Re: A Challenge for the Experts

From: Eugene Shubert (http://www.everythingimportant.org)
Date: 09/18/04 

Date: Sat, 18 Sep 2004 12:31:52 -0500

"Mike" <eleatis@yahoo.gr> wrote in message
news:9c1b39be.0409180824.6113078f@posting.google.com...
> "Eugene Shubert" <http://www.everythingimportant.org> wrote in
> message news:<414b602d@sys13.hou.wt.net>...
>
> The fact that two independent variables x and y are contraint
> to take the same value from the domain of a function f(x,y)
> does not inply x and y are dependent. ... In my counterexample,
> I demonstrated to you how a choice for the function k leads to
> a contraint of the form v = w.
> Does this mean v and w are dependent? Not at all.
> Example:
>
> F(x,y) = x+y, x,y in N
>
> If F(x,y) is contraint to take the value of 2 then one solution is x
> = y =1. Does that mean x,y are dependent. Of course not. They
> are independent but assume the same value.
>
> By the same toke, by contraining the relationshi of the variables
> v,w and funcgtion k by:
>
> k(v)/v = k(w)/w
>
> you can have several solutions that need not require k(v) = kv and
> k(w)=kw. Such solution is sufficient but not necessary. Depending on
> the form of the function, the two variables, although independent,
> maybe constrained to take the same value. That is what I have
> demonstrated.
>
> I'm sorry I cannot present my explanation in a simpler way. If you
> still do not understand it, that may be sue to the fact that you
> confuse values of variables with the dependency of variables. The
> equation:
>
> x = y
>
> does not imply the dependency of x and y. It is just the solution of
> the polynomial:
>
> f(x,y) = x-y = 0
>
> Thus, in my counterexample, I presented to you an equation that
> is a solution of your equation. that is v = w. You think that this
> contradicts the independency assumption. It does not. It's a
> solution only that states the two independent variable must have
> the same value.
>
> Mike

Thank you Mike. My paper was meant to challenge the experts around
here and judging from all the responses that I've received so far,
you're right up there at the top. Thanks for the polite reply.

> Further, that shows that your derivation leads to a parametric
> spacetime in which the Lorentz transform is only a possibility
> out of infinite possibilities.

Please don't attach my name to your discovery Mike. Please put
your name to all those other spacetimes.

Eugene Shubert
http://www.everythingimportant.org/relativity/special.pdf

Relevant Pages

  • Re: A Challenge for the Experts
    ... "Mike" wrote in message ... > The fact that two independent variables x and y are contraint ... > does not inply x and y are dependent. ... > spacetime in which the Lorentz transform is only a possibility ...
    (sci.physics.relativity)
  • Re: A Challenge for the Experts
    ... But I'd be happy to entertain a proper counterexample. ... that two independent variables x and y are contraint to take the same ... dependent. ... for the Lorentz transform in your concept of spacetime but only ...
    (sci.physics.relativity)
  • Re: A Challenge for the Experts
    ... But I'd be happy to entertain a proper counterexample. ... that two independent variables x and y are contraint to take the same ... dependent. ... for the Lorentz transform in your concept of spacetime but only ...
    (sci.physics)
  • Re: Optimised version works on dev machine but not on others
    ... Your .exe might be dependent on other DLL's being present in the system ... "Mike" wrote in message ... > My optimized build works without problems on my development machine. ...
    (microsoft.public.vc.mfc)
  • Re: doing my accpounts on excel
    ... Hi Mike, ... The solution you choose from those suppliedwould be dependent on if you want permanent ...
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