Saturday, December 11, 2004

Gravity Force Without Force and the Bohm Aharonov Effect

On Dec 10, 2004, at 8:44 PM, wrote:

Jack Sarfatti wrote:

On Dec 10, 2004, at 7:25 PM, wrote:

I think he understands my point (strongly affirmed by Alex Poltorak) that the choice of spacetime coordinates is a purely
mathematical issue and should be physically completely arbitrary -- which is simply not the case in Einstein GR, since a mere
change of coordinates can at least locally *annihilate* a sourced gravitational field -- without affecting the sources of this
field themselves in any way.


What you write

"which is simply not the case in Einstein GR, since a mere change of coordinates can at least locally *annihilate* a sourced gravitational field -- without affecting the sources of this field themselves in any way."

is really confused. You completely MISUNDERSTAND Einstein! You are starting from an absurd premise.

I am?

You do not understand the role of "locally" in your above sentence.

I don't?

Yes, you don't. If you did you would know that your GCT tensor gravity force T = 0 always!

You only pay lip service to it without grasping it.



You do not GENERALLY annihilate the {LC} field GLOBALLY everywhere in a small neighborhood in ONE GEODESIC COORDINATE PATCH. If the spacetime is REALLY CURVED you can only do it LOCALLY at a POINT P with the given geodesic coordinate patch.

But isn't that what I said?

You said it without understanding it.

You tell me I'm "confused" and that I don't understand anything -- and then you support this by basically repeating what I just said?

You do not see you are contradicting yourself. You cannot say what you said and also say T =/= 0. That's a contradiction. This proves you do not understand what you have written. Also you are not able to compute your T explicitly in the simplest possible problem of the Schwarzschild solution for r > 2GM/c^2r in ordinary vacuum with the conformal curvature tensor field for tidal stretch-squeeze positive and negative curvature components ~ GM/c^2r^3. It should be easy for you to construct your T from them, but you cannot do it!

There is no TENSOR gravity force in GR. This is Yilmaz's mistake. You do have a {LC} connection field in curved spacetime. {LC} = 0 in any LIF at any point of the manifold. {LC} =/= 0 in any LNIF at any point of the manifold. In particular, mc^2{LC}^i00, i = 1,2,3 is the Weight Inertial Force measured in the LNIF REST FRAME of m, where, in the Galilean limit of physical importance

mc^2{LC}^i00 + F^i(non-gravity) = 0

Here mc^2{LC}^i00 is the TRANSLATIONAL INERTIAL FORCE of gravity to use Newton's term.
The latter is bad to do because it mixes the two paradigms causing your confusion Paul.

Right now we neglect rotating frames. Stick only to translational case.

F^i(non-gravity) is the real non-gravity force, e.g. electrical forces creating our weight on surface of Earth where from Schwarzschild metric

mc^2{LC}^r00 = mg = mGM/r^2

We would not feel mg, the scale pointer would not move were it not for the real electrical reaction force and quantum pressure (Pauli Principle) from the rock of the Earth's crust pushing us off the geodesic of weightless free float inertial motion.

Jack, the point here is that in Einstein's model the "gravitational field" is completely "annihilated" at a spacetime point in any free-fall frame
of reference.

You do not understand what "annihilated" means. It means your T = 0 ALWAYS.

If you want to do it at another point P' you need a GCT OVERLAP FUNCTION or TRANSITION FUNCTION x^u -> x^u'to a different geodesic coordinate patch! If you can do it using only ONE coordinate patch then you are in Minkowski spacetime with a zero covariant curl of {LC) everywhere when in a finite region.

And just how is this supposed to conflict with what I said?

Because you say T =/= 0. You are saying 0 = 1.

You simply do not understand manifold theory and differential geometry. Neither does Hal Puthoff!


Are you denying that in the Einstein model, the gravitational field is completely annihilated at some spacetime point in any so-called "LIF"?

Red Herring. Obviously not. I am saying you do not understand what those words mean PHYSICALLY!

Otherwise I don't really see where you are actually disagreeing with anything I wrote above.

What arbitrary coordinate choices mean is that you can have a set of local detector/observers in arbitrary motion on geodesics and non-geodesics willy nilly and still get sensible data that can be compared among the observers. That is the physical meaning of the math. Physics is MORE than the math!

Which leads back to the interesting question as to why the mathematically arbitrary choice of spacetime coordinates should be linked to the motion of any particular observer, or set of detectors.

This proves my point that you do not understand general relativity because you think that trivial question is an interesting question! This means you do not at all understand the heuristic foundations of Einstein's Vision. You do not understand relativity period including special relativity! You also do not understand quantum theory.

Every choice of basis is a total experimental arrangement - a configuration of "ideal" detectors. This is true in general in both classical relativity and quantum measurement theory!

In quantum theory the basis (choice of coordinates) is in qubit "Hilbert space".

In relativity the basis (choice of coordinates) is in spacetime.

When spacetime is flat you can use global coordinates. When its curved you need local coordinate patches or charts weaved into an Atlas with overlap transition functions that cover the manifold. The overlaps are the GCTs. Puthoff does not understand this either. Matt Visser agreed with me on that at GR 17. Visser says Puthoff's K = e^2GM/c^2r is nonsense for this same reason I gave in my book Space-Time and Beyond II in 2002. Visser actually read Hal's paper and says its mathematical hogwash!

Given a locally flat metric

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

= (cdt)^2 - dr^2 - r^2(dtheta^2 + sin^2thetadphi^2) (1)

Make any GCT x^u -> x^u'(x^u) with Jacobian matrix X to

That is compute

dx^u = X^uu'dx^u'

and substitute into (1) to get

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


Problem, is (2) a curved spacetime or not?


Step 1. Compute the non-tensor connection {LC} from the metric guv.

Step 2. Compute the covariant curl of {LC}

If the covariant curl of {LC} is not identically zero in a finite region of the manifold, then that finite region is CURVED, i.e. with a real gravitational field in the general sense.

This property of the curl is INVARIANT under any further GCTs.

This is the complete solution to the problem you raised.

There is a simple EM analogy here. Think of {LC} like vector potential A.

Curvature is like magnetic field B = CurlA.

The difference is GR is like Yang-Mills, i.e. non-Abelian where the field carries charge. Here charge = mass. GCT in GR is like a gauge transformation in EM.

There is an analog to Bohm-Aharonov effect because the LOCAL MACRO-QUANTUM VACUUM COHERENCE of the micro-quantum false flat vacuum zero point fluctuations forming Einstein's gravity with residual zero point dark energy is a GIANT QUANTUM VACUUM WAVE FUNCTION that must be single-valued.

The flux through a closed loop is a quantum phase shift. In EM we have a magnetic flux quantum. In GR we have a conformal curvature flux quantum (in ordinary vacuum).

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