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Natural Science Forum / Physics / Relativity / August 2006And yet another...
Suppose we had a coordinate system on spacetime labeled u,v,w,t where t of course is time-like, u,v,w spacelike. Can we assert, as a matter of convention or otherwise, that increments in t are what would be measured by a standard clock at rest wrt the local spatial coordinates? Are there coordinate systems where other interpretations of "dt" are indicated? | Suppose we had a coordinate system on spacetime labeled u,v,w,t where t | of course is time-like, u,v,w spacelike. Can we assert, as a matter of | convention or otherwise, that increments in t are what would be | measured by a standard clock at rest wrt the local spatial coordinates? You can if you want to. This is PHYSICS, not math or logic, and "proof" is completely irrelevant. - Tom Roberts. news:P4Hqg.60105$Lm5.3167@newssvr12.news.prodigy.com | Are there coordinate systems where other interpretations of "dt" are | indicated? Sure, have as many as you like. Androcles spamspamspam > Suppose we had a coordinate system on spacetime labeled u,v,w,t where t > of course is time-like, u,v,w spacelike. Can we assert, as a matter of > convention or otherwise, that increments in t are what would be > measured by a standard clock at rest wrt the local spatial coordinates? No. For the simple reason that you could use as your t coordinate any function of the value of a standard clock. > Are there coordinate systems where other interpretations of "dt" are > indicated? Any coordinate system for which t is not the value of a standard clock has a different interpretation. Such as the Schw. coordinates on Schwarzschild spacetime. And, of course, there are coordiante systens for which there is no timelike coordinate at all; these have two spacelike and two null (lightlike) coordinates. Tom Roberts "Tom Roberts" <tjroberts137@sbcglobal.net> wrote in message news:P4Hqg.60105$Lm5.3167@newssvr12.news.prodigy.com... | This is PHYSICS, not math or logic, and "proof" is completely irrelevant. | Tom Roberts Silly Roberts. Androcles. Edward Green says... >Suppose we had a coordinate system on spacetime labeled u,v,w,t where t >of course is time-like, u,v,w spacelike. Can we assert, as a matter of >convention or otherwise, that increments in t are what would be >measured by a standard clock at rest wrt the local spatial coordinates? Definitely not. A clock doesn't measure t, it measures proper time along its path. If a clock is stationary in your coordinate system (that is, du/dtau = dv/dtau = dw/dtau = 0), then the elapsed time on the clock will be given by tau = integral of square-root(|g_tt|) dt That will be equal to t only in the special case g_tt = 1. >Are there coordinate systems where other interpretations of "dt" are >indicated? Every coordinate system in which g_tt is different from 1. For example, Schwarzchild coordinates in the vicinity of a black hole. -- Daryl McCullough Ithaca, NY > Edward Green says... > > [quoted text clipped - 11 lines] > > That will be equal to t only in the special case g_tt = 1. Ahhhhaaaaa...... It seems I had the interpretation of the metric rather bollixed up. So the _resultant_ of the metric contracted with the coordinate differentials is the time increment (proper time) a local clock would measure if the differential is time-like, and the length increment (proper length) a local meter stick would measure if the differential is space-like, and simply "spacetime interval" in the general case. A representation of the metric is a machine which outputs natural local structure (standard clock and rod measure) from the coordinates. That is an immense clarification: thank you. Now, persisting in crankery, I again ask what would happen to the metric if we "stretched" the manifold. Again taking a one-dimensional manifold equipped with the real coordinates and the obvious metric, and then taking a copy of this manifold and decreeing that we have dilated parts of it and the embedded coordinates, what happens to the metric? Evidently, nothing! If the local meter sticks have acquired a similar stretch, and the metric calculates what is measured by meter sticks, the new situation is indistinguishable from the old. In the one dimensional case the scoffers seem to have it: a "stretched" manifold is indistinguishable from an unstretched one, at least using the democratic metric which makes every clock and rod a king in its own neighborhood. I'm not so sure about the two dimensional case: in fact, I think the possibility of varying the stretch of one direction along another can be understood to lead to what is identified as curvature. > the _resultant_ of the metric