## Monday, November 15, 2004

I use here 1 -1 -1 -1 signature convention

ds^2 = nuvdx^udx^v

in globally flat spacetime using Cartesian coordinate.

n00 = + 1

n11 = n22 = n33 = -1

nuv = 0, u =/= v

Make a general coordinate transformation (GCT) for a fixed "point" physical event P, i.e. x^u(P) -> x^u'(P) in globally flat spacetime, the GCT is

X^u'u = x^u',u

,u = ordinary partial derivative

similarly for X^u,y'

Xu^w'Xw'^v = Unit Matrix etc.

The globally flat metric tensor in curvilinear coordinates is therefore

gu'v' = Xu'^uXv'^vnuv

The LC connection field in general is

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

Obviously in Cartesian coordinates in globally flat spacetime

{LC}^luv = 0

everywhere.

Therefore, the tidal force curvature tensor = 0 everywhere in any frame because tensors transform by multilinear transformation.

The LC connection field is not a tensor. It has no tensor part.

Under GCT, the {LC} transforms as

{LC}^luv -> {LC}^l'u'v' = {LC}^luvXl^l'X^uu'X^^vv' + Y^wv'u'X^l'w

Where Y^wv'u' = X^wv',u'

The LC connection field transformation has a homogeneous tensor part and an inhomogeneous non-tensor part. However, in no sense can you say that the LC connection field itself consists of a tensor part and a non-tensor part.

That is

{LC}^luv = Tensor^luv + (Non-Tensor)^luv

is false, except in the trivial case

Tensor^luv = 0

Back to globally flat spacetime, the LC connection field in curvilinear coordinates is simply

{LC FLAT}^l'u'v' = Y^wv'u'X^l'w

The 4th rank tidal curvature tensor computed from this will vanish.

The timelike geodesic equation in globally flat spacetime in Cartesian coordinates is simply

d^2x^l/ds^2 = 0

i.e. in Galilean limit

a = F(non-gravity)/m = 0

This same timelike geodesic equation in curvilinear coordinates is

d^2x^l'/ds^2 +{LC}^l'u'v'(dx^u'/ds)(dx^v'/ds) = 0

Imagine a test particle of mass m on such a timelike geodesic.

What is the rest frame of this test particle?

i = 1,2,3 space coordinates

In the geodesic LIF rest frame

d^2x^l'/ds^2 = 0

d^x^i/ds = 0

dx^0/ds = 1

Therefore, the geodesic equation in the LIF rest frame of the test particle on the geodesic is

{LC}^i00 = 0

All other components of LC are not measurable in the geodesic rest frame of the test particle.

In general the physically measurable g-force as a 3-vector is

g-force^i = {LC}^i00, i = 1, 2, 3

{LC}^i00 is an "inertial force field" that vanishes in the rest LIF of the test particle.

Therefore, any GCT in globally flat spacetime that has

{LC}^i00 =/= 0

at some event P is not an inertial GCT, but is a non-inertial GCT. Such a GCT is a physical description of a nongeodesic fleet of rocket ships in deep empty space whose rocket engines are firing in an arbitrary manner. This grid of rocket ships measures events P' and records the data. The original nuv in Cartesian coordinates describes an identical geodesic fleet of rocket ships with all their engines off.

The generalized Newtonian

F = ma

In Einstein's 1916 GR is

d^2x^l'/ds^2 +{LC}^l'u'v'(dx^u'/ds)(dx^v'/ds) = F^l(non-gravity)/mc^2

where guv is dimensionless and {LC} has dimension 1/(length) with tidal curvature dimension 1/Area.

The rest LNIF of the test particle obviously obeys

{LC}^i00 = F^i(non-gravity)/mc^2

Note that [Force] = [Energy/Distance].

* When doing physics always use dimensionally consistent equations even in intermediate steps. Mathematicians flout this rule thereby increasing the probability of making errors and missing important physical insights.

Therefore, the translational inertial g-force in globally flat spacetime is created by the non-gravity forces on the rocket ship nodes of the grid that is the end result of the GCT. Of course, any real set of rocket ships is only a coarse-grained finite lattice approximation to the formal continuum mathematics.

What the GCT mean physically are transformations between two sets of detectors in arbitrary relative motion. Call them the Red Fleet and the Blue Fleet.

Relativity theory is a generalized navigation theory from the British Navy's discovery of how to accurately measure latitude and longitude.

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