Mathematical Inconsistencies in Einstein's Derivation of the Lorentz Transformation
in [Relativity]
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From: Dirk Van de moortel on 10 Sep 2005 17:07 "shuba" <tim.shuba(a)lycos.ScPoAmM> wrote in message news:tim.shuba-95E2B6.14463310092005(a)corp.supernews.com... > Dirk Vdm wrote: > > [re: Thomas Smid] > > > Sometimes I think this guy is just playing a sordid little game. > > No one can be *that* stupid. > > Here's a link that ought to provoke your sense of humor. > > http://www.metaresearch.org/msgboard/topic.asp?TOPIC_ID=590 Yes, I had seen it. Even copied it for safekeeping. A fireworks chat between our friend and FlanderFace http://metaresearch.org/solar%20system/cydonia/proof_files/proof.asp Two idiots discussing their misunderstandings. Dirk Vdm
From: Thomas Smid on 11 Sep 2005 06:16 Daryl McCullough wrote: > Thomas Smid says... > > >It is not my mistake. > > Yes, it is certainly your mistake. > > >I am exactly applying the same rules > >you applied in your post #36 (by date) > >when you reduced your equation > > No, you are not. You are *mimicking* my derivation > without taking the care that I took to make sure that > I kept things straight. In particular, you keep forgetting > what's an independent variable and what's not, what's a > function of what. You don't keep of what is a constant > and what is a variable. > > If you thought that *I* was making that sort of mistake, > confusing independent and dependent variables, then you > should point it out in something I wrote. I did point it out to you above: Your original equations for the coefficients were x' - c t' = lambda (x - ct) + tau (x + ct) x' + c t' = mu (x + ct) + sigma (x - ct) Then you found that tau=0 under the provision that x' and x are positive and sigma=0 under the provision that x' and x are negative. But then you use x' - c t' = lambda (x - ct) x' + c t' = mu (x + ct) allowing x and x' to have the same sign in the upper and lower equation, whereas in fact you should have changed their sign in the lower equation, i.e. you should have x' - c t' = lambda (x - ct) x' - c t' = mu (x - ct) and hence lambda=mu or (since lambda=A-B and mu=A+B) B=0 (I obviously derived A=0 from your equations above as the latter had the signs not changed). (By the way, you could have arrived directly at this result by changing the conventions and have x1(t) and x2(t) both defined as positive i.e. x1(t)=ct and x2(t)=ct (and the same for the primed variables). Thomas
From: Thomas Smid on 11 Sep 2005 07:12 Daryl McCullough wrote: > Thomas Smid says... > > >It is not my mistake. > > Yes, it is certainly your mistake. > > >I am exactly applying the same rules > >you applied in your post #36 (by date) > >when you reduced your equation > > No, you are not. You are *mimicking* my derivation > without taking the care that I took to make sure that > I kept things straight. In particular, you keep forgetting > what's an independent variable and what's not, what's a > function of what. You don't keep of what is a constant > and what is a variable. > > If you thought that *I* was making that sort of mistake, > confusing independent and dependent variables, then you > should point it out in something I wrote. It doesn't > make a bit of sense for you to write something erroneous > and then blame its mistakes on me. (Unless you were just > *copying* my derivation.) > > You are being very sloppy, and your sloppiness is resulting > in an inconsistency. And then you are blaming the inconsistency > not on yourself but on me or Einstein, who were much > *more* careful than you, and who *didn't* produce an > inconsistency. > > The way to prove that someone made an error is by > redoing their derivation *more* carefully than they > did. You are repeating derivations in a way that's > much *less* careful. That doesn't prove anything. First of all, the issue of dependent variables is completely irrelevant here. I have pointed out before that the time dependences of x1(t), x2(t), x1'(t'), x2'(t') are understood and that I just do not write them here for convenience as they are obvious from the other equations. It does not make any difference at all for the algebra. Secondly, I did point out above the error you made in your original derivation: Your original equations for the coefficients were x' - c t' = lambda (x - ct) + tau (x + ct) x' + c t' = mu (x + ct) + sigma (x - ct) Then you found that tau=0 under the provision that x' and x are positive and sigma=0 under the provision that x' and x are negative. But then you use x' - c t' = lambda (x - ct) x' + c t' = mu (x + ct) allowing x and x' to have the same sign in the upper and lower equation, whereas in fact you should have changed their sign in the lower equation, i.e. you should have x' - c t' = lambda (x - ct) x' - c t' = mu (x - ct) and hence lambda=mu or (since lambda=A-B and mu=A+B) B=0 (I obviously derived A=0 from your equations above as the latter had the signs not changed). (By the way, you could have arrived directly at this result by changing the conventions and have x1(t) and x2(t) both defined as positive i.e. x1(t)=ct and x2(t)=ct (and the same for the primed variables). Thomas
