Is the electric field produced by uniformly moving charges non-conservative? And if it is, then why?

  • $\begingroup$ See the answer here $\endgroup$ Commented Sep 14, 2018 at 22:30

1 Answer 1


For a field to be conservative, you need $\nabla \times \mathbf{E} = 0$. See in my answer here how this relates to converservation of energy and path indepenedence.

The third Maxwell equation (Faraday's law + Lenz's law) states:

$$ \nabla \times \mathbf{E} = -\frac{\partial\mathbf{B}}{\partial t} \neq 0. $$

This is your first inkling that, when you have moving charges and hence electrodyanmics (as opposed to electrostatics, where $\partial/\partial t = 0$) entails a non-conservative electric field.

If you don't want to use Maxwell's equation, you can also start from the general expression for the electric field $$ \mathbf{E} = -\nabla\phi - \frac{\partial\mathbf{A}}{\partial t},$$

and left multiply by $\nabla \times$, obtaining the same result.

Now, to be a bit more pedantic.
In order to have $\nabla\times\mathbf{E} = 0$ from the equation above, you require:

  • either $\partial/\partial t = 0$, i.e. electrostatics
  • or $\mathbf{B} = 0$
  • or both.

This is why, I am assuming, knzhou asked you about accelerating charges in the comment.
Because, what if you have a moving charge which is not generating a magnetic field? Then $\partial/\partial t \neq 0$ but $\mathbf{B} = 0$, which guarantees an overall $\nabla\times\mathbf{E} = 0.$

However, you already know that a charge moving at constant velocity is basically a current (if you consider a very small length element $\mathrm{d}\mathbf{r}$), which you know creates a magnetic field $\mathbf{B} \neq 0$.

Indeed, the full expression for the electric and magnetic fields is given here, and an example of the expression is given in the answer linked by Alfred Centauri.

Let me add a final remark.

What if you travelling in a frame of reference that moves at the same speed as a constant-velocity moving charge? Then you would see the charge as stationary, with both $\mathbf{B} = 0$ (ignoring spin) and $\partial/\partial t = 0$.
This is because magnetic field naturally come out of electrostatics + special relativity, i.e. they are electric fields seen in a different reference frame. (magnetic fields from magnets are still relativistic because they come from spin, which is a "relativistic" effect in the sense that it comes from the Dirac equation).


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