electromagnetism - Faraday's law for a current loop being deformed

I'm working through Jackson's Classical Electrodynamics text (3rd version, chapter 5.15) about Faraday's law:

Faraday's law is pretty familiar:

$\int_c E \cdot dl = -\frac{d}{dt}(\int_s B \cdot n da)$

which states that the contour integral of the electric field around a bounded surface is equal to the negative time rate of change of the flux bounded by that surface. No problem. Here we have a total derivative and Jackson addresses 2 cases.

"The flux though the circuit may change because (a) the flux changes with time at a point, or (b) the translation of the circuit changes the location of the boundary"

In either of these cases, it is easy to show that:

$\frac{d}{dt}(\int_s B \cdot n da) = \int_s(\frac{\partial B}{\partial t} \cdot n da) + \int_c(B \times v)\cdot dl$

by using $\frac{d}{dt} = \frac{\partial}{\partial t} + v \cdot \nabla$ and then applying a "product rule", $\nabla \cdot B = 0$ and stokes theorem. i.e.:

$\frac{d}{dt}\int_s B \cdot n da = \frac{\partial}{\partial t}\int_s B \cdot n da + (v \cdot \nabla) \int_s B \cdot n da$

In the case where the surface is constant, we can exchange the order of differentiation and integration:

$\frac{d}{dt}\int_s B \cdot n da = \int_s \frac{\partial}{\partial t} B \cdot n da + \int_s (v \cdot \nabla) B \cdot n da$

I still have no problem with this. Now we can work on the second term on the right (apply a product rule, drop terms with $\nabla \cdot B$ and then apply stokes theorem) to get the $\int_c(B \times v)\cdot dl$ term.

which, when applied to faraday's law gives the nice result:

$\int_c [ E' - (v \times B)] \cdot dl = - \int_s \frac{\partial B}{\partial t} \cdot n da$

The problem arises when you allow the shape of the bounded surface to change. (Consider a circular wire loop in a magnetic field that you then hit with a hammer causing a dent in one side). How does changing the surface influence the result that we derived above? It seems to me that this would prevent us from being able to exchange the order of differentiation and integration which seems like it leaves us in a pretty tight spot ...

Answer

Faraday's Law can be given in differential form as follows in terms of the electric field $\mathbf E$ and the magnetic field $\mathbf B$ specified in an inertial frame: $$ \nabla\times\mathbf E = -\frac{\partial \mathbf B}{\partial t}. $$ As you note, this can be re-written as an integral equation, but the corresponding integral version takes different forms depending on how general one wants it to be. I'll address the fully general setting you're concerned with in which the loop around which the EMF is being computed has arbitrary time-dependence. This includes, for example, the corresponding surface that it is the boundary of being deformed in pretty odd ways, like being "dented."

For each time $t$, consider some surface $\Sigma_t$ with boundary $C_t$. We assume that all mathematical assumptions that physicist may not typically explicitly state hold so that the following computations hold true e.g. that $\Sigma_t$ is orientable. We can integrate both sides of the differential form of Faraday's law over the surface $\Sigma_t$ and use Stokes' theorem on the left hand side to obtain \begin{align} \int_{C_t}\mathbf E\cdot d\boldsymbol \ell = -\int_{\Sigma_t}\frac{\partial \mathbf B}{\partial t} \cdot d\mathbf a \end{align} Now, your question becomes

If $\Sigma_t$ constitues some sufficiently smoothly-varying family of surfaces but with otherwise arbitrary time dependence, what extra terms result when we attempt to take the time derivative outside of the integral on the right hand side of the above integral equation?

I now make the following

Claim. Let $\mathbf v$ denote the velocity of each point on the loop $C_t$, which will be a function of time, and the location of the point, then \begin{align} \int_{\Sigma_t}\frac{\partial\mathbf B}{\partial t}\cdot d\mathbf a &= \frac{d}{dt}\int_{\Sigma_t}\mathbf B\cdot d\mathbf a + \int_{C_t} \mathbf v\times\mathbf B \cdot d\boldsymbol\ell \end{align} As a result of this claim, the most general integral form of Faraday's Law for an arbitrarily time-dependent surface is as follows: \begin{align} \int_{C_t}\mathbf (\mathbf E+\mathbf v\times\mathbf B)\cdot d\boldsymbol \ell = - \frac{d}{dt}\int_{\Sigma_t}\mathbf B\cdot d\mathbf a. \end{align} The expression on the left hand side is the EMF around the loop $C_t$ (since by definition the EMF is the work done per unit charge by the electromagnetic fields when going around the loop, and the integrand written here is the Lorentz force per unit charge), and the expression on the right is the rate of change of the flux through the surface $\Sigma_t$ whose boundary is $C_t$.

Let's prove the claim above which allows us to take the time derivative outside of the integral even in the case of some arbitrarily time-varying surface/loop.

Proof of Claim.

