In Griffiths' Introduction to Electrodynamics, example 14 in chapter 7, "Electrodynamics." He makes an assumption/claim that I don't understand the reason behind, or don't agree with.

Example 7.14 Imagine two concentric metal spherical shells (Fig. 7.44).

The inner one (radius $a$) carries a charge $Q(t)$, and the outer one (radius $b$) an opposite charge $-Q(t)$. The space between them is filled with Ohmic material of conductivity $\sigma$, so a radial current flows:

$$ \textbf{J} = \sigma\textbf{E} = \sigma \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2} \hat{\textbf{r}}; \quad I = - \dot{Q} = \int\textbf{J}\cdot d\textbf{a} = \frac{\sigma Q}{\epsilon_0}. $$

This configuration is spherically symmetrical, so the magnetic field has to be zero (the only direction it could possibly point is radial, and $\nabla \cdot \textbf{B} = 0 \implies \oint \textbf{B} \cdot d\textbf{a} = \textbf{B}(4\pi r^2) = 0$, so $\textbf{B} = \textbf{0}$). What? I thought currents produce magnetic fields! Isn't that what Biot-Savart and Ampère taught us? How can there be a $\textbf{J}$ with no accompanying $\textbf{B}$?

Fig. 7.14

Solution. This is not a static configuration: $Q$, $\textbf{E}$, and $\textbf{J}$ are all functions of time; Ampère and Biot-Savart do not apply. The displacement current

$$ J_d = \epsilon_0 \frac{\partial\textbf{E}}{\partial t} = \frac{1}{4\pi} \frac{\dot{Q}}{r^2}\hat{\textbf{r}} = -\sigma \frac{Q}{4\pi\epsilon_0 r^2} \hat{\textbf{r}} $$

exactly cancels the conduction current (in Eq. 7.37), and the magnetic field (determined by $\boldsymbol{\nabla} \cdot \textbf{B} = 0, \boldsymbol{\nabla} \times \textbf{B} = \textbf{0}$) is indeed zero.

He writes $$\vec J=\sigma \vec E,$$ but in place of $\vec E$, he uses the electrostatic field produced by the shell of charge $Q$. However, we learned this formula is correct only if $E$ is the net electric field. The total $E$ could have, along with the conservative electrostatic component, a non-conservative component that could be caused by a changing magnetic field. Whether the magnetic fields are changing or not, we don't know beforehand.

What allows him to do this? Why is he neglecting possible non-conservative/tangential electric fields produced by changing magnetic fields, whose expressions may not look very neat/easily calculable? Why can he neglect the $B$ field, which could, in principle, be tangential whilst maintaining radial symmetry? I feel like, either there is some deeper symmetry that can be shown to allow us to do this, or this procedure is wrong.

  • $\begingroup$ Why did you just ignore meow in chat and repost this... You can prove that constant tangential fields will disagree with spherical symmetry. $\endgroup$ Sep 11, 2023 at 19:17

1 Answer 1


There are two ways of justifying what Griffiths concludes. The first approach—which is the one he adopts—is to argue that there is no possible magnetic field consistent with the spherical symmetry. You seem to be unconvinced of this, but the argument is correct; there is no tangential $\vec{B}$ that is radially symmetric.

To see why, consider a generic field written in spherical coordinates, $$\vec{B}=B_{r}(r)\,\hat{r}+B_{\theta}(r)\,\hat{\theta}+B_{\phi}(r)\,\hat{\phi}.$$ That the components only depend on $r$ is an immediate consequence of the symmetry; if they depended on the angles in a nontrivial way, then the magnitude $\left|\vec{B}\right|$ (a scalar) would be different at different angular positions—in other words, not invariant under rotations about the origin.

More subtle, but equally key to the argument in the textbook, is that both of the angular components of $\vec{B}$ must also vanish; $\vec{B}$ cannot point tangent to the surface of constant $r$, or even have a component in that direction. To see why, suppose that $\vec{B}$ does have a tangential component. We can choose our spherical coordinates so that we are looking at $\vec{B}$ at point $(x,y,z)=(r,0,0)$ and so that the tangential $\vec{B}$ points in the $\hat{\theta}$-direction, $\vec{B}=B_{\theta}\,\hat{\theta}$. Now ask yourself what happens if we rotate the whole system by $\pi$ around the $x$-axis. The rotation of the physical system will rotate the vector $\vec{B}$ to be $-B_{\theta}\,\hat{\theta}$. However, the configuration has not actually changed; the charge and current distributions are the same as they were to begin with, because they were spherically symmetric. Therefore, $\vec{B}$ cannot actually have changed either! This is only possible if $B_{\theta}=0$, meaning no tangential magnetic field.

The other way to justify the conclusion is self consistency. We can simply take $\vec{E}=E_{r}\,\hat{r}$ as an Ansatz and show that it provides a complete solution to the problem. Griffiths' calulations show that with a purely radial $\vec{E}$, the current and displacement current sources sum to zero; thus they combine to produce no net magnetic field. If there is no net $\vec{B}$, then the inductive part of $\vec{E}$—sourced by the time derivative of $\vec{B}$—is also zero. Since this is consistent with our original Ansatz, the uniqueness of the solutions to electrodynamics problems means that our Ansatz has supplied the one and only physical solution.

  • $\begingroup$ The first argument, I will completely agree with, (basically appealing to spherical symmetry) if you can show that, $E$ and $B$, being dependent on $\rho, \vec J, \dot{\rho}, \dot{J}$ obeys spherical symmetry. To show this, you need to show that $\rho, \vec J, \dot{\rho}, \dot{J}$ are themselves spherically symmetric, but these quantities themselves depend on $E$ and $B$. If you can show spherical symmetry somehow actually exists for all the variables, were done. But I think that calls for a $dt$ by $dt$ analysis. $\endgroup$ Sep 12, 2023 at 2:43
  • $\begingroup$ And for the uniqueness theorem for solutions to electrodynamic systems, could you please link me a source for this, I haven't seen this uniqueness theorem anywhere $\endgroup$ Sep 12, 2023 at 2:52
  • $\begingroup$ @nickbros123 The initial conditions are implicitly assumed to be spherically symmetric, and so the time evolution will continue to be. Strictly speaking, there is nothing, for example, the prevents the whole system from being immersed in a background magnetic field that breaks the symmetry at $t=0$ and thus for all subsequent times. However, it is implied that there is no such breaking of the symmetry; if there were such, it would have been an absolutely necessary part of the description of the system. $\endgroup$
    – Buzz
    Sep 12, 2023 at 3:06
  • $\begingroup$ This, by the way, makes is tricky to state precisely which mathematical uniqueness theory would apply, because there are many different ways that the boundary conditions (at spatial infinity, for example) could be stated that would all imply the same unique solution. Typically, it is not necessary to worry about about these situations; if you can find a self-consistent solution of the equations of motion for the electromagnetic field and its charged sources, it will be unique. The involvement of the matter means more than Helmholtz's Theory is required, but the essential idea is the same. $\endgroup$
    – Buzz
    Sep 12, 2023 at 3:10
  • $\begingroup$ I have thought for a while, and by observing infinitesimal changes, I can digest the use of spherical symmetry. I already knew about the spherical symmetry argument, but I like your answer on the grounds that I can actually, with simple considerations, solve the problem using the uniqueness theorem specified on the Wikipedia page. Thank you for introducing me to this theorem $\endgroup$ Sep 12, 2023 at 12:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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