Mathematical Inconsistencies in Einstein's Derivation of the Lorentz Transformation
in [Relativity]
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From: Mike on 7 Sep 2005 12:04 Bilge wrote: > Mike: > >If you were just a bit more careful reading my post, I do not claim any > >of this stuff. Dr. ALL did claim it. > > What you did was try to set up a strawman. I would have done that if I assumed: Bilge <---> Strawman But since I did not, you claiming I tried to set up a strawman is a strawman. Are you an idiot by the way? Mike
From: Thomas Smid on 7 Sep 2005 12:48 Daryl McCullough wrote: > Thomas Smid says... > > >I am sorry, but you still misinterprete the variables as specific > >coordinates. x1 and x2 are *not* coordinates but scalar variables for > >the two directions representing the vector variable x. Fully written it > >would be x1(t)=ct, x2(t)=-ct etc. I merely left out the argument (t) in > >my equations. > > That doesn't change my objections to your derivation. You wrote > > (3c) x1'+c t' = (A+B)(x1+ct) > (4b) x2'-c t' = (A-B)(x2-ct) > > The point is that x' and t' are *functions* of x and t. >.... >.... >.... > > Let's put in all the functional dependence explicitly. The > transformation equations are these > > x'(x,t) + c t'(x,t) = (A+B) (x + ct) > x'(x,t) - c t'(x,t) = (A-B) (x - ct) You *assumed* they are functions of x and t when writing down your transformation x'=Ax+Bct ct'=Bx+Act but this actually contradicts the use of t and t' as independent variables in x(t)=ct and x'(t')=ct'. As 'independent' means just this, you are actually not entitled to use t or t' with an argument like t'(x,t) (after all, you need some independent variables in any physical theory as a basis to work on, and it would be unphysical to assume that clocks are affected by the location of a light signal or by other clocks). But OK, let's just for the time being hypothetically assume that t' is not an independent variable. >Since x2(t) > is unequal to x1(t), the corresponding value of t' in (4b) > is *not* equal to the corresponding value of t' in (3c). > > Let's put in all the functional dependence explicitly. The > transformation equations are these > > x'(x,t) + c t'(x,t) = (A+B) (x + ct) > x'(x,t) - c t'(x,t) = (A-B) (x - ct) > > Now, you want to specialize to the case x1(t) = ct in > the first equation, and specialize to the case x2(t) = -ct > in the second equation. Fine. That gives us: > > x'(ct,t) + c t'(ct,t) = (A+B) (2ct) > x'(-ct,t) - c t'(-ct,t) = (A-B) (-2ct) > > Now, we use the fact that x' = ct' in the top > equation, and x' = -ct' in the bottome equation to get: > > > 2 c t'(ct,t) = (A+B) (2ct) > -2 c t'(-ct,t) = (A-B) (-2ct) > > Your mistake was assuming that t'(ct,t) = t'(-ct,t). Why should it be a mistake? There is nothing in our previous equations that would have prevented me from making this assumption, which, as mentioned above merely implies that t and t' are independent variables. But OK let's again consider the problem formally on the basis of the suggested transformation equations. For this, I use x1 and x2 again in order to avoid the x'=ct', x'=-ct' inconsistency. So the relevant equations are then (the time dependences x1(t), x2(t) and x1'(t'), x2'(t') are not explicity written) (1) x1'(x1,t) + c t'(x1,t) = (A+B) (x1 + ct) (2) x2'(x2,t) - c t'(x2,t) = (A-B) (x2 - ct) and additionally we have (3) x1'(x1,t) - c t'(x1,t) = 0 (4) x2'(x2,t) + c t'(x2,t) = 0 (5) x1=ct (6) x2=-ct Now by virtue of (3)-(6), (1) and (2) become (7) 2ct'(x1,t)=(A+B)2x1 (8) 2ct'(-x1,t)=(A-B)2x1 The question is how ct'(-x1,t) relates to ct'(x1,t) . For this we consider the original transformation for ct' (9) ct'(x1,t) = Bx1+Act which by virtue of Eq.(5) becomes (10) ct'(x1,t) = (B+A)x1 which obviously means that (11) ct'(-x1,t) = -(B+A)x1 =-ct'(x1,t) or in other words (from Eqs.(3) and (4)) (12) x2'(x2,t)=-x1'(x1,t). Now with (11), Eqs.(7) and (8) become (13) 2ct'(x1,t)=(A+B)2x1 (14) 2ct'(x1,t)=-(A-B)2x1 =(B-A)2x1 and thus (15) A=0. I have suggested B=0 before rather than A=0, but it does not make any difference at all for the transformation as x'=Ax+Bct ct'=Bx +Act yields either x'=x; ct'=ct or x'=ct; ct'=x (provided the remaining constant is set to 1) which is exactly identical because of the condition x=ct). Thomas
