Debunking Nimtz
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From: Tom Roberts on 10 Sep 2007 18:20 Tom Van Flandern wrote: > Tom Roberts writes: >> [Roberts]: Since you accept geometry for [straight lines], you ought >> to be able to accept geometry for the rest. Why don't you? > > Geometry cannot create 3-space forces. A purely geometric GR would > contain no dynamics. There is no necessity to "create" what you call "3-space forces". This is physics, and you cannot force your personal desires and prejudices onto nature. The requirement is that GR construct an accurate model of the world (including the behavior of objects in it), and it does. Do you seriously think physicists would accept GR if it did not accurately describe the behavior of a rock when one drops it? So using your terminology, the geometry of GR clearly DOES "create 3-space forces". The purely geometric GR most definitely does include dynamics. It seems that you keep thinking that geometry is 3-dimensional; that is false: in GR it is 4-dimensional, and the geometry of spaceTIME includes all the motions and (coordinate) accelerations you ascribe to "3-space forces". In the ADM formalism, one determines the time evolution of 3-d space, and by golly even in that formulation a dropped rock will fall. Of course this 3-d space need not be Euclidean as you insist.... Ask yourself this: if you were to ask any GR expert in the world the question "What is the essence of GR and how it describes gravity?", do you seriously think you would get an answer like "Gravity is a 3-space force"??? You know as well as I that the answer you would get is along the lines of "In GR the geometry is dynamic, and what we call gravity is merely an aspect of the geometry." Because that _IS_ the essence of GR. But what you are discussing is not GR. <shrug> Tom Roberts
From: Vern on 11 Sep 2007 13:23 On Sep 10, 6:20 pm, Tom Roberts <tjroberts...(a)sbcglobal.net> wrote: <snip> > Ask yourself this: if you were to ask any GR expert in the world the > question "What is the essence of GR and how it describes gravity?", do > you seriously think you would get an answer like "Gravity is a 3-space > force"??? You know as well as I that the answer you would get is along > the lines of "In GR the geometry is dynamic, and what we call gravity is > merely an aspect of the geometry." Because that _IS_ the essence of GR. > But what you are discussing is not GR. <shrug> If the field interpretation of GR is used, then gravity is a force and not just a function of the manifold as your geometric interpretation posits. Moreover, that force propagates faster than light, which is, of course, his point. In my opinion, you have never satisfactorily answered his claim that in the geometric interpretation a straight line may be a curve, but there is no reason for an object to follow that curve unless it is already moving. You still have to have a larger mass attracting the object to make it move along the geodesic and that's action at a distance. The field interpretation, on the other hand, requires a medium to constitute the field. Vern
From: Tom Roberts on 11 Sep 2007 17:48 Juan R. wrote: > On Sep 8, 4:55 pm, Tom Roberts <tjroberts...(a)sbcglobal.net> wrote: >> There is no doubt that the central equation of GR is the Einstein field >> equation. "What you were taught" does not describe all solutions of the >> EFE, and hence is not all of GR. > > It is specially ironic calling the EFE (Einstein FIELD equation), > whereas rejecting the field interpretation of gravitation in favor of > the more 'modern' geometric view where h_ab has not field > interpretation. In modern physics and mathematics, a field is a function on the manifold. The Einstein field equation relates the Einstein curvature tensor FIELD to the energy-momentum tensor FIELD, and it is usually interpreted as defining the metric tensor FIELD. And the metric on a manifold is what we mean by its geometry. I am not advocating anything unusual here, and my statements here are simply mainstream GR. <shrug> Note that your "h_ab" does not appear in the EFE. It does, however, appear in the small-field APPROXIMATION to the EFE. And in that approximation it is not truly a field, because it depends (implicitly) on the background metric to which it represents perturbations. So the "irony" is all in your head, not my writings. > That after many queries from my part you still did not provide a > simple paper favouring the geometric formulation over the field one is > not ironic but depresing from a scientific view. Hmmm. Your notion of "scientific view" does not correspond to mine, or to the mainstream of science. In science, we TEST THEORIES, and GR has been well tested within the solar system and in a few specific extra-solar systems. The words related to interpretations of theories are not directly testable. > Once again, when and where the geometric formulation > G_ab = k T_ab (e.g. Wald