Symmetry arguments in solving problems

Question

There was a question which involved calculation of final charges on two spheres when one uncharged and the other having charge $Q$ were brought in contact with each other. (radius same). If potential concept is applied charge will flow until potentials on the surfaces of the 2 spheres equalize and that will happen when both have $Q/2$. so $Q/2$ on each sphere. but in the book it was written as : "by symmetry, charges on the 2 will be $Q/2$". What logic does this symmetry argument provide that the charges on the 2 spheres will be $Q/2$ each and nothing else?? (it is irritating when problems are solved by symmetry)

Answer 1

I agree with you: if that's the whole answer the book gave, then it's a very poor answer.

Sure, you can easily deduce from a symmetry argument that the system of two (otherwise) identical charged spheres must have an equilibrium where the charges on the spheres are identical. However, on its own, this argument does nothing to establish:

1. that this equilibrium is stable, and

2. even if it is stable, that it will actually be reached from the given initial conditions.

For an example of where condition #2 fails, note that the same symmetry argument could equally well be used to argue that the symmetric situation, with both spheres charged to $Q/2$, is also an equilibrium before the spheres are brought into contact. This is perfectly correct, as far as it goes — but without a conductive path between the spheres, the system can't actually reach that equilibrium, unless it starts there.

For a striking example of condition #1 failing in electrostatics, see the Kelvin water dropper — a very simple symmetrical device that spontaneously (under the action of gravity, which also obeys the symmetry) breaks the symmetry and generates an increasingly asymmetrical static charge.

That said, symmetry arguments are hardly useless. In particular, if it can be shown that the symmetric equilibrium is the only possible one, that the dynamics of the system do not diverge beyond the range where our simplified model of it is valid (as happens in the Kelvin water dropper), and that the system cannot approach a cyclic or a chaotic attractor, then it can indeed be shown that the symmetric equilibrium must be stable and all dynamics must tend to it.

Now, these extra conditions may seem pretty hard to prove in general, but often they can be established a priori, just by noting that the system under consideration belongs to a class that doesn't exhibit such behavior. Also, even if we cannot (yet) rigorously show that the symmetric equilibrium indeed is the only possible attractor, just knowing that it exists can often guide our intuition, and can allow us to derive useful conditional results by assuming that, in absence of any external forces or constraints preventing it from doing so, the system will tend to this equilibrium.

Answer 2

I'm sure there are several lines of logic you could follow using the symmetry of spheres to arrive at that answer.

One that I thought of is, by the symmetry of the system, each sphere has the same surface area. Where the spheres are brought into contact we are allowing electrons to be exchanged freely between the two spheres. Whereas we previously had a surface area $A$ of one sphere that $Q$ amount of charge was distributed over, we now have a surface area $2A$ for the same $Q$ to distribute itself over; thus each area $A$ (each sphere) now has $Q/2$ charge distributed over it. This combines the logic you used regarding equalizing potentials with the symmetry of the two spheres.

So, in short, I suppose the logic used in the book's symmetry argument was the same logic you used to get to the same answer. However, bechira's answer in the comments explains how to use symmetry directly without considering directly what properties of the spheres are symmetric.