# How is the curl of the electric field possible?

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?

• 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. – Conifold Apr 19 '16 at 19:21
• This is a FAQ. Not all electrostatic formulas hold in electromagnetism. Related: physics.stackexchange.com/q/100028/2451 – Qmechanic Apr 19 '16 at 20:14
• You're missing the fact that electric field cannot always be expressed as the gradient of a scalar function! – DanielSank Apr 19 '16 at 22:24

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 (we suppose to be in $$\mathbb{R}^3$$, for simplicity) 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

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

• Just a slight correction: there's a minus sign missing in the last expression – binaryfunt Apr 19 '16 at 20:37
• "It is well-known that every divergenceless filed can be written a curl of another vector field (in a simply connected domain)." Actually, no: this is a common misconception. Divergence-free implies a vector potential in regions with vanishing second de Rham cohomology, NOT in simply connected domains. Take $\mathbb{R}^3$ minus the origin. That region is simply connected, and yet there exist divergenceless fields on it that do not, in fact, admit a vector potential. Simple connectedness is relevant for curl-free fields to be gradients, but not for divergence-free fields. – symplectomorphic Jan 14 at 2:32

When there is a time-varying magnetic field, the electric field is non-conservative and therefore cannot be written in the form $\mathbf{E}=-\nabla V$.

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 $\vec{\nabla} \cdot \vec{B} = 0$.