# Applications of group theory to classical mechanics

Today, a friend and I solved a classical mechanics problem using group theory. The problem was the following:

Around a circumference, there are $N$ children evenly spaced. In the center, there is a tire. Each children pulls the tire by a rope with equal forces. Is the resultant force always zero?

My friend associated each force vector with a complex root of unity, and using the fact that the group of the $N$-roots of unity is cyclic, showed the identity: $$1 + \zeta + \zeta^2 + \cdots + \zeta^{n-1} = 0$$ which equals saying that the resultant force is zero. I considered the set $\Omega$ of all permutations of the children, with a group action $\pi\colon \mathbb{Z}/N\mathbb{Z}\times \Omega \to \Omega$ by cyclic permutations. I considered the "force" function $f\colon \Omega \to \mathbb{R}^2$ which associated each system with it's resultant force vector. I also considered the group action $\varphi\colon \mathbb{Z}/N\mathbb{Z} \times \mathbb{R}^2 \to \mathbb{R}^2$ by rotations of the plane. Then, I argumented the following identities for all $S \in \Omega, x \in \mathbb{Z}/N\mathbb{Z}$: $$f(\pi(x, S)) = \varphi(x, f(S))$$ $$f(\pi(x, S)) = f(S)$$ which implied $f(S)$ is a fixed point of $\varphi$, and thus is always the zero vector.

I never expected a application of group theory to classical mechanics (though I know about it's uses in crystallography). Are there other well know examples of this?

• Ultimately, group theory is the study of symmetry. In many situations, the symmetry of a problem makes it easier to solve, or guarantees a certain outcome. In the easiest examples, one could intuitively figure out how to exploit that symmetry. However, in situations involving elaborate mathematical structures, group theory allows us to accomplish these ends systematically. Aug 18, 2016 at 13:42
• I would expect that you could find some interesting examples in chemistry. Aug 18, 2016 at 13:51