Top Document: Invariant Galilean Transformations On All Laws Previous Document: 4. The Encyclopedia Brittanica Incompetency. Next Document: 6. The data scale degradation absurdity. See reader questions & answers on this topic!  Help others by sharing your knowledge The traditional Gallilean transform is correct: t' = t x' = x  vt. But remember this: a transform of x doesn't effect just some values of x, but all of them, whether they are in the formula or not. This is important if you want to do things right. The crackpot position is strongly against this sci.math verified position, and the apparently standard coordinate pseudotransformation they suggest is perhaps the result. {See Table of Contents.] Let's use a simple equation: x^2 + y^2 = r^2, which is the formula for a circle with radius r, centered at a location where x=0. But what if the circle center isn't at x=0? Well, we'd want to use the form analytic geometry, vector algebra, and elementary measurement theory tells us to use, a form where we make explicit just where the circle center is, even if it is at x=x0=0: (xx0)^2 + (yy0)^2 = r^2. The circle center coordinate, x0, is an xaxis coordinate, just like all the xvalues of points on the circle. So, in proper generalized cartesian coordinate forms of laws/equations we want to transform every occurence of x and x0  by whatever name we call it: x.c, x_e, whatever. So, what is the transformed version of (xx0)? Why, (x'x0'); both x and x0 are xcoordinates, and every xcoordinate has a new value on the new axis. So, what is the value of (x'x0') in terms of the original x data? From the transform equations we see that x'=xvt, which is also true for x0'=x0vt: (x'x0')=[ (xvt)(x0vt) ]=(xx0). In other words, when we use the generalized coordinate form specified by analytic geometry, we find that the value of (x'x0') does not depend on either time or velocity in any way, shape, form, or fashion. Similarly for (yy0). We can treat time the same way if necessary: (tt0). The above is a proof that any equation in x,y,z,t is invariant under the galilean transforms. Just use the generalized coordinate form, with (xx0)/etc, in the transformation process, not the incompetently selected privileged form, with just x/etc. [The form is "privileged" because it assumes the circle center, point of emission, whatever, is at the origin of the axes instead at some less convenient point. After transform the coordinate(s) of the circle center/origin are also changed but the privileged form doesn't make this explicit and screws up the calculations, which should be based on (x'x0') but are calculated as (x'0).] The value of (x'x0') is the same as (xx0). That makes sense. Draw a circle on a piece of paper, maybe to the right side of the paper. On a transparent sheet, draw x and y coordinate axes, plus x to the right, plus y at the top. Place this axis sheet so the yaxis is at the left side of the circle sheet. Now answer two questions after noting the xcoordinate of the circle center and then moving the axis sheet to the right: (a) did the circle change in any way because you moved the axis sheet (ie because you transformed the coordin nate axis)? (b) did the coordinate of the circle center change? The circle didn't change [although SR will say it did]; that means that (x'x0') does indeed equal (xx0). The coordinate of the circle center did change, and it changed at the same rate (vt) as did every point on the circle. That means that x0'<>x0, and the fact the circle center didn't change wrt the circle, means that the relationship of x0' with x0 is the same as that of any x' on the circle with the corresponding x: x'=xvt; x0'=x0vt. This is to prepare you for the True Believer crackpots that say 'constant' coordinates can't be transformed; some even say they aren't coordinates. These crackpots include some that brag about how they were childhood geniuses, btw. QED: The galilean transformation for any law on generalized Cartesian coordinates is invariant under the Galilean transform. The use of the privileged form explains HOW the transformed equation can be messed up, the next Subject explains what the screwed up effect of the transform is, and how use of the generalized form corrects the screwup. User Contributions:Comment about this article, ask questions, or add new information about this topic:Top Document: Invariant Galilean Transformations On All Laws Previous Document: 4. The Encyclopedia Brittanica Incompetency. Next Document: 6. The data scale degradation absurdity. 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
