Mathematical Inconsistencies in Einstein's Derivation of the Lorentz Transformation
in [Relativity]
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From: Bilge on 5 Sep 2005 04:54 Mike: >If you were just a bit more careful reading my post, I do not claim any >of this stuff. Dr. ALL did claim it. What you did was try to set up a strawman.
From: Thomas Smid on 5 Sep 2005 16:33 Daryl McCullough wrote: > Thomas Smid says... > > I made a mistake in the equations relating A,B,D,E to > lambda,tau,mu,sigma. The correct equations are: > > lambda = 1/2(A-B-D+E) > tau = 1/2(A+B-D-E) > mu = 1/2(A+B+D+E) > sigma = 1/2(A-B+D-E) > > For reference, these constants are used as follows > > x' = A x + B ct > t' = D x + E ct > > x' - c t' = lambda (x-ct) + tau (x+ct) > x' + c t' = mu (x+ct) + sigma (x-ct) > > I'm claiming that tau = 0 and sigma = 0, > > [derivation deleted] > > >(7) A+B=D+E > >(8) A-B=E-D. > > Correct. Notice the first equation implies A+B-D-E=0, so tau=0. > The second equation implies A-B-E+D=0, so sigma=0. > > >If you add and subtract (7) and (8), you get therefore > > > >(9) A=E; B=D. > > Correct. > > >Now you have found previously that your constants tau and sigma are > >zero, so you from your defintions of tau and sigma: > > > >(10) A-D-B+E=0 > >(11) A+D-B-E=0 > > My mistake. Using the correct definitions of sigma and tau > show that > > tau = 1/2(A+B-D-E) > sigma = 1/2(A-B+D-E) > > So sigma=0 and tau=0 imply > > (10) A-D+B-E=0 > (11) A+D-B-E=0 > > Which again has the solution A=E, B=D. OK, with A=E and B=D we have then (1) lambda=A-B (2) mu=A+B This still leaves A and B to be determined. For this consider the resultant sets of equations for the two light signals travelling in opposite directions (3a) x(e1)=ct(e1) (3b) x'(e1)-c t'(e1) = (A-B)(x(e1)-ct(e1)) (= 0) (3c) x'(e1)+c t'(e1) = (A+B)(x(e1)+ct(e1)) and (4a) x(e2)=-ct(e2) (4b) x'(e2)-c t'(e2) = (A-B)(x(e2)-ct(e2)) (4c) x'(e2)+c t'(e2) = (A+B)(x(e2)+ct(e2)) (= 0) Now consider both wavefronts at the same time in each reference frame i.e. for t(e2)=t(e1) and t'(e2)=t'(e1). From Eqs.(3c) and (4b) (and using (3a),(3b),(4a),(4c)) one gets therefore the respective equations (5) ct'(e1)=(A+B)*ct(e1) (6) ct'(e1)=(A-B)*ct(e1). which obviously can only be true if (7) B=0. Now in order to determine A we have to make an additional assumption as both the systems of equations (3) and (4) are under-determined (we have 4 unknowns (A,B,x',t') but only 3 equations each). So as a further requirement we have to relate the independent variables (t and t') to each other. We assume therefore for simplity that we have identical and synchronized clocks in each system, i.e. (8) t'=t and with this we have thus from (5) and (7) (9) A=1. The correct transformation therefore simply is (10) x'=x ct'=ct. Thomas
From: Todd on 5 Sep 2005 17:27 "Thomas Smid" <thomas.smid(a)gmail.com> wrote in message news:1125952408.242594.268300(a)g43g2000cwa.googlegroups.com... > Now consider both wavefronts at the same time in each reference frame > i.e. for t(e2)=t(e1) and t'(e2)=t'(e1). If e1 and e2 are simultaneous in the unprimed frame, then you can't assume that they will also be simultaneous in the primed frame. Todd
From: Daryl McCullough on 5 Sep 2005 20:07 Thomas Smid says... >OK, with A=E and B=D we have then Wait a minute. Are you now in agreement that you were wrong about your *original* reason for saying that Einstein's derivation is inconsistent? You seem to be moving onto a completely different argument. We've discovered algebraic mistakes that you've made, and algebraic mistakes that I've made, but we have not yet found an algebraic mistake that Einstein made. Let's acknowledge at this point that you have not discovered any algebra mistakes in Einstein's derivation. >(1) lambda=A-B >(2) mu=A+B > >This still leaves A and B to be determined. > >For this consider the resultant sets of equations for the two light >signals travelling in opposite directions > >(3a) x(e1)=ct(e1) >(3b) x'(e1)-c t'(e1) = (A-B)(x(e1)-ct(e1)) (= 0) >(3c) x'(e1)+c t'(e1) = (A+B)(x(e1)+ct(e1)) > >and > >(4a) x(e2)=-ct(e2) >(4b) x'(e2)-c t'(e2) = (A-B)(x(e2)-ct(e2)) >(4c) x'(e2)+c t'(e2) = (A+B)(x(e2)+ct(e2)) (= 0) >Now consider both wavefronts at the same time in each reference frame >i.e. for t(e2)=t(e1) and t'(e2)=t'(e1). If t(e2)=t(e1), then you don't get t'(e2) = t'(e1). You get ct'(e2) = D x(e2) + E ct(e2) = -ct(e2) D + E ct(e2) = (A-B) ct(e2) ct'(e1) = D x(e1) + E ct(e1) = ct(e1) D + E ct(e1) = (A+B) ct(e1) So, if t(e1) = t(e2), we get t'(e1) = (A+B)/(A-B) t'(e2) So unless B=0, we don't get t'(e1) = t'(e2). -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 5 Sep 2005 20:09
Thomas Smid says... >Now consider both wavefronts at the same time in each reference frame >i.e. for t(e2)=t(e1) and t'(e2)=t'(e1). If you pick e1 and e2 such that t(e2) = t(e1), then t'(e2) will in general *not* be equal to t'(e1). -- Daryl McCullough Ithaca, NY |