Quantum Physics, CERN and Hawking radiation
post by MatthewBaker · 2011-06-16T20:44:33.885Z · LW · GW · Legacy · 66 commentshttp://lifeboat.com/blog/2011/06/dear-dr-hawking
Hey guys, my quantum physics is not powerful enough to understand this guy... Can anyone help me out with this one?
Thanks LW
66 comments
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comment by prase · 2011-06-17T12:57:40.029Z · LW(p) · GW(p)
Can anybody explain how a black hole with mass of few dozens of atoms can gravitationally attract a significant amount of matter to grow? Is there some paper which quantitatively analyses how fast would such a tiny stable BH grow under some plausible circumstances?
If I take 1000 TeV, the expected energy of two colliding Pb ions on LHC, to be the mass of the produced BH (which is certainly a gross overkill), its classical Schwarzschild radius will be of order 10^-48 m. For comparison, the charge diameter of a proton is about 10^-15 m, the interatomic distances are of order 10^-10 m.
Replies from: Douglas_Knight↑ comment by Douglas_Knight · 2011-06-21T22:40:10.718Z · LW(p) · GW(p)
A black hole small enough that Hawking radiation is relevant is too small to be a danger to anything. As far as I can tell, a quite large black hole could sit at the center of the earth without doing anything noticeable. I think my calculation was that a 10 micron radius black hole at the center of the earth would have a doubling time of a billion years. That link is to Scott Aaronson's blog where he asks that exact question. I recorded 10 microns there, but otherwise my calculations have been lost. Various people do similar calculations and seem to get similar numbers, but don't all make the qualitative conclusion. One of the commenters linked to this paper, but I don't recall extracting an answer from it.
comment by jimrandomh · 2011-06-16T21:23:55.730Z · LW(p) · GW(p)
The author of that article, Otto E. Rössler, believes that the Large Hadron Collider experiment could create a stable black hole and destroy the world. I have no knowledge of the relevant physics that would enable me to determine whether this claim is plausible. He has been writing about risk from the LHC on the Lifeboat blog for some time now. This particular post claims that a mechanism which is believed to cause black holes to lose mass, Hawking radiation, might not exist.
Replies from: gwern, timtyler↑ comment by gwern · 2011-06-16T21:36:05.950Z · LW(p) · GW(p)
I've been increasingly skeptical of Lifeboat Foundation, and have been meaning to write this up. Otto matches a number of items on the usual crank checklist, and his co-blogger Matt Funk matches even more. (Funk's writings are amazingly cranky; he doesn't mind posting rejections from ArXiv or including crank letters he's sent to prime ministers or Warren Buffet or Paul Allen, IIRC.) And then there's this stuff...
↑ comment by timtyler · 2011-06-16T21:46:24.075Z · LW(p) · GW(p)
Black holes not being able to lose mass would violate reversibility. If something can fall into a black hole, other things must be able to come out again - since physics exhibits microscopic reversibility, and the known laws of physics are symmetrical under T=-T.
Replies from: RolfAndreassen, prase↑ comment by RolfAndreassen · 2011-06-17T02:41:04.541Z · LW(p) · GW(p)
the known laws of physics are symmetrical under t=-t.
This is probably not correct. We know that CP symmetry is broken. There are good theoretical reasons to believe that CPT symmetry holds. Consequently T symmetry must be broken.
Replies from: timtyler↑ comment by timtyler · 2011-06-17T06:58:35.958Z · LW(p) · GW(p)
To quote from here:
Fredkin has suggested [here] that CPT symmetry is an immediate consequence of T symmetry - i.e. that inverting T has as its consequence inverting the parity and charge of all particles.
...for the reason to think that it is T symmetry, not CPT symmetry that holds.
Charge and Parity are likely to be implemented using internal rotation or sequences - and so will reverse automatically if T=>-T. We don't have to manually reverse particle momenta or spin if we are running things backwards. That happens automatically. The idea is that Charge and Parity function like that too - due to how these phenomena work. For instance, Charge seems likely to be implemented rather like a bi-directional pump. Reverse time, and such a pump runs backwards automatically. You do not have to go around reversing all the charges - they reverse themselves automatically. It works like spin does - and reverses automatically for much the same reason.
T-symmetry is neater and simpler. We should prefer it - unless there are good reasons not to do so.
Replies from: RolfAndreassen, gjm↑ comment by RolfAndreassen · 2011-06-17T17:43:18.767Z · LW(p) · GW(p)
that CPT symmetry is an immediate consequence of T symmetry
If so, then how can T symmetry hold? You seem to be saying that T symmetry implies CPT symmetry. But we know from experiment that CP symmetry is broken. If T symmetry holds, and CP symmetry does not hold, then CPT symmetry cannot hold.
Really, this looks pretty straightforward. The theory you quote has A->B. Experiment !B. Consequently, either !A or !(A->B).
Charge and Parity are likely to be implemented using internal rotation or sequences
Why do you think so?
We don't have to manually reverse particle momenta or spin if we are running things backwards.
Particle momenta, no; spin, yes. Although spin is angular momentum, it does not come about because particles are rotating about an internal axis, as you seem to have in mind. (To the best of anyone's knowledge, of course.) Consequently parity does not auto-reverse under time-reversal.
Replies from: timtyler↑ comment by timtyler · 2011-06-17T20:05:36.482Z · LW(p) · GW(p)
that CPT symmetry is an immediate consequence of T symmetry
If so, then how can T symmetry hold? You seem to be saying that T symmetry implies CPT symmetry. But we know from experiment that CP symmetry is broken. If T symmetry holds, and CP symmetry does not hold, then CPT symmetry cannot hold.
Really, this looks pretty straightforward. The theory you quote has A->B. Experiment !B. Consequently, either !A or !(A->B).
OK - so, you don't understand the idea. There is a much more detailed description of the associated model written by someone else here.
The punchline at the bottom reads:
There is a wonderful consequence of what we have just described. This DM model has T symmetry. However the T symmetry in this DM model is exactly equivalent to CPT symmetry in ordinary physics. If a model like this were to reflect the physics of the real world, then T symmetry would be restored to physics as consistent with all the laws of physics and all experimental evidence!
