# Relations between symmetries in sources and fields

I've edited this question way more times that I like to admit. I'll do my best

## Relation between highly symetric charge distributions and $$\nabla\times\mathbf{D}$$:

In electrostatics, for some charge distributions, we can use Gauss law in their integral from to compute the electric displacement field. This distributions have to be highly symmetric so that the field depends on one coordinte and has one direction (let's say $$\hat{f}$$). This coordinate must be the one you won't be integrating wrt at the surface where $$\hat{f}\cdot\mathbf{S}$$. So far so good. We are using the Gauss law and the charge distribution symetries to determine the field.

In the other hand, Helmholtz Theorem states that a field $$\mathbf{F}$$ is uniquelly determined by $$\nabla\cdot\mathbf{F} = d$$, $$\nabla\times\mathbf{F} = \mathbf{C}$$ if we have some particular boundary conditions (PBC) for $$\mathbf{C}$$ and $$\mathbf{F}$$.

1. $$\mathbf{C}$$ decay faster than $$\frac{1}{r^2}$$
2. $$\mathbf{F}$$ vanishes at infinity

It seems that the symetries are providing curl and PBC information. But they are doing more than that, they have so much information about the field that we can determine it by integrating with Gauss Law which is not possible even knowing curl and PBC. Even when $$\nabla\times\mathbf{D} = \vec{0}$$ (which seems to be the niciest curl condition for a field we can have) it isn't enough for using Gauss Law.

How can relate symetries with curl and PBC? A simpler question could be: Are there any cases where $$\nabla\times\mathbf{D}\neq\vec{0}$$ but we have enought symetries to determine the field with Gauss Law or having enought symetries implies $$\nabla\times\mathbf{D} = \vec{0}$$ (necesary but not sufficient condition)?

## Relation between spherical symetry of a punctual charge and $$\nabla\times\mathbf{D}$$

In Griffith's Introduction to Electrodynamics § 4.3.2, the author states that there isn't a Coulomb's Law analogue for electric displacement because,in general, $$\nabla\times\mathbf{D} = \nabla\times\mathbf{P} \neq \vec{0}$$. As far as I know, Coulomb's Law for charges densities can be deduced from Gauss law, the spherical symetry of a punctual charge and the Superposition Principle (here and here).

For $$\mathbf{D}$$, we have a Gauss Law and Superposition Principle. So what we must lack here is the spherical symetry for polarization charges. Thus, I conclude that $$\nabla\times\mathbf{P} \neq \vec{0}$$ implies that polarization charges have spherical symetry. I know that $$\rho_{p} = - \nabla\cdot\mathbf{P}$$ and $$\sigma_{p} = \mathbf{P}\cdot\hat{n}$$ but I can't relate it with my previous statement.

I have been taught that $$\nabla\cdot\mathbf{E} = \frac{\rho}{\epsilon_0}$$, $$\nabla\times\mathbf{E} = \vec{0}$$ and PBC imply Superposition Principle and Coulomb's law. This would also support that $$\nabla\times\mathbf{F} = \vec{0}$$ implies in some way that the punctual sources of a field have spherical symetry.

Those were some of the concerns I had, but any relation (as formal as posible) between Helmholtz's Theorem, Gauss Law integration and Coulomb's Law conditions or previous concepts is highly appreciated.

If my question made no sense or I'm missing huge points, I would appreciate any physics or mathematics reading. Thank you for your time.

P.D.: I was going to write the question in terms of magnetostatics but it didn't seem that analogous so I backed away.

• Are there any cases where $\nabla\times\mathbf{D}\neq\vec{0}$ but we have enought symetries to determine the field with Gauss Law or having enought symetries implies $\nabla\times\mathbf{D} = \vec{0}$ (necesary but not sufficient condition)? According to Griffiths, having symmetries implies $\nabla\times\mathbf{D} = \vec{0}$ . Jan 12, 2021 at 6:21