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Consider the class of electric fields given by

$$\mathbf{E}=\begin{cases} \ln (Cr)\hat{z} & 0\leq r < R\\ 0 & r> R \end{cases}$$

where $C$ is a constant and $r$ is the polar-distance from the $z$-axis (cylindrical coordinates). Maxwell's equations give us

$$\nabla\cdot\mathbf{E}=0$$

$$\nabla\times\mathbf{E}=\begin{cases} -\frac{1}{s}\hat{\phi} & 0\leq r < R\\ 0 & r> R\end{cases}$$

$$\mathbf{E}\rightarrow 0 \textrm{ as } r \rightarrow \infty$$

I thought that these relations would uniquely determine an electric field, but in this case there is still the free parameter/constant $C$. Moreover, if we had a magnetic field such that

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

then we wouldn't get a unique electric field (as I've just shown). For example, the magnetic field

$$\mathbf{B}=\begin{cases}\frac{t}{s} \hat{\phi} & 0\leq r < R\\ 0 & r>R\end{cases}$$

doesn't seem to produce a unique electric field by the argument presented above. Where is the error in this? I know this has to be wrong, because we can't have non-uniqueness in nature (we only measure one electric field).

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    $\begingroup$ Your statement $\vec\nabla \cdot \vec E = 0$ suggests you are generating nonuniform field along the $z$-axis, with a divergence at the origin, without any charges. This is not possible, and a hint that you may have postulated an unphysical field. There are lots of unphysical electric fields. Did you intend to have a $\hat z$ in the first equation? I think if so your curl of $E$ is incorrect. $\endgroup$ – rob Jan 26 '16 at 7:05
  • $\begingroup$ You're not allowed to set field to zero for r>R. It may become discontinuous. $\endgroup$ – LLlAMnYP Jan 26 '16 at 7:30
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    $\begingroup$ @LLlAMnYP Discontinuous $\vec E$ is fine if there's surface charge, just not if $\nabla\cdot\vec E=0$ everywhere. You can turn any weird local field into a monopole with a thick conducting spherical shell. Likewise discontinuous $\vec B$ at current sheets. $\endgroup$ – rob Jan 26 '16 at 8:00
  • $\begingroup$ @rob The tangent components have to match, otherwise there's an infinite B field somewhere $\endgroup$ – LLlAMnYP Jan 26 '16 at 9:13
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    $\begingroup$ This looks like a wire at r=0 with $I\propto t$ flowing along it and a cylinder of radius $R$ with the same total current flowing in the opposite direction. $\endgroup$ – LLlAMnYP Jan 26 '16 at 9:29
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The problem is that you calculated the curl wrong; you missed a delta function arising from the discontinuity. We can write $E_z$ as

$$ E_z = \ln(Cr) \Theta(R-r)$$

and, using that $\Theta'(x) = \delta(x)$, we get that

$$\nabla \times \mathbf{E} = -\frac{1}{r} \Theta(R-r) \hat{\phi} - \ln(CR)\delta(r-R)\hat{\phi}$$

so when giving $\nabla \cdot \mathbf{E}$ and $\nabla \times \mathbf{E}$ you do need to specify $C$: it's not a free constant.

This can be illustrated with a much simpler example. Suppose you demand $\nabla \cdot \mathbf{E} = 0$, $\nabla \times \mathbf{E} = 0$ and $\mathbf{E} \to 0$ at infinity. Then if you forget about the delta function, any field which is constant in a bounded region of space and zero elsewhere solves these equations.

The moral is that to get uniqueness out of Maxwell's equations you either need the fields to be differentiable (or maybe $C^1$), which your example is not, or you need to use distributions like I did.

Regarding your last point: there is no reason a given magnetic field should produce a unique electric field, because knowing $\mathbf{B}$ determines $\nabla \times \mathbf{E}$ and $\partial \mathbf{E} / \partial t$ but not $\nabla \cdot \mathbf{E}$ or the boundary conditions. You could add any static field with zero curl and still get a valid solution.


Let's see what kinds of sources could produce these fields. I rewrote this part because it was kind of a mess.

First observe that since $\nabla \cdot \mathbf{E} = 0$, there are no charges. We could have missed charges at $r=0$ because $\mathbf{E}$ is no defined there, but integrating along a cylindrical Gaussian surface reveals that there are no charges there either.

For later we'll need the following:

$$-\nabla^2 \mathbf{E} = \left[ \frac{\delta(r-R)}{r}(2+\ln(CR)) + \delta'(r-R)\ln(Cr)\right] \hat{z} \equiv \epsilon(r)\hat{z}$$

Using Faraday's law with zero magnetic field at $t=0$ we get that

$$\mathbf{B} = -(\nabla\times\mathbf{E})t = \left[\frac{1}{r}\Theta(R-r) + \ln(CR)\delta(r-R)\right] t \hat{\phi}$$

Now we can get the currents from Ampère's law. The differential version fails to reveal a wire of current at $r=0$ because $\mathbf{B}$ is undefined, so we need to use the integral version for that. The full result is

$$\mathbf{J} = \left[\frac{\delta(r)}{r}\hat{z}-\nabla^2 \mathbf{E}\right]t = \left[\frac{\delta(r)}{r}+\epsilon(r)\right]t\hat{z}$$

There are infinities everywhere, but as we know these aren't really a problem if we interpret them as approximations of smooth fields and currents. We have a wire at $r=0$ that carries a current that goes linearly with $t$; this is prefectly reasonable. The behavior at $r=R$ is a little weirder. There is a delta-function current sheet and a delta-function-derivative current sheet, which we might interpret as two currents going opposite ways very close together, not unlike a dipole sheet in electrostatics.