Please let me know if that fails to sort you out - and you are still interested.
Replies from: RolfAndreassen↑ comment by RolfAndreassen · 2011-06-18T03:10:16.029Z · LW(p) · GW(p)
First, the theory rests on the airy assertion that reversing T automatically causes the reversal of spin and other quantum numbers as well. I found the argument given for this unconvincing. Second, and more importantly, you do not seem to have grasped that you cannot possibly have both T symmetry and CPT symmetry, because CP symmetry is experimentally excluded. It does not matter if you invent a special form of T symmetry that is 'equivalent' to CPT symmetry.
Take a physical system that exhibits CP violation; assume it is described by the kind of theory outlined in your link. Now reverse time. By the argument in your link, this also reverses CP. Because the system is not symmetric under CP, it exhibits different behaviour. Bing, T symmetry has been broken: There is a measurement I can make that tells me which way time is flowing.
Replies from: timtyler↑ comment by timtyler · 2011-06-18T08:30:11.046Z · LW(p) · GW(p)
Well, I don't have a watertight argument for the first point. I think it is more likely than not, but if your intuition is the other way around, I won't argue too much. What I object to is the idea that T-symmetry is wrong. In fact, T-symmetry is pretty plausible, IMO.
From your second point, (from my perspective) you still don't get the logic of the whole idea - and you have exhausted most of my resources on the subject, so I am not sure what more to do with you.
Assuming that charge and parity quanta involve moving parts internally, then they would both reverse automatically if time is reversed - producing what appears to be CPT symmetry as a result. That would be consistent with all known experiments, and physics would then by time symmetric.
You said: "Because the system is not symmetric under CP, it exhibits different behaviour." No, because you have also reversed time, (you just said so yourself) - and if C,P and T are all reversed, then symmetry is restored. So, then there is no measurement you can make that tells you which way time is flowing.
Replies from: RolfAndreassen↑ comment by RolfAndreassen · 2011-06-18T20:22:36.947Z · LW(p) · GW(p)
Assuming that charge and parity quanta involve moving parts internally, then they would both reverse automatically if time is reversed - producing what appears to be CPT symmetry as a result.
No. Start with a left-handed neutrino. Reverse T under your assumption. It is now a right-handed antineutrino going the other way; reverse space as well to restore the original direction, if you like, although the argument does not depend on this. Because CP is broken, right-handed antineutrinos do not behave exactly as left-handed neutrinos do. Therefore you can tell how many times T has been reversed. You don't get the full symmetry back except by applying CP another time.
Replies from: timtyler↑ comment by timtyler · 2011-06-18T23:15:35.073Z · LW(p) · GW(p)
Assuming that charge and parity quanta involve moving parts internally, then they would both reverse automatically if time is reversed - producing what appears to be CPT symmetry as a result.
No. Start with a left-handed neutrino. Reverse T under your assumption. It is now a right-handed antineutrino going the other way;
Yes.
reverse space as well to restore the original direction, if you like, although the argument does not depend on this.
A parity) flip, I presume you mean.
Because CP is broken, right-handed antineutrinos do not behave exactly as left-handed neutrinos do.
That is indeed true.
Therefore you can tell how many times T has been reversed.
Well you only said you reversed it once - and then you flipped P, but not C, leaving things in a bit of a mess - and then you tried to make out the mess was something to do with me.
Reversing T an odd number of times changes everything. Reversing it an even number of times changes nothing. You can't distinguish between reversing T different numbers of times beyond that - under the hypothesis that reversing T automatically reverses C and P.
Replies from: RolfAndreassen↑ comment by RolfAndreassen · 2011-06-19T20:45:52.033Z · LW(p) · GW(p)
Ok, leave the parity flip out of it. If this is true:
Reversing T an odd number of times changes everything.
then you do not have T symmetry. Done.
Replies from: timtyler↑ comment by timtyler · 2011-06-19T21:13:26.371Z · LW(p) · GW(p)
It makes time run backwards. Those in charge may not think that this is such a null-op.
If you pressed the "rewind" button, you would normally expect to see some changes!
Replies from: RolfAndreassen↑ comment by RolfAndreassen · 2011-06-19T21:55:23.396Z · LW(p) · GW(p)
Ok, there's your problem: You don't understand what is meant by 'symmetry'.
Replies from: timtyler↑ comment by timtyler · 2011-06-19T23:31:21.330Z · LW(p) · GW(p)
At this stage, I don't really see why you are continuing to comment :-(
Replies from: RolfAndreassen↑ comment by RolfAndreassen · 2011-06-20T21:13:26.765Z · LW(p) · GW(p)
To convince you that you are wrong about CPT violation and T violation. Why are you posting?
Once more. Start with a left-handed antineutrino. T-reverse under your assumption that this also reverses CP. You now have a right-handed neutrino. Because of CP violation, it does not have the same physical properties that it started with. Therefore, T symmetry is broken. Which part of this argument do you disagree with?
Replies from: timtyler↑ comment by timtyler · 2011-06-20T21:44:31.739Z · LW(p) · GW(p)
The "Therefore". Reverse the universe, and a left-handed antineutrino turns into a right-handed neutrino travelling in the opposite direction. Everyone agrees about that. Its different properties don't prevent the universe from retracing its steps - rather they are essential for that to happen correctly.
Replies from: RolfAndreassen↑ comment by RolfAndreassen · 2011-06-21T01:59:58.389Z · LW(p) · GW(p)
No; wrong. Its different properties will, precisely, cause the universe not to retrace its steps exactly. The rate for X\to e^+ \nu_e is different from that for e^- \bar\nu_e \to X; this is what CP violation means. Therefore, when you have reversed time, the antineutrino will not precisely retrace the steps the neutrino took.
