I was reading Introduction to Electrodynamics by Griffiths, and I'm stuck on 4.3.2. He says:

If the problem exhibits spherical, cylindrical or plane symmetry, then you can get $\textbf{D}$ directly from $\oint \textbf{D} \cdot d \textbf{a} = Q_\mathrm{{free}_{enclosed}}$ by the usual Gauss's law methods. (Evidently in such cases $\nabla \times \textbf{P}$ is automatically zero, but since symmetry alone dictates the answer you're not really obliged to worry about the curl.)

I can't really see how $\nabla \times \textbf{P}$ is zero if the problem exhibits the symmetries; surely if the field goes round in a circle around the origin, that is cylindrically symmetric and the curl isn't zero?

For example:

enter image description here

Image Source: http://www2.math.umd.edu/~petersd/241/html/ex28.html


1 Answer 1


The situation you describe has not only translational & rotational symmetry but parity symmetry as well, which allows us to rule out a tangential (or longitudinal) component of $\vec{D}$. A completely arbitrary vector field $\vec{D}$ in cylindrical coordinates can be written as $$ \vec{D} = D_r(r,\theta,z) \hat{r} + D_\theta(r,\theta,z) \hat{\theta} + D_z(r,\theta,z) \hat{z}. $$ If the sources and media are invariant under cylindrical symmetry, we can argue that the components cannot depend on certain variables, and that some of them are zero:

  • Since the sources & media remain the same under a translation $z \to z + a$ for any $a$, none of the components can depend on $z$: $$ \vec{D} = D_r(r,\theta) \hat{r} + D_\theta(r,\theta) \hat{\theta} + D_z(r,\theta) \hat{z}. $$
  • Since the sources & media remain the same under a rotation $\theta \to \theta + a$ for any $a$, none of the components can depend on $\theta$: $$ \vec{D} = D_r(r) \hat{r} + D_\theta(r) \hat{\theta} + D_z(r) \hat{z}. $$

This much you had already figured out; including parity is what allows us to rule out a non-zero $D_\theta$ and $D_z$. Under reflection about the $xy$-plane, the source stays the same but $D_z \to - D_z$; thus, we must have $D_z(r) = - D_z(r) = 0$. Similarly, if we reflect about any plane containing the $z$-axis, the source remains the same but $D_\theta \to - D_\theta$, and thus $D_\theta(r) = - D_\theta(r) = 0$ as well. Thus, the only possible form for $\vec{D}$ can be $$ \vec{D} = D_r(r) \hat{r} $$ and from this assumption we can apply Gauss's Law to $\vec{D}$.

As you can see, the whole argument is somewhat subtle. It's also not necessary to invoke parity symmetry to apply Gauss's Law in the case of spherical symmetry, and it's a lot more obvious that you need to invoke it in the case of planar symmetry (to argue that $\vec{D}(z) = - \vec{D}(-z)$.) This is why the role of parity is usually elided in most undergraduate electromagnetism courses & texts, in favor of a statement to the effect that "it's obvious that the field must be radial".

  • $\begingroup$ Thank you Michael for taking the time to write this answer. My earlier answer completely trivialised the problem, ( so I deleted it) but I will study your answer carefully. $\endgroup$
    – user154420
    Jun 29, 2017 at 19:22

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