Mathematical Inconsistencies in Einstein's Derivation of the Lorentz Transformation
in [Relativity]
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From: Thomas Smid on 9 Sep 2005 13:18 Daryl McCullough wrote: > 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). I suggest we postpone the discussion of this point until we have sorted out the problem with the sign in the transformation equations (see below). > > >> 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. No they imply A=0 because t'(-x,t)=-t'(x,t). Let's make it again clearer by considering the variables for both light signals . The transformation equation for ct for both is (1) ct'(x1,t)=Bx1+Act (2) ct'(x2,t)=Bx2+Act and additionally we have the conditions (3) x1=ct (4) x2=-ct (5) x1'(x1,t)=ct'(x1,t) (6) x2'(x2,t)=-ct'(x2,t). These equations must hold together because otherwise it would not be justified to have simultaneously tau and sigma =0 when you derived the equations involving lambda and mu (which taught us that E=A and D=B in your original transformation). So let's just try to solve (1)-(6) purely mathematically without worrying what the terms actually means (i.e. we don't need to bother about the signs of t or t' at this stage): >From Eq.(1) and (3) we have (7) ct'(x1,t)=(B+A)x1 and from Eq.(2),(4),(3) (8) ct'(x2,t)=(B-A)x2=(A-B)x1. However from (3),(4),(7) it follows also that (9) ct'(x2,t)=ct'(-x1,t)=-(B+A)x1. By comparing (8) and (9) we find that (10) A=0, which means that (7) and (8) are now (11) ct'(x1,t)=Bx1 (12) ct'(x2,t)=-Bx1 and thus with (5) and (6) (11) x1'(x1,t)=Bx1 (12) x2'(x2,t)=Bx1. Now this is obviously a contradiction to our assumption that both signals travel in opposite directions in both frames and shows thus that the equations have no solution unless t=0 and x=0 (i.e. exactly the same situation as in Einstein's derivation addressed by me in the opening post of this thread.) Thomas
From: Thomas Smid on 9 Sep 2005 13:42 Daryl McCullough wrote: > 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). I suggest we postpone this point until we have sorted out the problem with the original transformation equations (see below). > > >> 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. No they imply A=0 because t'(-x,t)=-t'(x,t). Let's make it again clearer by considering the variables for both light signals . The transformation equation for ct for both is (1) ct'(x1,t)=Bx1+Act (2) ct'(x2,t)=Bx2+Act and additionally we have the conditions (3) x1=ct (4) x2=-ct (5) x1'(x1,t)=ct'(x1,t) (6) x2'(x2,t)=-ct'(x2,t). These equations must hold together because otherwise it would not be justified to have simultaneously tau and sigma =0 when you derived the equations involving lambda and mu (which taught us that E=A and D=B in your original transformation). So let's just try to solve (1)-(6) purely mathematically without worrying what the terms actually means (i.e. we don't need to bother about the signs of t or t' at this stage): >From Eq.(1) and (3) we have (7) ct'(x1,t)=(B+A)x1 and from Eq.(2),(4),(3) (8) ct'(x2,t)=(B-A)x2=(A-B)x1. However from (3),(4),(7) it follows also that (9) ct'(x2,t)=ct'(-x1,t)=-(B+A)x1. By comparing (8) and (9) we find that (10) A=0, which means that (7) and (8) are now (11) ct'(x1,t)=Bx1 (12) ct'(x2,t)=-Bx1 and thus with (5) and (6) (11) x1'(x1,t)=Bx1 (12) x2'(x2,t)=Bx1. Now this is obviously a contradiction to our assumption that both signals travel in opposite directions in both frames and shows thus that the equations have no solution unless t=0 and x=0 (i.e. exactly the same situation as in Einstein's derivation addressed by me in the opening post of this thread.) Thomas
