Preferred Frame Theory indistinguishable from SR
in [Relativity]
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From: Paul Stowe on 5 Jul 2010 14:26 On Jul 5, 10:00 am, harald <h...(a)swissonline.ch> wrote: > On Jul 5, 2:46 pm, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > > > harald says... > > > >Acceleration effects are not identified as gravitational fields in > > >Newtonian physics (which, as you now know, you didn't know); and > > >neither is that the case in SRT. In those theories acceleration is > > >"absolute", and no gravitational fields are caused by acceleration. > > > This is a topic for another discussion, but I'm talking about > > "pseudo-gravitational" fields, which crop up in both SR and Newtonian > > physics if you use accelerated coordinates. > > We agree on "pseudo", while Einstein rejected that. > > > >> I think it is because you have not made it very well. I still > > >> have no idea what your point is. > > > >Just study Einstein's paper carefully, > > > I want to know what *YOUR* point is. State it in your own words. > > I did, also in the part of my sentence that you exactly here snipped: > > *you'll know what theory the paradox challenges*. > > THAT (and only that) was my point: the clock paradox challenges the > General PoR. > > You certainly are aware that, despite Einstein's *suggestion* to that > effect in the introduction of his 1905 paper, SRT is *not* based on > the General PoR. Strictly speaking the 'domain' of the 'special' theory of relativity is limited to inertial states. The original principle of relativity as expressed in Einstein's 1905 work covered only that domain. So Harald is correct, the paradox is confined to the situation where, you have identical twins one remain in the original inertial frame, the other accelerated rapidly (nearly instantaneously) to speed ~c travels for x time wrt the original FOR, reverses comes to an equally rapid stop (wrt the original FOR) then returns the same way. Since SRT is based upon v^2 effects (second order quantities) the directionality of any asymmetry is lost in the expressions that quantify changes. However, there is NO! paradox, either in nature, or SRT, once one understands that limitation. The traveling twin, not the stay at home twin will be physically younger. On a one-way trip however, we can't say which one would be for an equal physical duration. That would depend upon the speeds of both FOR relative to the CMBR... Directionality does matter. > > >In this thread, you pretend no less than that: > > > >- Newton was mistaken with his Space postulate (while you evidently > > >didn't even bother to read his arguments) > > >- you know better than Einstein what his own theory is about (Cranky) > > > That isn't cranky. 100 years of relativity development does give > > one a more complete perspective. I would say that a typical physics > > graduate student understands Newtonian physics better than Newton, > > ?! You have just become aware (except if it hasn't sunk in yet) that a > typical physics student is clueless about Newtonian mechanics, since > he/she is taught something else instead. I find it a very illogical > statement that such a student would know Newton's physics better than > himself, after having been misinformed about it! > > > and understands relativity better than Einstein. > > That could very well be so (the word "relativity" isn't specifically > Einstein's theory). > Perhaps this is just about words? Do you call a plastic cup a "glass", > while I insist that it is a plastic cup, after which you insist that > plastic cups are much better glasses?! > > > That doesn't mean > > that they are smarter than Newton or Einstein---it's a lot harder > > to invent a new theory than it is to understand or expand on an > > existing theory. > > Yes. > > > Anyway, I don't care to argue with Newton or with Einstein. > > They are not around to argue with. > > > If you have a point to make, make it yourself. Don't > > hide behind Newton or Einstein. > > I made my point at the start (a factual statement) - and you jumped on > it. ;-) > > Harald
From: eric gisse on 5 Jul 2010 16:31 G. L. Bradford wrote: [...] > Exactly as I thought. I should pity you but I long ago quit pitying the > likes of you. I can read research literature in physics without getting lost. You whine on USENET. Exactly as I thought. [ snip babble ]
From: Androcles on 5 Jul 2010 17:21 "Esa Riihonen" <esa(a)riihonen.net.not.invalid> wrote in message news:pan.2010.07.05.21.17.07(a)riihonen.net.not.invalid... | Androcles kirjoitti: | | > "Esa Riihonen" <esa(a)riihonen.net.not.invalid> wrote in message | > news:pan.2010.07.05.13.36.28(a)riihonen.net.not.invalid... | Androcles | > kirjoitti: | > | | > | > "Esa Riihonen" <esa(a)riihonen.net.not.invalid> wrote in message | > | > news:pan.2010.07.05.09.35.48(a)riihonen.net.not.invalid... | > | Strange - | > | > the standard twin paradox uses at least three inertial | > frames: | | > > | > | > | > | > | | > | > | > | 1: The stay home frame 0 | > | > | > | > | > | > That's one. | > | > | > | > | > | > | > | > | > | 2: Outward frame 1, speed v_01 relative to the home frame | > | > | > | > | > | > | > Speed is neither a coordinate system nor a frame of reference. | | > | > | Indeed - why state the obvious. | > | > | > Because it is necessary when *you* blunder. | > | > | > | > | > What above means is that inertial systems | are moving with constant | > | > mutual velocities. This is basic stuff from | classical mechanics - | > so | > it is hard to believe you were really confused | by my wording. | | > > | > | > There are only two systems of coordinates in Einstein's SR, namely | > the | > "stationary frame", the "moving frame" and the "empty space" | > that light | > moves at speed c in. (Three always equals two.) | | > | Only two systems of coordinates - sheesh. Actually one can set as many | > | systems of coordinates in SR (inertial or not) as anyone care to | > consider. | > | > Hmmm... | > | > Well, there is the stationary frame K where the speed of light is c. | > That has coordinates (x,y,z) to which Einstein wrongfully adds t to give | > (x,y,z,t). | > | > Then there is the moving frame k where the speed of light is c-v one way | > and c+v back again. That has coordinates (xi,eta, zeta, tau) | > | > And then there is the kappa frame that Einstein didn't name, | | As I have said many times again - I am not at all interested how E | presented his theory over a century ago. Then this discussion is over, I have no interest in Esa Riihonen's theory of inertial frames.
