2
$\begingroup$

I have a spacetime $(M,g)$ and a set of discrete symmetries $S_\alpha: M\to M$. Given two fixed points $p_0,p_1\in M$ with $S_\alpha(p_i)=p_i$, I want to argue that a geodesic connecting the two points consists itself only of fixed points.

Obviously, this is not true in general: For instance, take rotations of the 2-sphere as symmetry that leaves the two poles invariant. The geodesics are the great circles and all of them have the same length and they get mapped into each other. However, if we consider the full 3d-space with rotations, then the shortest distance is actually invariant under rotations, because it is the straight line connecting the two poles.

In some sense, the symmetry is spontenously broken in the first case. Is there a way to distinguish the two cases? In particular, if I find a geodesic that is invariant under the symmetry transformations, is there a general argument that other ones that do not respect this symmetry should be longer (only locally shortest geodesics).

$\endgroup$
2
$\begingroup$

In classical mechanics spontaneous symmetry breaking corresponds to the case when the Hamiltonian is invariant under a symmetry but the initial conditions are not.

The examples given in the question indeed (and in a strict sense) correspond to spontaneously broken and unbroken rotation symmetry around the $z$ axis as stated in the question and as will be clarified in the following.

In the case of (non-relativistic) geodesic motion (of a massive particle), the Hamiltonian can be taken as: $$H = \frac{m}{2}g_{ij}(x)p^ip^j$$ ($g$ is the metric, $x$ is the position, $p$ is the momentum) This Hamiltonian is invariant under any rotation about the $z$ axis for both a round two-sphere and the embedding three space. (The question mentions a discrete symmetry which can be taken any discrete subgroup of the $U(1)$ group of rotations around the $z$ axis, but the discussion is valid for the whole continuous $U(1)$ group of rotations around the $z$ axis) . To see that, it is sufficient to prove that the Hamiltonian Poisson commutes with the generator of $z$ rotations: $$L_z = p_x y - p_y x$$ (In the sphere case, the components are not independent).

The geodesic equation is of second order, it needs 2 initial conditions: position and momentum. In order for the symmetry not to be spontaneously broken, both the initial position and the momentum need to be invariant under the symmetry. Now we can see the difference between the two cases: (Suppose that we start from the South Pole)

In the sphere case there is no possible initial momentum invariant under the rotation about the $z$ axis, since the momentum is tangent to the sphere, therefore parallel to the $x-y$ plane.

In the three space case, there is one possibility in which the initial momentum is parallel to the line connecting the South Pole to the North Pole. In this case the full set of initial conditions is invariant under the symmetry and the symmetry is not broken. Consequently every point along the geodesic is invariant under the symmetry. (All other options of the initial momenta will give rise to broke symmetry as in the case of the sphere).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.