Mathematical Inconsistencies in Einstein's Derivation of the Lorentz Transformation
in [Relativity]
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From: Daryl McCullough on 2 Sep 2005 14:07 Thomas Smid says... > >Daryl McCullough wrote: >> Einstein's equations were these >> >> (3) x' - ct' = lambda (x-ct) >> (4) x' + ct' = mu (x+ct) > >So how did he get then to (4) in your opinion? (Hint: he got to (3) >using his equations (1) and (2)) Yes, he used (1) and (2). Here's a more pains-taking explanation: For any event e, let x(e), t(e) be the location and time of e in the first frame, and let x'(e) and t'(e) be the location and time as measured in the other frame. We assume that these coordinates are linearly related: There is some parameters A,B,D,E that are functions of the relative velocity between the two frames such that for all events e x'(e) = A x(e) + B ct(e) ct'(e) = D x(e) + E ct(e) Now, these two equations can be rearranged into the equivalent equations (I'm not going to write the dependence on e, to simplify the appearance, but actually, x,t,x' and t' all depend on which event e you are talking about) (0.1) x' - c t' = lambda (x - ct) + tau (x + ct) (0.2) x' + c t' = mu (x + ct) + sigma (x - ct) where lambda, tau, mu, and sigma are linear combinations of A, B, D, and E: lambda = 1/2 (A-D+B-E) tau = 1/2 (A-D-B+E) mu = 1/2 (A+D+B+E) sigma = 1/2 (A+D-B-E) Okay, so what Einstein is arguing by considering light signals is that tau = 0 and sigma = 0. Why does that follow? Well, consider the following events: Let e0 be the event with coordinates x(e0) = 0, t(e0) = 0. Let a light signal travelling in the +x direction be sent from event e0 to some event e1. This event will have x(e1) > 0, t(e1) > 0. Because light travels at speed c, we know, in the first frame: x(e1) = c * t(e1) or (1) x(e1) - c t(e1) = 0 But light *also* travels at speed c in the second frame. So we have: x'(e1) = c * t'(e1) or (2) x'(e1) - c t(e1) = 0 By my equation (0.1) above, we know x'(e1) - c t'(e1) = lambda (x(e1) - c t(e1)) + tau (x(e1) + c t(e1)) Using (1) and (2) to simplify this, we get: 0 = 0 + tau (x(e1) + c t(e1)) Since x(e1) and t(e1) are both positive, it follows that tau = 0 Putting this together with my equation (0.1) gives (3) x' - c t' = lambda (x - ct) Now, we go through the same sort of thing for a light signal travelling in the -x direction: Let a light signal travelling in the -x direction be sent from event e0 to some event e2. This event will have x(e2) < 0, t(e2) > 0. Since light travels at speed c in both frames, we have (1') x(e2) + ct(e2) = 0 and similarly (2') x'(e2) + ct'(e2) = 0 My equation 0.2 gives: x'(e2) + c t'(e2) = mu (x(e2) + ct(e2)) + sigma (x(e2) - ct(e2)) Using (1') and (2') to simplify equation (0.2) gives: 0 = 0 + sigma (x(e2) - ct(e2)) or sigma (x(e2) - ct(e2)) = 0 Since x(e2) is negative, and so is -ct(e2), it follows that this is only possible if sigma = 0 Substituting this into my equation 0.2 gives (4) x' + c t' = mu (x + ct) -- Daryl McCullough Ithaca, NY
From: Perspicacious on 2 Sep 2005 16:13 Logically speaking, the Lorentz transformation may be derived with incomprehensible magic far above what is considered rational by mediocre physicists. See http://www.everythingimportant.org/relativity/special.pdf for example. Why do you insist that Einstein should have derived the Lorentz transformation correctly? Physicists aren't required to derive anything rigorously. The legitimacy or illegitimacy of Einstein's derivation has nothing to do with the question of consistency of special relativity. SR is only an interpretation of the Lorentz transformation.
