Preferred Frame Theory indistinguishable from SR
in [Math]
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From: Tom Roberts on 7 Jul 2010 10:05 Daryl McCullough wrote: > harald says... >> On Jul 6, 5:18=A0pm, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: >>> Here's an analogy: A flat Euclidean plane has no notion of a preferred >>> direction. Any direction is as good as any other. But it certainly has >>> a notion of a *change* of direction. If you draw a path on the Euclidean >>> plane, then you can unambiguously determine whether the line is >>> straight or curved, because a straight line connecting two points is >>> shorter than any curved line connecting the same two points. If you >>> measure the lengths of two curves, you can determine which one is >>> straight. >> Sorry but I can't resist pointing out the error of the above: a >> straight trajectory relatively to an Euclidean plane is *only* >> measured to be "straight" if that plane is part of what Einstein >> called the "privileged" group of inertial "spaces". > > I think you are confused about this point. Euclidean space has > an associated metric, which determines the lengths of paths. > A straight line is defined relative to that metric as the path > that minimizes the length between two points. It has nothing to > do with any "privileged space". harald is more confused than that. He did not realize that you were discussing "a flat Euclidean plane", and he thought you were still discussing relativity, despite your clear and unambiguous statement of this fact. Like so many around here, harald needs to learn how to read more accurately. > Having said that, we can define a special group of coordinate > systems for the Euclidean plane---the Cartesian coordinate systems, Right. They are the ANALOGY of the inertial frames in relativity. But they don't form a group, they form a set or a class. "Group" is a technical word with a different meaning than you intended. The transforms between pairs of such coordinates form a group. Tom Roberts
From: Daryl McCullough on 7 Jul 2010 12:02 harald says... > >On Jul 7, 1:46=A0pm, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: >> harald says... >> >> >> >> >On Jul 6, 5:18=3DA0pm, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrot= >e: >> >> harald says... >> >> >> >The twin scenario was presented by Langevin in 1911 to show that >> >> >physical acceleration is "absolute", even more so with SRT than with >> >> >Newton's mechanics. >> >> >> What does that mean? As I said, proper acceleration (as measured by >> >> an accelerometer) is absolute, but coordinate acceleration is >> >> certainly not. >> >> >It means that you agree on that point with Langevin. >> >> Well, it's hard for me to believe that Einstein was unaware of the >> fact that an accelerometer can measure accelerations. > >Einstein was as aware as most physicists that an accelerometer does >not distinguish between an acceleration and a gravitational field; >however, he pushed that idea to the extreme. Then I'm *not* disagreeing with Einstein. As I said, *proper* acceleration (acceleration relative to freefall) is certainly detectable, and Einstein agrees with that. >> >> Here's an analogy: A flat Euclidean plane has no notion of a preferred >> >> direction. Any direction is as good as any other. But it certainly has >> >> a notion of a *change* of direction. If you draw a path on the Euclide= >an >> >> plane, then you can unambiguously determine whether the line is >> >> straight or curved, because a straight line connecting two points is >> >> shorter than any curved line connecting the same two points. If you >> >> measure the lengths of two curves, you can determine which one is >> >> straight. >> >> >Sorry but I can't resist pointing out the error of the above: a >> >straight trajectory relatively to an Euclidean plane is *only* >> >measured to be "straight" if that plane is part of what Einstein >> >called the "privileged" group of inertial "spaces". >> >> I think you are confused about this point. Euclidean space has >> an associated metric, which determines the lengths of paths. >> A straight line is defined relative to that metric as the path >> that minimizes the length between two points. It has nothing to >> do with any "privileged space". > >I agree that a straight line relative to Euclidean space can be >clearly defined. Perhaps I misunderstood that you meant with straight >"path" a straight trajectory as defined in Newtonian mechanics and >SRT. If you did not mean that, I don't know what you tried to say. If you a space of points S, then a path is a 1-D subset of S that can be described as the image of a function from the reals (or an interval of reals) to S. So the position of an object as a function of time defines a path through Euclidean space, but a curve drawn on a piece of paper also is a path. >> Having said that, we can define a special group of coordinate >> systems for the Euclidean plane---the Cartesian coordinate systems, >> via the requirement: >> >> A line is straight >> <=> >> It can be parametrized so that (d/ds)^2 x = (d/ds)^2 y =3D 0. >> >> Being a straight line is independent of coordinate system. > >With that claim I wonder if I truly misunderstood you; for in >Newtonian mechanics as well as SRT, a path is very much dependent of >the kind of coordinate system that you use. That's not true. The path exists independent of the coordinates used to define it. If I have a road stretching across the surface of the Earth, that road defines a path (well, in the limit as the width of the road goes to zero, anyway). You don't need coordinates to give a path, and you don't need coordinates in order to say that a road is straight. Of course, you *can* describe a path with coordinates. You can describe a road by giving two functions lat(s) and long(s), which specifies the latitude and longitude as a function of the distance s along the road. >> The associated coordinate acceleration being zero is dependent >> on a choice of a special coordinate system. > >Yes. But what was your point? That the notion of "straight" versus "nonstraight" is *not* dependent on a coordinate system. Whether a path is straight (for Euclidean geometry) or inertial (for relativity) is an intrinsic property of the path, and a path doesn't change from straight to nonstraight when you change coordinate systems. As I said, there is a special set of coordinate systems (Cartesian coordinate systems, in the case of Euclidean geometry, inertial coordinate systems, in the case of relativity) such that straight paths or inertial paths are particular simple: In such a coordinate system, an inertial path can be written as: x(t) = x_0 + v_x t y(t) = y_0 + v_y t z(t) = z_0 + v_z t where x_0, y_0, z_0, v_x, v_y, and v_z are constants. Straight paths can *only* be written that way if you are using a Cartesian inertial coordinate system. >> >> >"The laws of physics must be of such a nature that they apply to >> >> >systems of reference in any kind of motion". >> >> >As a result, physical acceleration is, according to Einstein's GRT, >> >> >*relative* - which is just the contrary of what Langevin argued based >> >> >on his "twins" example of SRT. >> >> >> As I said, proper acceleration is definitely *not* relative, but >> >> coordinate acceleration trivially *is*. But proper acceleration is >> >> measuring acceleration relative to *freefall*. >> >> >Then we both disagree with Einstein; >> >> That's ridiculous. Einstein certainly knew that an accelerated >> observer feels "inertial forces", and an unaccelerated observer >> does not. > >Sure he did. :-) You agreed that he did, above. >> Whatever was meant by his generalized principle of relativity, > >You mean that you really did not know, and that you still don't - even >after reading all his explanations?! Well, it seems to me that you don't understand what Einstein meant. >> he certainly did *not* mean that what is now known >> as proper acceleration is undetectable. > >Indeed. Perhaps it helps to say it in other words than he did: he >meant *indistinguishable* from gravitation. When I say "proper acceleration", I mean acceleration *relative* to freefall. So that already takes into account gravity. In General Relativity, there *is* no "force of gravity". There are only inertial forces which appear whenever an observer is accelerating relative to freefall. That doesn't mean that gravitation is undetectable, just that a gravitational *force* is undetectable. Gravitation in GR is manifested through curvature, through the fact that the local standard for freefall (inertial motion) changes from location to location. Unlike Newtonian physics or Special Relativity, there is no longer a global notion of an inertial frame. >> The modern way of looking at it is that "inertial forces" are >> felt whenever the observer is accelerating *relative* to freefall. >> Einstein originally thought of the equivalence principle differently: >> He thought that an object accelerating in a gravitational field felt >> two different kinds of forces: (1) inertial forces due to acceleration, >> and (2) gravitational forces. These two forces canceled in the case >> of freefall. > >??? I strongly doubt that. Reference please! I cannot find an online reference, but it occurs in a discussion by Einstein of his "elevator" thought experiment. >> >According to his theory, we are entitled to say that such an object >> >is *not* (properly) accelerating but that instead a "real" >> >gravitational field is induced through the universe which accelerates >> >all the *other* objects. >> >> I think you are confusing the physical content of Einstein's theory >> with the way he chose to describe it. > >The purpose with which you and I try to describe things here is to >make the physical content of what think clear to the other. Do you >seriously believe that Einstein tried to do the opposite, to hide the >meaning of his words? No, what I'm saying is that in your case, Einstein failed to communicate (to you) what he meant. >Good, we are making progress. :-) >Einstein held that, as