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As far as I understand, for the field of a uniformly moving charge, curl of $\mathbf E$ is zero everywhere.

Since $\nabla \times \mathbf E = -\dfrac{\partial\mathbf B}{\partial t}$, magnetic field should be constant in every point in space.

This sounds wrong, since $\mathbf B$ is supposed to fall off proportionally to $r^2$, and $r$ is changing in time for a moving charge. What is wrong with this reasoning?

Even worse, $\nabla \times \mathbf B = \dfrac{\partial\mathbf E}{\partial t}$ , and since $\dfrac{\partial\mathbf E}{\partial t}$ is not constant (because $\dfrac{\partial^2\mathbf E}{\partial t^2}$ is not zero), curl of $\mathbf B$ keeps changing.

But how can $\nabla \times \mathbf B$ keep changing if $\mathbf B$ itself stays the same?

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  • $\begingroup$ Why is $\nabla \times {\bf E}$ zero? $\endgroup$
    – velut luna
    Commented Apr 24, 2016 at 14:29
  • $\begingroup$ The field lines are radial and symmetric, so I think it implies that curl is zero. Am I wrong? $\endgroup$
    – Vitaly Korzhik
    Commented Apr 24, 2016 at 14:31
  • $\begingroup$ The field is radial but not symmetric. $E_r$ depends on $\theta$. $\endgroup$
    – velut luna
    Commented Apr 24, 2016 at 14:33
  • $\begingroup$ The lines should be radial and symmetric. Then those are given by $\vec E \propto \vec r$. And this has no curl. In the rest frame, this certainly is the case. Boosting this might however also distort the field lines (length contraction). Not sure whether that will have a curl then, though. $\endgroup$
    – Martin Ueding
    Commented Apr 24, 2016 at 14:33
  • $\begingroup$ Am I missing something? A uniformly moving charge doesn't create a B-field, and, if you want to express it that way, a B-field of zero falls off with $\frac{1}{r}$. If/while you accelerate the charge, it creates a B-field: an electromagnetic wave, which "keeps switching between E and B" to satisfy maxwell equations. $\endgroup$
    – Solarflare
    Commented Apr 24, 2016 at 16:39

2 Answers 2

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For example, consider at $t=0$ the point charge be at the origin and moving in the $z$ direction with velocity ${\bf v}$. The electric field at this moment is $${\bf E}({\bf r})=kq\frac{1-v^2/c^2}{(1-v^2 \sin^2 \theta/c^2)^{3/2}}\frac{\hat{\bf r}}{r^2}$$ Then $$\nabla \times {\bf E}=-\frac{1}{r}\frac{\partial}{\partial \theta}kq\frac{1-v^2/c^2}{(1-v^2 \sin^2 \theta/c^2)^{3/2}}\frac{1}{r^2}\hat{{\bf \phi}}\ne {\bf 0}$$

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You only have the homogenious Maxwell equations there. The curl of the magnetic flux density $\vec B$ does also depend on the current density $\vec j$, see Wikipedia.

The current density for your uniformly moving point charge is $$ \vec j = e \vec v \, \delta^{(3)}(\vec vt - \vec x_0) \,.$$

This therefore also creates a curl in the magnetic flux density.

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  • $\begingroup$ Yes, I understand that there's certainly a curl of B. What I don't understand is that it looks like B itself isn't changing in time, and it makes no sense to me! $\endgroup$
    – Vitaly Korzhik
    Commented Apr 24, 2016 at 14:35
  • $\begingroup$ Non-mathematically, when the charge passes you it's field will be different when it is far away. $\endgroup$
    – jim
    Commented Apr 24, 2016 at 16:19

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