From: eric gisse on 5 Jul 2010 17:38 Esa Riihonen wrote: [...] Another poor fool tries to educate the androcles.
From: Edward Green on 5 Jul 2010 17:50
On Jul 4, 11:00 pm, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > Edward Green says... > > >There's a canard floating about here somewhere, which I see has very > >early origins. It runs as follows: "In SR, all inertial coordinate > >systems are shown to be equivalent, in GR _all_ coordinate systems are > >equivalent". It may not be false, but the two equivalences are not > >equivalent. One contains a strong physical insight, the other, more a > >tour of mathematical force, that all coordinate systems may be forced > >to be equivalent. I lack the skill to state this precisely, but there > >is something there. > > Can you say a little more about this difference? I might agree, or I might not. I'll try, but in the tradition of the Great Oz, I'm enjoying a passion fruit cocktail at this very moment -- if one can be said to enjoy something that tastes so bitter. There must be something wrong with my prep! > I've thought about it a little, and I realize that it's a little subtle to say > what it means to say that all inertial frames are equivalent. We can say roughly > that if there are two frames F and F', and you perform the same experiment in > both frames, you'll get the same results. But what, exactly, does "the same > experiment" mean? Well, you could use the self-same experimental apparatus, simply accelerated into your new frame. Presuming you're not doing the sort of experiment that consumes the apparatus. > What does "the same result" mean? Just one subtlety is system > of units: how do you know that the experimenters in F and F' are using the same > standards for length and time? You want to be able to say something like this: > Such and such a chemical reaction took T seconds in frame F, while it took T' > seconds in frame F'. To know whether this confirms or violates the relativity > principle, you have to know whether T and T' are being measured in the same > units. But how do you know that? Ultimately, you have to rely on some > reproducible experiment, such as chemical or mechanical reaction, in order to > establish that F and F' are using the same units of time. But then it becomes > tautological that those reactions work the same in both frames. One way to look at it is that everything comes along for the ride. With only one chemical experiment to choose from you might get a tautology. With an unlimited array of chemical, optical, acoustic, etc., experiments, you begin to believe that all really is behaving the same way: if there is a difference, it is a systemic one. > Another way to talk about invariance is in terms of coordinate transformations. > For a particular mathematical formulation of some law of physics, we can ask > whether its form is invariant under particular kinds of coordinate > transformations. Yes. > Newton's laws are invariant under Galilean transformations, but > not under Lorentz transformations, and not under nonlinear transformations. SR > is invariant under Lorentz transformations, but not under Galilean or nonlinear > transformations. GR is invariant under all possible transformations. Yes, and that's the nub of the question -- just how physical is this property of universal invariance, and how tautological. I think it is much more tautological than the invariance in form of Maxwell's equations under Lorentz transformation. It's more like casting physical law in the form " G(X) = 0 " and lo, and behold, noting that G'(X') = 0 also. There is less physical content to it, and more mathematical machinery. > But this characterization in terms of coordinate transformations has no real > physical content. You can *ALWAYS* write any law of physics in a way that is > invariant under arbitrary coordinate transformations. I see you anticipate me. > So what's the *physical* > sense in which SR satisfies a relativity principle, and GR satisfies a principle > of equivalence of all coordinate systems? > > Here's the completely non-obvious answer: It's in terms of nondynamic scalar, > vector and tensor fields! Let me explain this with the three theories under > discussion: Newtonian physics, Special Relativity, and General Relativity.. > > In Newtonian physics, there are no intrinsic spatial vectors. There can be > spatial vectors (such as the velocity vector of a particle), but they are always > *contingent*; they depend on initial conditions, and can be modified by applying > forces. How does the nonexistence of nondynamic spatial vectors satisfy the > principle of relativity? Well, if there were a preferred rest frame, then there > would be a preferred velocity vector---the velocity of the rest frame. It would > be an intrinsic spatial vector. > > Newtonian physics *does* have an intrinsic scalar field, namely universal time. > It's nondynamic in the sense that there are no forces that can ever change the > value of universal time. > > In Special Relativity, there is no intrinsic vector fields or scalar fields. But > there *is* an intrinsic tensor field, namely the metric tensor. It's nondynamic, > in the sense that there are no forces in SR that are capable of changing the > metric tensor. > > In General Relativity, there are no intrinsic, nondynamic scalars, vectors, or > tensors. There are important tensors (the metric tensor, the curvature tensor) > but they are dynamic, they evolve, and do not have god-given values. Not at all what I had in mind, but very deep nonetheless. Let me try to create a concrete example of what I had in mind. Say F = ma. We know this is invariant in form under Galilean transformation. But what might it look like under general non-linear transformations of coordinates? Perhaps F = m D x, where x are the generalized coordinates, and D is now a general differential operator. If we can always write our equation in this form, then we now have "invariance under arbitrary coordinate transformations". But we are playing some kind of a trick here. The "D" in general might be arbitrarily complicated, or at least more complicated than the d^2/ dx^2 implicit in the "a". We are camouflaging this complication under a single symbol, and we have obtained seemingly greater generality only at this price. There is no additional physical generality -- we already milked that out of the equations with the simple Galilean transformations. That is similar to what I think is going on between the invariance in form under Lorentz transformations and the invariance in form under general curvilinear transformations. |