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I am having trouble to understand what is going on with the Maxwell–Faraday equation: $$\nabla \times E = - \frac{\partial B}{\partial t},$$ where $E$ is the electric firld and $B$ the magnetic field. The equation is local, in the sense that any change at point $x$ will not affect what happens at another point $x'$, at least not instantaneously. That is, if there is a change in $B$ only at position $x$, then the change will need time to propagate to $x'$. But we can use the Kelvin–Stokes theorem and write the equation in integral form:
$$\int_{\partial \Sigma} E.d\ell = - \frac{\partial}{\partial t}\int_\Sigma B \cdot dS,$$ which is basically telling you that a change in $B$ at the center of the surface will affect instantaneously $E$ at the edge.

What is it wrong with my interpretation of these equations?

anomaly
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Wolphram jonny
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  • I think you should look up the concept of the retarded potential: EM affects travel at the speed of light. So no, information does not travel instantaneously. – Adrian Keister Dec 31 '18 at 18:08
  • @AdrianKeister thanks, I know that it does not, but it is what the second equation is saying, or not? E(t) at the edge depends on B(t) at the center, without any delays – Wolphram jonny Jan 01 '19 at 00:21
  • The integral version is not telling you what you wrote. Instead, think of it this way: an instantaneous change in the magnetic field in the center would affect the rate of change of the electric field on the surface, not the value of the electric field itself. – Adrian Keister Jan 01 '19 at 00:35
  • I am not sure I agree, let us say we have circular simmetry, then you have E=l*d/dt(int Bds), the E is at the edge of the surface, but the change in B can happen anywhere, like just at the center of the surface. – Wolphram jonny Jan 01 '19 at 01:40

2 Answers2

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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.

hypernova
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  • Great point, but I disagree with the conclusion, is you get diferent E's depending on how do you chose the surface, would not that mean that E is not defined, rather than unknown? – Wolphram jonny Jan 02 '19 at 20:42
  • @Wolphramjonny $\mathbf{E}$ has always been pre-determined before you choose any $\ell$ or $\Sigma$. While you may take different $\ell$ and $\Sigma$ to figure out the value of $\mathbf{E}$, your choice does not change its value. – hypernova Jan 03 '19 at 13:38
  • But is not your argument, which seems correct, that a good choice of $\Sigma$ will result in an area integral =0? in which case the computed fiels prediction is also zero? but with another surface choice the result would be different than zero? In such a case, regardless of physical reality(I agree E is well defined), the equations make inconsistent predictions, so the question is: what is it wrong with the equations? – Wolphram jonny Jan 03 '19 at 14:14
  • @Wolphramjonny: Not really. If you can find some $\Sigma_1$ such that $\oint_{\partial\Sigma_1}\mathbf{E}\cdot{\rm d}\mathbf{l}=0$, then for any other $\Sigma_2$ with $\partial\Sigma_2=\partial\Sigma_1$, it is a must that $\oint_{\partial\Sigma_2}\mathbf{E}\cdot{\rm d}\mathbf{l}=0$. My argument is thus as follows. If $\mathbf{B}$ observes a tiny change on $\Sigma_2$, it is hard to figure out $\oint_{\partial\Sigma_2}\mathbf{E}\cdot{\rm d}\mathbf{l}$. Nevertheless, if you can find some $\Sigma_1$ (with $\partial\Sigma_1=\partial\Sigma_2$) on which $\mathbf{B}$ does not change for the moment, – hypernova Jan 03 '19 at 15:09
  • @Wolphramjonny: (con't) then it is straightforward that $\oint_{\partial\Sigma_1}\mathbf{E}\cdot{\rm d}\mathbf{l}=0$. This result, in turn, implies $\oint_{\partial\Sigma_2}\mathbf{E}\cdot{\rm d}\mathbf{l}=\oint_{\partial\Sigma_1}\mathbf{E}\cdot{\rm d}\mathbf{l}=0$. This is a trick to help figure out $\mathbf{E}$. After all, your doubt to me lies in that, from the integral form, it is hard to tell if a tiny change in $\mathbf{B}$ influences the value of $\mathbf{E}$ at a remote location. The above trick is to clarify that there is no such influence. – hypernova Jan 03 '19 at 15:10
  • I wil read again your answer and the comments, i might have missinterpreted them – Wolphram jonny Jan 03 '19 at 15:13
  • @Wolphramjonny: Apologies. I updated my last comment with a few more words to help clarify my idea, in case you missed them. Don't worry. Kindly let me know if you have further questions :-) – hypernova Jan 03 '19 at 15:15
  • This confused me: "Since the change in B is tiny, you may want to find some Σ′, such that (1) ∂Σ′=ℓ, and that (2) B does not have any change on Σ′", it seems to say that the result of the surface integral depends on the choice of the surface, but perhaps you did not mean that. After thinking about the other answer, it seems clear to me now that any choice of surface will give the same result, because the magnetic lines will either be crosing the surface up and down (and be zero), or will cross the wire in which case any surface will result in a non zero result. Thanks, a lot! – Wolphram jonny Jan 03 '19 at 15:36
  • @Wolphramjonny: You've got it! You're right by putting "any choice of surface will give the same result". That is exactly what I meant. While a change in B seems (from the integral form) as if it leads to an immediate change in E at a remote location, you may show that this immediate change in E actually does not exist (by choosing different surfaces). By the way, this argument for locality does not need to make use of $\nabla\times B=0$ or other properties of magnetics; it is self-consistent in Maxwell-Faraday's law. – hypernova Jan 03 '19 at 17:16
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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 $E$.

wendy.krieger
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  • good point, I believe it might answer the question, but I need to think a little more a few issues. – Wolphram jonny Jan 02 '19 at 20:46
  • basically my doubt is how a field line that go outside the loop is formed: instantaneously across space or is slowly built, growing in diameter with time? I guess it is an electromagnetic wave that leaves loops of growing diameter behind the front? – Wolphram jonny Jan 02 '19 at 20:54