Replies from: timtyler↑ comment by timtyler · 2011-06-21T08:35:49.054Z · LW(p) · GW(p)
Do you realise that what you are claiming is pretty unconventional? Here is the conventional view:
The implication of CPT symmetry is that a "mirror-image" of our universe — with all objects having their positions reflected by an imaginary plane (corresponding to a parity inversion), all momenta reversed (corresponding to a time inversion) and with all matter replaced by antimatter (corresponding to a charge inversion)— would evolve under exactly our physical laws. The CPT transformation turns our universe into its "mirror image" and vice versa. CPT symmetry is recognized to be a fundamental property of physical laws.
Investingating to see if I could see what you are talking about found some claims that the symmetry between neutrinos and antineutrinos is violated:
Several experimental searches of such violations have been performed during the last few years and recently there has been some strong evidence for a violation of charge symmetry in that antineutrinos seem to have a different mass than neutrinos.
In the highly unlikely case of any such asymmetry being confirmed, that would break CPT symmetry - and serious revisions of fundamental physics would be needed.
Replies from: RolfAndreassen↑ comment by RolfAndreassen · 2011-06-21T18:25:51.996Z · LW(p) · GW(p)
Do you realise that what you are claiming is pretty unconventional?
No. I am giving you the conventional view, which you do not understand.
I do not wish to appeal to authority, but since we are now arguing in terms of what is the conventional view, perhaps I can legitimately mention that I have a PhD in experimental particle physics. True, I'm not a theorist, but I do feel I have a reasonable grounding in these matters.
In the highly unlikely case of any such asymmetry being confirmed,
Which part of "CP symmetry is broken" is unclear to you? If antineutrinos and neutrinos have different masses, that breaks C symmetry and its discoverer will certainly get a trip to Stockholm. But this is not required for the argument I gave above to be correct. The breaking of CP symmetry is already known, and has been known since the sixties. It has exactly the same consequences as if neutrino and antineutrino masses are different, it's just a bit more difficult to visualise.
Replies from: timtyler↑ comment by timtyler · 2011-06-21T19:56:43.621Z · LW(p) · GW(p)
I don't really see why you don't seem to understand what I am saying - and this message doesn't really help very much. Why do you think that I think that CP symmetry is not broken. What have I said that would lead you to that conclusion?
In an attempt to clarify, C P and T all need to fllp sign for proper reverse evolution to occur. From your above messages, it seems as though you doubt that - in which case you should probably say so clearly at this point. My messages just assume that the reader thinks that that is true.
The main issue is not whether that happens, but whether C and P flip themselves automatically if you just reverse T. Momenta flip automatically if you reverse T - because they are derivatives with respect to time. The hypothesis is that C and P would also behave like that - and probably for much the same reason.
Your messages so far seem to be concerned with whether C and P flip, and what happens if they don't. That is far from the issue under discussion - from my perspective.
Replies from: RolfAndreassen↑ comment by RolfAndreassen · 2011-06-21T20:49:25.952Z · LW(p) · GW(p)
Why do you think that I think that CP symmetry is not broken?
Because you apparently agree that breaking C symmetry would falsify your theory, but you do not understand that breaking CP symmetry has the same effect.
Your messages so far seem to be concerned with whether C and P flip, and what happens if they don't.
No. I have assumed that C and P flip. Hence my words: "Start with a left-handed neutrino, flip T under your assumption, end up with a right-handed antineutrino". A right-handed antineutrino is C and P flipped with respect to a left-handed neutrino. But, because CP is violated, it does not exactly retrace the steps its antiparticle took, unless T symmetry is violated so as to make up for the CP violation. Since you are asserting that T is a good symmetry, that cannot happen under your theory; consequently your theory is experimentally falsified. That your T flip carries with it the CP flip is not relevant, it has no effect on the argument.
Replies from: timtyler↑ comment by timtyler · 2011-06-21T21:15:35.824Z · LW(p) · GW(p)
Why do you think that I think that CP symmetry is not broken?
Because you apparently agree that breaking C symmetry would falsify your theory, but you do not understand that breaking CP symmetry has the same effect.
So: I am not clear on where you are getting that from either. I don't even recall discussing pure C symmetry in this thread.
Your messages so far seem to be concerned with whether C and P flip, and what happens if they don't.
No. I have assumed that C and P flip. Hence my words: "Start with a left-handed neutrino, flip T under your assumption, end up with a right-handed antineutrino". A right-handed antineutrino is C and P flipped with respect to a left-handed neutrino. But, because CP is violated, it does not exactly retrace the steps its antiparticle took, unless T symmetry is violated so as to make up for the CP violation.
Right, yes, so CPT all need to flip. I agree with that. You agree with that. Everyone agrees on that. At least that has nothing to do with our issue. The issue is not whether these things all need flipping, but whether they flip themselves automatically when T reverses (like momenta do). So far, you don't seem to have got as far as that issue - and are assuming that I am making the mistake that you just described. Which is not what is going on here at all.
Since you are asserting that T is a good symmetry, that cannot happen under your theory; consequently your theory is experimentally falsified. That your T flip carries with it the CP flip is not relevant, it has no effect on the argument.
If T flips, and consequently, C and P also flip, then C,P and T have all flipped - in which case we apparently just agreed that evolution proceeds backwards.
Here you can doubt the premise (sure, go ahead), but that's about the only option for criticism. I don't see how you can claim that flipping C, P and T does not lead to backwards evolution, having just agreed that it does.
Addendum, let me see if I can phrase this in language you are more likely to understand. Neither C, P or T symmetry hold alone. However, under the hypothesis that reversing T flips C and P, reversing T has the effect of reversing C, P and T, which then makes everything run backwards. That is what I am talking about - using your preferred terminology as best as I can manage.
Hopefully you can imagine why I prefer to describe that in terms of simple T-symmetry. Under the hypothesis, it makes much more sense to describe it that way.
Replies from: prase, RolfAndreassen↑ comment by prase · 2011-06-22T00:35:19.554Z · LW(p) · GW(p)
Edit: ignore the comment, it is wrong.
The issue is not whether these things all need flipping, but whether they flip themselves automatically when T reverses (like momenta do).
T-reversal isn't a physical process to be performed, it's a transformation of coordinates, t goes to -t. It "automatically reverses" momenta, because momenta are first derivatives of x with respect to t. It does not include reversal of spatial axes (P), but it includes charge change of particles into antiparticles, by definition.
