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Take the formula for Coloumb's law. It does not show in any case as to how the static field propagates. What I can feel is that the field of a static charge is ever prevading from the time the charge was created and other charges interact with it.

But consider a static charge kept in your room. Now another charge in Andromeda galaxy should take some time to respond to the field of this charge ( if the field travels at c).

Isn't Coloumb's law depicting an instantaneous action at a distance

because

But Coloumb's law is just like Newton's law, which Einstein desperately changed.

So how are Electrostatics and Magnetostatics in accord with the principle of relativity.

Their effect looks instantaneous.

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  • $\begingroup$ You first try to fix it up with retarded potentials, and then go full relativistic. $\endgroup$ – Jon Custer Jul 1 at 20:07
  • $\begingroup$ @JonCuster .. AI don't think so... First of all i read Retarded potentials for moving charges not static ones... Another - Panfosky in his book states the Field theoretical approach and retarded potentials are different $\endgroup$ – Shashaank Jul 1 at 20:10
  • $\begingroup$ Have you studied the full set of Maxwell’s equations and their Lorentz covariance? $\endgroup$ – G. Smith Jul 1 at 20:13
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    $\begingroup$ Because Newtonian gravity assumes that the inverse square law holds regardless of the motions of the masses. This is not the case with Coulomb’s inverse square law, which holds only when the charges are at rest. $\endgroup$ – G. Smith Jul 1 at 20:27
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    $\begingroup$ I don't understand how Electrostatics and Magnetostatics should be relativistically invariant They aren’t. For example, consider the electrostatic field of a point charge at rest. In a moving frame, there is no longer anything “electrostatic” about it. The particle is moving, and the field everywhere is dynamic. $\endgroup$ – G. Smith Jul 1 at 20:41
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I'm not sure what you mean by a "static" field "propagating"!

But you're right that Coulomb's law is an example of an "action-at-a-distance" law. However, the reason we usually use this law is that we are often only considering steady-state situations, where the charge distributions do not change. In such cases, there are no "changes" in the system, and so no instantaneous propagation occurs.

Your example of a charge being created goes against this assumption, and thus you're right that Coulomb's law cannot be used here.

This is why for a consistent Lorentz-invariant theory, Electromagnetism is actually formulated using fields, which satisfy Maxwell's Equations:

\begin{equation*} \begin{aligned} \nabla \cdot \mathbf{E} &= \frac{\rho}{\epsilon_0}\\ \nabla \cdot \mathbf{B} &= 0\\ \nabla \times \mathbf{B} &= \mu_0 \mathbf{j} + \frac{1}{c^2}\frac{\partial \mathbf{E}}{\partial t}\\ \nabla \times \mathbf{E} &= - \frac{\partial \mathbf{B}}{\partial t}\\ \end{aligned} \end{equation*}

Taking the curls of the last two equations, you can show that the Electric and Magnetic Fields satisfy two non-homogeneous wave equations with speed $c$:

\begin{equation*} \begin{aligned} \left( \nabla^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\right) \mathbf{E} &= \left(\frac{1}{\epsilon_0}\nabla\rho + \mu_0 \frac{\partial \mathbf{j}}{\partial t}\right),\\ \left( \nabla^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\right) \mathbf{B} &= -\mu_0(\nabla\times\mathbf{j}),\\ \end{aligned} \end{equation*}

where the right hand sides of both equations act as the sources for these fields. This shows that disturbances in the fields propagate at the speed of light $c$.

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