Top Document: Invariant Galilean Transformations On All Laws Previous Document: 6. The data scale degradation absurdity. Next Document: 8. What does sci.math have to say about x0'=x0vt? See reader questions & answers on this topic!  Help others by sharing your knowledge It has become apparent  whether misleading or not  that the crackpot responses to the obvious derive from a common source, whether it be bandwagoning or their SR instructors. Below, in the sci.math subject, we see that all sci.math respondents agree with the basic "controversial" position of this faq: every coordinate is transformed, whether a supposed "constant" or not. Think about it, the generalized coordinate of a circle center, x0, applies to infinities upon infinities of circle locations (given y and z, too); it is a constant only for a given circle, and even then only on a given coordinate axis. And even "variables" are often held 'constant' during either integration or differentiation. The utility of a "variable" is that you can discuss all possible particular values without having to single out just one. That utility does not make particular  singled out  values on the variable's axis not values of the variable just because they have become named values. In any case, all that is preamble to the incompetent idea they have proposed for a transform of coordinates. It is based on the idea that the circle center, point of emission, whatever, has coordinates that cannot be transformed. Let there be an equation, say (x)^2  (ict)^2 = 0. What is the transformed version of that equation? Answer: (x')^2  (ict')^2 = 0. That's the one thing the Brittanica got right. Note that the leading crackpot just criticized this faq for presuming to correct the Britt anica, but it then and before poses the incompetent pseudo transform we analyze here in this section. x to x' and t to t' are obviously coordinate transforms; the x and t coordinates have been replaced by the coord inates in the primed system. A tranform of an equation from one coordinate system to another is NOT a substitution of the/a definition of x for itself; that is not a coordinate transformation. The most that can said for such a substitution is that it is a change of variable. But the crackpots are calling this a coordinate trans form of the original equation: (x'+vt)^2  (ict')^2 = 0. It is not a coordinate transform, of course, except accidentally. (x'+vt) is not the primed system coordinate, it is another form/expression of x. They get that substitution by solving x'=xvt for x; x=x'+vt. So, by incompetent misnomer, they accomplish what they have been railing against all along. It has been the generalized coordinate form in question all this time: (xx0)^2  (ict)^2 = 0. Here they substitute for x instead of transforming to the primed frame: (x'+vtx0)^2  (ict')^2.  ^  ^  It is still x ^ but see what they have accomplished by their mis/malfeasance: [x'+vtx0]=[x'+(vtx0)]=[x'(x0vt)]. =[x'x0'] The crackpots have been bragging about how you don't have to transform the circle center's coordinate to transform the circle center's coordinate. Bragging that what they were doing was not what they said they were doing. This does give us insight as to some of the crackpot variations on their x0'<>x0vt theme, which in all the variations will be discussed in later sections.. They are used to seeing the mixed coordinate form, (x'+vtx0) without realizing what it respresented, so  accompanied with a lack of understanding of the term 'dependent'  they are used to seeing just the one vt term, and not the one hidden in the defi nition of x' and are used to imagining it makes the whole expression time dependent and thus not invariant. About which, let x=10, let, x0=20, v=10, and t variously 10 and 23: (xx0)=10. Using their (x'+vtx0): For t=10, we have (x'+vtx0) = [ (1010*10) + (10*10)  (20) ] = 90 + 100  20 = 10 = (xx0) For t=23, we have (x'+vtx0) = [ (1010*23) + (10*23)  (20) ] = 220 + 230  20 = 10 = (xx0) The result depends in no way on the value of time; we showed the obvious for a couple of instances of t just so you can see that the crackpots not only do not understand the obvious logic of the algebra { (x'x0')=[ (vt)(x0vt) ]=(xx0) }  which shows that the transform has no possible time term effect  but they don't understand even a simple arithmetic demonstration of the facts. Oh. Their (x'+vtx0) or (x'+vt'x0) reduces the same way since t'=t: (xvt+vtx0)=(xx0). Their process, which says (x'+vt') is the transform of x, says that (x'+vt') is the moving system location of x, but it can't be because x is moving further in the negative direction from the moving viewpoint. That formula will only work out with v<0 which is indeed the velocity the primed system sees the other moving at. However, that formula cannot be derived from x'=xvt, the formula for transformation of the coordinates from the unprimed to the primed, User Contributions:Top Document: Invariant Galilean Transformations On All Laws Previous Document: 6. The data scale degradation absurdity. Next Document: 8. What does sci.math have to say about x0'=x0vt? 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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