In our world we may observe a decay of a muon into an electron, a right-handed electron antineutrino and a left-handed muon neutrino. A T-reversed process would include a right-handed electron neutrino, a left-handed muon antineutrino and a positron merging into an antimuon (T includes turning particles into anti-particles, also by definition). But such a process contradicts the experimentally observed fact that right-handed neutrinos and left-handed antineutrinos don't interact (except gravitationally, if they exist at all). Therefore, T is not a symmetry.
Alternatively, T would be a symmetry iff the equations of motion (or Lagrangian) looked exactly the same (up to a total 4-divergence in case of the Lagrangian) after substituting -t for t. But it does not happen for the Lagrangian of the Standard Model, because T changes the left-handed weak interaction term to a right handed one.
Replies from: timtyler↑ comment by timtyler · 2011-06-22T07:08:23.389Z · LW(p) · GW(p)
T-reversal isn't a physical process to be performed, it's a transformation of coordinates, t goes to -t.
It depends on how you define a "physical process". Reversing time in a billiard ball machine is a physical process. Perhaps think about that to understand how time reversal could operate physically.
It "automatically reverses" momenta, because momenta are first derivatives of x with respect to t. It does not include reversal of spatial axes (P), but it includes charge change of particles into antiparticles, by definition.
No, it doesn't - that is simply incorrect. Perhaps read through the section of this, that I quote below - it should explain things:
The implication of CPT symmetry is that a "mirror-image" of our universe — with all objects having their positions reflected by an imaginary plane (corresponding to a parity inversion), all momenta reversed (corresponding to a time inversion) and with all matter replaced by antimatter (corresponding to a charge inversion)— would evolve under exactly our physical laws. The CPT transformation turns our universe into its "mirror image" and vice versa. CPT symmetry is recognized to be a fundamental property of physical laws.
C reversal is described here - and it is not conventionally included in T reversal.
Replies from: prase↑ comment by prase · 2011-06-22T12:02:25.871Z · LW(p) · GW(p)
You are right that T doesn't include particle-antiparticle mixing, of course. My previous comment was confused, I should never comment at 4AM.
But still, the interaction Lagrangian of the Standard model is not invariant with respect to T, since it is invariant with respect to CPT and CP invariance is violated by the CKM matrix.
↑ comment by RolfAndreassen · 2011-06-21T22:57:39.786Z · LW(p) · GW(p)
So: I am not clear on where you are getting that from either. I don't even recall discussing pure C symmetry in this thread.
In your post of 08:35, where you quoted someone saying there was evidence of charge violation, specifically neutrinos and antineutrinos having different masses, and said that if it was so, then CPT violation was broken. This is not actually true, because the P and T symmetries can be broken so as to exactly compensate. In fact this almost exactly happens in the weak force, where the C and P symmetries are separately almost-completely violated, but CP is almost a good symmetry. But all that is a separate point.
Neither C, P or T symmetry hold alone.
Ok, I'm glad we were finally able to agree on this, because that's what I've been saying all along: The laws of physics are not in fact T-symmetric.
Hopefully you can imagine why I prefer to describe that in terms of simple T-symmetry.
No, in fact I can't. You are confusing separate operators and introducing your own notation, and it has led to comments six deep because you refuse to distinguish T from CPT symmetry. Moreover, it leads to you contradicting yourself: In one paragraph you agree that T is not separately a good symmetry, and then in the next you say that you "prefer to describe [physics] in terms of simple T symmetry". If the symmetry is one that doesn't actually hold, then I suggest that it is not simple at all, and certainly not worth introducing nonstandard notation for.
Replies from: timtyler↑ comment by timtyler · 2011-06-22T00:12:42.797Z · LW(p) · GW(p)
You are confusing separate operators and introducing your own notation, and it has led to comments six deep because you refuse to distinguish T from CPT symmetry. Moreover, it leads to you contradicting yourself: In one paragraph you agree that T is not separately a good symmetry, and then in the next you say that you "prefer to describe [physics] in terms of simple T symmetry".
Paragraph 1 was me trying to "phrase this in language you are more likely to understand"
Paragraph 2 was me using the language I would normally use.
So: that was not a case of me "contradicting" myself at all.
If you simply reverse T, and the whole universe starts to run backwards, then it makes an awful lot of sense to call the universe "T symmetric", IMHO, conventional terminology or no. Then it is time to start saying "T" - instead of "CPT" - since the old "T" has turned out to be not a fundamental or interesting concept.
Anyway, I think a miscommunication.
↑ comment by gjm · 2011-06-17T10:15:22.786Z · LW(p) · GW(p)
So, which of the hypotheses of the CPT theorem do you find less compelling than, er, the fact that it would be kinda neat if T symmetry were correct? (If there's any reason beyond that in the page you linked to, I failed to see it.)
Replies from: timtyler↑ comment by timtyler · 2011-06-17T19:53:35.538Z · LW(p) · GW(p)
Well, T symmetry is favoured by Occam's razor. We have pruned away the momentum quanta from needing to be replaced by their anti-particles. The idea that T-symmetry is true just takes this a bit further, becoming more elegant and neat in the process.
Replies from: gjm↑ comment by gjm · 2011-06-17T23:15:27.274Z · LW(p) · GW(p)
You didn't answer the question, nor did you give any grounds for thinking it doesn't need answering.
Occam's razor applies to theories, not to individual propositions about them. CPT (or T) symmetry isn't something you build into a physical theory by having an axiom like "CPT-symmetry holds"; it arises from the structure of the theory. Do you have any actual reason for believing that theories with T-symmetry but not CPT-symmetry are simpler than theories with CPT-symmetry but not T-symmetry? The CPT theorem seems to me to give good reason not to believe that.
Replies from: timtyler↑ comment by timtyler · 2011-06-18T00:24:26.714Z · LW(p) · GW(p)
Do you have any actual reason for believing that theories with T-symmetry but not CPT-symmetry are simpler than theories with CPT-symmetry but not T-symmetry?
