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I have a question regarding, as the title says, this equation: $\nabla \times \textbf{E}=-\frac{\partial \textbf{B}}{\partial{t}}$

So, the above equation says that the curl of an electric field is proportional to the rate of change of a magnetic field. However, since electric fields have a potential associated with them (i.e. voltage), they should be conservative and thus have 0 curl.

Then this would mean that all magnetic fields are constant in time. This seems like a very strong result, though, and one that I never hear mentioned at that. So is my assumption above wrong? Or am I just misunderstanding the equation?

RubberDucky
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    The flaw is actually implicit in the statement that all electric fields are generated by a potential, $\mathbf{E}=\nabla\phi$. In fact, that's only valid for electrostatics. – DumpsterDoofus Apr 03 '14 at 20:05
  • Do you mean that only stationary electric fields have a voltage, then? – RubberDucky Apr 03 '14 at 20:07
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    Sort of? For example, when a decreasing magnetic flux is passing through a metal ring, the electrons will start cycling through the ring and the metal will heat up (slightly). If the electric field were truly conservative ($\mathbf{E}=\nabla\phi$), then by the fundamental theorem of calculus the path integral around the ring ought to vanish, and so no work would be done upon traversing the loop, and so the electrons would have no motivation to move. But experimentally, we find otherwise, and thus from that you have no choice but to conclude that $\mathbf{E}$ is not conservative. – DumpsterDoofus Apr 03 '14 at 20:15
  • Of course, that sounds reasonable. Thanks for the reply. – RubberDucky Apr 03 '14 at 20:18
  • @DumpsterDoofus: it is of course, easy enough to explain that away by saying that the heating is caused by friction between the electrons and the metal ring, just like you get heating when you run a DC current through a light bulb. – Zo the Relativist Apr 03 '14 at 22:03
  • @JerrySchirmer: Correct, and that was sort of the main point; if the path integral vanished, there would be no energy to move the electrons, and so no current would flow to cause resistive heating. But experiments show there's a current, and thus the path integral has to be be nonvanishing. – DumpsterDoofus Apr 03 '14 at 22:25
  • @DumpsterDoofus: but in a nice DC circuit like that, there are no time-dependent magnetic effects. It's just ordinary friction. There is a perfectly well defined potential at every point in the circuit. – Zo the Relativist Apr 04 '14 at 00:25

2 Answers2

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In fact, for dynamic fields, $\vec E$ is NOT generated by a scalar potential, and instead, is generated by the equation ${\vec E} = - {\vec\nabla}\phi - \frac{\partial{\vec A}}{\partial t}$, where $\vec A$ is the magnetic vector potential, which is related to the magnetic field by ${\vec B} = {\vec \nabla} \times {\vec A}$

Of course, these relations are just a clever way of capturing:

$$\begin{align} {\vec \nabla}\times {\vec E} &= - {\vec \nabla} \times \left(\vec \nabla \phi\right) - {\vec \nabla} \times \frac{\partial {\vec A} }{\partial t}\\ &= - 0 - \frac{\partial}{\partial t}{\vec \nabla}\times {\vec A}\\ &= -\frac{\partial {\vec B}}{\partial t} \end{align} $$

So, I've kind of just restated your question back at you.

Zo the Relativist
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(Originally posted as a comment):

The flaw is actually implicit in the statement that all electric fields are generated by a potential, $\mathbf{E}=-\nabla\phi$. In fact, that's only valid for electrostatics. Actual time-varying electric/magnetic fields are not necessarily conservative.

For example, when a decreasing magnetic flux is passing through a metal ring, the electrons will start cycling through the ring and the metal will heat up (slightly). If the electric field were truly conservative ($\mathbf{E}=-\nabla\phi$), then by the fundamental theorem of calculus the path integral around the ring ought to vanish, and so no work would be done upon traversing the loop, and so the electrons would have no motivation to move.

But experimentally, we find otherwise, and thus from that you have no choice but to conclude that $\mathbf{E}$ is not conservative.

See also this question.

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