Re: A Challenge for the Experts
From: Mike (eleatis_at_yahoo.gr)
Date: 09/17/04
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Date: 17 Sep 2004 12:51:02 -0700
"Eugene Shubert" <http://www.everythingimportant.org> wrote in message news:<41462d04@sys13.hou.wt.net>...
> I wrote up what I think is a very easy to understand derivation of
> the Lorentz transformation (6 pages) aimed at beginning students
> of math and physics. I believe that my approach to special relativity
> is sufficiently new, engaging and instructive and that it will charm
> and challenge the brightest students.
>
> Here's the challenge:
>
> http://www.everythingimportant.org/relativity/special.pdf
>
> If this derivation is faulty at any point, where's the first error in
> math or misstep in logic?
>
> Enjoy the charm.
>
> Eugene Shubert
> http://www.everythingimportant.org/relativity/special.pdf
There are at least two errors I can see. The first regards the fact
that the derivation assumes knowledge of the Lorentz transform in the
first place. It is thus a synthetic proof. Every formula, true or
false, implies a true formula. What you have done is not a deductive
proof but an abductive one.
Then,if you have used the standard procedure of numbering equations
every follows, it would be easire to point out the errors. At some
point you claim that since:
k(v)/v = k(w)/w = k and v and w are independent, k must be a constant.
This is not true in general. By counterexample:
Let k be the square function such as k(z) = z^2. Youe equation
becomes:
v^2/v = w^2/w or v = w
This does not imply that v = w = constant. By assuming the trivial
solution, you are implicitely assume the existence of globally
inertial reference frames, an axiom of SR. However, you derivation
implies in general a much richer spacetime than SR in which Lorentz
invariance is just a special case. The infinite number of other cases
are not Lorentz invariant. What kind of spacetime is that I cannot
say. I will call it Shubertian spacetime.:)
Mike
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