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Not really, no. What you are describing is a coordinate singularity; but whether a coordinate singularity has a physical meaning or is just an artifact of the c

by pdonis 4y ago

Not really, no. What you are describing is a coordinate singularity; but whether a coordinate singularity has a physical meaning or is just an artifact of the coordinates depends on the situation. In the case of the Earth, the coordinate singularity in longitude at the poles is an artifact of the coordinates: the Earth's poles are perfectly good parts of the Earth's surface and nothing physically goes wrong there. You just need to pick different coordinates if you want to cover the poles. The case of the horizon of a black hole in Schwarzschild coordinates is similar: there is a coordinate singularity, but it's an artifact of the coordinates. You can find other coordinate charts that have no coordinate singularity at the horizon. The black hole singularity at r = 0, however, is not an artifact of coordinates. It is a genuine physical singularity, where the spacetime curvature goes to infinity.

karmakaze 4y ago
Thanks for the explanation of the difference. To be fair we don't know the physical curvature goes to infinity--the math says it does.
- pdonis 4y ago
  > we don't know the physical curvature goes to infinity--the math says it does More precisely, the math of the theory that works everywhere we can test it, when extrapolated to r = 0 inside a black hole, where we cannot directly test it, says the curvature increases without bound as r = 0 is approached. The mathematical process of extrapolating a well tested theory into a domain where we cannot directly test it is used all the time in science; there's nothing inherently questionable or unreliable about it. The issue in this particular case is that the extrapolation gives us an answer that has a problem: it doesn't seem physically reasonable to have curvature increasing without bound in a finite time, along a finite arc length on the worldline of some observer, at which point spacetime just ends and the observer is destroyed. So most physicists believe that GR, the theory, will actually break down somewhere before this point is reached, and we will need a new theory, which most physicists believe will be a theory of quantum gravity, to tell us what actually happens in this regime. But you're correct that we won't know for sure unless and until we have some way of testing what happens in this regime.

by pdonis 4y ago

karmakaze 4y ago

Thanks for the explanation of the difference. To be fair we don't know the physical curvature goes to infinity--the math says it does.

pdonis 4y ago
> we don't know the physical curvature goes to infinity--the math says it does More precisely, the math of the theory that works everywhere we can test it, when extrapolated to r = 0 inside a black hole, where we cannot directly test it, says the curvature increases without bound as r = 0 is approached. The mathematical process of extrapolating a well tested theory into a domain where we cannot directly test it is used all the time in science; there's nothing inherently questionable or unreliable about it. The issue in this particular case is that the extrapolation gives us an answer that has a problem: it doesn't seem physically reasonable to have curvature increasing without bound in a finite time, along a finite arc length on the worldline of some observer, at which point spacetime just ends and the observer is destroyed. So most physicists believe that GR, the theory, will actually break down somewhere before this point is reached, and we will need a new theory, which most physicists believe will be a theory of quantum gravity, to tell us what actually happens in this regime. But you're correct that we won't know for sure unless and until we have some way of testing what happens in this regime.