Mathematical Inconsistencies in Einstein's Derivation of the Lorentz Transformation
in [Relativity]
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From: Daryl McCullough on 7 Sep 2005 17:04 Thomas Smid says... >> >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). But there are other criteria to consider. If someone is travelling at speed v relative to the (x,t) coordinate system, then he is at *rest* relative to the (x',t') coordinate system. This implies that for any event e such that x(e) = v t(e), x'(e) = 0. Going back to x' = A x + B t this implies that B = - vA Choosing B=0 is inconsistent. >> >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). You were the one who replaced x1 by -x1. Since x1 = ct, that is impossible, unless you are replacing t by -t. So don't blame me. [Aside: But actually, saying negative times are not defined is incorrect. The parameter t measures the time since some particular reference point. For example, in years, 2005 means 2005 years *after* the birth of Jesus. The year -600 means 600 years *before* the birth of Jesus. Negative times are just as meaningful as positive times.] Anyway, what you have are these: x'(x,t) + ct'(x,t) = (A+B) (x+ct) x'(x,t) - ct'(x,t) = (A-B) (x-ct) If you specialize the first equation to the case x=ct, and the second equation to the case x=-ct, then these become: 2ct'(ct,t) = (A+B) 2ct - 2ct'(-ct,t) = -(A-B) 2ct or t'(ct,t) = (A+B) t t'(-ct,t) = (A-B) t -- Daryl McCullough Ithaca, NY
From: Bilge on 8 Sep 2005 04:05 Mike: I'm going to skip your attempt to avoid the fallacies of your argument. Do you or do you not think the coordinate transformations for spatial rotations is valid? Is that a yes, no, or does the answer require a lot fallacious conditions to fit some ideology you like? >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? Not unless your thought processes are contagious.
From: Bilge on 8 Sep 2005 04:08 Perspicacious: >Igor wrote: >> Congratulations! You've just discovered that the average speed of two >> light rays moving in opposite directions vanishes. What this has to do >> with inconsistencies in the Lorentz transformation, which you didn't >> even get to, I have no idea. My first reaction was that this had to be >> a big joke, since no one could be that stupid. But I could be wrong. > >The greatest riddle, with as yet an undiscovered rationale, is why so >many kooks are drawn to special relativity. Since you're a kook and you believe you are the world's foremost authority on the subject, you should have a lot of insight into the answer. >I believe the answer is >that it special relativity is taught primarily by physicists who >strive for sensationalism and not by mathematicians who prefer >clarity and logical consistency. > >http://www.everythingimportant.org/relativity/special.pdf >
From: Thomas Smid on 8 Sep 2005 11:49 Daryl McCullough wrote: > Thomas Smid says... > > >> >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). > > But there are other criteria to consider. If someone is travelling > at speed v relative to the (x,t) coordinate system, then he is at > *rest* relative to the (x',t') coordinate system. Only if the two coordinate systems are also travelling with speed v relatively to each other. >This implies that > > for any event e such that x(e) = v t(e), > x'(e) = 0. > > Going back to > > x' = A x + B t > > this implies that > > B = - vA It should be x'=Ax+Bct i.e. B=-vA/c actually. > > Choosing B=0 is inconsistent. Not surprisingly because your new conditions x=vt x'=0 are inconsistent with the old ones x=ct x'=ct' It is with the latter conditions only that B=0 (or A=0) is consistent. > > >> >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). > > You were the one who replaced x1 by -x1. Since x1 = ct, that > is impossible, unless you are replacing t by -t. I replaced x2 by -x1. We have x1=ct and x2=-ct. x1 is always positive, x2 is always negative, hence t must in both cases always be positive. Thomas
From: Daryl McCullough on 8 Sep 2005 13:22
Thomas Smid says... > >Daryl McCullough wrote: >> Choosing B=0 is inconsistent. > >Not surprisingly because your new conditions >x=vt >x'=0 >are inconsistent with the old ones >x=ct >x'=ct' >It is with the latter conditions only that B=0 (or A=0) is consistent. I thought I explained to you already. These equations are for *different* events. The assumptions are these: 1. For all events e, if x(e) = 0, then x'(e) = -v t'(e). (The spatial origin of the (x,t) system is moving at velocity -v as measured in the (x',t') system.) 2. For all events e, if x'(e) = 0, then x(e) = v t(e). (The spatial origin of the (x',t') system is moving at velocity +v as measured in the (x,t) system.) 3. For all events e, if x(e) = c t(e), then x'(e) = c t'(e). (Anything that travels at the speed of light in one system also travels at the speed of light in the other system.) 4. For all events e, if x(e) = - c t(e), then x'(e) = - c t'(e). (Same as 3, but for the opposite direction.) To complete the transformation, we actually need one more fact, which is a symmetry condition between the two systems: 5. For all pairs of events e1 and e2 such that x(e1) = 0 and x'(e2) = 0 and t(e1) = t'(e2), then t'(e1) = t(e2). >> You were the one who replaced x1 by -x1. Since x1 = ct, that >> is impossible, unless you are replacing t by -t. > >I replaced x2 by -x1. We have x1=ct and x2=-ct. x1 is always positive, >x2 is always negative, hence t must in both cases always be positive. Either way, you made an error. The equations we had originally were x'(x,t) + c t'(x,t) = (A+B)(x + c t) x'(x,t) - c t'(x,t) = (A-B)(x - c t) In the special case in which x=ct, the first equation becomes x'(ct,t) + c t'(ct,t) = (A+B) 2ct Using x' = c t', this simplifies to (10) t'(ct,t) = (A+B) t In the special case in which x=-ct, the second equation becomes x'(-ct,t) - c t'(-ct,t) = -(A-B) 2ct Using x' = -c t', this simplifies to (11) t'(-ct,t) = (A-B) t These two equations don't imply B=0. -- Daryl McCullough Ithaca, NY |