textbook) > has proven to be superior to the field formulation > G_ab = k T_ab (proper EFE) > ? <giggle> Expecting clairvoyance of your readers is silly. But I repeat: my main disagreement with van Flandern is his CLAIM to be using GR, when he manifestly is not doing so. And I also disagree with his claims of propagation >>c that he thinks are general, but are actually theory specific and do not apply to GR. >> We observe strong, weak, and electromagnetic interactions, none of which >> propagate faster than c -- what makes gravity so special? >> Oh wait -- Coulomb forces ACT AS IF they propagate >> instantaneously, [...] > > Everything you say here is wrong. "Extraordinary claims require extraordinary proof." -- James "The Amazing" Randi You are claiming the every E&M textbook is wrong. That's extraordinary. (I'm looking at your references.) > DOGMA: "Gravity and EM are delayed by c". Amen. You rely on ambiguity in wording. When carefully phrased, this is certainly not mere "DOGMA" in GR, where it has been rigorously proven that no energy, momentum, or information can be transferred outside the light cone (i.e. locally faster than c). The observation that the "gravitational force" and the Coulomb force vectors point directly at the _current_ position of their source (rather than its retarded position), is not "proof" that these forces propagate instantaneously (or >>c). It is merely a measurement that is to be compared to the predictions of a theory. The predictions of GR and classical electrodynamics agree with the measurements, even though in both theories the changes to fields propagate at c (not >c). Note: Underlying my arguments is knowledge that at base the issue is about "propagating" and whether or not that really applies, especially as a model of some physical property of such systems. Whether or not the fields "propagate" is COORDINATE DEPENDENT, and that makes it inappropriate as a model of any physical phenomenon. (In the rest frame of the source, these fields are static and eternal, and no "propagation" is involved). Tom Roberts
From: Tom Roberts on 12 Sep 2007 11:40 Vern wrote: > If the field interpretation of GR is used, then gravity is a force and > not just a function of the manifold as your geometric interpretation > posits. Hmmm. Be careful calling this "the field interpretation", as that involves a PUN on "field". For instance: the connection is not a field, but the connection _is_ the "gravitational force" being discussed. So one must be ever vigilant about PUNs -- different people use the word "field" with significantly different meanings: I use it in the modern sense (function on the manifold), but van Flandern, Juan R., and others use it with an ancient meaning (as do you above). > Moreover, that force propagates faster than light, which is, > of course, his point. Hmmm. This is so only if "propagates" is an appropriate concept to use. In GR no energy, momentum, or information can be propagated faster than c, so I don't think it is an appropriate concept here. At least in GR. IOW: I think this elevates mere words and math to the status of "physical phenomena", and that's not sensible. Note, in particular, that this "faster than light propagation" occurs ONLY in certain coordinate systems, and as nature obviously uses no coordinates, such coordinate-dependent quantities cannot be valid models of physical phenomena. [E.g. in locally-inertial coordinates there is no "gravitational force", so how can it "propagate"???] Say, instead, that the force vector points toward the current position of the source, not at its retarded position. This avoids the inflammatory "propagates >>c", and is in any case a direct description of the situation. Now one can compare this to the predictions of GR, and one finds that those predictions agree (even though nothing propagates faster than c). That "propagates >>c" is a THEORY-DEPENDENT interpretation, not a direct experimental result, and it is invalid in GR. > In my opinion, you have never satisfactorily > answered his claim that in the geometric interpretation a straight > line may be a curve, but there is no reason for an object to follow > that curve unless it is already moving. You still have to have a > larger mass attracting the object to make it move along the geodesic > and that's action at a distance. I'm not sure what you are saying/asking, but here's an attempt at a response: In the Schw. manifold of GR external to the earth (ignoring its rotation, atmosphere, and all other massive objects), a small stone will fall when dropped. In coordinates fixed to the earth surface, that stone was initially at rest in my hand, but when I let go it immediately falls to the ground (both in the real world and in the predictions of GR). So the stone "followed that curve without already moving" -- see how bad it is to use COORDINATE-DEPENDENT quantities (like "moving") to attempt to describe physical