he put it, acceleration is "relative": >according to his theory we may just as well claim that the traveler is >*not* physically accelerated, contrary to Langevin's and your claim. No, you are confused. As I have said, there are two different notions of "acceleration": (1) proper acceleration (acceleration relative to the local standard for freefall) and (2) coordinate acceleration (acceleration relative to whatever coordinate system you are using). Einstein and I are in complete agreement that for the traveling twin, proper acceleration is nonzero, while coordinate acceleration is zero (using the appropriate noninertial coordinate system). So where is the disagreement? There is none. >He thought to solve the problem by saying that at the turnaround >(according to the stay-at-home), the traveler may consider himself as >remaining in place against an induced gravitational field that >appears. And certainly he may, in the sense that he may choose a coordinate system in which he is always at rest. The notion of being at rest is relative to a coordinate system in relativity. >> A lot of the confusion in physics discussions are because people are >> caught up in interpreting *words*, as if we are analyzing some holy >> text. I don't *care* what words Einstein, or anyone else, uses. > >In that case we have nothing to discuss, Are you saying that you had no point other than complaining about Einstein's way of describing his theory? >nor can you really discuss the clock paradox: I can discuss it perfectly well, from the point of view of physics. >it is foremost concerned with physical concepts that had been >expressed with words as well as with equations. >> His theories have physical content that are independent of the words used >> to describe them. > >Without definitions of the variables and their fields of application, >there is just mathematics without physical meaning. The physical meaning of the theory is defined by its predictions for *actual* experiments. General Relativity describes what happens when you take clocks and move them about, move them up and down in a gravitational field. It describes how mass affects gravitational fields, and how (indirectly) it affects the behavior of clocks. It describes how electromagnetic waves change frequency as they pass near massive bodies. It describes how massive bodies orbit one another. What other physical meaning could you possibly ask for???? If you are asking, not about General Relativity, but the General Principle of Relativity: that isn't a theory of physics, it is a heuristic, or a philosophical position, or metaphysics. It has no physical meaning, except to the extent that it guides us in coming up with better theories of physics. >> >> The bare statement "The laws of physics must be of >> >> such a nature that they apply to systems of reference in any kind of >> >> motion" is not a contradiction---on the contrary, it is nearly a >> >> tautology. You can always write the laws of physics so that you >> >> can use an arbitrary coordinate system. >> >> >If you think that a postulate of physics is a tautology, then probably >> >you misinterpret its meaning. >> >> Einstein didn't *realize* it was a tautology. He thought that the >> requirement that a physical theory be written in a way that had the >> same form in all coordinate systems would uniquely pin down the theory, >> or at least eliminate some candidate theories. > >There is also the key thought "simplest form" which is not written >there but implied: it is included in his special relativity >definition, of which the GPoR is an extension. > >Anyway, a theory is that what its author says it is. We disagree about this. The author is *irrelevant* except for historical purposes. The theory of relativity, or Newtonian mechanics, or electromagnetism, have all developed considerably since they were first invented by Einstein, Newton or Maxwell. What the original authors believed is interesting from a historical point of view, and for its insight into how great minds work, but has no significance in understanding modern theories. >Otherwise for example, I could claim that Ken Seto's Mechanics is right, >and that he doesn't know his theory well enough yet, but we will do >that for him! I couldn't care less what you call a theory, as long as you make it clear what theory you are talking about. Special Relativity and General Relativity are well-developed theories today. Any physicist knows what they are, and what their content is. That content is *not* determined by what Einstein believed in 1905 or 1916. >> I have read many of your posts, and I have *yet* to see you explain >> in what sense you think that the twin paradox is a consistency problem >> for any position that Einstein is likely to have believed. > >Einstein explains the cause of the distrust or criticism clearly >enough; nothing that I add can make clearer what the issue was. He makes it clear that there *is* no consistency problem. -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 7 Jul 2010 12:57 Tom Roberts