From: Russell BlackadarNewsgroups: sci.physics.research,sci.physics Subject: Re: Physics bitten by reverse Alan Sokal hoax? Date: Thu, 7 Nov 2002 22:52:43 +0000 (UTC) Organization: (none given) Message-ID: (aqeqrr$dpa$1@lfa222122.richmond.edu) "I/G.Bogdanoff" wrote: > baez@galaxy.ucr.edu (John Baez) wrote in message > news:(aq1uoo$sr2$1@glue.ucr.edu)... > > I know what it means for a process to conserve some quantity, > > e.g.: "nuclear fusion conserves charge". It means that the > > quantity doesn't change as the process happens. I don't > > know what it means for a plane to conserve something. > Comment : it simply means in our mind that the oscillation plane of > the pendulum conserves its initial orientation (whatever this initial > orientation is). This is the whole point of the well known 1851 > Foucault's experiment. Not really, since it's false! At least, false at the latitude of Paris's Pantheon. The whole point of the experiment is that the plane's orientation is not conserved in Earth-based coordinates, which are thereby proved to be non-inertial. But the converse -- that the plane *is* conserved in inertial coordinates -- does not follow and is in fact false. Grossly so, in Paris. Popular explanations of the F.P. often are quite wrong on this point. I am wondering if your whole introduction of *planes* here (in contrast to geodesic *paths*) is the result of this misconception. In any case the appearance of a plane in the case of the F.P. is a mere artifact of the experimental constraints -- a pendulum (which does define a plane) is the easiest means of allowing an object to move for long periods of time without any transverse external force. (Assuming that the suspension point is inertial -- which in Paris it is not, hence my remarks above.) But fundamentally, what's important is not the plane, but the *path* of the object. > > And > > I don't know what it means for something to conserve the > > initial singularity. > Comment : We admit that if we consider this expression ("...something > to conserve the initial singularity...") it does not make much sense. > What we would like to convey is simply that the more distant is the > inertial reference the smaller is the angular deviation of the > oscillation plane of the pendulum regarding this reference. I don't understand this. What is so special about the Foucault pendulum? Aren't *all* inertial paths equal in this regard? Or to use your terminology, aren't they *all* referenced to the initial singularity? JB's word "vacuous" seems apt.

From: baez@galaxy.ucr.edu (John Baez) Newsgroups: sci.physics.research,sci.physics Subject: Re: Physics bitten by reverse Alan Sokal hoax? Date: Thu, 21 Nov 2002 20:19:45 +0000 (UTC) Organization: UCR Message-ID: (arf6pq$5hh$1@glue.ucr.edu) In article (e8e077d9.0211101049.b640c0@posting.google.com), I/G.Bogdanoffwrote: >John Baez wrote: [some highly iterated quotation marks have been simplified here] >>For example, here's the beginning of [Igor's] paper "Topological Origin >>of Inertia: >>>We draw from the above that whatever the orientation, the plane of >>>oscillation of Foucault's pendulum is necessarily aligned with the >>>initial singularity marking the origin of physical space S^3, >>>that of Euclidean space E^4 (described by the family of instantons >>>I_beta of whatever radius beta), and, finally, that of Lorentzian >>>space-time M^4. >> Note that I am still waiting for an explanation of this sentence. >Answer: Let's be simple and heuristic : if we consider (as you wrote >yourself in your webpage) that Big Bang occured "everywhere", then FP >oscillation plane intersects the Initial singularity "everywhere" >(naively speaking). Okay. If this is what you're really saying above, it sounds like a complicated way of saying this: "Since the big bang happened everywhere, no matter which way a pendulum swings, the plane in which it swings can be said to `intersect the big bang'." Notice that this has nothing to do with with "instantons" and the like. In fact, it has nothing to do with the big bang! It really amounts to saying this: "No matter which way a pendulum swings, there is some point on the plane in which it swings." But now that we've simplified it this far, we can see this statement really has nothing to do with pendulums, either! The real content is simply: "Any plane contains a point." I don't think this can be too useful in "explaining the origin of inertia", as your paper hopes to do. If the paragraph quoted above says something more substantial than this, please say what it is. By the way, there's no need to be "simple and heuristic" or speak "naively". The audience here can absorb technical details, and we are actually very eager to see if such details exist. Anyway, let's continue: >> In particular, what does it mean to say the plane of oscillation >> of a pendulum is "aligned with the initial singularity"? This >> was your previous attempt at explanation: >>>in conjecture 4.9 (nothing more >>>that an conjecture, by