From: metares@well.sf.ca.us (Tom Van Flandern)
Subject: Re: Standard model of QSOsex
Organization: Whole Earth 'Lectronic Link
Distribution: sci
Date: Fri, 28 Aug 1992 13:38:20 GMT


njzy@vax5.cit.cornell.edu (T. Joseph Lazio, Cornell University) writes:

> I maintain that there the inertial mass enters into the problem.  If U is
> the potential energy of the system, e.g. for two point masses,
>
>                 U = -G*m1*m2/r,
>
> then o = U/m1, and
>
>                 A = d/dr[-G*m_g*m2/(m_i*r)] = (1/m_i)*dU/dr.
>
> Here, m_i is the inertial mass of particle 1, m_g is its gravitational
> mass, m2 is the other point mass, and G is the standard gravitational
> constant.  Substitute "potential energy" for "potential" in my earlier
> comments.

     Then your construction becomes perfectly parallel to my own.  If you
examine the definitions and terms in my equation, you will see where yours
departs from it.  You have not distinguished between active and passive
gravitational masses, which is crucial in this context.  m1 is most
definitely not equal to m_g, since m1 is a passive gravitational mass, and
m_g is an active gravitational mass.  (Please see definitions in previous
postings in this series.)

     You cannot do this analysis properly it you use only "gravitational" and
inertial masses.  The distinction between the field a body creates at a
distance (active gravitational mass), and the response of that body to an
existing field (passive gravitational mass = "weight") is part of the essence
of what we are examining.  You can see intuitively that passive mass <=>
response, and inertial mass <=> response, are similar types of measure of
mass; whereas the force acting at a distance is altogether different in kind.

     It is sometimes now said that the equivalence of passive gravitational
mass and inertial mass is the "weak equivalence principle", and the
equivalence of active gravitational mass with either of the others is the
"strong equivalence principle".  GR implies only the former.

     Once you equate m1 with the passive gravitational mass, then it cancels
with m_i.  Eotvos-type experiments have shown the equivalence of passive
gravitational mass and inertial mass to more than 10 significant figures.  If
there is any difference, it is not part of the Meta Model, and need not
concern us here.

>>>> consider the acceleration, A, of a small mass, m, in the field of a
>>>> larger mass, M, at a distance r:
>>>>
>>>>      m(i) A = G m(p) M(a) / r^2,
>>>>
>>>> where G = gravitational constant.

>>> The equation [above] only applies to a spherical system.

>> ??  The equation is quite general for any two-body problem with masses M
>> and m at distance r.  I don't understand your comment.

> The equation certainly applies to a two-body system.  By one of Newton's
> theorems it also applies to a perfectly spherical system for which M is the
> mass contained within a radius r.  That's the extent of it; if you wish to
> consider galaxies, which are almost certainly not spherically symmetric,
> then the above does not apply.

     The formula I cited comes directly from Newton's Universal Law of
Gravitation combined with F = m(i) A.  It applies to any two particles.  The
potential of our galaxy should be essentially that given by a point mass of 2
x 10^11 solar masses at its center.  The contribution of the disc to the
radial force should be unimportant at the Sun's distance from the center.
Galaxies should have spherically-symmetric fields to high accuracy.  The fact
that they appear not to has led to the postulate of "dark matter".  -|Tom|-

-- 
Tom Van Flandern / Washington, DC / metares@well.sf.ca.us
Meta Research was founded to foster research into ideas not otherwise
supported because they conflict with mainstream theories in Astronomy.