says... > >Daryl McCullough wrote: >> Having said that, we can define a special group of coordinate >> systems for the Euclidean plane---the Cartesian coordinate systems, > >Right. They are the ANALOGY of the inertial frames in relativity. > > But they don't form a group, they form a set or a class. > "Group" is a technical word with a different meaning than > you intended. The transforms between pairs of such coordinates > form a group. I was not meaning "group" in the technical sense, I was just meaning it in the sense of a collection. But actually, don't they form a group? The various Cartesian coordinate systems are related by operations such as (1) translations, (2) rotations, (3) scale transformations. Couldn't they form a group? -- Daryl McCullough Ithaca, NY
From: harald on 7 Jul 2010 14:32 On Jul 7, 6:02 pm, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > harald says... > > >On Jul 7, 1:46=A0pm, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > >> harald says... > > >> >On Jul 6, 5:18=3DA0pm, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrot= > >e: > >> >> harald says... > > >> >> >The twin scenario was presented by Langevin in 1911 to show that > >> >> >physical acceleration is "absolute", even more so with SRT than with > >> >> >Newton's mechanics. > > >> >> What does that mean? As I said, proper acceleration (as measured by > >> >> an accelerometer) is absolute, but coordinate acceleration is > >> >> certainly not. > > >> >It means that you agree on that point with Langevin. > > >> Well, it's hard for me to believe that Einstein was unaware of the > >> fact that an accelerometer can measure accelerations. > > >Einstein was as aware as most physicists that an accelerometer does > >not distinguish between an acceleration and a gravitational field; > >however, he pushed that idea to the extreme. > > Then I'm *not* disagreeing with Einstein. As I said, *proper* > acceleration (acceleration relative to freefall) is certainly > detectable, and Einstein agrees with that. OK - I understood "proper" as in SRT. [..] > Of course, you *can* describe a path with coordinates. You > can describe a road by giving two functions lat(s) and long(s), > which specifies the latitude and longitude as a function of the > distance s along the road. > > >> The associated coordinate acceleration being zero is dependent > >> on a choice of a special coordinate system. > > >Yes. But what was your point? > > That the notion of "straight" versus "nonstraight" is *not* > dependent on a coordinate system. It's definitely the case for "straight" trajectories, which are for example straight relative to an inertial system but not relative to a rotating system. > Whether a path is straight > (for Euclidean geometry) or inertial (for relativity) is an > intrinsic property of the path, and a path doesn't change from > straight to nonstraight when you change coordinate systems. See below. > As I said, there is a special set of coordinate systems > (Cartesian coordinate systems, in the case of Euclidean > geometry, inertial coordinate systems, in the case of relativity) > such that straight paths or inertial paths are particular > simple: In such a coordinate system, an inertial path can > be written as: > > x(t) = x_0 + v_x t > y(t) = y_0 + v_y t > z(t) = z_0 + v_z t > > where x_0, y_0, z_0, v_x, v_y, and v_z are constants. > Straight paths can *only* be written that way if you > are using a Cartesian inertial coordinate system. Ah - you used the right key words here; now we agree! :-) [..] > >> Whatever was meant by his generalized principle of relativity, > > >You mean that you really did not know, and that you still don't - even > >after reading all his explanations?! > > Well, it seems to me that you don't understand what Einstein > meant. I understand why he agreed to call the clock exercise a "paradox" and an "objection" against his theory, which required to be solved. It appears that you still don't understand why, and I don't think that adding more words will help. [...] > In General Relativity, there *is* no "force of gravity". There are > only inertial forces which appear whenever an observer is accelerating > relative to freefall. That doesn't mean that gravitation is undetectable, > just that a gravitational *force* is undetectable. Gravitation in GR > is manifested through curvature, through the fact that the local standard > for freefall (inertial motion) changes from location to location. Unlike > Newtonian physics or Special Relativity, there is no longer a global notion > of an inertial frame. > > >> The modern way of looking at it is that "inertial forces" are > >> felt whenever the observer is accelerating *relative* to freefall. > >> Einstein originally thought of the equivalence principle differently: > >> He thought that an object accelerating in a gravitational field felt > >> two different kinds of forces: (1) inertial forces due to acceleration, > >> and (2) gravitational forces. These two forces canceled in the case > >> of freefall. > > >??? I strongly doubt