From: Daryl McCullough on 7 Sep 2005 13:43 Thomas Smid says... >Daryl McCullough wrote: >> The point is that x' and t' are *functions* of x and t. >>.... >>.... >>.... >> >> Let's put in all the functional dependence explicitly. The >> transformation equations are these >> >> x'(x,t) + c t'(x,t) = (A+B) (x + ct) >> x'(x,t) - c t'(x,t) = (A-B) (x - ct) > >You *assumed* they are functions of x and t when writing down your >transformation > >x'=Ax+Bct >ct'=Bx+Act > >but this actually contradicts the use of t and t' as independent >variables in x(t)=ct and x'(t')=ct'. t' is not independent of x. t' is a *function* of x and t. You cannot choose x, t, x' and t' independently. You can choose *two* independently, but then the other two are determined. >As 'independent' means just this, t' is not independent of x and t. >you are actually not entitled to use t or t' with an argument like >t'(x,t) Yes, you certainly are. >But OK, let's just for the time being hypothetically assume that t' is >not an independent variable. > >>Since x2(t) >> is unequal to x1(t), the corresponding value of t' in (4b) >> is *not* equal to the corresponding value of t' in (3c). >> >> Let's put in all the functional dependence explicitly. The >> transformation equations are these >> >> x'(x,t) + c t'(x,t) = (A+B) (x + ct) >> x'(x,t) - c t'(x,t) = (A-B) (x - ct) >> >> Now, you want to specialize to the case x1(t) = ct in >> the first equation, and specialize to the case x2(t) = -ct >> in the second equation. Fine. That gives us: >> >> x'(ct,t) + c t'(ct,t) = (A+B) (2ct) >> x'(-ct,t) - c t'(-ct,t) = (A-B) (-2ct) >> >> Now, we use the fact that x' = ct' in the top >> equation, and x' = -ct' in the bottome equation to get: >> >> >> 2 c t'(ct,t) = (A+B) (2ct) >> -2 c t'(-ct,t) = (A-B) (-2ct) >> >> Your mistake was assuming that t'(ct,t) = t'(-ct,t). > >Why should it be a mistake? You are trying to show that Einstein's derivation led to an inconsistency. If you make an additional assumption that Einstein *didn't* make, then any inconsistency you derive is an inconsistency in the combination of *your* assumptions and *Einstein's* assumptions. It doesn't prove anything about Einstein's derivation. >There is nothing in our previous equations >that would have prevented me from making this assumption On the contrary, it leads to a contradiction. >which, as mentioned above merely implies that t and t' >are independent variables. They aren't independent variables. Given x and t, t' is determined. > (1) x1'(x1,t) + c t'(x1,t) = (A+B) (x1 + ct) > (2) x2'(x2,t) - c t'(x2,t) = (A-B) (x2 - ct) > >and additionally we have > >(3) x1'(x1,t) - c t'(x1,t) = 0 >(4) x2'(x2,t) + c t'(x2,t) = 0 >(5) x1=ct >(6) x2=-ct > >Now by virtue of (3)-(6), (1) and (2) become > >(7) 2ct'(x1,t)=(A+B)2x1 >(8) 2ct'(-x1,t)=(A-B)2x1 > >The question is how ct'(-x1,t) relates to ct'(x1,t). For this we >consider the original transformation for ct' > >(9) ct'(x1,t) = Bx1+Act > >which by virtue of Eq.(5) becomes > >(10) ct'(x1,t) = (B+A)x1 > >which obviously means that > >(11) ct'(-x1,t) = -(B+A)x1 =-ct'(x1,t) Thomas, you forget that x1 in equation 10 is *not* an independent variable, it is a *function* of t. In particular, x1 = ct, so (10) should be written as follows: (10) ct'(ct,t) = (B+A)ct Now, if you replace t by -t, you get the correct version of (11) (11) ct'(-ct,-t) = -(B+A)ct = -ct'(ct,t) [Rest deleted, since it relies on a false step] -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 7 Sep 2005 14:00 Thomas Smid says... [deleted] I just want to point out that, in spite of the title of this thread, you haven't actually discovered any mathematical inconsistencies in Einstein's derivation of the Lorentz Transformations. -- Daryl McCullough Ithaca, NY
From: Thomas Smid on 7 Sep 2005 15:47
