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
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