# The implications of symmetry + uniqueness in electromagnetism

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• EE18
In summary, the article "Symmetry, Uniqueness, and the Coulomb Law of Force" by Shaw (1965) discusses the use of symmetry arguments in electromagnetism problems. In the conversation, the speaker raises a question about the use of symmetry in a particular 1D problem with a charge distribution that is symmetric under a charge-flip and reflection about the xy plane. Through careful consideration of the intrinsic symmetries of the electric field, it is shown that the solution to this problem must have dipole characteristics.

#### EE18

I have tried to follow "Symmetry, Uniqueness, and the Coulomb Law of Force" by Shaw (1965) in both asking and solving this question, but to no avail. Some of the mathematical arguments there are a bit too quick for me but, it suffices to say, the paper tries to make the "by symmetry" arguments of introductory electromagnetism rigorous.

My question is the following: Consider a 1D situation in which I have a charge distribution which obeys 𝜌(𝑧)=−𝜌(−𝑧). Then I conclude that there must exist another solution obtained from my original solution via this symmetry: 𝐸′(𝑧)=−𝐸(−𝑧). But by the uniqueness of solutions to electromagnetism problems we have 𝐸′(𝑧)=𝐸(𝑧) so that we have 𝐸(𝑧)=−𝐸(−𝑧). But this is absurd, since it implies that the electric field everywhere points toward the origin which makes no sense for a distribution obeying 𝜌(𝑧)=−𝜌(−𝑧) (dipole) so that it should point in one direction everywhere. Where have I erred in "using symmetry"?

Edit: On reconsidering, I now have the following. We first observe that the symmetry ##\rho(z) = -\rho(-z)## is equivalent to saying that the system must be invariant under a reflection ##\rho(z) \to \rho(-z)## followed by a "flipping" of charge ##\rho(-z) \to -\rho(-z)##. These transformations being symmetries of the system mean that we can obtain another solution to the problem by performing ##E(z) \to E'(z') = E'(-z) = -E(-z)## (this step is analogous to Shaw equation (2)) followed by ##E'(z') \to E''(z') = -E'(z')= -(-E(-z)) = E(-z)##. Then, by uniqueness, it must be that the solution ##E''(z) = E(z)##...but from the above this just says ##E''(z) = E(-(-z)) = E(z)## which is useless (in the last step I have tried to use an analogue to Shaw equation (1)?

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I don't have the text you are following, but I think I can follow.

You have a charge density distribution that is symmetric under a charge-flip + reflection about the xy plane, as you have said.

First, you need to consider the intrinsic symmetries of the electric field. Intrinsically, no matter what charge distribution you have, the electric field will flip sign under a charge-flip (since it's proportional to source charge) and the z-component of the E-field will flip sign under reflection about the xy plane (since it is proportional to force and thus acceleration along z). The x and y components of the electric field will be unchanged by the reflection about the xy plane, but they will flip under a charge-flip transformation.

Putting all of that together, you find that your electric field must satisfy:
$$E_x(-z) = -E_x(z)$$ $$E_y(-z) = -E_y(z)$$ $$E_z(-z) = E_z(z)$$ This field has the dipole characteristic that you were looking for.

vanhees71