Apparent paradox when we use the Kelvin–Stokes theorem and there is a time dependency

Good question!

The answer, at least to me, lies in that the integral form holds for arbitrary surfaces $\Sigma$. This can be interpreted in two ways:

  • For a given closed loop $\ell\in\mathbb{R}^3$, there are infinitely many smooth surfaces $\Sigma$ such that $\partial\Sigma=\ell$;
  • The closed loop $\ell$ itself could also be arbitrarily specified.

Therefore, while the integral form appears non-local, it is actually local, as you may take a "small" closed loop $\ell$ (e.g., a circle with an infinitesimal radius).

Further, even if you take a "large" closed loop $\ell$, you may still choose different surface $\Sigma$, such that a local change of $\mathbf{B}$ in $\Sigma$ would not effect the value of $\mathbb{E}$ on $\ell=\partial\Sigma$.

With these arguments, your question could be interpreted as follows. Suppose you have chosen some $\ell$ and $\Sigma$ with $\ell=\partial\Sigma$. Suppose $\mathbf{B}$ observes a tiny change in the interior of $\Sigma$. Then according to $$ \oint_{\ell}\mathbf{E}\cdot{\rm d}\mathbf{l}=-\frac{\partial}{\partial t}\int_{\Sigma}\mathbf{B}\cdot{\rm d}\mathbf{S}, $$ it seems as if $\mathbf{E}$ also yields some changes along $\ell$. But wait! Since the change in $\mathbf{B}$ is tiny, you may want to find some $\Sigma'$, such that (1) $\partial\Sigma'=\ell$, and that (2) $\mathbf{B}$ does not have any change on $\Sigma'$. In this sense, you will obtain, at least for the moment, $$ \oint_{\ell}\mathbf{E}\cdot{\rm d}\mathbf{l}=-\frac{\partial}{\partial t}\int_{\Sigma'}\mathbf{B}\cdot{\rm d}\mathbf{S}=0, $$ with which you would have no idea whether or not $\mathbf{E}$ changes along $\ell$. For tiny changes in $\mathbf{B}$, you may apply the integral form around each point on $\ell$ with small closed loops $\ell'$ and surfaces $\Sigma''$ with $\partial\Sigma''=\ell'$ on which $\mathbf{B}$ does not find any change, and the arbitrariness of the choice of $\ell'$ and $\Sigma''$ would imply the free of change in $\mathbf{E}$. This trick fails only if the change in $\mathbf{B}$ hits $\ell$, which exactly indicates the locality of its physics.

Hope this could be helpful for you.


Since we have also $\nabla\times B = 0$, you can only change $B$ by adding an entire loop. In this case, it will either cross the surface $S$ once in each direction, so be 0, or it will actually go around the perimeter current, and induce a current, which will change $E$.

Tags:

Electromagnetism

Stokes Theorem

Related

Conjecture $\sum\limits_{n=0}^{\infty}{2n \choose n}^22^{-4n} \frac{n(6n-1)}{(2n-1)^2(2n+1)}=\frac{2C}{\pi} $ Ways to prove that $\int_0^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x = 0$. Find determinant and trace of product of non square matrices Are all functions that have a primitive differentiable? Probability of getting 6 heads in a row from 200 flips and intuition about this high value Derived series of Lie algebra in reverse Conjecture:$\int_0^{\pi/2} \left(\frac{\sin(2n+1)x}{\sin x}\right)^\beta dx$ is integer multiple of $\pi/2$,for integer $\beta>2$ Why do I keep getting this incorrect solution when trying to find all the real solutions for $\sqrt{2x-3}\ +x=3$. When is $p^2+1$ twice of a prime? Delta Epsilon Proof of $\lim_{x\to\infty}\frac{\ln{x}^2}{x}=0$ Let $f : \mathbb{R} \to \mathbb{R}$ be approximated arbitrarily well by polynomials of bounded degree. Prove $f$ is a polynomial. functor $\texttt{Nil}: Ring \longrightarrow Set$ is not representable

Recent Posts