Re: need help withb GR calculation
From: uzi kohavi (ekohavi_at_ics-iq.com)
Date: 03/01/05
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Date: Tue, 01 Mar 2005 19:43:58 GMT
after I thought again about it, I think Mr. Tompkins is right. if we have 2
massive stars, in order to calculate the force between them you need to find
the metric of 2 bodies system.
I do not know how you get to those equation?!
"uzi kohavi" <ekohavi@ics-iq.com> wrote in message
news:RS2Vd.16064$VD5.7416@twister.socal.rr.com...
> but we can look at that the the Mass M is going around the mass m and the
> curvature and the metric does not change so it has to be symetric?
> "Tom Capizzi" <etianshrdlu@verizon.net> wrote in message
> news:872Vd.41929$s16.26472@trndny02...
> >
> > "Mr. Tompkins" <mrtompkins@volcanomail.com> wrote in message
> > news:1109697134.329662.93500@g14g2000cwa.googlegroups.com...
> > >
> > > uzi kohavi wrote:
> > >> thanks, I did not know that veff can be depends on Etilda, I guess it
> > > could!
> > >> now for the last question about gravitation force:
> > >> if I take be and expand it in terms of GM/r like
> > >> B=1-2GM/r+a(2GM/r)^2+.....
> > >> the gravitational potential is the grad of -B/2,
> > >> gives fi=GM/r^2+a8G^2M^2/r^3+....
> > >> and the gravitation force is m*fi=GMm/r^2+a*8G^2M^2m/r^3
> > >> the second term depends on M^2*m which is not symmetric between the
> > > mass!
> > >> how can it be?
> > >> Uzi
> > >
> > > I don't think you should worry about that. The concept of
> > > gravitational "force" isn't very relevant here. The "big" mass M
> > > determines the metric structure of spacetime and the little test mass
m
> > > is assumed to move along a geodesic of this spacetime. I don't see why
> > > there should be any symmetry between m and M.
> > >
> > > Mr. Tompkins
> > >
> >
> > Mr.Tompkins,
> >
> > Doesn't GR also apply when m and M are comparable in size? Or
> > is it that M >> m is embedded in the derivation for this particular
> > formula, and some other derivation is necessary if that is not true?
> >
> >
>
>
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