Well, I think so, but maybe not in a format suitable for a short blog post. There are numerous small, simple CA with the property of being symmetrical under T=-T. For example, the BBM. My impression is that other means of reversal are correlated with automaton complexity. Then there's the idea of charge as a pump. That is appealing on other grounds - and pumps tend to have moving parts - which would then reverse automatically if T=>-T . Also, the possibility of simulationverse ideas would seem to favour ease of reversal, to some extent.
IMO, you should really not be counting the CPT theorem as evidence on the issue - one way or the other.
I certainly don't think people should be telling me that the known laws of physics are not symmetrical under T=>-T. IMO, it is more probable that they are symmetrical that way than that they are not. The idea that CPT symmetry illustrates that they are not is simply a popular misconception, with no basis in the facts of the matter.
Replies from: gjm↑ comment by gjm · 2011-06-18T16:44:44.060Z · LW(p) · GW(p)
The laws of physics as currently understood -- i.e., the laws in the best model we' ve got -- are in fact CPT-symmetric but not T-symmetric. (Because the best model we've got is a quantum field theory of the sort that the CPT theorem applies to; and because CP symmetry is violated (1) by that model and (2) in reality, according to the available evidence.)
Sure, there are plenty of small simple cellular automata with T-symmetry. And also with P-symmetry, which does not hold in the real world. So far as I know, CAs with PT symmetry are just about exactly as easy to make as ones with T symmetry. (And if you have CAs with a property corresponding to C, I bet CPT is as easy to arrange as T.) Why is any of this meant to mean that T-symmetry is simpler than CPT-symmetry?
You may find the idea of "charge as a pump" appealing; fair enough. I am at a loss to see why that is a reason for thinking that T symmetry is simpler than CPT symmetry.
Your argument, so far as you've provided one, seems to go like this: "I expect the universe to work like a cellular automaton. I have the feeling that T symmetry is simpler than CPT symmetry in cellular automata. Therefore the CPT theorem is irrelevant and we should expect T symmetry to prevail but not CPT symmetry". This strikes me as very strange since (1) we have very successful physical theories that are QFTs (to which the CPT theorem applies) and no successful physical theories based on cellular automata, and (2) the CPT theorem is an actual theorem governing QFTs, whereas your intuitions for relative simplicity are merely your intuitions.
Replies from: timtyler, timtyler↑ comment by timtyler · 2011-06-18T23:28:30.187Z · LW(p) · GW(p)
The laws of physics as currently understood -- i.e., the laws in the best model we' ve got -- are in fact CPT-symmetric but not T-symmetric. (Because the best model we've got is a quantum field theory of the sort that the CPT theorem applies to; and because CP symmetry is violated (1) by that model and (2) in reality, according to the available evidence.)
You understand that the claim is that that is just a historical accident about the way the model was built? The idea that C and P reverse automatically if T is reversed does not make any new predictions that the old model did not. The idea that CPT symmetry is favoured by some kind of experimental evidence seems completely wrong to me. Since the two models are totally equivalent experimentally, this is an issue for Occam.
Replies from: gjm↑ comment by gjm · 2011-06-19T09:25:29.578Z · LW(p) · GW(p)
You understand that the claim is that that is just a historical accident about the way the model was built?
There is a mathematical theorem that says that no model of the same kind can fail to have CPT symmetry. There are robust experimental results that say that the real world doesn't have CP symmetry. Therefore, it is not an accident that given the general kind of model it is it has CPT symmetry but not T symmetry.
If you are claiming that it's a historical accident that we have a QFT model and not a CA-type model, then show your evidence. Specifically, the obvious alternative hypotheses are things like "There is no CA-type model that actually fits the data" and "All CA-type models that fit the data as well as the current QFT model are much more complicated than the QFT model". Given the present state of the art as I understand it, these seem much more likely to me than your hypothesis that it's just a historical accident. Please feel free to convince me otherwise.
Since the two models are totally equivalent experimentally
What two models? I know of one model, the Standard Model, which fits the data extremely well and is reasonably simple (ha!), and which has CPT symmetry but not T symmetry (and cannot be fudged to have T symmetry without losing its agreement with experiment). What is the other model you propose, that supposedly is totally equivalent experimentally but has T symmetry?
Please be specific about this, because either you have some wonderful physics to show me that I haven't seen before or you're failing to distinguish between "some vague intuitions in Tim's head" and "a very thoroughly worked out and tested physical model".
Replies from: timtyler↑ comment by timtyler · 2011-06-19T09:43:06.987Z · LW(p) · GW(p)
There is a mathematical theorem that says that no model of the same kind can fail to have CPT symmetry. There are robust experimental results that say that the real world doesn't have CP symmetry. Therefore, it is not an accident that given the general kind of model it is it has CPT symmetry but not T symmetry.
This is all a misconception, though. Look carefully at the theorem you mention, and I expect that you will see that it is quite compatible with what I have been saying.
If you are claiming that it's a historical accident that we have a QFT model and not a CA-type model, then show your evidence.
This doesn't have anything to do with CAs, really. The two ideas are:
if T is reversed, you would have to manually swap each particle with its anti-particle and reverse parity as well to produce the correct backwards evolution.
T reversal has the effect of automatically swapping each particle with its anti-particle - and reversing its parity - due to these phenomena being implemented using rotating parts, cyclic phenomena, or similar things that do automatically run the other way if T is reversed.
These ideas do not produce different experimental predictions if running forwards normally. Standard physics describes both situations equally well. The second idea (due to Ed Fredkin) suggests that examining the nature of charge and parity might yield reasons why reversing T has that effect. Charge working like a pump is an example of how that might happen. Or maybe these systems will refuse to show us their internal workings.
In this kind of model, charge and parity are made of the same kind of stuff as everything else is. Their properties arise from how matter is patterned, not from them being somehow fundamental. As a result, charge doesn't even show up in Fredkin's scheme of fundamental units.
Replies from: gjm, gjm↑ comment by gjm · 2011-06-19T15:35:53.570Z · LW(p) · GW(p)
This is all a misconception, though. Look carefully at the theorem you mention, and I expect that you will see that it is quite compatible with what I have been saying.
Would you care to do me (and other readers) the courtesy of explaining what misconception you think I'm actually suffering from, and what about the CPT theorem you think I've failed to look at carefully enough, and what specific things it says that would make what you say look sensible?
Replies from: timtyler↑ comment by timtyler · 2011-06-19T16:41:16.575Z · LW(p) · GW(p)
Would you care to do me (and other readers) the courtesy of explaining what misconception you think I'm actually suffering from, and what about the CPT theorem you think I've failed to look at carefully enough, and what specific things it says that would make what you say look sensible?
You don't seem to understand how a model can lack CPT symmetry and be consistent with theory and observations. You should be aware that others besides Fredkin have seriously proposed this:
For example, according to the PDF of Spacetime symmetries and the CPT theorem Richard Feynman proposed much the same thing:
Replies from: gjmThe Feynman proposal has the consequence that the electric field flips sign under time reversal, and that the magnetic field does not but it, too, has the consequence that the theory is time reversal invariant.
↑ comment by gjm · 2011-06-19T17:07:51.951Z · LW(p) · GW(p)
You don't seem to understand how [...]
If I don't understand how that can happen, then perhaps the problem is either (1) that it can't or (2) that how it can is a subtle matter which hasn't been explained well enough for me to understand it. So far, in this discussion, you've offered no reason to think #2 more likely than #1, and in any case you haven't made any attempt to explain how it could happen.
The PDF does not appear to contain the sentence you purport to quote from it. (More specifically, it does not appear to contain the word "flips". (Neither does the abstract.) In any case, its proposal seems to amount simply to redefining "time reversal". If all you're saying is that if you use "T symmetry" to mean what everyone else calls "CPT symmetry" then physics is likely to be T-symmetric but not CPT-symmetric, then (duh!) I agree, but I'm not sure why that's supposed to be interesting.
Replies from: timtyler↑ comment by timtyler · 2011-06-19T17:15:53.209Z · LW(p) · GW(p)
The PDF does not appear to contain the sentence you purport to quote from it. (More specifically, it does not appear to contain the word "flips".
Section 2.4 page 32. Searching for "flips" doesn't work here either. Copy-n-paste into a text editor shows why - fi and fl are weird ligatures in this PDF.
In any case, its proposal seems to amount simply to redefining "time reversal". If all you're saying is that if you use "T symmetry" to mean what everyone else calls "CPT symmetry" then physics is likely to be T-symmetric but not CPT-symmetric, then (duh!) I agree, but I'm not sure why that's supposed to be interesting.
To quote from the article:
We have articulated the `geometric' notion of time reversal implicit in Malament's work, according to which time reversal consists in leaving all [other] fundamental quantities alone, and merely flipping the temporal orientation.
It's plainly proposing T symmetry. Does that help you to see how such a thing might be possible?
Replies from: gjm↑ comment by gjm · 2011-06-19T21:26:43.392Z · LW(p) · GW(p)
weird ligatures
Aha, of course. (I did search for some other substrings, though I forget what. Presumably they also contained ligatures. D'oh.)
plainly proposing T symmetry
... in Malament's proposal, which is not the same as the Feynman one you cite earlier. The purpose of the paper is to argue for a definitional change whereby we call "T" what is currently generally called "CT". Everything in the paper is concerned with classical, not quantum, electrodynamics. The paper does not argue that T symmetry (as generally understood or with a revised definition) is plausibly true in quantum electrodynamics.
Does that help you to see how such a thing might be possible?
It would make this discussion more pleasant for me if you'd be less patronizing. Whether you care about that is, of course, up to you.
↑ comment by gjm · 2011-06-19T15:31:04.891Z · LW(p) · GW(p)
The behaviour of Fredkin's model -- which he himself says "is grossly less comprehensive while far more inconsistent than conventional physics" really doesn't seem very important, in comparison with the behaviour of the actual models constructed by actual physicists that make actual predictions that actually fit actual physical data.
We have a very nice theory that seems to describe how the world works with great accuracy and precision. It does not have the property that when you simply T-reverse it C and P get reversed automatically. Nor can it be tweaked to have that property, without breaking its agreement with observation.
You have, so far as I can see, a bunch of handwaving that suggests that there possibly might be some sort of model of some of physics that has the property that T-reversing it brings C and P along automatically. You haven't actually produced such a model; no one has found one; no one seems to have much idea how to make one.
How on earth can it be reasonable to describe this situation by saying it's "just a historical accident" that one of the "two ideas" you describe happens to be dominant at the moment?
Replies from: timtyler↑ comment by timtyler · 2011-06-19T15:41:37.441Z · LW(p) · GW(p)
We have a very nice theory that seems to describe how the world works with great accuracy and precision. It does not have the property that when you simply T-reverse it C and P get reversed automatically.
Only because it doesn't say anything about that. It's a model of physics. In physics, you can't just reverse time, that is not a permitted operation.
Nor can it be tweaked to have that property, without breaking its agreement with observation.
That's incorrect. If P and C automatically reverse when you reverse T that breaks absolutely nothing.
You haven't actually produced such a model; no one has found one; no one seems to have much idea how to make one.
The model is as I already described: P and C automatically reverse when you reverse T. This is plausible since P and C might be physically implemented using moving parts. We can discuss how parsimonious that is. That is a discussion based around Occam's razor.
If I wanted to make a strong case that T symmetry was much more likely than CPT symmetry, then we would have to get into the possible details of hypotheses about why they might reverse. However, that was never my position. We don't know with much confidence that C and P reverse automatically, but equally we don't know with much confidence that they won't. The correct response to such a situation is not to declare CPT symmetry the winner, but to say that there's uncertainty, and that we don't know for sure.
Replies from: gjm↑ comment by gjm · 2011-06-19T17:18:25.493Z · LW(p) · GW(p)
Only because it doesn't say anything about that [...] you can't just reverse time, that is not a permitted operation.
WTF? Saying that a theory of physics has (say) T-symmetry just means: something is a possible history of the universe iff its time-reversal is.
If P and C automatically reverse when you reverse T
Could you please clarify whether you are saying anything about physics, or whether you are just making the content-free observation that by redefining T-reversal you can interchange the notions of T-symmetry and CPT-symmetry?
The model is as I already described
What model?
Replies from: timtyler↑ comment by timtyler · 2011-06-19T18:25:16.237Z · LW(p) · GW(p)
Only because it doesn't say anything about that [...] you can't just reverse time, that is not a permitted operation.
WTF? Saying that a theory of physics has (say) T-symmetry just means: something is a possible history of the universe iff its time-reversal is.
Uh, I am aware of what "T-symmetry" refers to.
Could you please clarify whether you are saying anything about physics, or whether you are just making the content-free observation that by redefining T-reversal you can interchange the notions of T-symmetry and CPT-symmetry?
As previously discussed, this is about how C and P work - and whether they reverse themselves if you reverse T. This does, inevitably, lead to time reversal not referring to the operation many people use it to refer to today. People think reversing T leaves P and C alone. The idea is that they are wrong about that. This is not a "content-free observation", it's about how the operation of parity and charge could depend on the direction of time.
The model is as I already described
What model?
You seem to be asking for more specifics than I, or anyone else, has. However, you snipped my explanation of why specific details are not needed to support my position. So, here it is again:
Replies from: gjmIf I wanted to make a strong case that T symmetry was much more likely than CPT symmetry, then we would have to get into the possible details of hypotheses about why they might reverse. However, that was never my position. We don't know with much confidence that C and P reverse automatically, but equally we don't know with much confidence that they won't. The correct response to such a situation is not to declare CPT symmetry the winner, but to say that there's uncertainty, and that we don't know for sure.
↑ comment by gjm · 2011-06-19T21:32:37.780Z · LW(p) · GW(p)
Uh, I am aware of what "T-symmetry" refers to.
Then perhaps you might care to clarify what your point was.
You seem to be asking for more specifics than I, or anyone else, has.
Then you should stop talking about "the model" as if, y'know, you actually have a model.
Replies from: timtyler↑ comment by timtyler · 2011-06-19T23:27:51.361Z · LW(p) · GW(p)
Uh, I am aware of what "T-symmetry" refers to.
Then perhaps you might care to clarify what your point was.
Well, this conversation is pretty tedious for me, and you seem to keep asking me to do more work. Well, OK. So, the context was:
We have a very nice theory that seems to describe how the world works with great accuracy and precision. It does not have the property that when you simply T-reverse it C and P get reversed automatically.
Only because it doesn't say anything about that. It's a model of physics. In physics, you can't just reverse time, that is not a permitted operation.
...and the idea was that the job of physics is mostly to tell us how the temporal evolution of the world works. It's main job is not to tell us what happens if an impossible physical event - like time running backwards - takes place. So, it is not a terribly big surprise that it doesn't have too much to say about the issue of whether charge reversal is an automatic consequence of time reversal - or not. That is not really an important part of its job description.
You seem to be asking for more specifics than I, or anyone else, has.
Then you should stop talking about "the model" as if, y'know, you actually have a model.
"Model" can be a pretty general term:
In the most general sense, a model is anything used in any way to represent anything else.
The problem is that you are not using the term in the same sense as me - which leads to communication problems. The results seem kind-of tedious to me.
Replies from: gjm↑ comment by timtyler · 2011-06-18T23:18:40.876Z · LW(p) · GW(p)
You may find the idea of "charge as a pump" appealing; fair enough. I am at a loss to see why that is a reason for thinking that T symmetry is simpler than CPT symmetry.
Well, if true, it is a reason to think T symmetry or PT symmetry holds - since then charge would reverse itself automatically if T was reversed.
Replies from: gjm↑ comment by gjm · 2011-06-19T09:26:30.129Z · LW(p) · GW(p)
Depending on the nature of the pump. For instance, a pump made out of Standard Model stuff would not exactly reverse itself if T were reversed. Presumably, then, you have something more specific in mind; would you care to say more exactly what it is?
Replies from: timtyler↑ comment by timtyler · 2011-06-19T09:30:06.024Z · LW(p) · GW(p)
a pump made out of Standard Model stuff would not exactly reverse itself if T were reversed.
Pumps usually reverse if T is reversed. Most pumps contain something like a rotating fan blade. Reverse T and the momenta all reverse, so the blade turns the other way - and the pump pushes in the opposite direction.
What about C and P you ask? Well, we are talking about a pump inside charged particles, whose action is responsible for charge. C is hardly likely to be relevant. If T symmetry holds, you would just have to reverse T. If PT symmetry holds (but not T symmetry alone), and the components of the pump are sensitive to the sign of P, then you might have to reverse both P and T to get the pump to run backwards.
Replies from: gjm↑ comment by gjm · 2011-06-19T15:33:56.383Z · LW(p) · GW(p)
Whether pumps "usually" reverse exactly when T is reversed depends on whether the universe is actually T-symmetric or not. If you assume that they do then you're begging the question.
Do you have any actual evidence that charged particles such as electrons and quarks are actually likely to contain pumps? This seems, on the face of it, monstrously implausible; of course it might be right -- anything might be right -- but why does the idea even deserve taking seriously, never mind using as the basis for your opinions about whether CPT symmetry is more likely than T symmetry?
Replies from: timtyler↑ comment by timtyler · 2011-06-19T16:14:50.751Z · LW(p) · GW(p)
Whether pumps "usually" reverse exactly when T is reversed depends on whether the universe is actually T-symmetric or not. If you assume that they do then you're begging the question.
It is not "begging the question" it is "explaining one way it could work". If charged particles contain pumps with moving parts, that would make simple T symmetry more plausible, since then there would be a clearer mechanism for explaining why reversing T would lead to all the charges in the universe reversing.
Do you have any actual evidence that charged particles such as electrons and quarks are actually likely to contain pumps? This seems, on the face of it, monstrously implausible; of course it might be right -- anything might be right -- but why does the idea even deserve taking seriously [...]
