Top Document: Invariant Galilean Transformations On All Laws Previous Document: 3. The Principle of Relativity and Transformation Next Document: 5. Transformations on Generalized Coordinate Laws See reader questions & answers on this topic!  Help others by sharing your knowledge One example of the traditional fallacious idea that an equation is not invariant under the galilean transformation comes from the Encyclopedia Brittanica: "Before Einstein's special theory of relativity was published in 1905, it was usually assumed that the time coordinates measured in all inertial frames were identical and equal to an 'absolute time'. Thus, t = t'. (97) "The position coordinates x and x' were then assumed to be related by x' = x  vt. (98) "The two formulas (97) and (98) are called a Galilean transformation. The laws of nonrelativ istic mechanics take the same form in all frames related by Galilean transformations. This is the restricted, or Galilean, principle of relativity. "The position of a light wave front speeding from the origin at time zero should satisfy x^2  (ct)^2 = 0 (99) in the frame (t,x) and (x')^2  (ct')^2 = 0 (100) in the frame (t',x'). Formula (100) does not transform into formula (99) using the transform ations (97) and (98), however." ................................................. Besides the trivially correct statement of what the Galilean 'transform' equations are, there is exactly one thing they got right. I. Eq100 is indeed the correct basis for discussing the question of invariance, given that eq99 is the correct 'stationary' (observer S) equation. [Let observer M be the 'moving'system observer.] In particular, eq100 is of exactly the same form [the square of argument one minus the square of argument two equals zero (argument three).] II. It is nonsense to say eq99 should be derivable from eq100; for one thing, the transforms are TO x' and t' from x and t, not the other way around, and the idea that either observer's equation should contain within itself the terms to simplify or rearrange to get to the other is ridiculous. As the transform equations say, the relationship of t', x' to t, x is based on the relative velocity between the two systems, but neither the original (eq99) equation nor the M observer equation is about a relationship between coordinate systems or observers. One might as well expect the two equations to contain banana export/import data; there is no relevancy. The 'transform' equations are the relationships between x' and x, t' and t and have nothing to do with what one equation or the other ought to 'say'. The equations' content is the rate at which light emitted along the xaxes moves. III. Most remarkable, the True Believer SR crackpots who most despise the consequences of measurement theory (demonstrable fact) contained in this document are those who want to argue against our saying the Britt anica got eq100 right; They insist that the correct equation is derived directly from x'=xvt and t'=t. Solve for x=x'+vt and replace t with t', then substitute the result in eq99: (x'+vt')^2  (ct')^2 = 0. Besides the fact that this results in an equation with arguments exactly equal to eq99, they will insist the transform is not invariant. IV. A major justification they have for their idea of the correct M system equation on which to base the the discussion of invariance, is that the variables are M system variables, never mind the fact that the arguments are S system values. That argument of theirs is arrant nonsense. The velocity v that S sees for the M system relative to herself is the negative of what the M system sees for the S system relative to himself. In other words, x'+vt' is a mixed frame expression and it is x'+(v)t' that would be strictly M frame notation, and that equation is far off base. [Work it out for yourself, but make sure you try out an S frame negative v so as not to mislead yourself.] V. In I. we said: "given that eq99 is the correct 'stationary' equation. Let's look at it closely: x^2  (ct)^2 = 0 (99) This whole matter is supposed to be about coordinate transforms. Is that what t is, just a coordinate? No. It isn't, in general. Suppose you and I are both modelling the same light event and you are using EST and I'm using PST. 'Just a time coordinate' is just a clock reading amd your t clock reading says the light has been moving three hours longer than my clock reading says. Well, that's what the idea that t is a coordinate means. Eq99 works if and only if t is a time interval, and in particular the elapsed time since the light was emitted. Thus, that equation works only if we understand just what t is, an elapsed time, with emissioon at t=0. However, we don't have to 'understand' anything if we use a more intelligent and insightful form of the equation: (x)^2  [ c(tt.e) ]^2 = 0, where t.e is anyone's clock reading at the time of light emission, and t is any subsequent time on the same clock. Similarly, x is not just a coordinate, but a distance since emission. (xx.e)^2  [ c(tt.e) ]^2 = 0 (99a) VI. In the spirit of 'there is exactly one thing they got right', the correct M system version of eq99a is eq100a: (x'x.e')^2  [ c(t't.e') ]^2 = 0 (100a) Every observer in the universe can derive their eq100a from eq99a and vice versa, not to mention to and from every other observer's eq99a. Now, THAT's invariance. [You do realize that every eq100a reduces to eq99a, when you back substitute from the transforms, right? t.e'=t.e, x.e'=x.evt.] User Contributions:Top Document: Invariant Galilean Transformations On All Laws Previous Document: 3. The Principle of Relativity and Transformation Next Document: 5. Transformations on Generalized Coordinate Laws Single Page [ Usenet FAQs  Web FAQs  Documents  RFC Index ] Send corrections/additions to the FAQ Maintainer: Thnktank@concentric.net (Eleaticus)
Last Update March 27 2014 @ 02:12 PM

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