Faraday's law of induction, $\mathrm{emf} = -d\Phi/dt$, can be expanded out to $$ \oint_{\partial S} \left(\mathbf{E} + \frac{d\mathbf{x}}{dt}\times \mathbf{B}\right)\cdot d\boldsymbol{\ell} = -\frac{d}{dt}\oint_S\mathbf{B}\cdot d\mathbf{A}, $$ where $d\mathbf{x}/dt$ refers to the velocity of the point on the boundary as the boundary changes in time. Meanwhile, the Maxwell equation often called Faraday's law is written in integral form as $$ \oint_{\partial S} \mathbf{E}\cdot d\boldsymbol \ell = -\oint_S \frac{\partial \mathbf B}{\partial t}\cdot d\mathbf{A}. $$ For these to be equivalent, the following identity needs to hold $$ \oint_{\partial S} \left(\frac{d\mathbf{x}}{dt}\times \mathbf{B}\right)\cdot d\boldsymbol \ell = -\frac{d}{dt}\oint_S \mathbf{B}\cdot d\mathbf{A} $$ where for the purposes of this identity $\mathbf{B}$ is considered to be constant in time. This is, of course, a purely geometric result, but I can't seem to prove it. Any pointers?


1 Answer 1


A rigorous derivation requires a form of the Reynolds Transport Theorem for moving two-dimensional manifolds.

In this form, if $\mathbf{B}(\mathbf{x},t)$ is a sufficiently smooth vector field and $S(t) \subset \mathbb{R}^3$ is such a moving surface then we have

$$\frac{d}{dt}\int_{S(t)} \mathbf{B} \cdot d\mathbf{A} = \frac{d}{dt}\int_{S(t)} \mathbf{B} \cdot \mathbf{n}\,dA = \int_{S(t)} \left(\frac{\partial \mathbf{B}}{\partial t} + (\nabla \cdot \mathbf{B})\mathbf{v} + \nabla \times (\mathbf{B} \times \mathbf{v})\right) \cdot \mathbf{n}\,dA $$

where the unit normal vector at the surface is $\mathbf{n}(\mathbf{x})$ and the velocity field is $\mathbf{v}(\mathbf{x})$. (Note that $\mathbf{v} = \frac{d \mathbf{x}}{dt}$ in your notation.)

A magnetic field satisfies $\nabla \cdot \mathbf{B} = \boldsymbol0$. Applying this condition along with Stokes' theorem we get your result

$$-\frac{d}{dt}\int_{S(t)} \mathbf{B} \cdot d\mathbf{A} = -\int_{S(t)} \left(\frac{ \partial \mathbf{B}}{\partial t} + \nabla \times (\mathbf{B} \times \mathbf{v}) \right)\cdot d\mathbf{A} =\oint_{C(t)} (\mathbf {E} + \mathbf {v} \times \mathbf{B}) \cdot d\boldsymbol\ell.$$

What is the basis for the transport theorem and why does this additional term appear as we interchange the derivative and the surface integral?

In one-dimension, the analogy is the Leibniz rule which gives

$$\frac{d}{dt} \int_a^{b(t)}f(x,t) \, dx = \int_a^{b(t)} \frac{\partial f}{\partial t} \, dx + f(b(t),t)b'(t),$$ and where a boundary term related to the "velocity" $b'(t)$ appears.

For a volume integral of a smooth scalar field $f(\mathbf{x},t)$ over a moving region $V(t) \subset \mathbb{R}^3$ with bounding surface $S(t)$,the transport theorem takes the form

$$\frac{d}{dt} \int_{V(t)}f(\mathbf{x},t) \, d\mathbf{x} = \int_{V(t)} \left(\frac{\partial f}{\partial t} + \nabla \cdot (f\mathbf{v}) \right) \, d\mathbf{x} = \int_{V(t)} \frac{\partial f}{\partial t} \, d\mathbf{x} + \oint_{S(t)} f\mathbf{v} \cdot \mathbf{n} \, dA.$$

The proof uses a change of variables with the family of smooth maps

$$\mathbf{\xi}(\cdot,t): \mathbf{x}_0 \mapsto \mathbf{\xi}(\mathbf{x}_0,t) = \mathbf{x}$$

to obtain an integral over a fixed region. Details of this proof are given here.

Proof of the theorem for surface integrals is analogous but more difficult and requires substantial background in differential geometry.

  • $\begingroup$ Interesting. I figured it was analogous to the Leibniz rule, but it sound like I need to look up more about differential geometry. I can see why this point is glossed over in introductory EM. And I guess I was wrong on it being purely geometric, since $\boldsymbol\nabla\cdot \mathbf B = 0$ is needed, too. $\endgroup$ Jul 15, 2017 at 14:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .