Mathematical Inconsistencies in Einstein's Derivation of the Lorentz Transformation
in [Relativity]
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From: Dirk Van de moortel on 11 Sep 2005 16:42 "Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote in message news:dg2319010b4(a)drn.newsguy.com... > 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? This is the point where he normally stops responding, just like he turned his back on http://groups.google.co.uk/group/sci.physics.relativity/msg/e0f827bae96207f4 Dirk Vdm
From: Dirk Van de moortel on 11 Sep 2005 16:43 "Thomas Smid" <thomas.smid(a)gmail.com> wrote in message news:1126471309.701446.277250(a)g43g2000cwa.googlegroups.com... > 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. You *really* haven't got a clue???? Smid, you are MEGA-stupid :-)) Dirk Vdm
From: Thomas Smid on 11 Sep 2005 17:08 Daryl McCullough wrote: > > 1' forall events e, (x'(e)-ct'(e)) = lambda (x(e)-ct(e)) OK let's choose then the 'event' x(e)=-ct(e) This yields -2ct'(e)=-lambda*2ct(e) > 2' forall events e, (x'(e)+ct'(e)) = mu (x(e)+ct(e)) OK let's choose then the 'event' x(e)=ct(e) This yields 2ct'(e)=mu*2ct(e) and hence lambda=mu. Also, if all the equations apply in any case, I should be entitled to make the transformation ct'(x2,t)=ct'(-x1,t)=-(B+A)x1. Not allowing me to do this means that you are applying double standards here. But anyway, 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: Dirk Van de moortel on 11 Sep 2005 17:30 "Thomas Smid" <thomas.smid(a)gmail.com> wrote in message news:1126472896.590382.127180(a)f14g2000cwb.googlegroups.com... > Daryl McCullough wrote: > > > > > 1' forall events e, (x'(e)-ct'(e)) = lambda (x(e)-ct(e)) > > OK let's choose then the 'event' x(e)=-ct(e) > This yields > -2ct'(e)=-lambda*2ct(e) Yes, but only for events that also satisfy x(e) = -c t(e) > > > 2' forall events e, (x'(e)+ct'(e)) = mu (x(e)+ct(e)) > > OK let's choose then the 'event' x(e)=ct(e) > This yields > 2ct'(e)=mu*2ct(e) Yes, but only for events that also satisfy x(e) = c t(e) > > and hence lambda=mu. No, stupid. If you want to combine the system of two equations { -2ct'(e)=-lambda*2ct(e) { 2ct'(e)=mu*2ct(e) you can do that only for events that satisfy { x(e) = -c t(e) { x(e) = +c t(e) in other words, for events that satisfy { x(e) = 0 { t(e) = 0 so the system of equations is { -2*0 = - lambda*2*0 { 2c*0 = mu*2c*0 or in other words { 0 = 0 { 0 = 0 from which you cannot deduce that lambda = mu Are you Marcel Luttgens in disguise? Dirk Vdm
From: Todd on 11 Sep 2005 17:31
"Thomas Smid" <thomas.smid(a)gmail.com> wrote in message news:1126472896.590382.127180(a)f14g2000cwb.googlegroups.com... > Daryl McCullough wrote: > >> >> 1' forall events e, (x'(e)-ct'(e)) = lambda (x(e)-ct(e)) > > OK let's choose then the 'event' x(e)=-ct(e) > This yields > -2ct'(e)=-lambda*2ct(e) > >> 2' forall events e, (x'(e)+ct'(e)) = mu (x(e)+ct(e)) > > OK let's choose then the 'event' x(e)=ct(e) > This yields > 2ct'(e)=mu*2ct(e) > > and hence lambda=mu. > No, you are talking about two *different* events here. So, you should let e1 be the event where x(e1)=-ct(e1) and let e2 be the different event where x(e2)=ct(e2). You then have 2ct'(e1)=lambda*2ct(e1) and 2ct'(e2)=mu*2ct(e2) Since event e1 is not the same event as e2, you cannot conclude that lambda = mu. Todd |