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Taking the curl of the electric field must be possible, because Faraday's law involves it: $$\nabla \times \mathbf{E} = - \partial \mathbf{B} / \partial t$$ But I've just looked on Wikipedia, where it says

The curl of the gradient of any twice-differentiable scalar field $\phi$ is always the zero vector: $$\nabla \times (\nabla \phi)=\mathbf{0}$$

Seeing as $\mathbf{E} = - \nabla V$, where $V$ is the electric potential, this would suggest $\nabla \times \mathbf{E} = \mathbf{0}$.

What presumably monumentally obvious thing am I missing?

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    $\begingroup$ As time derivative in the Faraday's law indicates the field there is not static, it does not therefore have to be conservative, i.e. a gradient of a scalar field. $\endgroup$ – Conifold Apr 19 '16 at 19:21
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    $\begingroup$ This is a FAQ. Not all electrostatic formulas hold in electromagnetism. Related: physics.stackexchange.com/q/100028/2451 $\endgroup$ – Qmechanic Apr 19 '16 at 20:14
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    $\begingroup$ You're missing the fact that electric field cannot always be expressed as the gradient of a scalar function! $\endgroup$ – DanielSank Apr 19 '16 at 22:24
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The fact is that, in the general case $$ \vec{E} = -\vec{\nabla}V - \frac{\partial\vec{A}}{\partial t}; $$ (signs depend on conventions used) where $\vec{A}$ is called vector potential. You can consult for example Wikipedia.

Let us consider homogeneous Maxwell equations:

$$ \begin{cases} \vec{\nabla}\cdot\vec{B} = 0,\\ \vec{\nabla}\times\vec{E} + \frac{\partial\vec{B}}{\partial t} = 0; \end{cases} $$

It is well-known that every divergenceless filed can be written a curl of another vector field (in a simply connected domain) just as we know that a curless field can be written as a gradient of a scalar function (always in a simply connected domain). Thus from the first equation,

$$ \vec{B} = \vec{\nabla}\times\vec{A}, $$

and substituting this in the second equation,

$$ \vec\nabla\times\left(\vec{E} + \frac{\partial\vec{A}}{\partial t}\right)=0, $$

since one can exchange the curl with the derivative w.r.t. time, and so one can set:

$$ \vec{E} + \frac{\partial\vec{A}}{\partial t} = -\vec\nabla V, $$

from which

$$ \vec{E} = -\vec{\nabla}V - \frac{\partial\vec{A}}{\partial t}. $$

Note that if your magnetic field is time-independent, you recover the well-know formula

$$ \vec{E} = -\vec\nabla V. $$

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    $\begingroup$ Just a slight correction: there's a minus sign missing in the last expression $\endgroup$ – binaryfunt Apr 19 '16 at 20:37
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When there is a time-varying magnetic field, the electric field is non-conservative and therefore cannot be written in the form $\mathbf{E}=-\nabla V$.

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For dynamic electric and magnetic fields, there is a piece of the electric field that depends on the vector potential: $$ \vec{E} = - \vec{\nabla} V - \frac{\partial \vec{A}}{\partial t}, \qquad \vec{B} = \vec{\nabla} \times \vec{A}. $$ Taking the curl of the first equation yields Faraday's Law (with the $V$-dependent term dropping out as you note); taking the divergence of the second one yields the "no monopole" law $\vec{\nabla} \cdot \vec{B} = 0$.

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protected by Qmechanic Apr 19 '16 at 20:02

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