Conclusion: as long as it is understood that infinite current densities are really approximations of very thin current sheets and that the solutions hold for a finite time interval (as they always do), this solution seems physical enough. At least it's not completely unrealistic. You might have some trouble with all the overlapping current sheets if you try to make it in the lab, though.

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  • $\begingroup$ Infinitesimally concentric cylindrical sheets of current are perfectly valid mathematical (and approximate physical) objects, but they aren't compatible with the discontinuities in ${\bf E}$ in this case. Can you give explicit expressions for ${\bf J}$ and/or ${\bf B}$? $\endgroup$ – tparker Jul 19 '16 at 3:13
  • $\begingroup$ @tparker: Why do you say they aren't compatible? All the equations seem to be satisfied, as far as I can see. $\endgroup$ – Javier Jul 19 '16 at 3:47
  • $\begingroup$ The problem is that placing two opposite sheets infinitesimally close together only creates a "sheet" magnetic field in between them, and we need a bulk magnetic field in order to induce the electric field, since there's no charge to source it. See my answer for details. $\endgroup$ – tparker Jul 19 '16 at 6:07
  • $\begingroup$ @tparker: but a delta function magnetic field works; the proof that the tangential component of $\mathbf{E}$ must be continuous assumes that $\mathbf{B}$ is finite. $\endgroup$ – Javier Jul 19 '16 at 14:10
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    $\begingroup$ @tparker: The expression for $\mathbf{B}$ is in the edited version of the answer (not sure if it was there before.) Both $\mathbf{B}$ and $\partial \mathbf{B}/\partial t$ are infinite at $r = R$, but they're both well-defined in a distributional sense. $\endgroup$ – Michael Seifert Jul 19 '16 at 15:31
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First of all, the tangential component of ${\bf E}$ (i.e. the component parallel to the interface) is always continuous across interfaces, so $E_z$ must be continuous at $r = R$, which fixes $C = 1/R$. There is no physical configuration of sources that could produce any other value for $C$.

${\bf E}$ does not depend on time, so ${\bf \nabla \times E} = -\partial {\bf B}/ \partial t$ gives $$ {\bf B} = -({\bf \nabla \times E})\, t = -\frac{1}{r} \Theta(R - r)\, t\, \hat{\phi} $$ (choosing for simplicity to set ${\bf B}(t = 0) = {\bf 0}$), so the magnetic field is azimuthal and increases steadily with time inside the cylinder of radius $R$.

Taking the curl gives that for $r > 0$, $$ {\bf \nabla} \times {\bf B} = {\bf J} = \frac{1}{r} \frac{d (r\, B_\phi)}{d r} \hat{z} = \frac{\delta(r-R)}{r}\, t\, \hat{z}.$$

We have to consider the axis $r = 0$ separately, because the fields diverge there. Noting that $\oint {\bf B} \cdot d{\bf l} = -2 \pi t$ for a circle around the $z$-axis inside radius $R$, the integral form of Ampere's law gives that there is a current ${\bf J} = -t\, \hat{z}$ at the axis.

The physical interpretation is that there is a line of electric current running down the $z$-axis and a cylindrical sheet of current with radius $R$ running up, both of them steadily increasing in time, with no net current parallel to $\hat{z}$. It's kind of like a solenoid with a square cross-section wound into a donut, then the hole in the middle shrunk down to a line, then the two flat faces taken very far away - and just like a solenoid, a steadily increasing current induces a steadily increasing ${\bf B}$ field but a constant ${\bf E}$ field.

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  • $\begingroup$ Hmmm. Is there a (changing) current density which could produce the required magnetic field? $\endgroup$ – rob Jul 19 '16 at 0:45
  • $\begingroup$ @rob You're right - edited my answer. $\endgroup$ – tparker Jul 19 '16 at 6:01
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Maxwell's equations only give a unique electric field subject to a set of boundary conditions and an initial condition for the field.

