Mathematical Inconsistencies in Einstein's Derivation of the Lorentz Transformation
in [Relativity]
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From: Thomas Smid on 12 Sep 2005 14:43 Daryl McCullough schrieb: > Thomas Smid says... > > > >Daryl McCullough wrote: > > >> 1. forall e, x'(e) = A x(e) + Bc t(e) > >> 2. forall e, ct'(e) = D x(e) + Ec t(e) > >> 3. forall e, if x(e) = ct(e), then x'(e) = ct'(e) > >> 4. forall e, if x(e) = -ct(e), then x'(e) = -ct'(e) > >> > >> imply that A=0 or B=0. Obviously, that's incorrect, because > >> one solution to these 4 equations is this: A = 5/3, B = -4/3 > >> > >> x' = 5/3 x - 4/3 ct > >> ct' = -4/3 x + 5/3 ct > > >You haven't actually solved your equations (1)-(4) as they contain D > >and E which you replaced by A and B (and it is here where the > >contradiction occurs (as I mentioned before)) > > I'm sorry. I thought it was clear what D and E were, but to be > explicit: The 4 equations above have a solution > > A=5/3 > > B=-4/3 > > D=-4/3 > > E=5/3 > > So it is incorrect to say that the 4 equations imply A=0 or B=0. > There is no contradiction. > > >> >(5) ct'(x1,t)=(B+A)x1 > >> > >> But that equation is *not* true for all x and t. It is only > >> true when x = ct. > >> > >> >(6) ct'(x2,t)=(B-A)x2 > >> > >> But that equation is *not* true for all x and t. It is only > >> true when x = -ct. > > > >These are conditions you interprete into the validity of the variables > >which aren't fixed mathematically anywhere. > > I don't know what you are talking about. > > As I said, these are > the four conditions we have talked about so far: > > >> 1. forall e, x'(e) = A x(e) + Bc t(e) > >> 2. forall e, ct'(e) = D x(e) + Ec t(e) > >> 3. forall e, if x(e) = ct(e), then x'(e) = ct'(e) > >> 4. forall e, if x(e) = -ct(e), then x'(e) = -ct'(e) > > Equations 1 and 2 say "forall e". Equations 3 and 4 > have conditions. You are ignoring these conditions. > > >There is only one function > >t'(x,t) defined here > > If you don't like events, then the same information can > be conveyed in terms of x and t: > > 1. forall x, forall t, x'(x,t) = A x + B c t > 2. forall x, forall t, t'(x,t) = D x + E c t > 3. forall x, forall t, if x=ct, then x'(x,t) = c t'(x,t) > 4. forall x, forall t, if x=-ct, then x'(x,t) = -c t'(x,t) > > >i.e. if you are serious about this issue you > >should also split t into t1 and t2, but I wonder how far you will be > >getting this way in the derivation of the Lorentz transformation. > > > >Also you still haven't answered my question why you are not changing > >the sign convention and let x1=ct and x2=ct. > > You were the one who said that x2 was -ct. Now you want to say > that x2 = +ct. That doesn't make any sense. On the contrary, it makes perfect sense if you have separate variables for the two directions. You only need negative numbers if you have one variable. Our calendar is for instance the best example for this: you have one scale named AD and another named BC, both measured positive. Obviously you could do the same for a spatial scale as well. If you want, you can define the location of the light signal by a spherical coordinate e.g. by r(phi,t) (where phi is a corresponding angle and thus r always positive). I am sure you are not suggesting that the derivation of the Lorentz transformation depends on the use of a Cartesian coordinate system. Thomas
From: Daryl McCullough on 12 Sep 2005 15:00 Thomas Smid says... > >Daryl McCullough wrote: >> >> 1. forall e, x'(e) = A x(e) + Bc t(e) >> >> 2. forall e, ct'(e) = D x(e) + Ec t(e) >> >> 3. forall e, if x(e) = ct(e), then x'(e) = ct'(e) >> >> 4. forall e, if x(e) = -ct(e), then x'(e) = -ct'(e) >> The 4 equations above have a solution >> >> A=5/3 >> >> B=-4/3 >> >> D=-4/3 >> >> E=5/3 >No there isn't a contradiction as you restricted again the validity of >the equations: > >if you have two arbitrary events e1 and e2 , the transformation >equations are > >x'(e1) = A x(e1) + Bc t(e1) >ct'(e1)= B x(e1) + Ac t(e1) > >x'(e2) = A x(e2) + Bc t(e2) >ct'(e2)= B x(e2) + Ac t(e2) > >These equations must hold for all events e1 and e2 simultaneously, >whether x(e1) and x(e2) are positive or negative, and thus also for >further events at -x(e1) and -x(e2), What does "further events at -x(e1) and -x(e2)" mean? I think that maybe what you mean is this: let e3 and e4 be events such that x(e3) = -x(e1) x(e4) = -x(e2) I'll assume that's what you mean. >so we can actually double up the equations to > >(1a) x'(e1) = A x(e1) + Bc t(e1) >(1b) ct'(e1)= B x(e1) + Ac t(e1) > >(2a) -x'(e1) = -A x(e1) + Bc t(e1) That doesn't make any sense. If you want to consider some event e3 such that x(e3) = - x(e1), and t(e3) = t(e1), then you have (2a) x'(e3) = A x(e3) + Bc t(e3) = - A x(e1) + B c t(e1) Note: the left side is in terms of e3, not e1. You seem to be assuming that x'(e3) = - x'(e1) but there is no reason to assume that. >> You were the one who said that x2 was -ct. Now you want to say >> that x2 = +ct. That doesn't make any sense. > >On the contrary, it makes perfect sense if you have separate variables >for the two directions. That doesn't make any sense. I have no idea what you mean. -- Daryl McCullough Ithaca, NY
