Mathematical Inconsistencies in Einstein's Derivation of the Lorentz Transformation
in [Relativity]
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From: Daryl McCullough on 11 Sep 2005 17:53 Thomas Smid says... > >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) Okay, but you need to distinguish between variables and constants. Let e1 represent some particular event such that x(e1) = -ct(e1) >This yields >-2ct'(e)=-lambda*2ct(e) Right. For that particular event. So for that event, we have t'(e1) = lambda t(e1) >> 2' forall events e, (x'(e)+ct'(e)) = mu (x(e)+ct(e)) > >OK let's choose then the 'event' x(e)=ct(e) Okay, that's some new event, so don't use the same label. So let e2 be some event such that x(e2)=ct(e2). >This yields >2ct'(e)=mu*2ct(e) For that particular event e2, we have 2ct'(e2) = mu * 2 c t(e2) or t'(e2) = mu t(e2) >and hence lambda=mu. No, there is one event e1 satisfying x(e1)=-ct(e1). There is a *different* event e2 satisfying x(e2) = +ct(e2). They aren't the same event (unless t(e1) = 0). Thomas, you just keep making algebraic mistake after algebraic mistake. >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. As I have told you several times, x2 is not a variable. It is (by assumption) equal to -ct. x1 is not a variable. It is equal to +ct. So you can't go from ct'(x1,t) = (B+A) x1 to ct'(-x1,t) = -(B+A)x1 That amounts to going from ct'(ct,t) = (B+A) ct to ct'(-ct,t) = - (B+A) ct That's clearly incorrect, and you keep doing it. -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 11 Sep 2005 17:59 Thomas Smid says... > >Daryl McCullough wrote: >> 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. Thomas, B is a *constant*. A is a *constant*. If you can prove that A=0 for some value of x (for instance, x=-1), then A is *always* 0. A doesn't change values. That's why it's called a constant. If A is always 0, then in particular, A is 0 when x=+1. Okay, so when x=+1, we have A=0. (Because A is *always* equal to 0). But when x=+1, we also have (by the first statement) A*(+1) = 2*B*(+1) which simplifies to A = 2*B Now, we've already established that A is always equal to zero. Therefore, we have B = 0 Even though we got this result in the special case x=+1, since B is a constant, B must *always* be equal to 0. -- Daryl McCullough Ithaca, NY
From: Thomas Smid on 12 Sep 2005 05:10 Daryl McCullough wrote: > Thomas Smid says... > > > >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) > > Okay, but you need to distinguish between variables > and constants. Let e1 represent some particular event > such that > > x(e1) = -ct(e1) > > >This yields > >-2ct'(e)=-lambda*2ct(e) > > Right. For that particular event. So for that event, we have > > t'(e1) = lambda t(e1) > > >> 2' forall events e, (x'(e)+ct'(e)) = mu (x(e)+ct(e)) > > > >OK let's choose then the 'event' x(e)=ct(e) > > Okay, that's some new event, so don't use the same label. > So let e2 be some event such that x(e2)=ct(e2). > > >This yields > >2ct'(e)=mu*2ct(e) > > For that particular event e2, we have > > 2ct'(e2) = mu * 2 c t(e2) > > or > > t'(e2) = mu t(e2) > > >and hence lambda=mu. > > No, there is one event e1 satisfying x(e1)=-ct(e1). There is a > *different* event e2 satisfying x(e2) = +ct(e2). They aren't the > same event (unless t(e1) = 0). > > Thomas, you just keep making algebraic mistake after algebraic > mistake. > > >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. > > As I have told you several times, x2 is not a variable. It is > (by assumption) equal to -ct. x1 is not a variable. It is equal > to +ct. So you can't go from > > ct'(x1,t) = (B+A) x1 > to > ct'(-x1,t) = -(B+A)x1 > > That amounts to going from > > ct'(ct,t) = (B+A) ct > > to > > ct'(-ct,t) = - (B+A) ct > > That's clearly incorrect, and you keep doing it. We have gone through all this already. Each time the algebra is not working in your sense, you are trying to change the formalism by adding arguments to the variables. This does not change anything about the algebraic conclusions (as we have seen before when we did the split into x1 and x2). If you have ever written a computer program to solve mathematical equations, then you know that variables are characterized by simple names that have no arguments (unless they are arrays). Arguments are only introduced in analytical maths in order to better keep track of things when doing operations like differentiation and integration etc. Again, they do not change anything about the algebra and the solution of the equations (all the algebraic connections are given through the equations between the variables, whether you denote the latter with arguments or not). Assume you have a function f(x) which is defined for all numbers x (positive as well as negative) and you have the two equations (1) f(x)=(A+B)x (2) f(-x)=(A-B)x then you can *logically* conclude from (1) that (3) f(-x)=-(A+B)x and thus by comparison with (2) (4) A=0. In the same sense, if your function t'(x,t) is defined on all numbers x (positive and negative) and you have the relationships (5) ct'(x1,t)=(B+A)x1 (6) ct'(x2,t)=(B-A)x2 (7) x2=-x1 then you can *logically* conclude from (5) (8) ct'(x2,t)=ct'(-x1,t)=-(B+A)x1 and from (6) (9) ct'(x2,t)=(B-A)x2=(A-B)x1 and thus by comparing (8) and (9) (10) A=0. This is straightforward algebra and I can not see how you can arrive at any other conclusion. Thomas