phenomena? This falling is not action "at a distance" -- at each and every point along its trajectory, the stone simply continues along its geodesic path, and that is COMPLETELY determined by the geometry at each point. In GR there is no need to invoke "action at a distance" at all, because while falling the trajectory of the stone is not affected by the earth at all (after all, the earth is not located where the stone is located). The geometry of the manifold is of course affected by the earth, and the geometry is what determines the geodesics, but the stone is completely oblivious to the presence of the earth, and only knows the geometry WHERE IT IS LOCATED. Do not be confused by van Flandern's PUNs -- he uses "geometry" to mean 3-space, while I use it in the normal way of GR to refer to the full 4-d spacetime manifold with metric. A better description is that while I'm holding it the stone does not follow a geodesic path because of the upward force exerted by my hand; when I release it that force is no longer applied to it, and the stone naturally follows the geodesic path determined by its position and 4-velocity at the moment of release (remember its 4-velocity is the tangent 4-vector to its worldline). That geodesic, of course, accelerates downward (using those same coordinates fixed to the surface). > The field interpretation, on the > other hand, requires a medium to constitute the field. Not necessarily. But, of course, no such medium has been observed, and it must have quite remarkable and counter-intuitive properties. For instance, with a medium involved it's not clear how to "propagate instantaneous action at a distance".... Nor is it clear how such a medium could exert gravitational forces without itself being affected, and without impeding the motion of planets (i.e. it has no viscous drag and yet exerts "force" on objects). No real fluid or material comes anywhere close to having the requisite properties.... Tom Roberts
From: Vern on 13 Sep 2007 08:26
On Sep 12, 11:40 am, Tom Roberts <tjroberts...(a)sbcglobal.net> wrote: > Vernwrote: <snip> > > In my opinion, you have never satisfactorily > > answered his claim that in the geometric interpretation a straight > > line may be a curve, but there is no reason for an object to follow > > that curve unless it is already moving. You still have to have a > > larger mass attracting the object to make it move along the geodesic > > and that's action at a distance. > > I'm not sure what you are saying/asking, but here's an attempt at a > response: > > In the Schw. manifold of GR external to the earth (ignoring its > rotation, atmosphere, and all other massive objects), a small stone will > fall when dropped. In coordinates fixed to the earth surface, that stone > was initially at rest in my hand, but when I let go it immediately falls > to the ground (both in the real world and in the predictions of GR). So > the stone "followed that curve without already moving" -- see how bad it > is to use COORDINATE-DEPENDENT quantities (like "moving") to attempt to > describe physical phenomena? > > This falling is not action "at a distance" -- at each and every point > along its trajectory, the stone simply continues along its geodesic > path, and that is COMPLETELY determined by the geometry at each point. > In GR there is no need to invoke "action at a distance" at all, because > while falling the trajectory of the stone is not affected by the earth > at all (after all, the earth is not located where the stone is located). > The geometry of the manifold is of course affected by the earth, and the > geometry is what determines the geodesics, but the stone is completely > oblivious to the presence of the earth, and only knows the geometry > WHERE IT IS LOCATED. > > Do not be confused by van Flandern's PUNs -- he > uses "geometry" to mean 3-space, while I use it in > the normal way of GR to refer to the full 4-d spacetime > manifold with metric. > > A better description is that while I'm holding it the stone does not > follow a geodesic path because of the upward force exerted by my hand; > when I release it that force is no longer applied to it, and the stone > naturally follows the geodesic path determined by its position and > 4-velocity at the moment of release (remember its 4-velocity is the > tangent 4-vector to its worldline). That geodesic, of course, > accelerates downward (using those same coordinates fixed to the surface). First Tom, thank you for the respectful reply. In your last sentence above, you say that the geodesic accelerates downward (wrt the surface of the Earth). So then it sounds like you are saying that the fact that the geodesic is accelerating downward would be the cause of an object affected by that geodesic moving downward. In the absence of gravity, does this principle mean that an object is still being moved by the geodesic, or would an object placed in the geodesic remain motionless? Vern |