LTE
in [Relativity]
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From: Jack Campin - bogus address on 17 Jul 2010 18:57 > Does anyone out there know how velocity addition works to describe > how we measure it besides a math? What is the physical reason? Whatever you're talking about in that subliterate gibberish has nothing to do with logic. ----------------------------------------------------------------------------- e m a i l : j a c k @ c a m p i n . m e . u k Jack Campin, 11 Third Street, Newtongrange, Midlothian EH22 4PU, Scotland mobile: 07800 739 557 <http://www.campin.me.uk> Twitter: JackCampin
From: Tom Roberts on 18 Jul 2010 11:31 Jonathan Doolin wrote: > I was wondering about something related. Since if you take the unit > circle, x=cos(theta) y=sin(theta), the arc length of the unit circle > is the same as the angle in radians. > > Is the arc-length of the unit hyperbola x=cosh(theta), y= sinh(theta) > also equal to its angle? It seems like it ought to be, but I can't > figure out the path-integral. > > I just wondered, because then you can say that the rapidity change is > equal to the arc-length of the unit hyperbola, just the same as > rotation angle is the arc-length of the unit circle. I'm not interested enough to work it out. But I point out that a key part of doing that integral in Euclidean space is the identity sin^2(x) + cos^2(x) = 1 There is a related identity cosh^2(x) - sinh^2(x) = 1 The minus sign is precisely what is needed, because of the minus sign in the metric components. Tom Roberts
From: Jonathan Doolin on 18 Jul 2010 14:51 On Jul 18, 10:31 am, Tom Roberts <tjroberts...(a)sbcglobal.net> wrote: > Jonathan Doolin wrote: > > I was wondering about something related. Since if you take the unit > > circle, x=cos(theta) y=sin(theta), the arc length of the unit circle > > is the same as the angle in radians. > > > Is the arc-length of the unit hyperbola x=cosh(theta), y= sinh(theta) > > also equal to its angle? It seems like it ought to be, but I can't > > figure out the path-integral. > > > I just wondered, because then you can say that the rapidity change is > > equal to the arc-length of the unit hyperbola, just the same as > > rotation angle is the arc-length of the unit circle. > > I'm not interested enough to work it out. But I point out that a key part of > doing that integral in Euclidean space is the identity > sin^2(x) + cos^2(x) = 1 > > There is a related identity > cosh^2(x) - sinh^2(x) = 1 > > The minus sign is precisely what is needed, because of the minus sign in the > metric components. > > Tom Roberts Thanks for looking. I reposted the question at sci.math, but I think I'm wrong, anyway. Rapidity is not linear with the arc-length of the unit hyperbola, after all. path-length = int (cosh^2(t) + sinh^2(t)) = int (sqrt(cosh 2t)) dt ~ int e^t when t is large which is definitely not equal to t.
From: harald on 18 Jul 2010 18:16 On Jul 17, 12:39 am, xxein <xxxx...(a)gmail.com> wrote: > On Jul 16, 2:53 pm, glird <gl...(a)aol.com> wrote: > > > It is said that the Lorentz Transformation Equations (LTE) have been > > experimentally confirmed by many different experiments. I would > > appreciate it if someone would provide a list of all the experiments > > that did so. > > > glird > > xxein: Don't worry about it. Experiments are measuremental > observations and affects. They are put into a math form without > regard to really understanding the physic that caused them to be > observed and effect in this way. > > Oh wow! There is a velocity addition formula. Does it explain what > is really happening or is it just a math wysiwyg? > > The Lorentz transormations are fine, but when you math-shortcut them > to SR (and its postulates), you strip the essense of the physic out of > it. I told you this before (in some fashion) but you decided that it > was too much for you to try to understand. > > And now an appeal. Does anyone out there know how velocity addition > works to describe how we measure it besides a math? What is the > physical reason? I know what it is but I doubt that anyone else does. > > Geez! Doesn't anybody know how to think logically of the physic > beyond the archaic sceintific method? Yes of course - there are at least several people who contribute to this group and who understand this very well. For sure Lorentz and Poincare understood it! The later generation of "geometers" lost grip with reality but physical modeling is certainly part of science. Harald
From: glird on 19 Jul 2010 16:49
On Jul 18, 6:16 pm, harald <h...(a)swissonline.ch> wrote: > On Jul 17, 12:39 am, xxein <xxxx...(a)gmail.com> wrote: > > > Geez! Doesn't anybody know how to think logically of the physic > > beyond the archaic sceintific method? > >< Yes of course - there are at least several people who contribute to this group and who understand this very well. For sure Lorentz and Poincare understood it! The later generation of "geometers" lost grip with reality but physical modeling [i.e. METAPHYSICS] is certainly part of science. > Metaphysics, which is the study of the things that exist and how their mechanisms physically work, is not only "part of science" it was and -- despite being ignored by today's physicists -- remains the most important part! glird |