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Home Jack Sarfatti's Blog Is Jim Woodward's Mach Effect Engine Theory Consistent? 1-15-12 v2

*On Jan 15, 2012, at 7:43 AM, This email address is being protected from spambots. You need JavaScript enabled to view it. wrote: I just remembered something. I think in GR, M in the below equations is the rest mass and a constant (i.e., DM/ds = 0 and F = MDV/ds whether EM forces are present or not). So if I'm not mistaken if M is not constant in these equations, by experiment or by use of Mach's principle, then that fact would be an inconsistency with GR. Is Woodward claiming this quantity M is not constant?*

Yes I think that's correct. Somehow his whole idea is to change rest mass m0 using action at a distance from distant matter/horizon, but it makes no sense to me. I don't see any coherent narrative there. But perhaps I misunderstand Jim's idea. He has not been clear about it enough for me.

Now, the geodesic equation is essentially Newton's First Law (zero acceleration in absence of a real force) for a test particle in a local accelerating frame LNIF (leaving off indices)

D^2X/ds^2 = 0

remember in general

ds^2 = g00(cdt)^2 + g0idx^i(cdt) + gijdx^idx^j

= (cdt)^2[g00 + g0iV^i/c + gijV^iV^j/c^2]

therefore

cdt = ds/[g00 + g0iV^i/c + gijV^iV^j/c^2]^1/2

= dsGAMMA(LNIF)

GAMMA = [g00 + g0iV^i/c + gijV^iV^j/c^2]^-1/2

ds = cdt/GAMMA

d/ds =( GAMMA/c)(d/dt)

Note that

GAMMA(LNIF) ----> gamma(LIF) = (1 - V^2/c^2)^-1/2 > 0

GAMMA maybe less than 1 in the presence of a strong gravimagnetic field?

Woodward's phi and A?

g00 = 1 + phi/c^2

g0i = Ai

g0iV^i/c =

However, we can't assume U1 gauge transformations and Maxwell type equations for them in the general case. That seems to be limited to the weak field post-Newtonian first order perturbation theory on a non-dynamical Minkowski background that precludes any HORIZON effect. Hence, the Sciama-Bernstein-Woodward model seems inconsistent to me.

Recall that (supressing indices)

DX/ds = dX/ds

(D/ds)(dX/ds) = d^2X/ds^2 + {LC}(dX/ds)(dX/ds)

However, in general Newton's first law is for the dynamical KINETIC MOMENTUM MV allowing charge

http://en.wikipedia.org/wiki/Kinetic_momentum

If there is an EM field then we add the Lorentz force and the radiation reaction jerk force and we now have Newton's 2nd Law of Motion.

So, in general we must have instead of the pseudo-Riemannian geodesic equation

D(MV)/ds = 0 when e = 0

i.e. MDV/ds + VdM/ds = 0

M = m0GAMMA

dM/ds = GAMMAdm0/ds + m0dGAMMA/ds

Category: MyBlog

Written by Jack Sarfatti

Published on Sunday, 15 January 2012 14:12

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