From: Daryl McCullough on 12 Sep 2005 06:57 Thomas Smid says... > >Daryl McCullough wrote: >> As I have told you several times, x2 is not a variable. It is >> (by assumption) equal to -ct. x1 is not a variable. It is equal >> to +ct. So you can't go from >> >> ct'(x1,t) = (B+A) x1 >> to >> ct'(-x1,t) = -(B+A)x1 >> >> That amounts to going from >> >> ct'(ct,t) = (B+A) ct >> >> to >> >> ct'(-ct,t) = - (B+A) ct >> >> That's clearly incorrect, and you keep doing it. > >We have gone through all this already. Each time the algebra is not >working in your sense, Each time, you fail to keep track of dependencies between variables, and it gets you into trouble. As I have said before, if you want to show that *I'm* making a mistake, then you need to be *more* careful than I am. But you are consistently *less* careful. >you are trying to change the formalism by adding >arguments to the variables. I'm saying that that is the *correct* way to reason about these things. If you want to show *my* reasoning leads to a contradiction, then you need to use *my* reasoning. If you want to show that *your* reasoning leads to a contradiction, then by all means, use your own reasoning. You are the one who introduced x1 and said that it was equal to ct. You are the one who introduced x2 and said that it was equal to -ct. Thus it makes no sense to go from >> ct'(x1,t) = (B+A) x1 >> to >> ct'(-x1,t) = -(B+A)x1 How can it make any sense, when x1 is just another name for ct? >This does not change anything about the >algebraic conclusions The correct algebraic conclusion is that you are having trouble with algebra. Have you ever heard of *checking* your results? You say that the equations 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 Let's check: Suppose x=ct. Then x' = 5/3 ct - 4/3 ct = 1/3 ct Then t' = -4/3 ct + 5/3 ct = 1/3 ct. So x' = ct'. Check. Suppose x = -ct. Then x' = -5/3 ct - 4/3 ct = -3 ct. Then t' = +4/3 ct + 5/3 ct = +3 ct. So x' = -ct'. Check. So, obviously my 4 equations above are consistent with both A and B being nonzero. So if you come up with A=0 or B=0, then you are adding something new. If that something new leads to a contradiction, then that's *your* contradiction, not mine. >Assume you have a function f(x) which is defined for all numbers x >(positive as well as negative) and you have the two equations > >(1) f(x)=(A+B)x >(2) f(-x)=(A-B)x >then you can *logically* conclude from (1) that >(3) f(-x)=-(A+B)x >and thus by comparison with (2) >(4) A=0. Right. But we are not in that situation. We have a function of *two* variables x and t. >In the same sense, if your function t'(x,t) is defined on all numbers x >(positive and negative) and you have the relationships > >(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. >(7) x2=-x1 >then you can *logically* conclude from (5) (5) only applies when x=ct. (6) only applies when x=-ct. So for what values of x and t do *both* 5 and 6 hold? That's a simple question, Thomas, and it has a simple answer: If x=ct, and x=-ct, then x=0 and ct=0. That's the *only* case in which both 5 and 6 hold. >(8) ct'(x2,t)=ct'(-x1,t)=-(B+A)x1 Right. In the special case x=0 and t=0, it follows that ct'(x2,t) = ct'(-x1,t) = -(B+A)x1 because in that special case, x1 = 0, x2 = 0, t = 0, t' = 0. >and from (6) > >(9) ct'(x2,t)=(B-A)x2=(A-B)x1 Only in the special case x1 = x2 = t = t' = 0. Plug in these values for x1,x2,t,t', and you get 0 = (B-A) * 0 = (A-B) 0 You cannot conclude that B-A = A-B except by dividing by zero. >and thus by comparing (8) and (9) > >(10) A=0. > >This is straightforward algebra Yes, it is a straight-forward example of division by zero. >and I can not see how you can arrive at >any other conclusion. By actually taking care that I not divide by zero. -- Daryl McCullough Ithaca, NY
From: Thomas Smid on 12 Sep 2005 10:34