the way) we have considered that the >>>2-dimensional plane of oscillation of the pendulum conserves the initial >>>singularity S for inertial reference, whatever the orientation of this >>>plane in physical space R3." >> This attempt did not help at all. >> >> I still don't what it means for a plane to "conserve the initial >> singularity S for inertial reference". What does it mean? >Answer : It means that (at the pole) whatever the intial direction of >the oscillation plane P is, there is always a geodesic that belongs >to P up to the "everywhere". I don't know what `up to the "everywhere"' means. This expression is not used by the mathematicians and physicists that I know. >Initial singularity (that is the reason >why the pendulum oscillation plane does not "rotate" and conserves its >average initial direction). Is this supposed to be a new sentence or a continuation of the previous one? Anyway, I guess you are saying that "conserve the initial singularity S for inertial reference" is taken by you to mean "does not rotate" and you are offering some explanation of why the plane of a Foucault pendulum does not rotate. Unfortunately, this explanation is quite unclear - even neglecting the fact, noted by other posters, that the plane of a Foucault pendulum *does* rotate. >In (slightly) more technical terms, there >is a theorem (Hawking/Penrose) showing that in the standard Roberton >Walker model, all past directed timelike as well as null geodesics >from a point P converge within a compact region (initial >singularity). I'm not sure this is quite the right way to state it, but you appear to be attempting to state one of the singularity theorems which shows that an initial singularity is to be expected in a large class of cosmological models. But in fact, we don't need this theorem to prove the existence of an initial singularity in the standard Robertson-Walker model, because this solution can be written down in closed form, so one can work out the existence of the initial singularity directly. In other words, here you seem to be roughly sketching how one might use a high-powered theorem result to prove a known fact: there is a big bang in the standard big bang cosmology. Okay. >>In this case any 2-dimensional spacelike surface will >>intersect the cosmological past lightcone at the origin. Okay, so now you are returning to your theme: any 2d spacelike surface "hits the big bang singularity". Right? The problem is, this does not prove the nonrotation of the plane in which a Foucault pendulum swings - which is, in any case, false. Anyway, let's continue. >> The trouble starts here: >>>As far as our own model is concerned >> Which model is that??? >Answer : The model based on what we call the "quantum superposition" >of Lorentzian and Riemannian metrics. Please describe this model in detail, so we can see how you establish the following result: >>>we have established that at the vincinity of this "singular point" >>>(0 scale) the 4 dimensional metric must be considered as positive >>>definite (euclidean signature ++++). >> Ah! You say you have *established* this. That means you have >> some sort of definite proof or calculation, at least at the physical >> level of rigor. I would like to understand what you actually did. >> So: >> >> Firstly, tell me what model you are talking about. >Answer : The model of superposition (as quoted above). Again, at the >Planck scale, we consider that the spacetime metric and its signature >are subject to some "quantum fluctuations". The quantum fluctuation of >the signature implies that the lorentzian signature +++- should be >extended to +++(+,-). >The "space of oscillation" between the lorentzian and the euclidean >metrics can be described -as we did- by the separated quotient >topological space sigma = R3,1 cross R4 diagonally quotiented by >SO(3) (the slicing being here given by the orbits of the action of >SO(3) on R3,1 cross R4). This of course suppose an extention to a >5-dimensional underlying space. By the way : the topological space >sigma is constructed from the symmetric homogeneous space SO(3,1) >cross SO(4) diagonally quotiented by SO(3). In terms of generators, >this homogeneous space is 9-dimensional; but sigma (considering the >orbits of the action of SO(3) is 5-dimensional. I understand the rather simple homogeneous spaces you're describing here, but the problem is, you aren't describing your physical model in any detail, so I can't imagine how you would begin to: "establish that at the vincinity of this "singular point" (0 scale) the 4 dimensional metric must be considered as positive definite (euclidean signature ++++)." You see, when you discuss R^{3,1} x R^4 modulo the diagonal copy of SO(3), you are basically telling me about a certain well-defined 5-dimensional space - but you aren't saying what this space has to do with the *physics of the big bang*. Thus, I can't use your remarks to infer that the spacetime metric right at the big bang is positive definite. >> Secondly, tell me what mean by saying the 4-dimensional metric >> "must be considered as" positive definite. Are you saying it >> *is* positive definite? That would be very odd if you were >> talking about the FRLW cosmology, because here the metric is >> never positive definite, not even in any vicinity (neighborhood) >> of the initial singularity. So, what's going on? >Answer : The superposition model is NOT anymore the FRLW model (which >only valid up to the Planck scale). In the framework of superposition >space of the metric it is easy to show (as proved in prop. 2.2.4 of >Grichka's thesis) that sigma top has the structure of a convex cone >endowed with a singular origin. On this origin (and only on this >singular point), the metric is euclidean. Please describe this in more detail. You are putting a metric on this space "sigma" now, and saying that *that* metric is positive definite at a certain point of sigma? What does this have to do with whether the spacetime metric is positive definite? >> Thirdly, tell me how you *established* that the metric must be >> considered as positive definite. >Answer : There are different kinds of arguments (possible >demonstrations). I would be very happy to see one. >Some belongs to algebraic topology or Lee groups >theory, some others to quantum groups theory or van Neumann algebra, >and still some others have a more physical content. But all are very >long to expose. That's okay - I have lots of time. >The best is to see chap 2, 3, 4, 5, 7 of Grichka's thesis. Unfortunately, I can't read French, so I'm hoping one of you could explain your work in English, as your published papers also seek to do. By the way, since I'm trying to understand your published papers, it would be a bit odd if I had to read some unpublished *other* work, written by your brother, to do this. >In one sentence and on physical basis, we could say that KMS >condition applied to spacetime system at the Planck scale necesseraly >implies complexification of the timelike direction of the metric. >This involves (in terms of factors) a III-lambda factor able to be >decomposed into a semi-direct product of R and a II-infinite factor (M >0,1). One can then show that at the origin t goes to 0, the >one-parameter automorsphism group of the "residual" II-infinite >factor M 0,1 is an euclidean automorphism semi-group of the form : >exp - beta h M O,1 exp +beta h. This gives a "pseudo evolution" in >imaginary time (euclidean metric). > >The question here : do we have a Lagrangian corresponding to thie >"superposed state" of the metric at the Planck scale? The answer is >yes, but only in the framework of an extension of classical gravity. A >good toy model of this extended lagrangian is : > >L sugrav = beta hat R + 1 over g squared times R squared + alpha R R >dual. > >Here, beta is the compactification radius of the theory, 1 over g >squared the dilaton (g being the coupling constant) and alpha the >axion (super partner of the dilaton field). We have shown that this >lagrangian has an infrared and an ultraviolet limit. For g goes to >arbitrary large values (infrared limit corresponding to beta > to >Planck scale) the extended lagragian is reduced to the Einstein term >beta R. In this case, of course, the underlying metric is >4-dimensional lorentzian. On the contrary, for g goes to 0. (ie.beta >goes to 0, ultraviolet limit ) the Einstein term becomes neglectible. >Moreover, as the theory becomes self dual on this limit, the only >relevant term is the topological term alpha R R dual. This limit >represents the instantonic pole of the theory. In this case, the >timelike direction is compactified on the circle S1 of radius beta=0. > Dually, the spacelike direction (whose the compactification radius in >an inverse function of beta) is decompactified on the straight line R >for beta=0. The underlying theory is 4-dimensional endowed with an >euclidean metric. This all sounds very familiar from your papers, but unfortunately it is quite vague, so it doesn't answer my question. I'm hoping for something considerably more precise. And by the way, I'm apparently not the only one! In the November 17th issue of the New York Times, George Johnson wrote: "I do think it is possible to tell good work from bad," said Dr. Steve Giddings, a string theorist at the University of California, Santa Barbara. "Even when researchers are confused, and only have a partial understanding of a puzzle, it is important that their explanations have some element of logic and consistency." This is where experts say that sincere or otherwise, the Bogdanovs' papers fall flat. Reading through an Internet debate between them and the physicist John Baez of the University of California at Riverside is like watching someone trying to nail Jell-O to a wall. Regardless of whether this is a fair description of what's going on, it captures the dissatisfaction that some people feel with the explanations thus far.

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