Daryl McCullough wrote: > Thomas Smid says... > > >Daryl McCullough wrote: > > >> The point is that x' and t' are *functions* of x and t. > >>.... > >>.... > >>.... > >> > >> Let's put in all the functional dependence explicitly. The > >> transformation equations are these > >> > >> x'(x,t) + c t'(x,t) = (A+B) (x + ct) > >> x'(x,t) - c t'(x,t) = (A-B) (x - ct) > > > >You *assumed* they are functions of x and t when writing down your > >transformation > > > >x'=Ax+Bct > >ct'=Bx+Act > > > >but this actually contradicts the use of t and t' as independent > >variables in x(t)=ct and x'(t')=ct'. > > t' is not independent of x. t' is a *function* of x and t. You > cannot choose x, t, x' and t' independently. You can choose *two* > independently, but then the other two are determined. > > >As 'independent' means just this, > > t' is not independent of x and t. > > >you are actually not entitled to use t or t' with an argument like > >t'(x,t) > > Yes, you certainly are. > > >But OK, let's just for the time being hypothetically assume that t' is > >not an independent variable. > > > >>Since x2(t) > >> is unequal to x1(t), the corresponding value of t' in (4b) > >> is *not* equal to the corresponding value of t' in (3c). > >> > >> Let's put in all the functional dependence explicitly. The > >> transformation equations are these > >> > >> x'(x,t) + c t'(x,t) = (A+B) (x + ct) > >> x'(x,t) - c t'(x,t) = (A-B) (x - ct) > >> > >> Now, you want to specialize to the case x1(t) = ct in > >> the first equation, and specialize to the case x2(t) = -ct > >> in the second equation. Fine. That gives us: > >> > >> x'(ct,t) + c t'(ct,t) = (A+B) (2ct) > >> x'(-ct,t) - c t'(-ct,t) = (A-B) (-2ct) > >> > >> Now, we use the fact that x' = ct' in the top > >> equation, and x' = -ct' in the bottome equation to get: > >> > >> > >> 2 c t'(ct,t) = (A+B) (2ct) > >> -2 c t'(-ct,t) = (A-B) (-2ct) > >> > >> Your mistake was assuming that t'(ct,t) = t'(-ct,t). > > > >Why should it be a mistake? > > You are trying to show that Einstein's derivation led > to an inconsistency. If you make an additional assumption > that Einstein *didn't* make, then any inconsistency you > derive is an inconsistency in the combination of *your* > assumptions and *Einstein's* assumptions. It doesn't > prove anything about Einstein's derivation. > > >There is nothing in our previous equations > >that would have prevented me from making this assumption > > On the contrary, it leads to a contradiction. Which would be what? Setting B=0 and A=1 (or vice versa) is fully consistent with the assumptions x=ct and x'=ct' (which are the only relevant equations regarding the propagation of light). > > >which, as mentioned above merely implies that t and t' > >are independent variables. > > They aren't independent variables. Given x and t, t' is > determined. > > > (1) x1'(x1,t) + c t'(x1,t) = (A+B) (x1 + ct) > > (2) x2'(x2,t) - c t'(x2,t) = (A-B) (x2 - ct) > > > >and additionally we have > > > >(3) x1'(x1,t) - c t'(x1,t) = 0 > >(4) x2'(x2,t) + c t'(x2,t) = 0 > >(5) x1=ct > >(6) x2=-ct > > > >Now by virtue of (3)-(6), (1) and (2) become > > > >(7) 2ct'(x1,t)=(A+B)2x1 > >(8) 2ct'(-x1,t)=(A-B)2x1 > > > >The question is how ct'(-x1,t) relates to ct'(x1,t). For this we > >consider the original transformation for ct' > > > >(9) ct'(x1,t) = Bx1+Act > > > >which by virtue of Eq.(5) becomes > > > >(10) ct'(x1,t) = (B+A)x1 > > > >which obviously means that > > > >(11) ct'(-x1,t) = -(B+A)x1 =-ct'(x1,t) > > Thomas, you forget that x1 in equation 10 is *not* > an independent variable, it is a *function* of t. > In particular, x1 = ct, so (10) should be written > as follows: > > (10) ct'(ct,t) = (B+A)ct > > Now, if you replace t by -t, you get the correct > version of (11) > > (11) ct'(-ct,-t) = -(B+A)ct = -ct'(ct,t) > > [Rest deleted, since it relies on a false step] I am sorry, but I don't think you are right . We can't change the sign of the explicit t-dependence here (after all negative times are not defined). The only change of sign comes when we switch from x2 to x1 when formulating equation (8) i.e. ct'(x2,t)=ct'(-x1,t) (according to (5) and (6), which fully written imply x2(t)=-x1(t) but not x2(t)=-x1(-t)). Thomas |