Re: Special Relativity is Dead! (second proof)


<jan.verheul@xxxxxxxxxx> wrote in message news:1163925792.739814.224840@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Dirk,

Thanks for your thorough treatment of the experiment, and for the
calculations, that seems to be correct on a first glance. The problem
with this calculation is that the calculation rules of S.R. are only
applied from the viewpoint of K and not from the viewpoint of K'. The
problem is that anomalies and conflicts start to occur if you apply the
calculation rules of S.R. to the viewpoints of BOTH clocks. Proponents
of S.R. then usually say: that's not allowed because an accelerating
clock "is not in one inertial frame".

Ha, yes, perhaps you misunderstood what these proponents
meant to say.
In S.R. the calculation can be made from the viewpoint of K'
as well, but it is much more difficult, since K' is not an inertial
frame.
Since K' is not inertial, we don't have to calculate an integral
like
int{ sqrt( 1 - [v(t)/c]^2 ) dt }
where v(t) is the velocity of clock K' at time t according to
clock K, but this time we have to calculate something like
int{ sqrt( 1 - [v(t')/c]^2 ) ( 1+a x'(t') ) dt' }
where x'(t') and v'(t') are the distance and velocity of clock K
at time t' as seen by clock K'.
Needless to say this is a bit tricky, and *maybe* later I'll show
you how it can be done.

But first let's get back to your statement about "the problem.
with this calculation".
When you want to calculate the number of times some kid's
jo-jo goes up and down on a merry-go-round on one of
Saturn's satelites, you need the equation of motion of the jo-jo,
and calculate an integral. When I give you these equations in
the kid's reference frame, this is rather simple.
Are you going to say that the problem with this calculation is
that it has not been made from the viewpoint of yourself, riding
a rollercoaster in Amsterdam? Are you going to insist that
someone goes through the trouble of finding the equations of
motion of the jo-jo in your frame, and verify whether the rather
difficult integral will produce the same number? If the person
is not prepared to do that for you, will you do the job yourself?

Dirk Vdm


.

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