that. Reference please! > > I cannot find an online reference, but it occurs in a discussion > by Einstein of his "elevator" thought experiment. As far as I remember, he held that an object accelerating in a gravitational field feels no force at all; does it make a difference? > >> >According to his theory, we are entitled to say that such an object > >> >is *not* (properly) accelerating but that instead a "real" > >> >gravitational field is induced through the universe which accelerates > >> >all the *other* objects. > > >> I think you are confusing the physical content of Einstein's theory > >> with the way he chose to describe it. > > >The purpose with which you and I try to describe things here is to > >make the physical content of what think clear to the other. Do you > >seriously believe that Einstein tried to do the opposite, to hide the > >meaning of his words? > > No, what I'm saying is that in your case, Einstein failed to > communicate (to you) what he meant. Not Einstein, but we to each other. However, it just got better! > >Good, we are making progress. :-) > >Einstein held that, as he put it, acceleration is "relative": > >according to his theory we may just as well claim that the traveler is > >*not* physically accelerated, contrary to Langevin's and your claim. > > No, you are confused. As I have said, there are two different notions > of "acceleration": (1) proper acceleration (acceleration relative to > the local standard for freefall) and (2) coordinate acceleration > (acceleration relative to whatever coordinate system you are using). > Einstein and I are in complete agreement that for the traveling > twin, proper acceleration is nonzero, while coordinate acceleration > is zero (using the appropriate noninertial coordinate system). So > where is the disagreement? There is none. There is no disagreement on that point. What about the induced gravitational field? > >He thought to solve the problem by saying that at the turnaround > >(according to the stay-at-home), the traveler may consider himself as > >remaining in place against an induced gravitational field that > >appears. > > And certainly he may, in the sense that he may choose a coordinate > system in which he is always at rest. The notion of being at rest > is relative to a coordinate system in relativity. He only may do so if his induced gravitational field can be held to be, as his theory claims, "physical", and propagating according to the same laws of physics as all other gravitational fields. > >> A lot of the confusion in physics discussions are because people are > >> caught up in interpreting *words*, as if we are analyzing some holy > >> text. I don't *care* what words Einstein, or anyone else, uses. > > >In that case we have nothing to discuss, > > Are you saying that you had no point other than complaining > about Einstein's way of describing his theory? ?! I have no complaints at all. My point, about which *you* "complained", was the simple fact that the "clock paradox" concerns the General PoR; that is irrelevant for SRT. [..] > General Relativity describes what happens when > you take clocks and move them about, move them up and down in a > gravitational field. It describes how mass affects gravitational > fields, and how (indirectly) it affects the behavior of clocks. > It describes how electromagnetic waves change frequency as they > pass near massive bodies. It describes how massive bodies orbit > one another. What other physical meaning could you possibly ask for???? I don't ask for anything; Einstein provided more! > If you are asking, not about General Relativity, but the General > Principle of Relativity: that isn't a theory of physics, it is > a heuristic, or a philosophical position, or metaphysics. It has > no physical meaning, except to the extent that it guides us in > coming up with better theories of physics. I rarely saw a more aggressive criticism against Einstein's theory. :-) Thanks for the discussion. Harald
From: Daryl McCullough on 7 Jul 2010 16:05
harald says... >On Jul 7, 6:02=A0pm, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: >> That the notion of "straight" versus "nonstraight" is *not* >> dependent on a coordinate system. > >It's definitely the case for "straight" trajectories, which are for >example straight relative to an inertial system but not relative to a >rotating system. That's mistaken. Whether a trajectory is straight (or unaccelerated) is *not* relative to a coordinate system. If it is straight, it is straight in all coordinate systems. What varies from coordinate system to coordinate system is the *equation* describing a straight path. For Cartesian coordinates, the path x(s), y(s) satisfies: (d/ds)^2 x = (d/ds)^2 y = 0 For non-cartesian coordinates, the equation of a straight path is more complicated. >I understand why he agreed to call the clock exercise a "paradox" and >an "objection" against his theory, His dialog was a response to critics. The fact that he responded doesn't amount to admitting the critics were right. He's explaining