Dirk Van de moortel wrote:

> "Pentcho Valev" <pvalev@yahoo.com> wrote in message 
> news:1120133321.965731.68580@o13g2000cwo.googlegroups.com...
> >
> >
> > Dirk Van de moortel wrote:
> >
> > > "Pentcho Valev" <pvalev@yahoo.com> wrote in message 
> > > news:1120121299.933446.13420@z14g2000cwz.googlegroups.com...
> > > >
> > > > All Einsteinians, both profane and initiated, should immediately take
> > > > their heads out of the sand and resolve the following problem. In
> > > > Chapter 23 in his "Relativity" Einstein claims that a clock at rest
> > > > undergoes time contraction, that is, runs fast by a factor of gamma
> > > > relative to a clock situated at the edge of a rotating disk. The
> > > > Juggler also claims that this follows from Lorentz transforms. However
> > > > Lorentz transforms predict no time contraction - rather, the only time
> > > > distortion they predict is symmetrical time dilation for two inertial
> > > > systems. So Einsteinians should prove that their god is not a Juggler
> > > > and derive, RIGOROUSLY, the time contraction factor (gamma) from
> > > > Lorentz transforms.
> > >
> > > Since this has been explained to you before (at least
> > > 10 times), you can stop reading here.
> > >
> > > For the others:
> > >
> > > The symmetry of the Lorentz transformation exists
> > > between two inertially moving clocks. The clock at the
> > > edge of the disk is not inertial.
> > >
> > > If you want to know the total elapsed time of a non
> > > inertially moving clock, you must integrate.
> > > The total proper time T of a non-inertially moving clock
> > > as calculated form the point of view of an inertial clock is
> > > given by
> > >         T = Integral{ t1...t2; sqrt( 1 - [v(t)/c]^2 ) dt }
> > > where v(t) is the speed of the non-inertial clock as a
> > > function of the time t on the inertial clock.
> > > You immediately see that
> > >         T <= t2 - t1
> > >
> > > You can *not* turn this around to something like
> > >         t = Integral{ T1...T2; sqrt( 1 - [v(T)/c]^2 ) dT }
> > > since the T-clock is not moving inertially, so there is no
> > > symmetry.
> >
> >
> > You should start from Lorentz transforms and obtain, rigorously and by
> > giving a detailed explanation of each step, the time contraction factor
> > gamma. Again (for a profane): LORENTZ TRANSFORMS -> INFERENCE -> GAMMA.
> > It is this inference that Einstein mentions in Chapter 23 in his
> > "Relativity" but forgets to give the steps. So try again. Your
> > arguments above are irrelevant.
> >
> > Pentcho Valev
>
> Weren't you in write-only mode?
> I explicitly said "For the others".
> So that seems to have worked.
>
> Matrix form of Lorentz transformation with movement in arbitrary
> direction between momentarily co-moving inertial frame with
> coordinates (T,X,Y,Z) of non-inertial clock and coordinates (t,x,y,z)
> of inertial clock:
>   ( dT )   ( g         -g*v*nx        -g*v*ny        -g*v*nz     ) ( dt )
>   ( dX )   ( -g*v*nx   (g-1)*nx^2+1   (g-1)*ny*nx   (g-1)*nz*nx  ) ( dx )
>   ( dY ) = ( -g*v*ny   (g-1)*nx*ny    (g-1)*ny^2+1  (g-1)*nz*ny  ) ( dy )
>   ( dZ )   ( -g*v*nz   (g-1)*nx*nz    (g-1)*ny*nz   (g-1)*nz^2+1 ) ( dz )
> where:
>    we better look at this in a fixed font,
>    we work in units where c = 1,
>    (nx,ny,nz) = unit vector in direction of velocity (constantly changing) ,
>    v = amplitude of velocity (constant in this case)
>      = sqrt( [v nx]^2 + [v ny]^2 + [v nz]^2 )
>      = sqrt( [dx/dt]^2 + [dy/dt]^2 + [dz/dt]^2 ) ,
>    g = 1/sqrt(1-v^2) = gamma.
>
> Since we are only intersted in events on the rotating clock, we
> always have dX=dY=dZ=0 and we can concentrate on the time
> equation only
>  (and now writing multiplication with spaces in stead of with "*"):
>    dT = g dt - g v nx dx - g v ny dy - g v nz dz
>       = g ( 1 - v nx dx/dt - v ny dy/dt - v nz dz/dt ) dt
>       = g ( 1 - v nx v nx - v ny v ny - v nz v nz ) dt
>       = g ( 1 - (v nx)^2 - (v ny)^2 - (v nz)^2 ) dt
>       = g ( 1 - v^2 ( nx^2 + ny^2 + nz^2 ) ) dt
>       = g ( 1 - v^2 ) dt
>       = sqrt( 1 - v^2 ) dt
>       = 1/g dt
>
> So
>    T = Integral{ t1...t2; sqrt( 1 - [v(t)]^2 ) dt }
>      = Integral{ t1...t2; sqrt( 1 - [v]^2 ) dt }
>      = sqrt( 1 - [v]^2 ) Integral{ t1...t2; dt }
>      = sqrt( 1 - [v]^2 ) ( t2-t1 )
>      = 1/g ( t2-t1 )
> and turning it around
>    t2-t1 = g T
> so
>    t2-t1 > T
> For two events on the rotating clock, the clock at rest shows
> more elapsed time t2-t1 than the rotating clock (T), i.o.w.
> the rotating clock "runs slow" as compared to the other.
>
> Exercise:
> Find the the expression starting from the invariant interval:
>     (dT)^2 = (dt)^2 - (dx)^2 - (dy)^2 - (dz)^2
> 
> Dirk Vdm

[reply #1]

No progress, although I should admit your diligence. Discuss the
problem with the initiated and try again.

Pentcho Valev

[reply #2]

Moortel, I would pay more attention to your subnormal "integration" if
it were part of the official zombi methodology as presented e.g. in
textbooks. However I have the impression that it is part of what may be
called profane folklor popular on internet forums. So far this
"integration" has been performed by intellects like sal, Andersen, you
and nobody cleverer. I am prejudiced perhaps but nothing could be done.

Pentcho Valev

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