From: Thomas Smid on 13 Sep 2005 07:47 Daryl McCullough wrote: > Thomas Smid says... > > > >Daryl McCullough wrote: > > >> >> 1. forall e, x'(e) = A x(e) + Bc t(e) > >> >> 2. forall e, ct'(e) = D x(e) + Ec t(e) > >> >> 3. forall e, if x(e) = ct(e), then x'(e) = ct'(e) > >> >> 4. forall e, if x(e) = -ct(e), then x'(e) = -ct'(e) > > >> The 4 equations above have a solution > >> > >> A=5/3 > >> > >> B=-4/3 > >> > >> D=-4/3 > >> > >> E=5/3 > > >No there isn't a contradiction as you restricted again the validity of > >the equations: > > > >if you have two arbitrary events e1 and e2 , the transformation > >equations are > > > >x'(e1) = A x(e1) + Bc t(e1) > >ct'(e1)= B x(e1) + Ac t(e1) > > > >x'(e2) = A x(e2) + Bc t(e2) > >ct'(e2)= B x(e2) + Ac t(e2) > > > >These equations must hold for all events e1 and e2 simultaneously, > >whether x(e1) and x(e2) are positive or negative, and thus also for > >further events at -x(e1) and -x(e2), > > What does "further events at -x(e1) and -x(e2)" mean? > > I think that maybe what you mean is this: let e3 and e4 > be events such that > > x(e3) = -x(e1) > x(e4) = -x(e2) > > I'll assume that's what you mean. > > >so we can actually double up the equations to > > > >(1a) x'(e1) = A x(e1) + Bc t(e1) > >(1b) ct'(e1)= B x(e1) + Ac t(e1) > > > >(2a) -x'(e1) = -A x(e1) + Bc t(e1) > > That doesn't make any sense. If you want to consider > some event e3 such that x(e3) = - x(e1), > and t(e3) = t(e1), then you have > > (2a) x'(e3) = A x(e3) + Bc t(e3) > = - A x(e1) + B c t(e1) > > Note: the left side is in terms of e3, not e1. You seem > to be assuming that > > x'(e3) = - x'(e1) > > but there is no reason to assume that. > Sorry, but I deleted the section you commented on before you posted your reply. I noticed that it might not improve my argument but that the introduction of further events may actually confuse things further rather than clarify them. So I want to get back to the issue of not writing the arguments of variables here (which you used in order to prevent me from making certain substitutions in the equations). As I indicated before, the arguments of variables are irrelevant for the algebra. The value of a variable on the left hand side of an equation is only determined by the expression on the right hand side and is not in any way affected by writing any arguments or not. In this sense, if I have the identies (1) x2=-x1; x2'=-x1' and the further relationships (2) x1'=Ax1+Bct (3) x2'=Ax2+Bct then there is nothing to prevent me from substituting the mathematical identies (1) into (3) which thus yields (4) -x1'=-Ax1+Bct and thus (5) x1'=Ax1-Bct and hence by a comparison with (2) (6) B=0. Every other conclusion is in my opinion mathematically incorrect and would involve making further assumptions or restrictions that are not inherent in the equations. Thomas
From: Thomas Smid on 13 Sep 2005 07:53 Daryl McCullough wrote: > Thomas Smid says... > > > >Daryl McCullough wrote: > > >> >> 1. forall e, x'(e) = A x(e) + Bc t(e) > >> >> 2. forall e, ct'(e) = D x(e) + Ec t(e) > >> >> 3. forall e, if x(e) = ct(e), then x'(e) = ct'(e) > >> >> 4. forall e, if x(e) = -ct(e), then x'(e) = -ct'(e) > > >> The 4 equations above have a solution > >> > >> A=5/3 > >> > >> B=-4/3 > >> > >> D=-4/3 > >> > >> E=5/3 > > >No there isn't a contradiction as you restricted again the validity of > >the equations: > > > >if you have two arbitrary events e1 and e2 , the transformation > >equations are > > > >x'(e1) = A x(e1) + Bc t(e1) > >ct'(e1)= B x(e1) + Ac t(e1) > > > >x'(e2) = A x(e2) + Bc t(e2) > >ct'(e2)= B x(e2) + Ac t(e2) > > > >These equations must hold for all events e1 and e2 simultaneously, > >whether x(e1) and x(e2) are positive or negative, and thus also for > >further events at -x(e1) and -x(e2), > > What does "further events at -x(e1) and -x(e2)" mean? > > I think that maybe what you mean is this: let e3 and e4 > be events such that > > x(e3) = -x(e1) > x(e4) = -x(e2) > > I'll assume that's what you mean. > > >so we can