why they are *not* right. >which required to be solved. It appears that you still don't >understand why, And the fact that you can't give a coherent answer to the question: why is the twin paradox a consistency challenge for Einstein's generalized principle of relativity seems to me to mean that you don't understand why, either. >> >> The modern way of looking at it is that "inertial forces" are >> >> felt whenever the observer is accelerating *relative* to freefall. >> >> Einstein originally thought of the equivalence principle differently: >> >> He thought that an object accelerating in a gravitational field felt >> >> two different kinds of forces: (1) inertial forces due to acceleration= >, >> >> and (2) gravitational forces. These two forces canceled in the case >> >> of freefall. >> >> >??? I strongly doubt that. Reference please! >> >> I cannot find an online reference, but it occurs in a discussion >> by Einstein of his "elevator" thought experiment. > >As far as I remember, he held that an object accelerating in a >gravitational field feels no force at all; does it make a difference? Right. The modern explanation is that an object in freefall is *not* accelerating; it is moving inertially. Einstein's original explanation (if I'm remembering it correctly) was that the object feels *two* forces that cancel each other: A downward force due to gravity, and an upward "inertial" force. >> >Good, we are making progress. :-) >> >Einstein held that, as he put it, acceleration is "relative": >> >according to his theory we may just as well claim that the traveler is >> >*not* physically accelerated, contrary to Langevin's and your claim. >> >> No, you are confused. As I have said, there are two different notions >> of "acceleration": (1) proper acceleration (acceleration relative to >> the local standard for freefall) and (2) coordinate acceleration >> (acceleration relative to whatever coordinate system you are using). >> Einstein and I are in complete agreement that for the traveling >> twin, proper acceleration is nonzero, while coordinate acceleration >> is zero (using the appropriate noninertial coordinate system). So >> where is the disagreement? There is none. > >There is no disagreement on that point. What about the induced >gravitational field? That's just the ordinary inertial forces associated with an accelerated observer. Calling them a gravitational field is to remind you that in Einstein's theory, there is no difference between a gravitational force and inertial forces. They are both manifestations of accelerating relative to the local notion of freefall. >> >He thought to solve the problem by saying that at the turnaround >> >(according to the stay-at-home), the traveler may consider himself as >> >remaining in place against an induced gravitational field that >> >appears. >> >> And certainly he may, in the sense that he may choose a coordinate >> system in which he is always at rest. The notion of being at rest >> is relative to a coordinate system in relativity. > >He only may do so if his induced gravitational field can be held to >be, as his theory claims, "physical", and propagating according to the >same laws of physics as all other gravitational fields. And that is the case. It's important to distinguish "gravitational field" from "gravity". They aren't the same thing. There are two different phenomena at work in the modern view of gravity: (1) Spacetime is *curved* by matter. What this means is that at each point in spacetime, there is a local notion of "freefall" or "inertial motion". Curvature means that this notion varies from point to point, rather than there being a global notion of an inertial frame. (2) Acceleration relative to the local notion of freefall results in inertial forces. This effect is exactly like Newtonian physics, where acceleration results in inertial forces. The difference is that in Newtonian physics, there is a consistent *global* notion of freefall or inertial motion, while in General Relativity, freefall varies from place to place. For effect (2), there is no distinction between "gravitational force" and any other inertial force. They're all inertial forces due to acceleration relative to the local notion of freefall. There is no "propagation" of effect (2). If you start accelerating, you instantly feel inertial forces. Inertial forces don't propagate in any physical sense. On the other hand, effect (1) has a very definite dynamic to it, which is describe by Einstein's field equations. Curvature is influenced by the presence of mass/energy/momentum. >> If you are asking, not about General Relativity, but the General >> Principle of Relativity: that isn't a theory of physics, it is >> a heuristic, or a philosophical position, or metaphysics. It has >> no physical meaning, except to the extent that it guides us in >> coming up with better theories of physics. > >I rarely saw a more aggressive criticism against Einstein's >theory. :-) The generalized principle of relativity is not a theory. -- Daryl McCullough Ithaca, NY |