Mathematical Inconsistencies in Einstein's Derivation of the Lorentz Transformation
in [Relativity]
|
From: Thomas Smid on 4 Sep 2005 05:31 Daryl McCullough wrote: > 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) > OK, but you should note that fully written your equations (3) and (4) read (1) x'(e1) - c t'(e1) = lambda (x(e1) - ct(e1)) (2) x'(e2) + c t'(e2) = mu (x(e2) + ct(e2)) so you can't use them to determine lambda and mu (and hence the Lorentz transformation) like Einstein did. But you can determine them in fact from your original equations (3) x'(e1) = A x(e1) + B ct(e1) ct'(e1) = D x(e1) + E ct(e1) (4) x'(e2) = A x(e2) + B ct(e2) ct'(e2) = D x(e2) + E ct(e2) which using your conditions (5) x(e1)=ct(e1); x'(e1)=ct'(e1) (6) x(e2)=-ct(e2); x'(e2)=-ct'(e2) yields (7) A+B=D+E (8) A-B=E-D. If you add and subtract (7) and (8), you get therefore (9) A=E; B=D. 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 from which you get (12) A-B=D-E (13) A-B=E-D. Because the left hand sides of (12) and (13) are identical but the right hand side merely has the sign inverted, this means that (14) A=B; D=E. Now together with (9) you have therefore (15) A=B=D=E, which, going back to the definitions of your constants, tells you that not only tau and sigma but also lambda is zero. The only constant left to determine is mu , but from (2) and (6) we know that (16) 0=mu*0, so mu (and hence the numerical value of (15)) is arbitrary, which makes sense because the original equations (3) and (4) read now (17) x'(e1) = A x(e1) + A ct(e1) ct'(e1) = A x(e1) + A ct(e1) (18) x'(e2) = A x(e2) + A ct(e2) ct'(e2) = A x(e2) + A ct(e2) which using (5) and (6) yields (19) x'(e1)-ct'(e1)=0 (20) x'(e2)+ct'(e2)=0, which are just your separate equations for the two light signals all over again. This shows that a consistent set of equations (like yours) can not result in the Lorentz transformation. Thomas
From: Dirk Van de moortel on 4 Sep 2005 05:41 "Thomas Smid" <thomas.smid(a)gmail.com> wrote in message news:1125826285.035521.262840(a)o13g2000cwo.googlegroups.com... > Daryl McCullough wrote: > > 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: > > [snip pains-staking explanation] > > OK, but you should note that fully written your equations (3) and (4) > read > > (1) x'(e1) - c t'(e1) = lambda (x(e1) - ct(e1)) > (2) x'(e2) + c t'(e2) = mu (x(e2) + ct(e2)) > > > so you can't use them to determine lambda and mu (and hence the Lorentz > transformation) like Einstein did. Einstein assumed that his readers would master the basics of linear algebra. You clearly have no clue. [snip pathetic demonstration of ignorance] > This shows that a consistent set of equations (like yours) can not > result in the Lorentz transformation. Actually, it perfectly demonstrates that you haven't got a clue about linear algebra. Doesn't that hurt? Honestly? Dirk Vdm
From: Mike on 4 Sep 2005 12:00 If you were just a bit more careful reading my post, I do not claim any of this stuff. Dr. ALL did claim it. Mike
From: Daryl McCullough on 4 Sep 2005 11:46 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. -- Daryl McCullough Ithaca, NY
From: Todd on 4 Sep 2005 12:22
----- Original Message ----- From: "Thomas Smid" <thomas.smid(a)gmail.com> Newsgroups: sci.physics.relativity Sent: Sunday, September 04, 2005 4:31 AM Subject: Re: Mathematical Inconsistencies in Einstein's Derivation of the Lorentz Transformation > Daryl McCullough wrote: >> Thomas Smid says... >> > >> >Daryl McCullough wrote: >> >> >> Einstein's equations were these >> >> >> >> (3) x' - ct' = lambda (x-ct) >> >> (4) x' + ct' = mu (x+ct) >> > [snip] > OK, but you should note that fully written your equations (3) and (4) > read > > (1) x'(e1) - c t'(e1) = lambda (x(e1) - ct(e1)) > (2) x'(e2) + c t'(e2) = mu (x(e2) + ct(e2)) > > > so you can't use them to determine lambda and mu (and hence the Lorentz > transformation) like Einstein did. Einstein assumes that equation (3) holds for _any_ event, not just for those events for which x-ct = 0. Please note where Einstein says that (3) ''is fulfilled in general'', which just means that it is assumed to hold for _arbitrary_ events. Likewise (4) is assumed to hold for _any_ event. Therefore, you may apply (3) and (4) to the _same_ event and add and subtract them as Einstein did. What may be confusing you is that Einstein does not make any effort to show why (3) and (4) hold for _arbitrary_ events. David McCullough has been kind enough to do that for you using the underlying assumption that the _general_ transformation equations are linear. [snip] Todd |