CORRECTED 2nd Draft

On Nov 13, 2004, at 1:05 PM, Jack Sarfatti wrote:

From the published papers of STAIF and NASA BPP (now defunct) it is apparent that many aerospace engineers do not understand the essentials of Einstein's theory of gravity in their attempts to come up with exotic advanced propulsion systems. Hence the silly proposal to change "rest mass" (part of Hal Puthoff's PV program) for more efficient space-craft. Indeed, you might be able to do that, but it would be a disaster WMD to do so as a reading of Sir Martin Rees's "Just Six Numbers" shows. (Also Tipler & Barrow "The Anthropic Cosmological Principle).

I only use 1916 plain vanilla GR with the Levi-Civita connection. No extended "Affine" connection etc.

Start with

ds^2 = guvdx^udx^v

Summation convention over repeated upper and lower indices.

Equations in physics must be dimensionally consistent. Mathematicians sometimes flout this in differential geometry. One can do this virtually in intermediate steps as long as one checks that the final real output equations are physically consistent dimensionally. Similarly theoretical physicists have a pure mathematician's Cargo Cult fetish and like to be "elegant" setting everything in sight = 1, e.g. h = c = G = 1 that causes a lot of confusion and leads to them missing interesting connections between seemingly disparate quantities. As Ludwig Boltzmann said "Elegance is for tailors."

ds is the local frame invariant differential length, dx^u is a frame-dependent coordinate differential length, therefore, the metric tensor components guv must be dimensionless pure numbers and the L-C connection has dimension 1/(length). Therefore, c^2{L-C} is an "acceleration".

Example 1. Global Inertial Frame GIF in really flat space-time.

In Cartesian coordinates

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

goo = -1

g11 = g22 = g33 = 1

all off-diagonal components = 0.

Look at (see below for proof)

{LC}i00 = (1/2)(g0i,0 + gi0,0 - g00,i) = goi,0 - (1/2)goo,i

Obviously

{LC}i00 = 0

in GIF Cartesian coordinates.

1a. Do a static orthogonal coordinate transformation to spherical polar coordinates. This is not a local general coordinate transformation but a rigid global coordinate transformation.

z = rcos(theta)

x = rsin(theta)cos(phi)

y = rsin(theta)sin(phi)

dt = dt'

dz = drcos(theta) - rsin(theta)dtheta

dx = drsin(theta)cos(phi) + rcos(theta)cos(phi)dtheta - rsin(theta)sin(phi)dphi

dy = drsin(theta)sin(phi) + rcos(theta)sin(phi)dtheta + rsin(theta)cos(phi)dphi

Substitute to get

ds^2 = -(cdt)^2 + dr^2 + r^2[dtheta^2 + sin^2(theta)dphi^2]

The infinitesimal coordinate differentials in the little orthonormal triad at (r,theta, phi) in space are

der = erdr

detheta = ethetardtheta

dephi = ephirsin(theta)dphi

where er, etheta and ephi are an orthonormal 3-vector basis or LOCAL ORTHOGONAL TRIAD FRAME in 3D space.

Obviously, the dimensionless metric tensor components in this static orthogonal coordinate transformation are for

0 = t, 1 = x, 2 = y, 3 = z

0' = t, 1' = r, 2' = theta, 3' = phi

ds^2 = gu'v'(de^u')(de^v')

g0'0' = g00 = -1

g1'1' = g11 = +1

g2'2' = g22 = +1

g3'3' = g33 = +1

That is, the individual metric tensor components under this restricted set of global static orthogonal coordinate transformations in globally flat space-time are invariants.

Therefore {L-C}i'0'0' = 0 as well in GIF static orthogonal spherical polar coordinates.

This result is intuitively obvious, but needs to be shown formally as well.

Let's call these trivial coordinate transformations.

1b. The simplest non-trivial coordinate transformation to a non-inertial frame with the appearance of an "inertial force" (mis-named "non-inertial force" is more precise) is the global Galilean transformation along the z axis

t -> t' = t

x -> x' = x

y -> y' = y

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

where gt/v << 1

therefore,

dz = dz' + gt'dt'

Substitute into

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

to get

ds^2 = -(cdt')^2 + dx'^2 + dy'^2 + (dz' + gt'dt')^2

= -(cdt')^2 + dx'^2 + dy'^2 + dz'^2 + (gt')^2dt'^2 + 2gt'dt'dz'

= [(gt')^2 - c^2]dt'^2 + 2gt'dt'dz' + dx'^2 + dy'^2 + dz'^2

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

Therefore, the dimensionless GLOBAL NONINERTIAL FRAME (GNIF) metric tensor components are

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

g0'3' = g3'0' = 2(gt'/c)

Note that this off-diagonal space-time metric tensor component is a 3-vector Lense-Thirring "frame drag" gravimagnetic field in the sense of Ray Chiao's "gravity radio" superconducting transducer proposal.

Finally notice that in this GNIF as the rest frame for object of rest mass m

(1/2)g0'^3',0' = g/c^2

Therefore, the translational "weight" as an inertial force in this particular GNIF is simply

(Weight 3-Vector)3' = mc^2(1/2)g0'^3',0'= mc^2{LC}3'0'0'/2 since goo,3' = 0 in this case.

Therefore, this particular case simplifies to

W = mg along the z axis.

Note that a non-gravity force is causing the frame to accelerate.

Locally this is equivalent to a gravity "g-force", but globally it does not fall off to zero as does a real gravity field from a compact source. See Chap 10 of Landau & Lifshitz "Classical Theory of Fields".

Note that there are no inertial forces in inertial frames (either global or local).

The plain vanilla L-C connection has no GCT tensor part.

That is the L-C connection in 1916 GR is 100% inertial force, no GCT tensor force component.

Extended connections with torsion and non-metricity dynamical fields do have GCT tensor pieces, but they are not part of 1916 GR.

The not-GCT tensor Levi-Civita connection for parallel transport of geometric objects in curved space-time is

{LC}wuv = (1/2)(gvw,u + gwu,v - guv,w)

u,v,w,l = 0,1,2,3

Note that indices ij,k = 1,2,3 confined to the 3D spacelike surface in a given 3+1 foliation of the manifold.

Therefore,

{LC}i00 = (1/2)(g0i,0 + gi0,0 - g00,i) = goi,0 - (1/2)goo,i

First look only at the ordinary partial time derivative of the gravimagnetic field

(1/2)gi0,0

Comma means ordinary partial derivative.

Here I prove that the "translational" "Weight" actually measured in the "rest LNIF" of the object of rest mass m being weighed is

(Weight 3-Vector)^i = mc^2{LC}^i00/2 -> mc^2(1/2)gi0,0 when g00,i = 0

This comes from the "ma = F" of Einstein's 1916 GR:

d^2xl/ds^2 + (1/2){LC}lvw(dx^v/ds)(dx^w/ds) = Fl(non-gravity)/mc^2

In the special LNIF REST FRAME of the center of mass of object of rest mass m where

d^2x^i/ds^2 = 0, i = 1,2,3

dx^i/ds = 0

dx^0/ds = 1

d^2x0/ds^2 = 0

(1/2){LC}i00 = Fi(non-gravity)/mc^2

(1/2){LC}000 = F0(non-gravity)/mc^2

Homework Problem

Show what combination of {LC} connection field components give the centrifugal and Coriolis forces under what kind of non-trivial coordinate transformations of the globally flat metric field.

## Saturday, November 13, 2004

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