Re: Bile awski the Janitor.

Androcles wrote:
> "JanPB" <filmart@xxxxxxxxx> wrote in message
> news:1126737794.538043.17850@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
> | I've sent the edited DVDs out so I have more time now for a bit of
> fun.
> | So let me start by stating clearly my understanding of what it is I'm
> | supposed to do.
> |
> | -------------------------------------------
> | I'll prove by *merely going over Einstein's own argument in more
> | detail* that the "tau-equation" together with the usual assumptions
> | (Einstein's two postulates, the clock synchronisation definition,
> etc.)
> | plus the assumption of transform linearity IMPLIES that the transform
> | is of the Lorentz type:
> |
> | tau = phi(v)beta(v)(t - xv/c^2)
> | xi = phi(v)beta(v)(x - vt)
> | eta = phi(v)y
> | zeta = phi(v)z
> |
> | ...where phi(v) is a velocity-dependent constant still to be
> | determined, and:
> |
> | beta(v) = 1/sqrt(1 - v^2/c^2)
> | -------------------------------------------
> |
> | Am I correct in assuming that this is your objection?
>
> No, you are not correct.
> My objection is to the half in
>
> ½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v))
>
> which comes from
>
> [quote]
> we establish by definition that the "time" required by a turtle to
> travel
> from A to B equals the "time" it requires to travel from B to A.
> [end quote]

Well, it does come from it. So what is it you claim is incorrect then?
(You did claim before that the Lorentz transform could not be derived
from the tau-equation, that's why I assumed that's what your objection
was.)

--
Jan Bielawski

.

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