Mathematical Inconsistencies in Einstein's Derivation of the Lorentz Transformation
in [Relativity]
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From: Daryl McCullough on 13 Sep 2005 10:37 Thomas Smid says... Thomas, this is pretty silly. We have the following constraints: 1. forall x, forall t, x'(x,t) = A x + Bc t 2. forall x, forall t, ct'(x,t) = D x + Ec t 3. forall x, forall t, if x=ct, then x'(x,t) = c t'(x,t) 4. forall x, forall t, if x=-ct, then x'(x,t) = -c t'(x,t) Obviously, these constraints are not inconsistent, since they have at least one solution, A = 5/3, B = -4/3 D = -4/3 E = 5/3 If the equations are inconsistent, then there is no solution. Also, you claim that B=0. That is obviously *not* implied by equations 1-4, since we have a solution with B nonzero. So there is something wrong with your reasoning. If we have a solution with B nonzero, and you prove that the only solution is B=0, then obviously you made a mistake. B=0 is *not* the only solution. -- Daryl McCullough Ithaca, NY
From: Thomas Smid on 13 Sep 2005 12:40 Daryl McCullough wrote: > Thomas Smid says... > > Thomas, this is pretty silly. We have the following > constraints: > > 1. forall x, forall t, x'(x,t) = A x + Bc t > 2. forall x, forall t, ct'(x,t) = D x + Ec t > 3. forall x, forall t, if x=ct, then x'(x,t) = c t'(x,t) > 4. forall x, forall t, if x=-ct, then x'(x,t) = -c t'(x,t) > > Obviously, these constraints are not inconsistent, since > they have at least one solution, > > A = 5/3, > B = -4/3 > D = -4/3 > E = 5/3 > > If the equations are inconsistent, then there is no solution. > Also, you claim that B=0. That is obviously *not* implied by > equations 1-4, since we have a solution with B nonzero. So > there is something wrong with your reasoning. > > If we have a solution with B nonzero, and you prove that > the only solution is B=0, then obviously you made a mistake. > B=0 is *not* the only solution. Then insert A=5/3 and B=-4/3 into x1'=Ax1+Bct x2'=Ax2+Bct=-Ax1+Bct which gives x1'=5/3*x1-4/3*ct x2'=-5/3*x1-4/3*ct and thus x1'+x2'=-8/3*ct whereas it should be zero. Other combinations of the transformation equations may be consistent, but if a certain subset of them is not consistent, the whole set is obviously not consistent either. Thomas
From: Daryl McCullough on 13 Sep 2005 13:35 Thomas Smid says... > >Daryl McCullough wrote: >> 1. forall x, forall t, x'(x,t) = A x + Bc t >> 2. forall x, forall t, ct'(x,t) = D x + Ec t >> 3. forall x, forall t, if x=ct, then x'(x,t) = c t'(x,t) >> 4. forall x, forall t, if x=-ct, then x'(x,t) = -c t'(x,t) >> >> Obviously, these constraints are not inconsistent, since >> they have at least one solution, >> >> A = 5/3, >> B = -4/3 >> D = -4/3 >> E = 5/3 Can you please acknowledge that you understand that equations 1-4 have the solution A=E=5/3, B=D=-4/3? Plug it in. Check your work. Okay, now if you agree that B=-4/3 is a correct solution, then it follows that your proof that B=0 is an *incorrect* proof. B is *not* necessarily 0. >Then insert A=5/3 and B=-4/3 into > >x1'=Ax1+Bct >x2'=Ax2+Bct=-Ax1+Bct > >which gives >x1'=5/3*x1-4/3*ct >x2'=-5/3*x1-4/3*ct > >and thus > >x1'+x2'=-8/3*ct > >whereas it should be zero. No, it shouldn't be zero. Show how x1' + x2' = 0 follows from equations 1-4 above. It doesn't. You are making the same mistakes over and over and over. Look, we are considering two functions x'(x,t) and t'(x,t). You keep introducing new functions and then forgetting that they are functions. You want to introduce a new function x1(t) = ct. Fine. You want to introduce a new function x2(t) = -ct. Fine. Then you can define functions x2'(t) = x'(x1(t),t) = x'(ct,t) and x1'(t) = x'(x2(t),t) = x'(-ct,t). Why in the world do you think you are justified in saying x2'(t) = - x1'(t)? Show me which of my equations 1-4 imply that. Where you are getting confused is this: If e1 and e2 are two events such that x(e1) = ct(e1) x(e2) = -ct(e2) t'(e1) = t'(e2) then in that circumstance x'(e1) = -x'(e2) What this implies for our functions x2' and x1' is the following: x2'(t(e1)) = -x1'(t(e2)) But t(e1) is not equal to t(e2). You are getting confused because you are failing to keep track of the functional dependencies. You keep thinking that if t(e1) = t(e2) then t'(e1) = t'(e2) and vice-versa. That's not true. -- Daryl McCullough Ithaca, NY
From: Dirk Van de moortel on 13 Sep 2005 14:00 "Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote in message news:dg72lc0tca(a)drn.newsguy.com... > Thomas Smid says... > > > >Daryl McCullough wrote: > > >> 1. forall x, forall t, x'(x,t) = A x + Bc t > >> 2. forall x, forall t, ct'(x,t) = D x + Ec t > >> 3. forall x, forall t, if x=ct, then x'(x,t) = c t'(x,t) > >> 4. forall x, forall t, if x=-ct, then x'(x,t) = -c t'(x,t) > >> > >> Obviously, these constraints are not inconsistent, since > >> they have at least one solution, > >> > >> A = 5/3, > >> B = -4/3 > >> D = -4/3 > >> E = 5/3 > > Can you please acknowledge that you understand > that equations 1-4 have the solution A=E=5/3, B=D=-4/3? > > Plug it in. Check your work. > > Okay, now if you agree that B=-4/3 is a correct solution, > then it follows that your proof that B=0 is an *incorrect* > proof. B is *not* necessarily 0. > > >Then insert A=5/3 and B=-4/3 into > > > >x1'=Ax1+Bct > >x2'=Ax2+Bct=-Ax1+Bct > > > >which gives > >x1'=5/3*x1-4/3*ct > >x2'=-5/3*x1-4/3*ct > > > >and thus > > > >x1'+x2'=-8/3*ct > > > >whereas it should be zero. > > No, it shouldn't be zero. Show how x1' + x2' = 0 follows > from equations 1-4 above. It doesn't. > > You are making the same mistakes over and over and over. Deliberately. Until *you* give it up. Bet? Remember that I won this kind of bet *many* times before ;-) Dirk Vdm
From: Daryl McCullough on 13 Sep 2005 14:36
Dirk Van de moortel says... >> You are making the same mistakes over and over and over. > >Deliberately. >Until *you* give it up. Bet? > >Remember that I won this kind of bet *many* times before ;-) No, you haven't. Androcles gave up first. He plonked *me*. So I win. -- Daryl McCullough Ithaca, NY |