Re: Teaching logic is urgent

[pvalev <pvalev@BAS.BG>]

--- Bob LaMontagne <rlamont@POSTOFFICE.PROVIDENCE.EDU> wrote:
> pvalev wrote:
>
> > > You say (A) and (C) are compatible and lead to the Lorentz
> > transformations. If I take
> > > one of those transformations, namely
> > >
> > > x = (x' + vt')/sqrt(1-v^2/c^2)
> > >
> > > and substitute x' = -vt', I get x=0, which is your statement
(B).
> > You have chosen a version of Lorentz transformations that agrees
with
> > (B) but does not agree with (A). Substitute x=vt and you will not
> > obtain x'=0.
>
> ???? The standard complementary transformation back from x,t to
x',t',
> namely x'=(x+vt)/sqrt(1-v^2/c^2), immediately gives x'=0.
>
> Even if one restricts oneself to the transformations from x',t' to
x,t,
> namely x=(x'+vt')/sqrt(1-v^2/c^2) and t=(t'+vx'/c^2)/sqrt(1-
v^2/c^2) and
> then substitute x=vt, one still gets x'=0. At this very elementary
level,
> your statements (A), (B) and (C) appear logically compatible when
the
> Lorentz transformations are used.
>
> You proposed this as a question to be presented to students. Could
you
> please present what you would consider an acceptable student
solution -
> something more detailed than simple assertions that (A), (B) and
(C) are
> incompatible.

You are right - (A), (B) and (C) are indeed compatible within the
Lorentz transforations. For the moment, I am not quite sure how I
made this naive mistake. I remember starting from

x' = ax + by + cz + dt

x' = 0  <->  x = vt

x = 0  <->  x' = -vt'

and then I obtained transformations different from Lorentz
transformations. So I made the conclusion that the last two
conditions are incompatible within the Lorentz transformations but I
did not check this. I am sorry. Tomorrow I will have a more detailed
explanation.

Thanks,
Pentcho