2
$\begingroup$

In Electrodynamics the law of induction for a moving integration contour (not necessarily a conducting wire) is, according to my books, e.g. Jackson, Classical Electrodynamics

$\oint_C \vec E' \cdot d \vec l = -\frac{d}{dt}\int_S \vec B \cdot \vec n da$

Here, notably, $\vec E' = \vec E + \vec v \times \vec B$ is the field in the rest system of the element $dl$

Just to mention, this approach is taken also in the Wikipedia article.

With the relation

$\frac{d}{dt}\int_S \vec B \cdot \vec n da = \int_S \frac{\partial }{\partial t} \vec B \cdot \vec n da + \oint (\vec B \times \vec v)\cdot d \vec l $

this is consistent with the well known Maxwell Eqation

$\vec \nabla \times \vec E = -\frac{\partial \vec B}{\partial t}$

Now, there is a book which I dont't want to mention in detail, where the law of induction is stated like

$\oint_C \vec E \cdot d \vec l = -\frac{d}{dt}\int_S \vec B \cdot \vec n da$

taking E as the field in the laboratory frame. By combining this with the same vector identity above they derive a new equation

$\vec \nabla \times \vec E = -\frac{\partial \vec B}{\partial t}+\vec \nabla \times (\vec v \times \vec B) $

The author says: The first two terms are identical to Maxwell's equation, the third term is due the Lorentz force.

Now, w.t.f. is this? The equation doesn't give any sense for me, because how can the relation between B and E depend on some velocity of an imaginary integration contour??? However, the author insists, that this equation is a "better" generalization of Maxwell's equation and that this was argued also by Richard.P. Feynman in his Lectures, Volume II. Looking into this book I cannot find anything like this...

I'm quite confused now...

$\endgroup$
2
3
$\begingroup$

To me, Maxwell's equations are synonymous with the work of god, so much so that I'll even take these equations as definitions for the fields. So, what I call the electric and magnetic fields are respectively the two vector fields $\mathbf{E}$ and $\mathbf{B}$ which (in the presence of charges and currents $\rho,\mathbf{J}$) satisfy

  • $\nabla \cdot \mathbf{E}=\dfrac{\rho}{\epsilon_0}$
  • $\nabla\cdot{\mathbf{B}}=0$
  • $\nabla \times \mathbf{E}=-\dfrac{\partial \mathbf{B}}{\partial t}$
  • $\nabla \times \mathbf{B}=\mu_0\mathbf{J}+\mu_0\epsilon_0\dfrac{\partial \mathbf{E}}{\partial t}$

To me calling anything other than these two fields to be electric/magnetic field is just weird. Also the third equation is what I always refer to as "Faraday's Law". Once we have agreed on this terminology for what the electric and magnetic fields are, then the rest is a simple matter of calculus, there's no more physics involved. More precisely, we have the following theorem:

Let $\Sigma\subset \Bbb{R}^3$ be a compact, oriented, smooth two-dimensional manifold with boundary (i.e a nice surface for which Stokes' theorem is valid). Let $t\mapsto \psi_t$ be a one-parameter group of diffeomorphisms with infinitesimal generator $\mathbf{v}$ (i.e for every $p\in\Bbb{R}^3$, we define $\mathbf{v}(p):=\frac{d}{dt}|_{t=0}\psi_t(p)$). For notational purposes, denote $\Sigma_t:= \psi_t[\Sigma]$. For ANY smooth vector field $\mathbf{B}$, define its time-varying flux relative to the family of surfaces $\Sigma_t$ as \begin{align} \Phi_{\mathbf{B}}(t):=\int_{\Sigma_t}\mathbf{B}(\mathbf{r},t)\cdot \mathbf{n}(\mathbf{r},t)\,da \end{align} Then, (using Leibniz's integral rule for differentiating under the integral sign and Stokes theorem) \begin{align} \Phi_{\mathbf{B}}'(t)&=\int_{\partial\Sigma_t}-(\mathbf{v}\times \mathbf{B})\cdot\, d\mathbf{l} + \int_{\Sigma_t}\frac{\partial \mathbf{B}}{\partial t}\cdot\mathbf{n}\,da + \int_{\Sigma_t}(\nabla \cdot \mathbf{B})\,\mathbf{v}\cdot \mathbf{n}\,da \end{align}

Thus far, this is a purely mathematical result, with zero appeal to any physics.

Now if we appeal to physics (or rather just some basic terminology established in the first paragraph on what constitutes as an electric/magnetic field) then we find that by taking $\mathbf{E}$ and $\mathbf{B}$ to satisfy Maxwell's equations, the second term on the RHS becomes the curl of $\mathbf{E}$ which can be transformed via Stokes' theorem. The third term vanishes. Therefore, we obtain \begin{align} \Phi_{\mathbf{B}}'(t)&=-\int_{\partial\Sigma_t}\left((\mathbf{v}\times \mathbf{B}) + \mathbf{E}\right)\cdot d\mathbf{l}. \end{align}

At this point, we can define the integral on the RHS as the EMF in the loop $\partial \Sigma_t$, so that the integral form of Faradays law reads "the rate of change of magnetic flux through a surface is minus the EMF in the loop enclosing the surface" (with all the appropriate caveats regarding orientation).


Also you haven't explained why the author claims your proposed equation is a "better generalization" of Maxwell's equations. In any case, I would find this pretty weird because first of all, the last equation you wrote is not a generalization; to me it just looks like a confusing re-definition. Also, I took a brief look at Feynman's book and I don't see anything which says that Maxwell's equations aren't general enough.

$\endgroup$
6
  • $\begingroup$ This is and was exactly my point of view on the topic. I just wanted to ask whether I'm completely wrong or not. $\endgroup$
    – MichaelW
    Jun 10 at 19:30
  • $\begingroup$ @MichaelW as written, I find the authors statements very confusing. I have never encountered such statements when reading Griffiths (which I read thoroughly), Jackson (briefly, so I don't recall too much), or Feynman (just now briefly). To me it seems like a completely confusing redefinition of terms (like renaming what we call left and right... just weird). I'm no authority figure, but the math is unambiguous here on this issue, so if you're still having issues with that book, I suggest finding another one. So, no I don't think your objections are wrong. $\endgroup$
    – peek-a-boo
    Jun 10 at 19:33
  • $\begingroup$ But really, this isn't a physics/math issue, it's an issue about terminology (and it seems you and I and every book I've read are in agreement). $\endgroup$
    – peek-a-boo
    Jun 10 at 19:35
  • $\begingroup$ The "generalized Maxwell Equation" at the end of my posting is in my opinion not only a redefinition, but simply wrong. The author insists, that he is right and even Jackson is wrong! So I'm a bit shocked... $\endgroup$
    – MichaelW
    Jun 10 at 19:37
  • $\begingroup$ @MichaelW oh right I actually misread it. I thought you wrote $\nabla \times E'$ on the left (hence I called it a re-definition of the electric field). But yes, $\nabla \times E=-\frac{\partial B}{\partial t} + \text{non-zero stuff}$ is just plain wrong. Maxwell's equations are already perfectly general, and you're right that trying to ascribe a relation between $E$ and $B$ which depends on the velocity of an arbitrary contour of integration is absurd. $\endgroup$
    – peek-a-boo
    Jun 10 at 19:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.