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For any given electric field $\mathbf{E}(\mathbf{r},t)$ in some region of space, and assuming no constraints on the magnetic field, there is always a set of charges and currents (both possibly time-dependent) in that region of space that produce that particular electric field. What's more, while the charges required are uniquely determined, the currents are not. Here's the construction:

  1. Let $\rho(\mathbf{r},t) = \epsilon_0 \mathbf{\nabla \cdot E}$. This uniquely defines the charge density required in the region of space of interest.
  2. We now wish to find a magnetic field $\mathbf{B}(\mathbf{r},t)$ such that $\mathbf{\nabla \cdot B} = 0$ and $\partial \mathbf{B} /\partial t = - \mathbf{\nabla \times E}$. This is easy to do: let $$ \mathbf{B}(\mathbf{r},t) = - \int \mathbf{\nabla \times E} \; dt + \mathbf{B}_0(\mathbf{r}), $$ where $\mathbf{B}_0(\mathbf{r})$ is any function of $\mathbf{r}$ satisfying $\mathbf{\nabla \cdot B}_0 = 0$. The freedom in $\mathbf{B}_0(\mathbf{r})$ implies that we will not have a unique $\mathbf{B}(\mathbf{r}, t)$; more on this later.
  3. Finally, define $\mathbf{J}(\mathbf{r},t)$ to be $$ \mathbf{J}(\mathbf{r},t) = \frac{1}{\mu_0} \mathbf{\nabla \times B} - \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}. $$ By explicit construction, we have $\mathbf{\nabla \cdot J} = - \partial \rho / \partial t$ (take the divergence of Ampère's Law and the time derivative of Gauss's Law to prove this.) Any $\rho$ and $\mathbf{J}$ thus constructed will therefore yield the desired electric field $\mathbf{E}(\mathbf{r},t)$.

Since $\mathbf{B}$ is not unique in this construction, $\mathbf{J}$ will not be unique either. In particular, $\mathbf{J}$ splits up naturally into a time-dependent piece $\mathbf{J}_E$ that is determined by $\mathbf{E}$ and a static piece $\mathbf{J}_0$ that is independent of $\mathbf{E}$. Combining our definitions for $\mathbf{B}$ and $\mathbf{J}$ above, we get $$ \mathbf{J}(\mathbf{r},t) = \underbrace{\frac{1}{\mu_0} \int \left[ \nabla^2 \mathbf{E} - \mathbf{\nabla (\nabla \cdot E)} \right] dt - \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}}_{{} \equiv \, \mathbf{J}_E} + \underbrace{\frac{1}{\mu_0} \mathbf{\nabla \times B}_0}_{\equiv \, \mathbf{J}_0}. $$ Since $\mathbf{J}_0$ and $\mathbf{B}_0$ satisfy $\mathbf{\nabla \times B}_0 = \mu_0 \mathbf{J}_0$ and $\mathbf{\nabla \cdot B}_0 = 0$, we can interpret them as a stationary current distribution (satisfying $\mathbf{\nabla \cdot J}_0 = 0$ by definition) and the magnetic field produced by this current. Thus, our freedom in choosing $\mathbf{J}$ above corresponds to our freedom to take the superposition of a set of sources $\rho$ & $\mathbf{J}_E$ (which produce a given electric field $\mathbf{E}$) and a stationary current configuration $\mathbf{J}_0$; this superposition will still yield the same electric field by definition.

Note that even if we impose spatial boundary conditions for $\mathbf{B}$, this does not eliminate the freedom we have to add any $\mathbf{J}_0$ to our solution; there will still be an infinite number of stationary current configurations we can add to our solution.

Finally, note that the charge, current, and magnetic field so defined may not be terribly well-behaved if we have a discontinuous function for $\mathbf{E}$, as was chosen in the original question. They will therefore not correspond to physically well-defined fields, but they can be viewed as the limit of some family of more realistic configurations as some parameter goes to zero. For example, the solution found by Javier (which basically uses the above construction to find $\mathbf{J}$, with $\mathbf{B}_0$ set equal to zero) involves the limit of two infinitesimally thin current sheets with zero separation; we can view this as a limit of two thick cylindrical shells of current separated by a finite distance, and then take the shell thickness and separation to zero simultaneously (while keeping the shell's current per length constant.)

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  • $\begingroup$ -1. I disagree with your first paragraph. The method of images is an example where 2 different charge distributions yield the same potential and thus same fields. So a given electric field does not uniquely determine the charge distribution. $\endgroup$ – thermomagnetic condensed boson May 2 '18 at 21:10
  • $\begingroup$ @lobotomized_sheep_99: In a given region of space, knowing the electric field allows you to say what the charge distribution is in that region of space. So in the example of (say) a point charge above a conducting plane, knowing what $\vec{E}(\vec{r})$ is for $\vec{r}$ above the plane tells you what $\rho(\vec{r})$ is above the plane. If I don't tell you what $\vec{E}$ is below the plane, then you're correct that $\rho$ below the plane is not uniquely determined; it could be a conductor, or it could be another point charge. I'll edit my answer to clarify this distinction. $\endgroup$ – Michael Seifert May 3 '18 at 14:52
  • $\begingroup$ I should add that to get the electric field of a point charge above a conducting plane, you do have to have the point charge there; it's not the case that the field above the plane is solely produced by induced charges on the conducting plane. If you were to freeze the induced charges in place and remove the point charge, you'd have a different electric field above the plane. $\endgroup$ – Michael Seifert May 3 '18 at 15:04
  • $\begingroup$ Fair enough, downvote removed! $\endgroup$ – thermomagnetic condensed boson May 3 '18 at 20:51

protected by Qmechanic Jul 19 '16 at 15:43

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