## Tuesday, November 16, 2004

Last time I wrote (typo-corrected here with missing ^):

The only non-vanishing terms in this rest LNIF case are

{LC}^1'0'0' = (1/2)g^1^'0'g0'0',0' + (1/2)g^1'^1'g0'1',0'

= -g^1'^0'(g^2/c^3)t' + g^1'^1'g/c^2

Note that there appears to be an additional non-Newtonian term here. We need to calculate the inverse matrix to gu'v' of course.

OK I did the calculation and lo and behold g^1'0' = 0 exactly! There is NO "additional Non-Newtonian term" in this simple example. Such a term would mar the beauty of Einstein's general theory in this simple Galilean limit analogous to doing the simple hydrogen atom in the quantum theory of atoms.

Note that in this simple problem the inertial artificial gravity force felt by the non-inertial observer in globally flat spacetime is a pure "gravimagnetic" effect in terms of Einstein's local geometrodynamics in which there are no Newtonian gravity forces acting at a distance.

On Nov 15, 2004, at 3:59 PM, Jack Sarfatti wrote:

Example - uniformly accelerating non-inertial frame in globally flat space-time in the Galilean limit as a consistency check.

On Nov 15, 2004, at 2:45 PM, Jack Sarfatti wrote:

The LC connection field in general is

{LC}^luv = (1/2)g^lw(gvw,u + gwu,v - guv,w)

Therefore,

{LC}^i00 = (1/2)g^iw(g0w,0 + gw0,0 - g00,w) = (1/2)g^iw(2g0w,0 - g00,w)

In the case

x -> x' = x - (1/2)gt^2

t = t'

y = y'

z = z'

gt/c << 1

dx -> dx' = dx - gtdt

dx^2 = (dx' + gt'dt')^2 = dx'^2 + 2gt'dx'dt' + (gt')^2dt'^2

= dx'^2 + 2(gt'/c)dx'cdt' + (gt'/c)^2c^2dt'^2

ds^2 = (cdt)^2 - dx^2 - dy^2 - dz^2

= [1 - (gt'/c)^2](cdt')^2 - 2(gt'/c)(cdt')(dx') - dx'^2 - dy'^2 - dz'^2

g0'0' = [1 - (gct'/c^2)^2]

g0'1' = H1' = 2(gt'/c) gravimagnetic field

g0'0',0' = -2(g^2/c^3)t

g0'1',0' = 2g/c^2

The only non-vanishing terms in this rest LNIF case are

{LC}^1'0'0' = (1/2)g^1^'0'g0'0',0' + (1/2)g^1'^1'g0'1',0'

= -g^1'^0'(g^2/c^3)t' + g^1^'1'g/c^2

Note that there appears to be an additional non-Newtonian term here. We need to calculate the inverse matrix to gu'v' of course.

The Galilean approximation here is gt'/c << 1

Therefore,

{LC}^1'0'0'~ g^1'1'g/c^2

Where obviously

g^1'1' = -1

Therefore

{LC}^1'0'0'~ -g/c^2

But the REST LNIF non-geodesic equation for the test particle is, in this case

{LC}^1'0'0' = F(non-gravity reaction force)/mc^2

Therefore

-g ~ F(non-gravity reaction force)/m

QED!