# Lorentz symmetry approaching a black hole event horizon

• I
Hello people,
I have been thinking about a concept that I was taught whilst learning GR, If I understand correctly it is that Lorentz symmetry becomes local when we consider GR. This makes sense to me as then the metric is generally speaking not Minkowski, only for a sufficiently small area with no mass/energy present.

So my question is this, as you approach a black hole is it correct to say that the local area in which lorentz symmetry holds gets smaller?

If so can it then be inferred that the symmetry is broken at event horizon?

PeterDonis
Mentor
If I understand correctly it is that Lorentz symmetry becomes local when we consider GR.

This is true, but you have to be clear about what that means. It means that Lorentz symmetry is not a symmetry of spacetime at all. It's a symmetry of the tangent space at each point.

This makes sense to me as then the metric is generally speaking not Minkowski, only for a sufficiently small area with no mass/energy present.

Even if there is no stress-energy present, the metric will not be Minkowski in general. For example, Schwarzschild spacetime is a vacuum solution, but the metric is not Minkowski anywhere; it's curved everywhere.

as you approach a black hole is it correct to say that the local area in which lorentz symmetry holds gets smaller?

You are confusing local Lorentz symmetry with the equivalence principle. In general, as tidal gravity increases (which it will as you get closer to any gravitating body, black hole or otherwise), given a fixed measurement accuracy, the area over which tidal effects will be negligible gets smaller. But that only tells you the area over which you can approximate the actual curved spacetime with a small patch of flat spacetime. It doesn't change anything about the Lorentz symmetry of the tangent space at a point.

If so can it then be inferred that the symmetry is broken at event horizon?

No. See above.

Ibix