From: Dirk Van de moortel on 11 Sep 2005 08:00 "Thomas Smid" <thomas.smid(a)gmail.com> wrote in message news:1126437123.716979.80550(a)z14g2000cwz.googlegroups.com... > Daryl McCullough wrote: > > Thomas Smid says... > > > > >It is not my mistake. > > > > Yes, it is certainly your mistake. > > > > >I am exactly applying the same rules > > >you applied in your post #36 (by date) > > >when you reduced your equation > > > > No, you are not. You are *mimicking* my derivation > > without taking the care that I took to make sure that > > I kept things straight. In particular, you keep forgetting > > what's an independent variable and what's not, what's a > > function of what. You don't keep of what is a constant > > and what is a variable. > > > > If you thought that *I* was making that sort of mistake, > > confusing independent and dependent variables, then you > > should point it out in something I wrote. It doesn't > > make a bit of sense for you to write something erroneous > > and then blame its mistakes on me. (Unless you were just > > *copying* my derivation.) > > > > You are being very sloppy, and your sloppiness is resulting > > in an inconsistency. And then you are blaming the inconsistency > > not on yourself but on me or Einstein, who were much > > *more* careful than you, and who *didn't* produce an > > inconsistency. > > > > The way to prove that someone made an error is by > > redoing their derivation *more* carefully than they > > did. You are repeating derivations in a way that's > > much *less* careful. That doesn't prove anything. > > First of all, the issue of dependent variables is completely irrelevant > here. > I have pointed out before that the time dependences of x1(t), x2(t), > x1'(t'), x2'(t') are understood and that I just do not write them here > for convenience as they are obvious from the other equations. It does > not make any difference at all for the algebra. > > > Secondly, I did point out above the error you made in your original > derivation: > > Your original equations for the coefficients were > > x' - c t' = lambda (x - ct) + tau (x + ct) > x' + c t' = mu (x + ct) + sigma (x - ct) > > Then you found that tau=0 under the provision that x' and x are > positive and sigma=0 under the provision that x' and x are negative. tau is not 0 "under the provision that x' and x are positive". tau is 0 under the provision that: for all the special events that take place on a light ray going in the positive x and x'-direction, the coordinate pairs (x,t) transform to pairs (x',t') according to the above equations that are valid for *all* possible events. sigma is not 0 "under the provision that x' and x are negative". sigma is 0 under the provision that: for all the special events that take place on a light ray going in the negative x and x'-direction, the coordinate pairs (x,t) transform to pairs (x',t') according to the above equations that are valid for *all* possible events. Did you ever have linear algebra in high school? Didn't they tell you how to find the coefficients of a linear transformation by using the known transformation of the coordinates of a few linearly independent points? You are fighting (and severly raping) linear algebra. Dirk Vdm
From: Daryl McCullough on 11 Sep 2005 11:01
Thomas Smid says... >Your original equations for the coefficients were > >x' - c t' = lambda (x - ct) + tau (x + ct) >x' + c t' = mu (x + ct) + sigma (x - ct) > >Then you found that tau=0 under the provision that x' and x are >positive and sigma=0 under the provision that x' and x are negative. tau and sigma are *constants*, independent of x and t. So if they can be proved to be 0 in the special cases x=ct and x=-ct, then they must be zero in *all* cases. Suppose A and B are constants, and I tell you 1. forall x, A*x = 2*B*x 2. forall x, if x is a prime number, then A*x = 0. Do you come to the conclusion that A=0 only when x is prime? Let's make everything more explicit. 1. For all events e, x'(e) - c t'(e) = lambda (x(e) - ct(e)) + tau (x(e) + ct(e)) 2. For all events e, x'(e) + c t'(e) = mu (x(e) + ct(e)) + sigma (x(e) - ct(e)) 3. For all events e such that x(e) = c t(e), then x'(e) = c t'(e). 4. For all events e such that x(e) = - c t(e), then x'(e) = -c t'(e). Putting 3 and 1 together, we get For all events e such that x(e) = ct(e), then x'(e) = ct'(e) (from 3) and x'(e) - c t'(e) = lambda (x(e) - ct(e)) + tau (x(e) + ct(e)) (from 1) which simplifies to For all events e such that x(e) = ct(e), then tau = 0. But since tau is a *constant*, independent of e, then if tau = 0 in a particular case, it is *always* 0. Similarly, we prove that sigma is always 0. Therefore, 1 and 2 simplify to 1' forall events e, (x'(e)-ct'(e)) = lambda (x(e)-ct(e)) 2' forall events e, (x'(e)+ct'(e)) = mu (x(e)+ct(e)) Which step do you disagree with? -- Daryl McCullough Ithaca, NY |