5
$\begingroup$

Suppose a homogenous Magnetic Field $\vec{B}$ in vacuum that varies with time, but always points in the z-direction. This induces a curl in the Electric Field $\vec{\nabla} \times \vec{E} = -\frac{\partial B}{\partial t} $, which is also uniform in space and points in the z-direction. If we calculate the integral of this on a horizontal loop we get a non zero EMF through the loop, which means that the horizontal Electric Field must be non-zero at least in a portion of space. Because of translational symmetry one could argue that if $\vec{E} $ is non zero in one point, it must be non zero everywhere. Moreover, it should have the same value everywhere, which is an absurd, since this would mean the curl is zero everywhere, and so would be the EMF.

Where is the mistake in the argument?

Is it that a perfectly uniform time-varying magnetic field is inconsistent with Maxwell's Equations? Or does it have something to do with Lorentz/Poicarré Invariance being the proper symmetry of the system?

My first thought was that the field can't be uniform and time-dependant at the same time because it takes time for the change in the field to propagate, but I would like to have a more elaborate and/or mathematical answer, if this reasoning is correct.

$\endgroup$
  • 1
    $\begingroup$ Related questions: 1, 2 $\endgroup$ – Puk Jul 31 at 2:09
  • $\begingroup$ The main point in these other questions seems to be boundary conditions and constant charge density implying constant electric field-constant electric field implying in zero charge density contradiction. Is the problem here also boundaries? If we set that $\vec{B}(t) = f(t) \hat{z} $ for some $f$, and chose the boundaries $ curl(\vec{E}) = -\dot{f(t)} \hat{z}$ at infinity, does this not have enough conditions for a solution to the differential eqs.? $\endgroup$ – Lucas Baldo Jul 31 at 2:52
2
$\begingroup$

The main reason this argument is giving counter-intuitive results is, as the comments suggest, boundary conditions. Generally speaking, taking domains to be infinite in the study of differential equations gives rise to 'badly behaved' functions - discontinuities, Dirac deltas and other non-differentiable objects are readily obtained when differentiating analytic functions that live on infinite domains. (A classic example is $\nabla^2 \frac{1}{r} = -4\pi \delta^3(\vec{r})$).

Essentially, the issue is that the solution of a differential equation is not really a function. They do not in general have th eproperty that you can actually evaluate them at a point, so trying to think about what the solutions's value is will always give confusing results. In this case, to solve your problem, we want an electric field such that $\nabla \times E = -f'(t)\hat{z}$, or rather

$$\partial_y E_x - \partial_x E_y = f'(t)$$

Firstly, note that the solution for $E$ is not unique - adding any irrotational field $\vec{F}$ to the electric field globally does not change this equation. One solution is $\vec{E}^{(1)} = \hat{x}yf'(t)$ - there's nothing wrong with that, but remember that $\vec{E}^{(2)} = -\hat{y}xf'(t)$. is just as good.

This is a bit of an issue, since electric fields can be directly measured by e.g. a test charge. In order to decide which of this infinite array of solutions is physical, you would need to specify a boundary condition on the electric field.

However, you can still get physical results without specifying the boundary conditions. Look at the integral form of Maxwell's equations, where all of the scalar divergences and Dirac deltas have been implicitly integrated out.

$$\oint_{\partial S} \vec{E} \cdot d\vec{l} = -\frac{\partial}{\partial t}\iint_S \vec{B} \cdot d\vec{A} = -f'(t) A_\perp$$

Where $A_\perp$ defines the cross-sectional area of the surface $S$ that is 'facing' the $z$ axis.

Then we have a clear result- The AC signal that a wire loop will pick up is a direct measure of the time derivative of $f$, amplified by the area $A_\perp$. This is the relevant physics.

None of these problems arise if you keep your system of charges and currents finite, such that the solutions you get are well-defined and make physical sense.

$\endgroup$

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.