Mathematical Inconsistencies in Einstein's Derivation of the Lorentz Transformation
in [Relativity]
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From: Dirk Van de moortel on 13 Sep 2005 15:12 "Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote in message news:dg768a017r0(a)drn.newsguy.com... > 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. If you look at it as like to a game, no argument on that one :-) But I was talking more generally. There's more retards than Androcles. And there's more people who try to help (or play with) them. We're not alone... Anyway, enjoy :-) Dirk Vdm
From: Thomas Smid on 14 Sep 2005 06:43 Daryl McCullough wrote: > 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. I have algebraically proved above that it is true and you haven't given any valid objection against it (as I pointed out already, your 'functional dependences' argument is algebraically completely irrelevant). You don't seem to realize that the original equations (1) x'=Ax+Bct (2) ct'=Bx+Act are inconsistent upon changing the signs of x and x' (i.e. when applying the equations to a light signal travelling in the opposite direction) because from (1) and (2) we have (3) x'+ct'=(A+B)(x+ct) , but by changing the signs of x and x' we have (4) -x'=-Ax+Bct (5) ct'=-Bx+Act and by multiplying (4) by -1 (6) x'=Ax-Bct (7)ct'=-Bx+Act , so from (6) and(7) (8) x'+ct'=(A-B)(x+ct) , which is inconsistent with (3) unless B=0 (or x=x'=0 and t=t'=0) Thomas
From: Daryl McCullough on 14 Sep 2005 07:09 Thomas Smid says... >I have algebraically proved above that it is true Every single one of your proofs starts off with an error. You can *see* that it is an error by checking your work. You claim to have proved that B=0 is the only solution to the equations 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) You can *see* that that is incorrect by trying the solution A=5/3 B=-4/3 D=-4/3 E=5/3. If you prove something that is false, then that is evidence that you made a mistake. I explicitly pointed out what your mistake was, but you don't need me to point it out in order for you to admit that you made a mistake: You proved that B=0 was the only solution. B=0 is *not* the only solution. Therefore, your proof is wrong. That is the simplest, and most airtight rule for mathematics. If you prove something that turns out to be false, then you've made a mistake. You proved something that turned out to be false. Therefore, you made a mistake. -- Daryl McCullough Ithaca, NY
From: Dirk Van de moortel on 14 Sep 2005 08:45 "Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote in message news:dg90d00371(a)drn.newsguy.com... > Thomas Smid says... > > >I have algebraically proved above that it is true > > Every single one of your proofs starts off with an > error. You can *see* that it is an error by checking > your work. You claim to have proved that B=0 is the > only solution to the equations > > 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) > > You can *see* that that is incorrect by trying > the solution A=5/3 B=-4/3 D=-4/3 E=5/3. If you > prove something that is false, then that is > evidence that you made a mistake. I explicitly > pointed out what your mistake was, but you > don't need me to point it out in order for you > to admit that you made a mistake: > > You proved that B=0 was the only solution. > B=0 is *not* the only solution. > Therefore, your proof is wrong. > > That is the simplest, and most airtight rule > for mathematics. If you prove something that > turns out to be false, then you've made a mistake. > You proved something that turned out to be false. > Therefore, you made a mistake. Actually that is debatable. Deliberately writing something completely wrong in order to annoy someone, isn't really making a mistake. It's more like lying and deceiving, don't you agree? But I do agree that the very act of thinking that he can get away with this kind of deceptive behaviour, can be called a mistake ;-) Dirk Vdm
From: Daryl McCullough on 14 Sep 2005 09:16
Dirk Van de moortel says... >Deliberately writing something completely wrong in order >to annoy someone, isn't really making a mistake. It's more >like lying and deceiving, don't you agree? I don't know about that explanation. It seems to give Thomas too much credit to assume that he *understands* his errors, but is making them intentionally in order to win some argument. I don't see any evidence of such understanding. But maybe that's part of his deviousness---that he carefully hides his own understanding? The question is: what possible motivation could someone have to lie about the Lorentz transformations? Are you thinking that some people (Thomas and maybe Androcles) actually know that Einstein's derivations were correct, but are trying to cast doubt on them anyway? Okay, I guess I have a hypothesis as to what may ge going on, that's almost the same as your dishonesty hypothesis, but doesn't credit them with understanding. Thomas and Androcles and the other relativity-bashers *sincerely* believe that relativity is wrong, and that there is something wrong with Einstein's derivation. However, they also know that they don't have the energy or mathematical ability to figure out exactly where the mistake is. But they reason: It doesn't matter exactly what the mistake is---if it's mistaken, then people shouldn't be using relativity. So, in the spirit of "The ends justify the means" they are using methods that they know are incorrect to try to convince people of a *correct* (to them, anyway) conclusion: that relativity is nonsense. If the conclusion is correct, who cares about the picky details? -- Daryl McCullough Ithaca, NY |