We note that since $\nabla\cdot\mathbf B = 0$, there exists (under suitable topological assumptions which we assume hold true here) some vector field $\mathbf A$ for which $\mathbf B = \nabla\times\mathbf A$. It follows that \begin{align} \int_{\Sigma_t}\mathbf B\cdot d\mathbf a &= \int_{\Sigma_t}\nabla\times\mathbf A\cdot d\mathbf a = \int_{C_t}\mathbf A\cdot d\boldsymbol \ell \end{align} Now, we are after the time derivative of the flux integral on the left which, by this computation, equals the time derivative of the line integral of the vector potential around $C_t$. To facilitate performing this integral, we parameterize the curve $C_t$ in order to remove the time-dependence from the bounds of integration. In particular, for each time $t$, let $\boldsymbol \gamma(t,\lambda)$ denote a parameterization of $C_t$ that traverses $C_t$ once as gamma ranges through the interval $[0,1]$. Then the line integral of the vector potential can be written as \begin{align} \int_{C_t}\mathbf A\cdot d\boldsymbol\ell = \int_0^1 \mathbf A(t, \boldsymbol\gamma(t,\lambda))\cdot\frac{\partial\boldsymbol\gamma}{\partial t}(t,\lambda) d\lambda \end{align} putting these last two equations together, and taking the time derivative of both sides, we get \begin{align} \frac{d}{dt}\int_{\Sigma_t}\mathbf B\cdot d\mathbf a &= \int_0^1 \left(\frac{\partial A_i}{\partial t}\cdot\frac{\partial\gamma_i}{\partial\lambda}+\frac{\partial\boldsymbol\gamma}{\partial t}\cdot \nabla A_i\frac{\partial\gamma_i}{\partial \lambda}+A_i\frac{\partial^2\gamma_i}{\partial t\partial\lambda}\right)d\lambda\\ &= \int_0^1 \frac{\partial A_i}{\partial t}\cdot\frac{\partial\gamma_i}{\partial\lambda}d\lambda + \int_0^1 \left(\frac{\partial\boldsymbol\gamma}{\partial t}\cdot \nabla A_i\frac{\partial\gamma_i}{\partial \lambda}+A_i\frac{\partial^2\gamma_i}{\partial t\partial\lambda}\right)d\lambda \\ &=\int_{C_t} \frac{\partial\mathbf A}{\partial t}\cdot d\boldsymbol\ell + \int_0^1 \left(\frac{\partial\boldsymbol\gamma}{\partial t}\cdot \nabla A_i\frac{\partial\gamma_i}{\partial \lambda}+A_i\frac{\partial^2\gamma_i}{\partial t\partial\lambda}\right)d\lambda\\ &= \int_{\Sigma_t}\frac{\partial\mathbf B}{\partial t}\cdot d\mathbf a + \int_0^1 \frac{\partial\boldsymbol\gamma}{\partial t}\cdot \nabla A_i\frac{\partial\gamma_i}{\partial \lambda} d\lambda + \int_0^1 A_i\frac{\partial^2\gamma_i}{\partial t\partial\lambda}d\lambda \end{align} Let's focus on the last integral on the right. Using integration by parts on that integral, and using the fact that the boundary term vanishes because the curve over which we are doing the line integral is closed, and using the chain rule \begin{align} \frac{\partial}{\partial\lambda} A_i(t, \boldsymbol\gamma(t,\lambda)) &= \nabla A_i(t,\boldsymbol\gamma(t,\lambda))\cdot \frac{\partial\boldsymbol\gamma}{\partial\lambda}(t,\lambda) \end{align} we obtain \begin{align} \int_0^1 A_i\frac{\partial^2\gamma_i}{\partial t\partial\lambda}d\lambda &= -\int_0^1 \frac{\partial\gamma_i}{\partial t} \nabla A_i\cdot \frac{\partial\boldsymbol\gamma}{\partial\lambda}d\lambda \end{align} so that \begin{align} \frac{d}{dt}\int_{\Sigma_t} \mathbf B\cdot d\mathbf a &= \int_{\Sigma_t} \frac{\partial\mathbf B}{\partial t}\cdot d\mathbf a + \int_0^1 \left(\frac{\partial\boldsymbol\gamma}{\partial t}\cdot \nabla A_i\frac{\partial\gamma_i}{\partial \lambda} - \frac{\partial\gamma_i}{\partial t} \nabla A_i\cdot \frac{\partial\boldsymbol\gamma}{\partial\lambda}\right) d\lambda \end{align} Now, if we make the appropriate notational identifications \begin{align} \mathbf v = \frac{\partial\boldsymbol\gamma}{\partial t}, \qquad d\boldsymbol \ell = \frac{\partial\boldsymbol\gamma}{\partial\lambda} d\lambda \end{align} and if we take note of the following vector identity \begin{align} \mathbf v\times\mathbf B = (v_j\partial_i A_j - v_j\partial_jA_i)\mathbf e_i \end{align} then we obtain \begin{align} \frac{d}{dt}\int_{\Sigma_t} \mathbf B\cdot d\mathbf a &= \int_{\Sigma_t} \frac{\partial\mathbf B}{\partial t}\cdot d\mathbf a -\int_{C_t} \mathbf v\times\mathbf B\cdot d\boldsymbol \ell \end{align} which is simply a rearrangement of the claim.

Note. The claim just mentioned is really just a special case of the Leibniz integral rule (See Art's answer to this question) when the vector field under consideration has vanishing divergence. If you'd like, I could probably give a proof of the general rule as well, but it's really unnecessary since the magnetic field does, in fact, have zero divergence.