G. 't Hooft
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 Sep 6 comment Does any particle ever reach any singularity inside the black hole? Your theory about leaving the black hole alive is not going to hold up at all, it's wild guesses that would create totally unnecessary conflicts with everything we do know about black holes. Simple thermodynamical arguments tell us that, in the real world, the only thing that can get out of a black hole is Hawking radiation. Astronauts would be suppressed by gigantic Boltzmann exponentials. The classical story would have you come out in some other universe; that's against any thermodynamical law so with hbar, it won't happen, even there. Sep 6 comment Does any particle ever reach any singularity inside the black hole? Mass is defined asymptotically, yes, but the mass term always comes in the combination $Mr$, so when $r\rightarrow-\infty$, where the angular part of the metric goes as $r^2$, you have to replace $r$ by $|r|$ to see what happens. So $M$ goes to $-M$. Look up the Penrose diagrams in Hawking and Ellis for example. And look up the Kerr metric (too long for these "comments"). If you do AdS/CFT you are doing quantum, and you won't avoid thermalization. Only the classical theory would seem to allow you to get out, God knows in which universe. Sep 6 comment Does any particle ever reach any singularity inside the black hole? You were talking of rotating black holes. If they don't rotate the negative $r$ region is closed off by a singularity. But if the holes rotates, you can get at $r<0$ because the singularity is only at the equator - you need to take a northern or southern route, also to avoid the region with closed timelike curves (at small negative $r$ and large $\sin^2\theta$. I'm talking of the Kerr and Kerr-Newmann metric (necessary for rotation). Sep 5 comment Does any particle ever reach any singularity inside the black hole? And then, sorry, but the location of a horizon is a function of the $r$ variable, not $t$, so after passing that second horizon you will enter into the negative $r$ regime. The coordinates $r$ and $t$ do interchange, in the sense that, between the two horizons, $t$ becomes spacelike and $r$ timelike. But does the BH get opposite spin? Please think: how did you define the spin direction, in which coordinates? The statement is empty. Sep 5 comment Does any particle ever reach any singularity inside the black hole? @Ron I'm afraid you were saying some more incorrect things: you "believe" that the observer comes out in the same universe. But what will he (it) look like? Quite likely is that the observer will not be noticed by the other inhabitants in his (former) universe since he turned into Hawking radiation. All modern theories say that the best he could do is give some subtle twists in the Hawking particles of the same BH that he entered. He (it) turned into a ghost. Sep 5 comment Why do people categorically dismiss some simple quantum models? For many physicists, this is all that matters, also in the 1960s. I presume Einstein did think of something like hidden variables. 2: WITH Hidden variables, Bell claims that the hidden variables are non-local, but what he really means is that the Ansatz equation that he starts with cannot be satisfied. My claim is that in a superdeterministic scenario that equation cannot hold, even if the evolution laws of a CA are local. Sep 5 comment Why do people categorically dismiss some simple quantum models? @user 7348: 1: if you don't care about hidden variables, quantum mechanics as it is, or more precisely quantum field theory, is entirely local. Locality means that if we have, in the Heisenberg notation, two field operators depending on space-time: $Op_1(x_1,t_1)$ and $Op_2(x_2,t_2)$, then they must commute if $(x_1,t_1)$ and $(x_2,t_2)$ are completely space-like separated. This holds for QFT and even (in spite of claims to the contrary) for string theory (if two points are spacelike separated in target space, they also are so in the world sheet - assuming we may ignore certain projections). Aug 30 comment Why do people categorically dismiss some simple quantum models? OK let me correct that last statement. Some pseudo-non-locality can enter in two ways: 1: the description of the vacuum state as a superposition of CA states. The vacuum makes discussion of Bell's inequalities very difficult in CA. 2: There are good reasons for imagining information loss to occur in the CA. You can still map that onto a quantum system (with full CPT invariance), but that mapping leads to holography and apparent non-locality. Aug 30 comment Why do people categorically dismiss some simple quantum models? They even don't violate locality, since the QFT has commutators that always vanish outside the light cone. It could be that there is some