actually double up the equations to > > > >(1a) x'(e1) = A x(e1) + Bc t(e1) > >(1b) ct'(e1)= B x(e1) + Ac t(e1) > > > >(2a) -x'(e1) = -A x(e1) + Bc t(e1) > > That doesn't make any sense. If you want to consider > some event e3 such that x(e3) = - x(e1), > and t(e3) = t(e1), then you have > > (2a) x'(e3) = A x(e3) + Bc t(e3) > = - A x(e1) + B c t(e1) > > Note: the left side is in terms of e3, not e1. You seem > to be assuming that > > x'(e3) = - x'(e1) > > but there is no reason to assume that. Sorry, but I deleted the section you commented on before you posted your reply. I noticed that it might not improve my argument but that the introduction of further events may actually confuse things further rather than clarify them. So I want to get back to the issue of not writing the arguments of variables here (which you used in order to prevent me from making certain substitutions in the equations). As I indicated before, the arguments of variables are irrelevant for the algebra. The value of a variable on the left hand side of an equation is only determined by the expression on the right hand side and is not in any way affected by writing any arguments or not. In this sense, if I have the identities (1) x2=-x1; x2'=-x1' and the further relationships (2) x1'=Ax1+Bct (3) x2'=Ax2+Bct then there is nothing to prevent me from substituting the mathematical identities (1) into (3) which thus yields (4) -x1'=-Ax1+Bct and thus (5) x1'=Ax1-Bct and hence by a comparison with (2) (6) B=0. Every other conclusion is in my opinion mathematically incorrect and would involve making further assumptions or restrictions that are not inherent in the equations. Thomas
From: Thomas Smid on 13 Sep 2005 08:03
Daryl McCullough wrote: > Thomas Smid says... > > > >Daryl McCullough wrote: > > >> >> 1. forall e, x'(e) = A x(e) + Bc t(e) > >> >> 2. forall e, ct'(e) = D x(e) + Ec t(e) > >> >> 3. forall e, if x(e) = ct(e), then x'(e) = ct'(e) > >> >> 4. forall e, if x(e) = -ct(e), then x'(e) = -ct'(e) > > >> The 4 equations above have a solution > >> > >> A=5/3 > >> > >> B=-4/3 > >> > >> D=-4/3 > >> > >> E=5/3 > > >No there isn't a contradiction as you restricted again the validity of > >the equations: > > > >if you have two arbitrary events e1 and e2 , the transformation > >equations are > > > >x'(e1) = A x(e1) + Bc t(e1) > >ct'(e1)= B x(e1) + Ac t(e1) > > > >x'(e2) = A x(e2) + Bc t(e2) > >ct'(e2)= B x(e2) + Ac t(e2) > > > >These equations must hold for all events e1 and e2 simultaneously, > >whether x(e1) and x(e2) are positive or negative, and thus also for > >further events at -x(e1) and -x(e2), > > What does "further events at -x(e1) and -x(e2)" mean? > > I think that maybe what you mean is this: let e3 and e4 > be events such that > > x(e3) = -x(e1) > x(e4) = -x(e2) > > I'll assume that's what you mean. > > >so we can actually double up the equations to > > > >(1a) x'(e1) = A x(e1) + Bc t(e1) > >(1b) ct'(e1)= B x(e1) + Ac t(e1) > > > >(2a) -x'(e1) = -A x(e1) + Bc t(e1) > > That doesn't make any sense. If you want to consider > some event e3 such that x(e3) = - x(e1), > and t(e3) = t(e1), then you have > > (2a) x'(e3) = A x(e3) + Bc t(e3) > = - A x(e1) + B c t(e1) > > Note: the left side is in terms of e3, not e1. You seem > to be assuming that > > x'(e3) = - x'(e1) > > but there is no reason to assume that. > Sorry, but I deleted the section you commented on before you posted your reply. I noticed that it might not improve my argument but that the introduction of further events may actually confuse things further rather than clarify them. So I want to get back to the issue of not writing the arguments of variables here (which you used in order to prevent me from making certain substitutions in the equations). As I indicated before, the arguments of variables are irrelevant for the algebra. The value of a variable on the left hand side of an equation is only determined by the expression on the right hand side and is not in any way affected by writing any arguments or not. In this sense, if I have the identities (1) x2=-x1; x2'=-x1' and the further relationships (2) x1'=Ax1+Bct (3) x2'=Ax2+Bct then there is nothing to prevent me from substituting the mathematical identities (1) into (3) which thus yields (4) -x1'=-Ax1+Bct and thus (5) x1'=Ax1-Bct and hence by a comparison with (2) (6) B=0. Any other conclusion from the transformation equations would only indicate that they are in fact algebraically inconsistent. So either way it shows that they don't make any sense as they would only be consistent if all variables are identically zero. Thomas |