Re: Hierarchical Inertial System (HIS) and GPS function.
From: Dirk Van de moortel (dirkvandemoortel_at_ThankS-NO-SperM.hotmail.com)
Date: 03/12/05
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Date: Sat, 12 Mar 2005 16:24:09 +0100
"Rafael Valls Hidalgo-Gato" <valls@icmf.inf.cu> wrote in message news:33d06fe2.0503111129.66b66bcf@posting.google.com...
> "Dirk Van de moortel" <dirkvandemoortel@ThankS-NO-SperM.hotmail.com> wrote in message
news:<75%Xd.1428$rq3.1285@news.cpqcorp.net>...
I just noticed your reply on the binaries news server of my ISP.
Your post was much too large for the text based news server.
I will snip heavily.
[snip]
> > > > In our example exercise of yesterday with A, B and C moving all
> > > > inertially along one line, where C was "in the middle" and therefore
> > > > the perfect candidate for the center of mass of A and B, but also for
> > > > the center of mass of A, B and C itself, we had
> > > > Va = (1-sqrt(1-u^2)) / u
> > > > Vb = (-1+sqrt(1-u^2)) / u
> > > > and therefore
> > > > Fab = sqrt(1-Va^2) / sqrt(1-Vb^2) = 1
> > > > How do you get to sqrt(1-u^2) with the Fab factor with value 1?
> > > >
> > > You are right! I have a big error. What I obtain with my proposed
> > > place for C is an equal to sqrt(1-u^2) time dilation factor for both,
> > > A and B, as viewed in the frame C. But the relative time dilation
> > > factor Fab of A respect B, as viewed in the frame C, is equal to 1 as
> > > you found correctly.
> >
> > Yes, and so it is of course in the case of the equilateral
> > triangle. Symmetry dictates that Fab = 1 as well in that
> > case. Problem!
> >
> Which problem are you referring?
Just like above where you said
"You are right! I have a big error.".
In the setup with the equilateral triangle between A, B and C,
your Fab = 1 as well, but there *is* time dilation between A
and B, so your scheme does not work either in that case.
And furthermore, when you put C on one of the points of the
triangle, C is *not* at the center of mass of A and B to
begin with.
So your scheme does not work for inertial A and B
in the follwing cases:
(1) When A and B move along their common connecting
line. You have aknowledged this.
(2) When A and B move in a plane. You can reduce this
to the triangle case by picking the right C (analogous to
the first case but with the "angle equation" for velocity
composition). You have not understood this yet. But I
will not insist.
(3) When A and B move in arbitrary directions. I already
gave a hint, but I won't bother to try convince you.
[snip]
> You seem making a list of problems. Can you specify to what theory are
> you charging them? We are yet in an orthodox SR environment (by your
> decision since already many posts ago). Are you suggesting that those
> are SR problems?
No, they are not problems with SR. They are problems with
your entire approach. It doesn't work.
I made 100% sure that we have sufficient common ground
of SR, and it took me quite some work to convince you that
you have a "big error" in the first case already.
[major snip]
> > These are all rather serious problems with your approach...
> >
> You have not convinced me yet about it, but I appreciate a lot your
> decision to address my HIS approach. Make a search with (HIS) -with
> the ( ) included- in this group and you will see how many attempts I
> had done to share it with others.
I have tried to help you understand where your approach fails.
I also have shown you that it actually fails everywhere, but
taking into account the amount of energy you have invested
in it, I understand that you have great difficulty opening your
mind for that. So I will not address the remarks you made
to my previous message. I don't have the energy for that.
Dirk Vdm
- Next message: Lefty: "Re: What really is "Light" (assuming SR is correct)??"
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