From: Daryl McCullough on 9 Sep 2005 16:06 Thomas Smid says... > >Daryl McCullough wrote: >> (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. > >No they imply A=0 because t'(-x,t)=-t'(x,t). No, that's not true. >The transformation equation for ct for both is > >(1) ct'(x1,t)=Bx1+Act >(2) ct'(x2,t)=Bx2+Act > > >and additionally we have the conditions > >(3) x1=ct >(4) x2=-ct >(5) x1'(x1,t)=ct'(x1,t) >(6) x2'(x2,t)=-ct'(x2,t). >>From Eq.(1) and (3) we have > >(7) ct'(x1,t)=(B+A)x1 Note: (7) is not true in *general*. It is only true in the special case in which x1 = ct. Under those circumstances, x1 is *not* an independent variable, but is is determined by the value of t. So you should write it as (7) ct'(x1(t),t) = (B+A)x1(t) Actually, it is silly to even *have* x1, since it in this case just means ct. You should really write (7) as (7') ct'(ct,t) = (B+A)ct >and from Eq.(2),(4),(3) > >(8) ct'(x2,t)=(B-A)x2=(A-B)x1. Once again, (8) is not true in general, it is only true in the special circumstances x1=ct x2=-ct. So it really should be written as ct'(x2(t),t) = (B-A)x2(t) = (A-B)x1(t) or even better: (8') ct'(-ct,t) = (B-A)(-ct) = (A-B)ct >However from (3),(4),(7) it follows also that > >(9) ct'(x2,t)=ct'(-x1,t)=-(B+A)x1. No, no, no. We've already been through this. The equation (7) is not valid *except* in the case x1 = ct. x1 is not an independent variable in (7), it's just an obscure way of writing ct. You cannot replace x1 by -x1 in (7) This argument at this point has nothing to do with Einstein. It is just algebraic manipulation, and you keep making mistakes that have nothing to do with Einstein. -- Daryl McCullough Ithaca, NY
From: Thomas Smid on 10 Sep 2005 10:32 Daryl McCullough wrote: > Thomas Smid says... > > > >Daryl McCullough wrote: > > >> (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. > > > >No they imply A=0 because t'(-x,t)=-t'(x,t). > > No, that's not true. > > >The transformation equation for ct for both is > > > >(1) ct'(x1,t)=Bx1+Act > >(2) ct'(x2,t)=Bx2+Act > > > > > >and additionally we have the conditions > > > >(3) x1=ct > >(4) x2=-ct > >(5) x1'(x1,t)=ct'(x1,t) > >(6) x2'(x2,t)=-ct'(x2,t). > > >>From Eq.(1) and (3) we have > > > >(7) ct'(x1,t)=(B+A)x1 > > Note: (7) is not true in *general*. It is only > true in the special case in which x1 = ct. Under > those circumstances, x1 is *not* an independent > variable, but is is determined by the value of t. > So you should write it as > > (7) ct'(x1(t),t) = (B+A)x1(t) > > Actually, it is silly to even *have* x1, since it > in this case just means ct. You should really write > (7) as > > (7') ct'(ct,t) = (B+A)ct > > >and from Eq.(2),(4),(3) > > > >(8) ct'(x2,t)=(B-A)x2=(A-B)x1. > > Once again, (8) is not true in general, it is > only true in the special circumstances x1=ct > x2=-ct. So it really should be written as > > ct'(x2(t),t) = (B-A)x2(t) = (A-B)x1(t) > > or even better: > > (8') ct'(-ct,t) = (B-A)(-ct) = (A-B)ct > > >However from (3),(4),(7) it follows also that > > > >(9) ct'(x2,t)=ct'(-x1,t)=-(B+A)x1. > > No, no, no. We've already been through this. > The equation (7) is not valid *except* in the > case x1 = ct. x1 is not an independent variable > in (7), it's just an obscure way of writing ct. > You cannot replace x1 by -x1 in (7) > > This argument at this point has nothing to do with > Einstein. It is just algebraic manipulation, and you > keep making mistakes that have nothing to do with > Einstein. It is not my mistake. I am exactly applying the same rules you applied in your post #36 (by date) when you reduced your equations x' - c t' = lambda (x - ct) + tau (x + ct) x' + c t' = mu (x + ct) + sigma (x - ct) to x' - c t' = lambda (x - ct) x' + c t' = mu (x + ct) . My equations (1)-(6) are fully consistent with this and you are merely interpreting something into them which is not there and which contradicts your own procedure. Thomas
From: Daryl McCullough on 10 Sep 2005 14:09
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. -- Daryl McCullough Ithaca, NY |