneously change at very large distances?

• That's a very weird way to say something. Basically he said "The force propagates instantly, except not". – Javier Jul 4 '15 at 14:25
• To echo @Javier: the only thing that changes instantaneously is the result of the formula $F= kq_1q_2/r^2$, which shows that that law isn't good if stuff is moving. – Andrea Jul 4 '15 at 15:15
• It is called electrostatics for a reason. The formulas are truly valid only if nothing varies with time. You can use some of those formulas as an approximation when the variations are sufficiently slow in a sufficiently small region of space, but in general they are not valid any longer. – Lorenzo Donati -- Codidact.org Jul 4 '15 at 17:30
• I think the point your teacher was trying to make is that Newtonian physics contains forces that instantaneously change, but the real world is not like that (a la Einstein). – CharlieB Jul 4 '15 at 23:02
• @CharlieB Yes, I agree. Admirable motives, but the point needs to be made very carefully and not on the fly, separating the two paradigms clearly. I would devote a whole lecture to this one point alone. Otherwise, it comes across exactly as Javier says: "the force propagates instantly, but not"! – WetSavannaAnimal Jul 5 '15 at 1:21

The force does not change instantaneously, the correct way the electromagnetic field of (and thus the force exerted by) a moving electric charge is given by the Liénard-Wiechert potential, where one can see that the effect of the charge does not travel faster than light.

• It would be groovy to point out the features of the LW potentials which make this clear. Nice answer. – DanielSank Jul 5 '15 at 0:57

To add to ACuriousMind's answer on the Liénard-Weichert potentials, you can put these formulas into an even more wonderfully descriptive form since you can derive Feynman's formula from them for the radiation from a moving charge:

$$\vec{E} = -\frac{q}{4\,\pi\,\epsilon_0}\left(\frac{\vec{R}}{R^3}+\frac{R}{c}\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\vec{R}}{R^3}\right)+\frac{1}{c^2}\frac{\mathrm{d}^2}{\mathrm{d}t^2}\left(\frac{\vec{R}}{R}\right)\right)$$

Here $\vec{R}$ is the vector $\vec{R}_2-\vec{R}_q([t])$ joining the moving charge's position, $\vec{R}_q([t])$, evaluated at the retarded time, $t-\frac{R}{c}$, to the fixed position, $\vec{R}_2$, of the observer. So this explicitly shows the finite propagation speed of the effects, as do the Liénard-Weichert potentials, which are also functions of $\vec{R}_2-\vec{R}_q([t])$ alone. The first term is simply a time delayed version of Coulomb's law, the delay travelling at the speed of light. The last term's magnitude varies like $|R|^{-1}$ and represents a radiation field: energy permanently shed from the charge and propagating off into space. The middle term falls off more swiftly than $|R|^{-1}$ and is called the near field. It represents energy that shuttles to and fro in space, but ultimately is not lost from the charge. If the charge makes a simple harmonic motion, this term represents a reaction force of the EM field on the charge that is in phase quadrature to the oscillation, and thus the charge does no net work over a cycle on the force.

Feynman used this formula extensively in the 1950s thinking about synchrotron radiation.

Incidentally, two nice derivations of the Liénard-Weichert potentials are firstly in section 21-5 in Vol 2 of the Feynman Lectures on Physics. Section 25-5 derives them again, in a relativistic discussion making their Lorentz covariance more explicit.

The force is not propagated instantly. It takes time for the information to get from one point to another.

You can treat that as an instant if you are working with small enough distances and velocities, but it's not. If you'll ever study field theory you'll meet retarded potentials that are just this: the field propagates at the speed of light and it's no longer seen as instant.

To reach the Lienard-Wiechert potentials or to prove Feynman's equation (exposed in his lectures without proof), it's necessary to begin with the so-called retarded potentials expressed here conveniently by the following.

$$\phi\left(\mathbf{r},t\right)=\dfrac{1}{4\pi\varepsilon_{o}}\iiint d^{3}\mathbf{r}^{\prime}\dfrac{\rho\left(\mathbf{r}^{\prime},t-\dfrac{\|\mathbf{r}^{\prime}-\mathbf{r}\|}{c}\right)}{\|\mathbf{r}^{\prime}-\mathbf{r}\|}\:, \quad \text{scalar potential} \tag{01a}$$

$$\mathbf{A}\left(\mathbf{r},t\right)=\dfrac{\mu_{o}}{4\pi}\iiint d^{3}\mathbf{r}^{\prime}\dfrac{\mathbf{j}\left(\mathbf{r}^{\prime},t-\dfrac{\|\mathbf{r}^{\prime}-\mathbf{r}\|}{c}\right)}{\|\mathbf{r}^{\prime}-\mathbf{r}\|}\:, \quad \text{vector potential} \tag{01b}$$

Now, what explains the non-instantaneous action are the terms in parentheses. For example, if the action was instantaneous then the scalar potential $\:\phi\:$ at point $\:\mathbf{r}\:$ in time $\:t\:$ would be that produced electro-statically by the charge density $\:\rho\:$ from various points $\:\mathbf{r}^{\prime}\:$ in that SAME MOMENT t:

$$\dfrac{1}{4\pi\varepsilon_{o}}\iiint d^{3}\mathbf{r}^{\prime}\dfrac{\rho\left(\mathbf{r}^{\prime},t\right)}{\|\mathbf{r}^{\prime}-\mathbf{r}\|}\ne\phi\left(\mathbf{r},t\right) \tag{02}$$

But the term $$t-\dfrac{\|\mathbf{r}^{\prime}-\mathbf{r}\|}{c}= t^{\prime} \tag{03}$$ "repairs" exactly this, taking into account that the electromagnetic disturbance from charges at $\:\mathbf{r}^{\prime} \:$ needs a time interval

$$\dfrac{\|\mathbf{r}^{\prime}-\mathbf{r}\|}{c} \tag{04}$$ to arrive at $\:\mathbf{r}\:$ "running" with velocity $\:c\:$.