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Assume we have a (3-dimensional) resistor network that is mirror symmetric with respect to a plane $p$. The symmetry is only broken by a battery that creates a potential difference between points $A$ and $A'$ where $A'$ is the mirror image of $A$. Let $P$ be any point in the plane $p$. I would expect that for any point $Q$ on the network the potential difference between $Q$ and $P$ is identical to the potential difference between $P$ and $Q'$ (where $Q'$ is the mirror image of $Q$).

However, just mumbling "symmetry ..." does not really cut it for me. After all the symmetry is broken by the battery: there is definitely a net current from one side of the plane to the other.

Can you provide a general argument (or maybe some good intuition) that supports my expectation? Or disprove it?

And is it still true if there are multiple batteries (either connecting mirror images or occurring in symmetric pairs)?

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To formalize a symmetry argument, you should consider the different transformations of the system. Figure out how each transformation affects the physical properties of the system, and which combinations of transformations are really symmetries, i.e. leave the system physically indistinguishable from the original state.

For this problem, consider the following two transformations on the system $S$:

  • $T_1$: reflect in plane $p$
  • $T_2$: multiply all voltage sources by $-1$.

Neither $T_1$ nor $T_2$ are symmetries of $S$. However, $T_1T_2$ is a symmetry of $S$, i.e. $T_1T_2S$ is indistinguishable from $S$.

Therefore $T_1 S$ is indistinguishable from $T_2S$, and these two configurations have the same physical properties, including the same potential differences.

Let $V_S (X_1, X_2)$ denote the potential difference function between points $X_1$ and $X_2$ for a system $S$. The symmetry we identified above means that $V_{T_1 S} (X_1, X_2) = V_{T_2 S} (X_1, X_2)$ for any $X_1$, $X_2$. In particular, $V_{T_1 S} (P, Q) = V_{T_2 S} (P, Q)$ for a point $P$ on $p$.

Due to the nature of reflections, $V_{T_1 S} (P, Q) = V_{S} (P, Q')$. Due to the nature of voltage sign flips, $V_{T_2 S} (P, Q) = - V_{S} (P, Q)$. Putting this together, $V_{S} (P, Q') = - V_{S} (P, Q)$, which is what you wanted to prove.

To figure out whether a complicated system with multiple batteries qualifies for the symmetry argument, just ask yourself whether $T_1 T_2$ is still a symmetry of the system.

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