Mathematical Inconsistencies in Einstein's Derivation of the Lorentz Transformation
in [Relativity]
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From: Daryl McCullough on 11 Sep 2005 11:03 Thomas Smid says... >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. And I pointed out to you that tau and sigma are *constants*. If they are equal to 0 in some particular case, then they are *always* equal to zero. So, no, you have not pointed out any mistakes. -- Daryl McCullough Ithaca, NY
From: Thomas Smid on 11 Sep 2005 14:54 Daryl McCullough wrote: > 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? Your example is inappropriate as the prime numbers are contained in all numbers, but positive and negative numbers are mutually exclusive. Your example should read in fact like this: For all x positive A*x=2*B*x For all x negative A*x=0 So can you conclude that B=0? Not as far as I am concerned (it's not about the mere appearance of constants here but about the validity of equations). > > 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)) If all the equations always apply, I should also have been entitled to make the transformation ct'(x2,t)=ct'(-x1,t)=-(B+A)x1 above. Not allowing me to do this means that you are applying double standards here. But why don't you make life easier for yourself and avoid negative x-coordinates in the first place (as I suggested above already). If you split x into x1 and x2 there is no point anyway having x2 negative. As the sign of the measuring unit is only a convention, you can just take x1(t)=ct and x2(t)=ct (imagine you have two rulers going into opposite directions, one with a scale entitled x1 and one with x2). Thomas
From: Daryl McCullough on 11 Sep 2005 16:11 Thomas Smid says... >Your example should read in fact like this: > >For all x positive A*x=2*B*x >For all x negative A*x=0 > >So can you conclude that B=0? Of course you can. >Not as far as I am concerned Well, I don't know what to say, except that this is elementary algebra. If you're having trouble with it, it isn't Einstein's fault. Why don't you tell me what value of the constants A and B can make both of these statements true: For all x positive A*x=2*B*x For all x negative A*x=0 other than the choice A=0, B=0? -- Daryl McCullough Ithaca, NY
From: Dirk Van de moortel on 11 Sep 2005 16:39 "Thomas Smid" <thomas.smid(a)gmail.com> wrote in message news:1126464875.087754.309620(a)o13g2000cwo.googlegroups.com... > Daryl McCullough wrote: > > 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? > > Your example is inappropriate as the prime numbers are contained in all > numbers, but positive and negative numbers are mutually exclusive. > > Your example should read in fact like this: > > For all x positive A*x=2*B*x > For all x negative A*x=0 > > So can you conclude that B=0? Not as far as I am concerned (it's not > about the mere appearance of constants here but about the validity of > equations). For all x negative A*x=0 ==> A*(-1) = 0 (taking x = -1) ==> A = 0 For all x positive A*x=2*B*x ==> For all x positive 0=2*B*x (since A = 0) ==> 0 = 2*B*1 (taking x = 1) ==> B = 0 Dirk Vdm
From: Thomas Smid on 11 Sep 2005 16:41
Daryl McCullough wrote: > Thomas Smid says... > > >Your example should read in fact like this: > > > >For all x positive A*x=2*B*x > >For all x negative A*x=0 > > > >So can you conclude that B=0? > > Of course you can. > > >Not as far as I am concerned > > Well, I don't know what to say, except that > this is elementary algebra. If you're having > trouble with it, it isn't Einstein's fault. > > Why don't you tell me what value of the constants > A and B can make both of these statements true: > > For all x positive A*x=2*B*x > For all x negative A*x=0 > > other than the choice A=0, B=0? > Sorry, I had already deleted my corresponding post as I intended to re-edit it. I wanted to write originally 'So can you conclude that B=0 *if x<0*?'. In this case the first equation is simply not defined and B could have any value. Thomas |