contracted with the coordinate > differentials is the time increment (proper time) a local clock would [quoted text clipped - 3 lines] > representation of the metric is a machine which outputs natural local > structure (standard clock and rod measure) from the coordinates. Yes. > Now, persisting in crankery, I again ask what would happen to the > metric if we "stretched" the manifold. Again taking a one-dimensional > manifold equipped with the real coordinates and the obvious metric, and > then taking a copy of this manifold and decreeing that we have dilated > parts of it and the embedded coordinates, what happens to the metric? This depends upon how you stretch the manifold. First consider a change of coordinates: the coordinates of a given point P will change, reflecting the stretching, but the components of the metric will change correspondingly, and the distance between any pair of points remains unchanged. Instead consider a second 1-d manifold identical to the first, and call the first manifold M and the second N; consider a diffeomorphism f(.) from M to N such that the image point P' in N of a point P in M is P'=f(P) for all points P in M (and therefore all P' in N). Let the metric be mapped appropriately so the distance between any two points P and Q in M is the same as the distance between their image points P' and Q' in N. Note I have said nothing about how M and N might look if one laid them down next to each other. Do that however you like (stretching and/or shrinking in any manner, but keeping the two manifolds adjacent to each other); now compare points that are adjacent, instead of points related by f(.) -- this is now a "stretching", because the distance between pairs of adjacent points can have a different value in M than in N. Do this smoothly and it is a diffeomorphism from M to N I'll call g(.): P"=g(P) for all P in M (and therefore all P" in N). Note that g(.) does NOT carry the metric, but f(.) does -- that is essential to the notion of "stretching": the distance between P' and Q' is the same as the distance between P and Q, but the distance between P" and Q" can be different (each distance in the respective manifold M or N). Now to consider a "stretching" of a single manifold rather than diffeos. between manifolds, simply apply either f(.) or g(.) from M to M (not to N!). it should be clear that using f(.) will preserve the distances between points, but using g(.) will not. > Evidently, nothing! If the local meter sticks have acquired a similar > stretch, and the metric calculates what is measured by meter sticks, [quoted text clipped - 3 lines] > democratic metric which makes every clock and rod a king in its own > neighborhood. See above. This depends on how you "stretch" the manifold. If you carry the metric with the diffeo. (the usual case, like f(.) above) then you are correct. But if you don't carry the metric (as in g(.) above) then this is wrong. > I'm not so sure about the two dimensional case: in fact, I think the > possibility of varying the stretch of one direction along another can > be understood to lead to what is identified as curvature. The basic idea generalizes to N dimensions. Yes indeed, varying diffeomorphisms like g(.) appropriately can generate curvature; varying diffeos like f(.) cannot. Note that the standard mathematical notion of curvature of a manifold is identically zero for any 1-d manifold, even though in everyday speech we often talk of "curved lines". Tom Roberts > > Edward Green says... > > > [quoted text clipped - 38 lines] > democratic metric which makes every clock and rod a king in its own > neighborhood. Right. And the only reason you can even talk about "stretching" (undetectable in the 1D manifold itself) is that you have an extra ingredient: the ambient context in the form of the enclosing space containing the 1D manifold. Spacetime in GR is considered intrinsically: there is no concept of an ambient super-duper-space containing spacetime. This is the usual situation in differential geometry so whenever one uses terms like curvature, it's assumed to be intrinsic. > I'm not so sure about the two dimensional case: in fact, I think the > possibility of varying the stretch of one direction along another can > be understood to lead to what is identified as curvature. It's the same thing in the sense that this sort of distorting is really meaningless without some extrinsic reference like the ambient space. Mathematically this corresponds to taking any diffeomorphism of the manifold ("distortion" except it's going to be undetectable) and carrying the metric over using this diffeomorphism. The result is an isometry. -- Jan Bielawski |
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