Daryl McCullough wrote: > Thomas Smid says... > > > >Daryl McCullough wrote: > > >> As I have told you several times, x2 is not a variable. It is > >> (by assumption) equal to -ct. x1 is not a variable. It is equal > >> to +ct. So you can't go from > >> > >> ct'(x1,t) = (B+A) x1 > >> to > >> ct'(-x1,t) = -(B+A)x1 > >> > >> That amounts to going from > >> > >> ct'(ct,t) = (B+A) ct > >> > >> to > >> > >> ct'(-ct,t) = - (B+A) ct > >> > >> That's clearly incorrect, and you keep doing it. > > > >We have gone through all this already. Each time the algebra is not > >working in your sense, > > Each time, you fail to keep track of dependencies between > variables, and it gets you into trouble. As I have said > before, if you want to show that *I'm* making a mistake, > then you need to be *more* careful than I am. But you are > consistently *less* careful. > > >you are trying to change the formalism by adding > >arguments to the variables. > > I'm saying that that is the *correct* way to reason > about these things. If you want to show *my* reasoning > leads to a contradiction, then you need to use *my* > reasoning. If you want to show that *your* reasoning > leads to a contradiction, then by all means, use your > own reasoning. > > You are the one who introduced x1 and said that it was > equal to ct. You are the one who introduced x2 and said > that it was equal to -ct. Thus it makes no sense to go > from > > >> ct'(x1,t) = (B+A) x1 > >> to > >> ct'(-x1,t) = -(B+A)x1 > > How can it make any sense, when x1 is just another > name for ct? > > >This does not change anything about the > >algebraic conclusions > > The correct algebraic conclusion is that you are having > trouble with algebra. Have you ever heard of *checking* > your results? You say that the equations > > 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 > > Let's check: Suppose x=ct. Then x' = 5/3 ct - 4/3 ct = 1/3 ct > Then t' = -4/3 ct + 5/3 ct = 1/3 ct. So x' = ct'. Check. > > Suppose x = -ct. Then x' = -5/3 ct - 4/3 ct = -3 ct. > Then t' = +4/3 ct + 5/3 ct = +3 ct. So x' = -ct'. Check. > > So, obviously my 4 equations above are consistent with both > A and B being nonzero. So if you come up with A=0 or B=0, > then you are adding something new. If that something new > leads to a contradiction, then that's *your* contradiction, > not mine. 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)) > > >Assume you have a function f(x) which is defined for all numbers x > >(positive as well as negative) and you have the two equations > > > >(1) f(x)=(A+B)x > >(2) f(-x)=(A-B)x > >then you can *logically* conclude from (1) that > >(3) f(-x)=-(A+B)x > >and thus by comparison with (2) > >(4) A=0. > > Right. But we are not in that situation. We have a function > of *two* variables x and t. > > >In the same sense, if your function t'(x,t) is defined on all numbers x > >(positive and negative) and you have the relationships > > > >(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. > > >(7) x2=-x1 > > >then you can *logically* conclude from (5) > > (5) only applies when x=ct. (6) only applies when x=-ct. > So for what values of x and t do *both* 5 and 6 hold? > That's a simple question, Thomas, and it has a simple > answer: > > If x=ct, and x=-ct, then x=0 and ct=0. > > That's the *only* case in which both 5 and 6 hold. > > >(8) ct'(x2,t)=ct'(-x1,t)=-(B+A)x1 > > Right. In the special case x=0 and t=0, it follows that > > ct'(x2,t) = ct'(-x1,t) = -(B+A)x1 > > because in that special case, x1 = 0, x2 = 0, t = 0, t' = 0. > > >and from (6) > > > >(9) ct'(x2,t)=(B-A)x2=(A-B)x1 > > Only in the special case x1 = x2 = t = t' = 0. Plug in these > values for x1,x2,t,t', and you get Let's make things clearer by considering again your equations involving lambda and mu for the two directions x1 and x2 (1)(x1'-ct')=lambda*(x1-ct) (2)(x1'+ct')=mu*(x1+ct) (3)(x2'-ct')=lambda*(x2-ct) (4)(x2'+ct')=mu*(x2+ct) Now since both signals are supposed to travel in opposite directions, we have in addition the global condition (5) x2=-x1; x2'=-x1' Note that these are mathematical identities i.e. we can replace x2 and x2' by -x1 and -x1' respectively in (3) or (4) as we please without changing anything at all about the equations (whether we replace the variables themselves or the (not written) arguments) So (3) therefore becomes (6) (-x1'-ct')=lambda*(-x1-ct) or (7) (x1'+ct')=lambda*(x1+ct) Now we know that equations(1)-(4) must hold simultaneously (after all they are all functions of the same variable t,t'), so from (7) and (2) we have therefore (8) lambda=mu or (9) A-B=A+B and hence (10) B=0. Previously, I had A=0, but I think this is due to the fact that I changed the sign of one equation for the ct-transformation, whereas it should have been ct'(x1,t)=(A+B)x1 ct'(x2,t)=ct'(-x1,t)=-(B-A)x1=(A-B)x1 i.e. also B=0 from a comparison of both equations.I reckon this is the correct result, but it does not really make a difference anyway as in both cases the transformation is only consistent if x=x'=t=t'=0. Thomas |