We don't know the details of how electromagnetic and gravitational fields are generated. Something generates fields following an inverse-square law. A pump would be likely to have that effect - and seems about as plausible as anything else. Why do you describe the idea as being "monstrously implausible"?
Replies from: gjm↑ comment by gjm · 2011-06-19T17:14:38.037Z · LW(p) · GW(p)
I think it's monstrously implausible because it requires charged particles to have intricate internal structure of a very curious sort, a thing for which we have no evidence whatever. (Assuming you actually mean a pump, rather than (e.g.) "a source of some substance that can flow", in which case I see no reason whatever for thinking it should be T-symmetric.)
Your justification seems to be that "charged particles contain pumps" would somehow explain the inverse square law, but I don't see that at all. The idea that they are sources/sinks of some substance that spreads out geometrically might (though I don't know how you're going to make that work in the quantum context) but what you're suggesting is both more specific (pumps as such are not required for an inverse square law) and less specific (pumps as such do not imply an inverse square law).
Replies from: timtyler↑ comment by timtyler · 2011-06-19T18:05:05.475Z · LW(p) · GW(p)
I think it's monstrously implausible because it requires charged particles to have intricate internal structure of a very curious sort, a thing for which we have no evidence whatever.
So: charged particles are very tiny, we have limited knowledge about how they do what they do. To reverse under T=>-T you need something that rotates, or something that cycles through more than two states, or something like that - it need not be particularly intricate. Sources could reverse their operation under T=>-T too - it depends on the details of how they are constructed. It doesn't take very much to reverse under T=>-T. Rotating is enough to do it, for example.
Sources make slightly more sense for gravity than electromagnetism, IMO. With electromagnetism you typically need sources and sinks - and a pump does both of those jobs pretty neatly.
Replies from: gjm↑ comment by gjm · 2011-06-19T21:30:36.028Z · LW(p) · GW(p)
I am entirely unable to understand how you can say "a pump does both of those jobs pretty neatly" without actually having at your disposal anything remotely resembling a theory that does any sort of job of matching observation in which charged particles are pumps.
You might as well point to some property of elementary particles and say "a billiard ball does this pretty neatly". Except that we do at least have models of some of physics in which particles are a bit like billiard balls, which appears to put that proposal ahead of explaining electric charge via pumps.
Actual working models trump handwaving, for me, because the problem with handwaving is that without working out the details you have nothing remotely resembling an upper bound on the actual complexity of the theory -- if any even exists -- that the handwaving might be gesturing towards.
Replies from: timtyler↑ comment by timtyler · 2011-06-19T23:53:54.741Z · LW(p) · GW(p)
I am entirely unable to understand how you can say "a pump does both of those jobs pretty neatly" without actually having at your disposal anything remotely resembling a theory that does any sort of job of matching observation in which charged particles are pumps.
I just meant that a pump can act as a source - or a sink - depending on which way around you use it.
A specific model would probably not help much. My position is that there are a whole class of models which are isomorphic to conventional physics and exhibit T symmetry. We don't yet know which of those models are correct, or indeed if any of them are.
The Wheeler–Feynman Time-Symmetric theory appears to be one such idea.
↑ comment by prase · 2011-06-17T12:31:21.668Z · LW(p) · GW(p)
That a time-inverted process doesn't contradict the fundamental laws doesn't mean that we could observe it with frequency basically comparable to that of the uninverted process. Think of thermodynamical irreversibility, for an example of class of processes which practically don't have inverted counterparts, even if these are perfectly compatible with microscopic physics.
Also, losing mass is not clearly defined process. If by mass of a BH one means the mass included below the horizon (or calculated from the diameter of the horizon) observed from constant distance, then black holes never lose or gain mass, since such an observer would never see anything cross the horizon in any direction: for an outside observer it takes infinite time for any falling object to reach the horizon.
If mass is calculated from the gravitational force measured in a constant distance, it may grow as mass gets attracted towards the horizon, or it may shrink if the mass near the horizon has enough outward momentum to overcome the gravity of the black hole. The latter scenario is quite unlikely to happen during typical black hole formation, at least I think so.
comment by XiXiDu · 2011-06-17T09:39:23.658Z · LW(p) · GW(p)
This is not related to your question, but I thought it is important (and interesting) to note that predictions in physics can turn out not to match reality very well:
The vacuum catastrophe is sometimes cited as the biggest disagreement between theory and experiment ever. They disagree by a factor of at least 10^107.
In cosmology the vacuum catastrophe refers to the disagreement of 107 orders of magnitude between the upper bound upon the vacuum energy density as inferred from data obtained from the Voyager spacecraft of less than 10^14 GeV/m³ and the zero-point energy of 10^121 GeV/m³ calculated using quantum field theory. This discrepancy has been termed "the worst theoretical prediction in the history of physics!"
More: http://en.wikipedia.org/wiki/Vacuum_catastrophe
Replies from: khafracomment by JoshuaZ · 2011-06-16T22:32:19.698Z · LW(p) · GW(p)
Essentially he's claiming that Hawking radiation either does not exist or will be at such small levels that it won't prevent black holes from evaporating. My impression from physicists is that we'd have to be pretty wrong for it not to exist or to exist at much lower than predicted levels. But there's also a more serious problem with this whole approach: the physics that predicts that black holes will go poof is very much connected to the physics that would suggest that it might be possible to make mini black holes in the LHC. If Hawking radiation doesn't exist then the chance that a black hole could form in the LHC goes down, not up. So the total danger attached to "The LHC will make black holes that won't go away" is very small.
comment by MrMind · 2011-06-20T15:24:15.509Z · LW(p) · GW(p)
He is a crackpot, I'm 99.99% sure. Hawking radiation stems from the second law of thermodynamics, in a general relativity setting one cannot have the first without the second. Besides this, collisions between atoms of the energy produced at the LHC happens every time in baseball, between a ball and a bat.
comment by Dreaded_Anomaly · 2011-06-16T22:47:39.275Z · LW(p) · GW(p)
It's not a reasonable concern. CERN addresses various safety claims here.