Even in the Newtonian limit of

GR where v/c << 1 for test particles

guv ~

nuv(flat) +

huvhuv <<

nuv
and radii of curvature >> measurement scales

the

Levi-Civita connection ~

huv,w will still vanish in

LIFs via

EEP
so if you try to make a

Tuv(gravity) quadratic in

huv,w it will vanish in an

LIF, therefore the gravity energy is still nonlocal even in this Newtonian limit

h00 ~

2VNewton/c^2 << 1

{LC}^

rtt ~

-VNewton,r

i.e.

g = -GM/r^2 etc.

On the other hand the

LIF gravity field tetrad

Cartan 1-forms e^I are

GCT scalar

invariants. Their torsion 2-forms

T^I = De^I = de^I + (spin connection)^

IKe^K

have local tensor components

Their Yang-Mills

Lagrangian density is ~

TraceT^I*T^J yields a local stress-energy gravity field tensor.

Using

teleparallelism you can get the

Einstein-Hilbert Lagrangian from the torsion tensor.

So this approach seems to solve the problem.

On Sep 2, 2011, at 12:15 PM, Woodward, James wrote:

*The difference between Newtonian gravity and GRT is that in Newtonian theory you can localize the gravitational potential energy of objects. In GRT you can't. They aren't the same at all. One is framed in absolute space and time, the other is framed in absolute spacetime. Apples and oranges.*
From: Paul

Zielinski [

iksnileiz@gmail.com]

Sent: Friday, September 02, 2011 12:42 PM

To: JACK

SARFATTI
Subject: Re: let's be more precise in the use of informal language

On 9/1/2011 10:38 PM, JACK

SARFATTI wrote:

On Sep 1, 2011, at 9:50 PM, Paul

Zielinski wrote:

*Doesn't precisely the same equivalence principle also hold in a generally covariant spacetime formulation of Newtonian theory?*
Jack wrote: Einstein's

GR limits to Newtonian theory in the two limits v/c << 1 and weak curvature

i.e.

guv ~

nuv(flat) +

huvhuv <<

nuv
and radii of curvature >> measurement scales

for a spherically symmetric static source

g00(static

LNIF) = 1 +

2V/c^2

V = Newton's gravity potential per unit test mass

e.g
V = - c^

2rs/r for finite source

r (observer) >

rsrs for Earth is ~ 1 cm

V = - /\r^2 de Sitter space

observer located at r = 0

/\ = Einstein's cosmological constant ~ dark energy density accelerating universe ~ (area-entropy of our future event horizon)^-1

Z:

* I think the correct answer is "yes".*

How can one "locally" distinguish the effects of the dynamical acceleration of a test object observed in its own kinematical rest frame from the non-tidal effects of a Newtonian gravitational force field acting in a non-accelerating kinematical frame?JS: I have no idea what your sentence means without seeing some equations.

Z: I

*n fact it's a very clearly stated question. *JS: Not to me.

*Z: The correct answer is that in Newtonian theory you cannot locally distinguish the effects of physically accelerating a test object, as observed from the object's non-inertial rest frame, from the non-tidal effects of a Newtonian gravitational force observed in an inertial frame of reference.*JS: No that is completely confusing apples with oranges.

The non-tidal effects in the

Ruvwl curvature tensor are measured with pairs of neighboring test particles each on geodesics.

In contrast, the g-force of Newton's gravity {LC}^

rtt ~

-dV/

dr is measured with a single-test particle pushed off a geodesic by an electrical-based force (including

Van-der Walls forces).

Z:

*The point here is that precisely the same heuristic argument about "local equivalence" holds in Newtonian theory.*JS: Not even wrong.

Given any object, if an accelerometer clamped to it moves off zero then that object is really accelerating.

http://

en.wikipedia.org/wiki/Accelerometer