Re: Mathematical Inconsistencies in Einstein's Derivation of the Lorentz Transformation

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)
> >>
> >> 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.

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), 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)
(2b) ct'(e1)= -B x(e1) + Ac t(e1)

(3a) x'(e2) = A x(e2) + Bc t(e2)
(3b) ct'(e2)= B x(e2) + Ac t(e2)

(4a) -x'(e2) = -A x(e2) + Bc t(e2)
(4b) ct'(e2)= -B x(e2) + Ac t(e2)

Note that at this point e1 and e2 are arbitray and we have not defined
a relationship x(e2)=-x(e1) yet. Still it is obvious from (1b) and (2b)
for instance that B=0.


>
> >> >(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

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