OTHERwise EMPTY fields.

Bilge wrote: sal: On Tue, 07 Feb 2006 01:22:37 +0000, Bill Hobba
wrote: wrote in message
Wondering if there is a simple test for being in curved space.

GR says that gravity is space-time curvature..

Bill, please be careful of how you say this. "Gravity" -- the observable
gravitational "force" -- is _not_ due to the curvature of spacetime, as
I'm sure you know perfectly well; it's due to the nonzero connection
coefficients. The Christoffel symbols, from which we can find the local
gravitational acceleration, are functions of the _first_ derivatives of
the metric. The curvature -- the Riemann tensor -- is a function of the
_second_ derivatives of the metric.

That is true, but... there is a catch.

In a region with a uniform gravitational field, spacetime is _flat_. If
you throw a ball in a uniform gravitational field, it will follow a
parabolic path, just like it will in a uniform "acceleration field". In
both cases there is no curvature involved.

The catch is that such a field is unobservable.[1] All that changes is
ones definition of inertial, which is defined by the fields _other_ than
gravity.

[1] It's likely that such a field is unobservable, however, if there were
such a thing as a ``gravitational aharonov-bohm'' effect, then the con-
nection coefficients themselves would lead to observable consequences. As
a comparison with E&M, the gauge covariant derivative,

D_u\phi == (d_u + ieA_u)\phi

is the electromagnetic analog of

D_u V^v == d_u V^v + C^v_bu V^b.

The field, A_u is the electromagnetic ``connection.'' In general relativity,
the riemann tensor is obtained by the commtator of the generally covariant
derivatives,

[D_u, D_v] V^a = R^a_buv V^b

and the electromagnetic ``curvature'' temsor is the faraday tensor, which
is obtained from the commutator of the gauge covariant derivatives,

[D_u, D_v] = (1/ie)F_uv

[by the way, since maxwell's equations in terms of the faraday tensor
are just d_u F^uv = j^v, that relation works to derive the equivalent
set of ``maxwell equations'' for the weak and strong force using the
appropriate covariant derivative and the fact that the fields for
the weak and strong interaction don't commute, so that there is an
additional term, [B_u, B_v] that goes with the G_uv.]

Now, _classically_ (i.e., maxwell's equations), if F_uv vanishes, (i.e.,
E and B are zero), then there is no physical effect. Maxwell's equations
are insufficient to give any physical meaning to the potentials (other
than as a vector identity). However, quantum mechanically, that isn't the
case. The presence of a vector potential, even where E and B vanish, leads
to a change in phase over a closed path of (e/har)\integral A_u dx^u,
where the integral is the line integral over the path.

Similarly, general relativity is a classical theory and the only
gravitational effects which are observable are those for which the
curature tensor doesn't vanish. However, if there were some quantum
phenomenon which depended directly on the connection coefficients, then it
might be possible to find some physical phenomenon that could distinguish
between a universe with and without a uniform gravitational field.
Otherwise, the difference amounts to an arbitrary definition of
coordinates, which would be determined by observation.

$$ OTHERwise EMPTY fields.

$$ You are employing the ol' "objects in EMPTY space" trick, Bilge.
Semantically speaking, say "objects in [OTHERwise] EMPTY space".

$$ WHAT is intended is, "a BALL in an [OTHERwise] UNiFORM ..field".
So semantically say, "a BALL in an [OTHERwise] UNiFORM ..field".

$$ Such a CLARiFiCATiON from YOU ought help, with SOME credibility.

```Brian.
Simple test needed for being in curved space
OTHERwise EMPTY fields.