Re: little gee and its sign

[John Mallinckrodt <ajmallinckro@CSUPOMONA.EDU>]

On Sat, 29 Sep 2001, John S. Denker wrote:

> ... Operationally, g(x) is the acceleration of a free particle
> (at location x) relative to the chosen reference frame.

Agreed.

>   -- The person who _uses_ g  or _measures_ g doesn't know and doesn't care
> how much of the local g value is due to purely-gravitational effects and
> how much is due to acceleration of the reference frame.

Hmm.  I'd say "... can't know and, thus, mustn't waste time caring
..."

> It's impossible in principle to tell the difference.

Yes.  Just so and worth repeating:  It is impossible *in
principle* (and obviously all the more so in practice) to tell the
difference ... locally.

Another (and, as long time phys-l'ers know all too well, in my
opinion better) way of saying the same thing is to repudiate the
very existence of a "local purely-gravitational effect."  "Local
gravity" is always entirely caused by acceleration relative to
local frames in which there is no "local gravity."  Furthermore, I
would claim that it is impossible in practice to calculate an
accurate local value for g from first principles. That is to say,
the only practical way to determine one's acceleration relative to
local freely falling frames (i.e., g) is to measure it.

It turns out that when we ride on the surface of a large dense
planet that is far away from large concentrations of mass in the
universe, we can do a passable job of calculating an approximate
value for g.

First, the fact that the planet is a freely falling object almost
eliminates our need to know about the distribution of mass
throughout the rest of the universe.  I say "almost" because we
*would* need to know that distribution *in detail* if we cared to
take into account its "tidal effect."  The tidal effect does make
a difference in the local value of g on the surface of the planet
but we have kept it small by riding "a large dense planet that is
far away from all other large concentrations of mass in the
universe."

Having almost eliminated the combined effect of all the rest of
the mass in the universe, we need only take into account the
effect of the mass distribution and rotational motion of the Earth
itself.  We can do that part pretty well using the Newtonian
approximation which is appropriate given the other approximations
we have already made.

John Mallinckrodt      mailto:ajm@csupomona.edu
Cal Poly Pomona          http://www.csupomona.edu/~ajm