Re: Nitpicking: gravity is not a force???

[Hugh Logan <hlogan@ix.netcom.com>]

William Beaty wrote:
> On Sat, 8 Aug 1998, Jerome Epstein wrote:
>
> > You cannot feel a (uniform) gravitational "force" (i.e. field). If a
> > gravitational force is the only force acting on you, you feel
> > weightless. This causes no end of confusion to students in elementary
> > classes, and I have seen teachers with Ph.D's make a total shambles of
> > this.
> >
> > The only exception would be an extremely non-uniform gravitional field
> > as in falling towards the singularity of a black hole. In that case, the
> > weaker field at your head as compared with your feet, causes you to be
> > pulled apart by tidal forces.
>
> Another exception: when we take a large-scale viewpoint, Earth's radial
> "field" is obvious.  For example, if my body was as large as the moon's
> orbit, then Earth's field would be unmistakably radial, and I would find
> it hard to remove the "field" by adopting a non-intertial reference
> frame.
You could only remove the field locally by using a sufficiently small
free-falling ("free float") reference frame. But such a frame would be
an inertial frame.
 
> Another idea (don't know if it jibes exactly with GR): a positive atomic
> nucleus in a parallel e-field would not experience the electrostatic
> acceleration, and would feel "weightless", yet we still feel confident
> in talking about electrostatic "fields".

As Jerry states, you cannot feel a uniform gravitational field. Any
massive object in such a field has the same acceleration, whatever its
mass if gravity is the only force. If you are in a capsule falling in
such a field, the capsule and any objects inside it have the same
acceleration due to gravity. This is because of the equivalence of
gravitational mass (the mass in F=G*(m*m')/(r^2) ) and inertial mass
(the mass in F=m*a ). If the mass, and hence the inertia, of an object
doubles, so does the gravitational force on the object, so the
acceleration remains the same. This is illustrated by the heavy and
light stones dropped at the same time from the Leaning Tower of Pisa.
Except for some extraneous effects, both will strike the ground at the
same time.

As I see it, the electrostatic field is not completely analogous. It
takes more than a uniform field for one not to "feel" the field. There
is no principle of equivalence between the charge q in F=k(q*q')/(r^2)
(or F=q*E from the field point of view) and the inertial mass m of a
charged particle). The acceleration of a charged particle in a uniform
electric field E is a = F/m = (q*E)/m = (q/m)*E . Unless all particles
had the same charge-to-mass ratio, their accelerations would be
different for a given uniform E-field.

In a gravitational field g (not necessarily the value at the surface of
the Earth), the acceleration 

   a =(force of gravity)/(inertial mass)
     = [(gravitational mass)*g]/(inertial mass)
     = [(gravitational mass)/(inertial mass)]*g = g

since the principle of equivalence of gravitational and inertial mass
tells us that (gravitational mass)/(inertial mass) = 1 (assuming that
the same object has been used as a standard of both gravitational and
inertial mass).

Hugh Logan