From: Dirk Van de moortel on 2 Sep 2005 16:42 "Perspicacious" <iperspicacious(a)yahoo.com> wrote in message news:1125691994.385246.36950(a)g14g2000cwa.googlegroups.com... > Logically speaking, the Lorentz transformation may > be derived with incomprehensible magic far above what > is considered rational by mediocre physicists. See > http://www.everythingimportant.org/relativity/special.pdf where Eugene Shubert hits the Windshield :-) > for example. Why do you insist that Einstein should have > derived the Lorentz transformation correctly? Physicists > aren't required to derive anything rigorously. > > The legitimacy or illegitimacy of Einstein's derivation > has nothing to do with the question of consistency of > special relativity. SR is only an interpretation of the > Lorentz transformation. Careful with that axe, Eugene. Dirk Vdm
From: Bill Hobba on 2 Sep 2005 18:32 "Thomas Smid" <thomas.smid(a)gmail.com> wrote in message news:1125667659.658642.57840(a)g44g2000cwa.googlegroups.com... > Todd wrote: > >> As Bill Hobba says, you must think about what the symbols denote. It >> might >> help to write (1) and (3) as >> >> (1) x1 - ct = 0 >> >> (3) x2 + ct = 0 >> >> where x1 is the position of the light pulse that's traveling in the >> positive >> x direction and x2 is the position of the other pulse traveling in the >> negative x direction. Note that x1 never equals x2 except at time t = 0. >> >> When you subtract them you get an equation that may be written as >> >> x1 - x2 = 2ct >> >> This just says that the distance between the pulses is increasing at the >> rate of 2c, which makes sense. > > Yes, it would make sense if x2=-x1 i.e. 2x1=2ct, but evidently > Einstein's derivation would then not 'work' anymore as it relies on > x1=x2=x i.e. 2ct=0. > > Thomas It does not matter how you cut and dice it solving equations simultaneously gives you what happens when both hold true. For a light ray in the x direction and one in the opposite direction the only time they both hold true is at time t = 0 which occurs at the origin. And guess what, when you solve them simultaneously you get t=0 and hence x = 0. As usual you show an appalling understanding of basic math - but that is correctable with the right attitude. You problem is you sucking attitude - or as Dirk says 'Sucking Logic, Sucking Algebra, Sucking Attitude, Sucking Thumbs, the deadly combination of IGNORANCE and ARROGANCE, topped with a coulis of SELF-RESPECT as from now known under the name ARROGNORANCE'. When ending up with contradictory answers most people examine there premises and sort out what is going on. Your trouble, and the trouble with a lot of cranks around here, is they assume they are infallible and something must be wrong with the subject that they are the only ones smart enough to see. Such an attitude has 'Unskilled and Unaware of It: How Difficulties in Recognizing One's Own Incompetence Lead to Inflated Self-Assessments' written all over it http://www.phule.net/mirrors/unskilled-and-unaware.html Bill
From: Jon Bell on 3 Sep 2005 01:33
In article <1125652015.288928.309540(a)z14g2000cwz.googlegroups.com>, Thomas Smid <thomas.smid(a)gmail.com> wrote: > >It is only necessary here to examine the initial equations for this, >which describe the 'equations of motion of a light signal' in the >unprimed and primed reference frames, i.e. > >(1) x-ct=0 >(2) x'-ct'=0 >where c is the speed of light (which obviously has to be a constant >0) > >In the same way, the propagation of a signal in the opposite direction >yields >(3) x+ct=0 >(4) x'+ct'=0 >(note that these equations are not written explicitly in Einstein's >derivation). > >>From equations (1)-(4), the Lorentz transformation is then derived by >some algebraic manipulations. > >But are the above equations mathematically consistent at all? Let's >subtract equation (1) from (3), which yields >(5) 2ct=0 >which means that for any time t>0 >(6) c=0, >in contradiction to the requirement that c>0. Is there any reason why I should not apply your argument to a slightly different situation, as follows? It is only necessary here to examine the initial equations for this, which describe the 'equations of motion of an automobile' in the unprimed and primed reference frames, i.e. (1) x-vt=0 [snip equation (2) because it's irrelevant to your argument] where v is the speed of the automobile, which must be >0 In the same way, the propagation of the automobile in the opposite direction yields (3) x+vt=0 [snip equation (4) because it's irrelevant to your argument] But are the above equations mathematically consistent at all? Let's subtract equation (1) from (3), which yields (5) 2vt=0 which means that for any time t>0 (6) v=0, which means that the automobile must be at rest. -- Jon Bell <jtbell(a)presby.edu> Presbyterian College Dept. of Physics and Computer Science Clinton, South Carolina USA |