rudimentary non-locality in the mapping, but I have not really encountered that yet. Aug 30 comment Why do people categorically dismiss some simple quantum models? Superdeterminism is obviously there, if you care to give it some thought. Now, when people talk of "conspiracy" they really mean that they don't understand the result. But you can understand it this way: the CA can be treated as a fully grown quantum system, as a QFT. This QFT leads to correlations that look like conspiracy. Hence this apparent conspiracy is there. Don't be afraid of spooks. Mathematically, there's nothing wrong with them, just a bit difficult. Aug 29 comment Why do people categorically dismiss some simple quantum models? @user7348, Do you mean those pilot waves? I think these are ugly concoctions, but I do agree that they show that there is a possibility in principle. I think that elegance and plausibility will be important assets of a healthy theory. I can't make working field theories using Bohm. I am talking about much more fundamental principles. And most importantly: my theory IS quantum mechanics, not an "alternative". Aug 28 comment Can superdeterminism resolve contextuality, entanglement and Shor's algorithm in quantum mechanics? It's essential to add interactions, making the CA non-trivial. Only then my arguments make sense. Without that, they also work but it's just formalities and semantics. Interactions turn your CA into a universal computer. You can still "solve" the CA equations but you can't speed them up using, say, renormalization group (RG) techniques, to go to larger time and distance cales. Therefore you would need a computer with Planckian dimensions. That does not exist today. Aug 28 comment Can superdeterminism resolve contextuality, entanglement and Shor's algorithm in quantum mechanics? I'd be happy to continue in "chat" but I am not going to waste my time trying to figure out how to do this. Aug 27 revised In 't Hooft beable models, do measurements keep states classical? added 698 characters in body Aug 27 revised In 't Hooft beable models, do measurements keep states classical? added 698 characters in body Aug 27 comment Can superdeterminism resolve contextuality, entanglement and Shor's algorithm in quantum mechanics? That these correlations are "absurd" or "ridiculous" is not a good enough argument to me. The mouse gut, the brain, a flipper machine, they all obey conservation laws such as energy and angular momentum, and the laws of thermodynamics. So they also obey unitarity. There's nothing absurd or ridiculous about that. Just do the CA - quantum mapping. Aug 27 comment Can superdeterminism resolve contextuality, entanglement and Shor's algorithm in quantum mechanics? @ Mitchel: The CA is a universal computer, as such non integrable. I think my CA - quantum mapping just proves that the CA generates "miraculous finetuning". It does not exactly randomise. It computes more accurately than a mouse so, yes, it controls the mouse's gut. Now remember that the QFT generated by the CA still obeys quantum causality (i.e. commutators vanish outside the light cone), so the finetuning is still causal, which makes it a bit less "miraculous". We can never short circuit this universal computer, but we can compute its correlations, using the mapping on QM. Aug 26 comment Can superdeterminism resolve contextuality, entanglement and Shor's algorithm in quantum mechanics? @ Mitchell: Do you agree with the following? Your "arbitrarily complicated macroscopic processes" do have a past; to change the setting of polarization filter $a$, you have to change that past. By resetting the past, the "local vacuum data", both at $a$ and at $b$, will have changed completely. My point is that, whether a photon at $a$ or at $b$ gets through, is difficult or impossible to compute in the deterministic theory. Only the chances can be calculated by using QM as a tool. Acceptable or not? Aug 24 comment Can superdeterminism resolve contextuality, entanglement and Shor's algorithm in quantum mechanics? The sad thing is that I can't get it into people's heads that counterfactual measurements are not meaningful, even in CA theories. Operators that are not diagonal in the CA basis cannot be measured. The point that seems to escape all the time is that a CA can be so complex that no counterfactual measurements are needed to describe the outcome of a macroscopic observation. This is why non-diagonal operators may nevertheless be extremely useful to describe intermediate results of a quantum calculation. Aug 24 comment Can superdeterminism resolve contextuality, entanglement and Shor's algorithm in quantum mechanics? @ Mitchell: even if the logical chain does not pass through the mouse's gut, it does reset the data for the EPR experiment. So the photons are affected by the past data.