4 ms·
> 1. I must admit to a lack of sufficient GR education to feel confident in this, but I think that one of the issues that made physicists unwilling to accept th
by senderista 2mo ago
> 1. I must admit to a lack of sufficient GR education to feel confident in this, but I think that one of the issues that made physicists unwilling to accept the idea of black holes when they were first postulated was that there ended up being a division by zero in the mathematics.
Well, the Ricci curvature scalar blows up to infinity, which is obviously unphysical.
- zmgsabst 2mo agoWhy is it unphysical?
- senderista 2mo agoThe curvature scalar can be physically evaluated by measuring the volume of a small ball of freely falling test particles and comparing to its volume in flat spacetime. https://en.wikipedia.org/wiki/Scalar_curvature#Relation_between_volume_and_Riemannian_scalar_curvature https://en.wikipedia.org/wiki/Scalar_curvature#Relation_betw...
- zmgsabst 2mo agoOkay — why is it unphysical such a ball has unbounded curvature?
- senderista 2mo agoIf this unbounded state were asymptotic then it wouldn't be unphysical, but a freely falling particle reaches the singularity in finite time. A freely falling observer crossing the event horizon could in principle perform this measurement as they approached the singularity, and general relativity could no longer describe their measurements.
- zmgsabst 2mo agoOkay — you’re still not explaining what actually breaks, just repeating they could measure over and over. Please say what you specifically believe is unphysical about the situation — what trajectory reaches the singularity in finite time and why specifically is that unphysical? My understanding is that you have a cusp singularity that is actually an infinite spike, ie, distance to the singularity is unbounded; that is, no matter how small a circle/sphere around the singularity, you have an infinite diameter. And so you will need to be much more explicit about where the problem lies.