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Aug
24
comment Does any particle ever reach any singularity inside the black hole?
The metric contains terms $(1-{2M\over r})$, so that if $r<0$ one must replace $r=-r',\ M=-M'$ to describe the extended region there. In Kerr, positive and negative r are smoothly connected. The singularity is only along the equator. We do have another pathology in that region: there is a region with closed timelike curves. Just look at the complete metric expression carefully. Now where is this wrong wrong wrong?
Aug
24
comment Does any particle ever reach any singularity inside the black hole?
I said that the SECOND horizon is unstable. It really means that objects falling in long after the observer himself, will get in his way, with infinite blue shift. There is also an issue about infalling objects in the new universe where your observer left the bh. But I thought we agreed there. What you did wrong was that you did not notice that, in the Kerr case, ignoring the problems with the second horizon, the observer can get out in two directions: $r\rightarrow\pm\infty$. The negative r direction corresponds with negative mass BH. Just look at the equs for metric, and Penrose diagram.
Aug
23
comment In 't Hooft beable models, do measurements keep states classical?
It's not a question of "thinking that ..." or "intuition", I am talking of simple mathematical facts. As explained in step 3 above, all that needs to be computed in a CA, with states $|Q_i(t)\rangle$ is the amplitudes $\langle Q_i(t_{\,1})|Q_j(t_{\,2})\rangle$. They are 0 or 1 (mostly 0). I can compute this in any basis I like. These are unitary transformations. There's nothing more to it than that. But I am done. I terminate this discussion.
Aug
22
comment Does any particle ever reach any singularity inside the black hole?
@Anixx: The answer by Gordon is correct; this is the paradoxical thing about Hawking radiation. It is not observed by the observer falling in.
Aug
22
comment Does any particle ever reach any singularity inside the black hole?
@John, you are right, but only in a formal sense: the infalling particle will close in to the horizon so rapidly, and the evaporation is so slow, that the external observer will not be able to distinguish the signals he receives from Hawking radiation. The signals from the particle will red-shift way too fast.
Aug
22
comment Does any particle ever reach any singularity inside the black hole?
@Ron and Killercam: sorry, you both got it wrong: in the Penrose diagram there are several spots where you might hope to cross the second horizon, at $r\rightarrow+$ or $-\infty$. The negative $r$ choice will land you in a negative mass black hole. But we agree that that horizon is unstable and will kill the astronaut.
Aug
22
comment In 't Hooft beable models, do measurements keep states classical?
From your previous comments, I only now realize what's bothering you. You seem to think that, also in the CA basis, one should look at inner products $\langle\psi_1|\psi_2\rangle$ and the absolute values of these inner products cannot directly be interpreted as probabilities in the CA. True, but those inner products will never occur in the final expressions. I insist on the postulate that macroscopic measurements also refer only to the CA states. The pointer on a measuring device is described by a CA state, since it is ontic. and Ron, then apply step 3 as I said above.
Aug
22
comment Can superdeterminism resolve contextuality, entanglement and Shor's algorithm in quantum mechanics?
How do you imagine a change in the distant past that affects the gut of a mouse but not the EPR photons? Even if the EPR photons are outside the lightcone, I can simply go further to the past ... all the way to the inflationary early universe, if needed.
Aug
22
awarded  Necromancer
Aug
22
comment Discreteness of Spacetime and Violation of Lorentz symmetry
I don't think one can write down field theories that break the continuous Lorentz group but keep the discrete subgroup intact. Therefore my guess is: none. But it's a guess.
Aug
22
comment In 't Hooft beable models, do measurements keep states classical?
@Ron: The same holds for the planets. I can try to figure out how the Earth-Mars exchange operator evolves, use Schroedinger equations. But in the very end, these Schroedinger equations simply permute probability states of the planets, so here it is easy to understand what the "quantum calculation" does: for the planets, it does nothing that you couldn't do the conventional way. For atoms, the conventional way (which now would amount to calculate the CA evolution) would be too difficult.
Aug
22
comment In 't Hooft beable models, do measurements keep states classical?
@Ron: Sorry for these repeats, but they seem to be needed ...
Aug
22
comment In 't Hooft beable models, do measurements keep states classical?
3) And now look at a quantum experiment. Atoms, photons, molecules are too small to affect collective behavior of a CA. My math tells me that one can map the CA states onto wave functions describing these atoms. The theory says: do that math. Solve those Schroedinger equs. You will end up with a wave function describing the final state. THEN go back and project that on the CA states in your detector. Voila, you did quantum physics to explain what your detector did. IT'S THE ONLY WAY TO EXPLAIN THAT!
Aug
22
comment In 't Hooft beable models, do measurements keep states classical?
2) That means that classical observations all commute, which of course we already knew. The fact that they commute with the CA positions, as a bonus, explains why we perceive what some think is a "collapse of the wave function". Probabilities, Born's rule, they now all make sense.
Aug
22
comment In 't Hooft beable models, do measurements keep states classical?
@Ron: The idea that (macroscopic) measurements are classical means that macroscopically you can measure only beables, the states of the automaton. You may view this as a new "postulate". It makes sense: 1) it assumes that macroscopic objects, planets, people, a pointer on a device, etc., have at least some minute effect on the collective modes of the CA. It is hard to believe that there should be absolutely no such effect. As soon as that's the case, you need to look only at the CA to see that you have a planet or a person somewhere.
Aug
22
awarded  Notable Question
Aug
22
awarded  Nice Answer
Aug
21
awarded  Good Question
Aug
21
comment Why do people rule out local hidden variables?
The strange thing is that this point of resetting the experiment, both for Alice and for Bob, should be obvious from the rules everyone uses in QM; it is the reason why QM disobeys the Bell inequalities. All you have to say is that a hidden variable theory must have the same property, and this is possible, even if it is a local theory, but you have to accept its deterministic nature all the way. Don't dilute that with "free will".
Aug
21
comment Why do people rule out local hidden variables?
What this really means is that, strictly speaking, you can't make a "counterfactual" observation. You choose the operator you want to measure, and you can't change your mind anymore. The usual objections people make to this are based on ideas they claim to have about "free will". As you might infer from a previous answer, it has something to do with religion...