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Its a short and elegant proof set in pure theoretical general relativity (GR). The idea is that if there’s a sphere of space where if you try to emit light rays

by zachf 6y ago

Its a short and elegant proof set in pure theoretical general relativity (GR). The idea is that if there’s a sphere of space where if you try to emit light rays and the light rays don’t initially start separating, then they can’t start separating due to gravity, because in classical GR, gravity is always attractive. You can then show that this implies that inside that sphere, spacetime must end, basically because you can’t outrun light. The proof is important because it was previously believed that black holes are not interesting because they require very special perfect conditions to create, like balancing a pencil on its tip is physically possible but requires perfect aim. But these aforementioned spheres are very common and easy to find so it turns out black holes are common too. If you think (as almost every physicist does) that GR is approximately correct to describe reality, but needs fixes at very tiny lengths because of poorly understood quantum effects, the proof does not directly carry over. One immediate problem is that the proof assumes that energy densities are positive, implying that that gravity is universally attractive, which for quantum matter can never be always true for every quantum state (this is a consequence of Reeh-Schlieder, that every QFT contains states with negative energy density). None of this invalidates Penrose’s work. Physicists have always used different physics to describe different scales. Newtonian physics is great to describe most physics on a human scale, but it’s “wrong” in the sense that GR supersedes it. Similarly GR is “wrong” but still approximately right for a ton of questions of cosmology. But if you fall into a black hole, once you wait long enough, we don’t know what will happen. In string theory, there are objects that are black hole-like. It is generally believed that the singularity is “resolved” (not truly present) in string theory but the details are very tricky to work out. It still is true that geometry breaks down near the singularity and whats left is some stringy stuff, something very new and confusing. Of course it might turn out that string theory does not describe our reality either...

andrewon 6y ago
Thanks for your reply. Sounds like the proof is an interesting piece to study.
ssivark 6y ago
Could you elaborate why Reeh-Schlieder implies negative energy states? Typically in QM we only care about the spectrum being lower bounded (existence of vacua), and ignoring additive energy constants.
- zachf 6y ago
  Good question! You can definitely add a constant to the Hamiltonian without changing the local physics, but I was referring to the stress-energy tensor (i.e. energy density), not the Hamiltonian (total energy). The stress tensor can be locally negative but still integrate to a positive or zero value of total energy. So let's talk about energy density, I'll call it T. I need two facts: 1. The Reeh-Schlieder theorem implies that any operator that the vacuum state of QFT is "separating", meaning that the only operator A that satisfies A|0> = 0 is the trivial solution A = 0. (Here |0> means vacuum.) 2. The energy density operator has zero expectation value in the vacuum state, <0|T|0> = 0. Why? Explicitly: it has to be constant because of translational symmetry, and it has to because 0 is the only value that is constant and integrates to 0 total energy. From fact 1, we know that T|0> is not 0 in any nontrivial theory (otherwise T = 0 identically, and no state has energy density at all). From fact 2, it follows that T|0> is orthogonal to |0>. Therefore |0> and T|0> span a 2D subspace of the Hilbert space. Now write down the matrix T in that space. It looks like [0, b*] [b, c ] where b is nonzero because of the way the subspace is defined, and c is real because T is Hermitian. Such a matrix is never positive definite, therefore there exists a state with negative expectation value in this subspace. Explicitly, one of the eigenvalues of this matrix is c - sqrt(|b|^2 + c^2) which is negative because b is nonzero. The corresponding eigenvector therefore has negative energy density. (Source for this argument: [0]) An example of such a state is the Casimir state, see [1]. [0] https://arxiv.org/abs/1803.04993 https://arxiv.org/abs/1803.04993 [1] https://en.wikipedia.org/wiki/Casimir_effect https://en.wikipedia.org/wiki/Casimir_effect
  - ssivark 6y ago
    Thanks for the nice explanation, and the Witten ref. So it seems that in QFT we must have at least one negative energy state (should we consider such a state “below” the vacuum? Maybe the message is that we should give up on a strict ordering once we have many local degrees of freedom). It’s interesting to interpret the Casimir effect in this language — is it that the location of negative energy density is specified by spontaneously broken translational symmetry (negative between plates and zero/positive outside). What property decides whether the system is sitting in the “vacuum” or the “Casimir” state? (If we’ve given up on “lowest energy”)

by zachf 6y ago

andrewon 6y ago

Thanks for your reply. Sounds like the proof is an interesting piece to study.

ssivark 6y ago

Could you elaborate why Reeh-Schlieder implies negative energy states? Typically in QM we only care about the spectrum being lower bounded (existence of vacua), and ignoring additive energy constants.

zachf 6y ago
Good question! You can definitely add a constant to the Hamiltonian without changing the local physics, but I was referring to the stress-energy tensor (i.e. energy density), not the Hamiltonian (total energy). The stress tensor can be locally negative but still integrate to a positive or zero value of total energy. So let's talk about energy density, I'll call it T. I need two facts: 1. The Reeh-Schlieder theorem implies that any operator that the vacuum state of QFT is "separating", meaning that the only operator A that satisfies A|0> = 0 is the trivial solution A = 0. (Here |0> means vacuum.) 2. The energy density operator has zero expectation value in the vacuum state, <0|T|0> = 0. Why? Explicitly: it has to be constant because of translational symmetry, and it has to because 0 is the only value that is constant and integrates to 0 total energy. From fact 1, we know that T|0> is not 0 in any nontrivial theory (otherwise T = 0 identically, and no state has energy density at all). From fact 2, it follows that T|0> is orthogonal to |0>. Therefore |0> and T|0> span a 2D subspace of the Hilbert space. Now write down the matrix T in that space. It looks like [0, b*] [b, c ] where b is nonzero because of the way the subspace is defined, and c is real because T is Hermitian. Such a matrix is never positive definite, therefore there exists a state with negative expectation value in this subspace. Explicitly, one of the eigenvalues of this matrix is c - sqrt(|b|^2 + c^2) which is negative because b is nonzero. The corresponding eigenvector therefore has negative energy density. (Source for this argument: [0]) An example of such a state is the Casimir state, see [1]. [0] https://arxiv.org/abs/1803.04993 https://arxiv.org/abs/1803.04993 [1] https://en.wikipedia.org/wiki/Casimir_effect https://en.wikipedia.org/wiki/Casimir_effect
- ssivark 6y ago
  Thanks for the nice explanation, and the Witten ref. So it seems that in QFT we must have at least one negative energy state (should we consider such a state “below” the vacuum? Maybe the message is that we should give up on a strict ordering once we have many local degrees of freedom). It’s interesting to interpret the Casimir effect in this language — is it that the location of negative energy density is specified by spontaneously broken translational symmetry (negative between plates and zero/positive outside). What property decides whether the system is sitting in the “vacuum” or the “Casimir” state? (If we’ve given up on “lowest energy”)