2,324 reputation
1122
bio website staff.science.uu.nl/~hooft101
location Utrecht, Netherlands
age
visits member for 1 year, 11 months
seen Feb 4 at 15:15

Theoretical physicist, Utrecht University.

University Professor. Elementary particle physics, quantum gravity, black holes, quantum mechanics.

Won some prizes (such as the nobel prize 1999), but please don't hold that against me.


Nov
4
awarded  Citizen Patrol
Nov
4
answered Will Determinism be ever possible?
Nov
1
comment Is flying really easier on smaller scales?
Apologies to AlanSE, my comments and questions were to be directed to @miceterminator.
Nov
1
revised Is flying really easier on smaller scales?
added 6 characters in body
Oct
30
comment Is flying really easier on smaller scales?
But this is what I read on my screen, signed by your name. If you didn't mean $\sqrt{\text{diameter}}$, then what are you trying to say?
Oct
28
comment Is flying really easier on smaller scales?
@AlanSE Please explain why you write $\sqrt{\text{diameter}}$ where I would have expected $\text{diameter}^2$, assuming that strength of muscle tissue is more or less size independent. Of course, the energy produced by a muscle scales with its volume, times frequency.
Oct
28
revised Is flying really easier on smaller scales?
added 2 characters in body
Oct
27
answered Is flying really easier on smaller scales?
Sep
15
revised Why do people categorically dismiss some simple quantum models?
added 282 characters in body
Sep
12
comment Why do people categorically dismiss some simple quantum models?
In principle, yes, one should be able to construct a Bohmian field theory, but I think it would be inelegant. To my taste, Bohmian mechanics adds far too many "unobservable observables" in the form of pilot waves. This would be awful for field theories, where the pilot wave would be a field functional, or a function of infinitely many particle positions.
Sep
7
comment Why do people categorically dismiss some simple quantum models?
Think of qft as a large set of quantum harmonic oscillators, each oscillating at isolated points in space. Then assume that each oscillator shows interactions only with its direct neighbors. In qft, these are quantum interactions. To most theorists, this looks sufficiently local, no spooky signals.
Sep
7
comment Why do people categorically dismiss some simple quantum models?
@user7348: No, quantum field theory as it stands has causality built in; for that, it is sufficient to demand that all commutators vanish outside the light cone. This guarantees that no signal ever will go faster than light. So qft obeys relativity and has ordinary qm particles in its non-relativistic limit. Everything is fine, no problem with relativity, until you try to understand what the ontology is. You have to remember that spacelike correlations are fine if you can explain them in terms of intial states in the past.
Sep
6
comment Why do people categorically dismiss some simple quantum models?
If you repeat an experiment, or do it many times, you therefore can't modify one observable without affecting an other one, somewhere, somehow. It is difficult to understand how this happens, you have to remember that the vacuum surrounding us is a very complicated entangled state. All this is the real reason why Bell may be violated in the CA. So I ignore Bell, and in that case qm (rather: qft) is local.
Sep
6
comment Why do people categorically dismiss some simple quantum models?
@ user7348: I do not accept Bell's theorem that easily. It is difficult to see exactly what happens, but it is crucial that all CA observables at all times commute. After my unitary mapping, therefore, only observables that are orthogonal to each other are uniquely defined in terms of CA variables. Now, since the mapping is complicated, these observables are different every time you do an experiment. Therefore we can have counterfactual observables that do not commute.
Sep
6
comment Why do people categorically dismiss some simple quantum models?
@ drake: I don't know what theories you talk about. In my theory, obtained from a local mapping from a local CA, the only non-locality is over a small number of lattice sites. Further away, all commutators outside the light cone do vanish.
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.