From: metares@well.sf.ca.us (Tom Van Flandern)
Subject: Re: Standard model of QSOsex
Organization: Whole Earth 'Lectronic Link
Distribution: sci
Date: Thu, 20 Aug 1992 02:40:11 GMT


Earlier, I wrote:

>> You have a confusion here in the distinction between types of masses. In
>> dynamics, *all* effective masses are active gravitational masses only.
>> Inertial masses of stars do not affect their velocities in the
>> gravitational potential.  To see this, recall that the acceleration of a
>> body in response to a gravitational force is independent of its own mass.
>> [Lead and aluminum balls drop at the same rate.]

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

> Somebody else has made a similar comment.  Fearing that I was messing up
> the distinction between grav. and inertial masses, I went back to Halliday
> & Resnick.  After reading H & R, I disagree with the above.
>
> The acceleration of a star in a grav. potential is
>
>              A = (1/m_i)*(do/dr)
>
> where A is the acceleration, m_i is the inertial mass, and o is the grav.
> potential.  I would argue that the inertial mass quite clearly enters into
> the acceleration, and hence the velocity of a star.

     Let's sort this out carefully.  The term do/dr is the gradient of the
potential of the large, distant mass, while m_i is the inertial mass of the
accelerating body.  There are two problems with this formula.  (1) do/dr
implicitly assumes that the local body has negligible mass, since it has
acceleration units, not force units.  (2) The mass implicit in do/dr should
have cancelled m_i.  Since it did not, you are left with an incorrect
formula.  In your formula, acceleration approaches infinity as the local
"test" mass m_i approaches zero.  This is obviously non-physical.

>> In more technical language, let m(i) be inertial mass [response to
>> collisions], m(p) be passive gravitational mass [weight], and m(a) be
>> active gravitational mass [pull on other bodies].  Then 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.  Experiments place only mild constraints
>> on the relationship between active gravitational masses and either of the
>> other two types.  However Eotvos-type experiments have shown the equality
>> of passive gravitational masses and inertial masses to more than 10
>> significant figures.  So m(i) and m(p) in the above equation cancel,
>> leaving the acceleration A dependent upon the active gravitational mass in
>> the system only.

> 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.

> I'm afraid I don't know what the active gravitational mass is.

     The active gravitational mass creates the potential at a distance, or
the force on other bodies.  The passive gravitational mass is the response of
a body to a potential; i.e. it is the "weight" of a body.  The inertial mass
measures the resistance to acceleration, and is the mass effective in
collision experiments.

> you clearly (or so I thought) indicated in an earlier post that m(i) > m(g)
> resulting in A being less than if m(i) = m(g).  Of course, it was later
> pointed out to me that the difference was very small except for extremely
> massive stars, so the effect probably could not be seen.

     In the Meta Model, m(i) = m(p) > m(a).  But the difference is usually
too small to be detectible, as you correctly note.

     Granted, in many contexts the distinction between the two types of
gravitational mass is not often made.  But in this context it is critical.
As I remarked at the outset, gravitational dynamics depend on active
gravitational masses only.

>>> It is still not clear how extra interior matter increases stability.  The
>>> stability of a star depends upon its surface gravitational field.

>> Since the extra matter lowers the gravity, temperature, and pressure at
>> the surface, it increases stability and lifetime.

> How does the extra matter affect the temperature?  The stability of a
> massive star is a balance between radiation pressure and gravity.  If the
> gravity drops, the luminosity must drop by the same factor or the star will
> become less stable;

     Blue giants and red giants can have similar luminosities but vastly
different temperatures.  Massive size compensates for lower temperatures in
the red giants.  The same factors continue in effect in the red and blue
extensions of the H-R diagram in the Meta Model.  Both classes of stars get
cooler and bigger than they would be in the standard model.  This is most
easily seen by asking what would happen to an MM supermasive star if all its
interior matter started to affect the surface gravity.  The surface gravity
would suddenly increase, causing the star to contract and heat up until a new
equilibrium was established.  This would be a less stable state.  -|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.
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