Tuesday, August 01, 2006

Understanding "Gravity Force"
In a nut shell Zielinski makes the common amateur error of pushing the superficial analogy of Newton's gravity potential -GM/r with Coulomb's electrical potential Q/r too far. See details below.

Franz Kafka, feeling like a cockroach under a magnifying glass focusing a blinding light, wakes up having been abducted by Gray cyborgs. He is in a closed room with no windows. He manages to undo the straps of the table bolted to the floor and he begins to float off the table. He holds on to one of the straps. Being a good physicist as well as story writer, Kafka knows instantly he is on a timelike geodesic out in space. Warp drive is the same experience the difference is that the Captain can control his ship's geodesic glide path to point in any direction and speed he likes without feeling any g-forces at all because no propellant is ejected in the maneuvers including sharp turns in space even instant U-turns. This is metric engineering the dark energy-matter envelope around the ship.

Kafka manages to find a port hole cover. He opens it. To his horror he sees that the ship is falling toward the Sun! That explains why it's been getting hotter. He sees a control panel on the side of the table he had been strapped to. He pulls on the strap to get closer to it. In desperation he starts playing with a joystick and suddenly he is pulled to the table. He feels weight returning. The ship's normal impulse swivel rocket engines have been switched on. He plays with the stick and some of the knobs and finally through his psychic intuition he finds a setting in which the ship stands still relative to the distant Sun that fortunately is still as far away as Venus. Indeed looking in a different direction he can see the planet Venus. The zero point energy powered rocket engines must continually fire to keep the ship hovering at a fixed distance from the Sun and from Venus. This hovering fixed point is a non-geodesic path through spacetime. Having wiped the sweat off his brow, he sees a spigot on the table side and is able to get some sweet-tasting water that instantly gives him a sense of well-being and relaxation in addition to a feeling of great muscular strength and youthful vitality. At that moment three Grays enter the room. The delicate one in the middle who seems more feminine than the other two says to him telepathically without sound "That was nicely done Professor Kafka. We apologize for putting you under stress, but you were in no real danger. Would you like us to explain to you how our ship works." Kafka answers - thinking telepathically "Is The Pope Catholic?"

On Aug 1, 2006, at 3:23 PM, Paul Zielinski wrote:

Jack Sarfatti wrote:

Paul you are fundamentally wrong. I need to run now. More later tonite.
You are not the only one confused by this. It's subtle. There is no intrinsic gravity force. It's an inertial force created by nongravity forces pushing test particles off geodesics.

Yes it is very confusing owing to the nature of the GR terminology, but I can assure you I'm not the one who is
confused here.

It's also confusing because gravity really is a force in Newton's theory, though not in Einstein's. Einstein eliminates gravity as an intrinsic force like electromagnetic force replacing g-force with curved space-time.

Note the "curvature" in internal fiber space is the gauge "force". Therefore, the correct analogy between gravity and electromagnetism

space-time curvature analogous to electromagnetic force

d^2(x^u - x'^u)/ds^2 = R^uvwl(dx^v/ds)(dx'^w/ds)(x^l - x'^l) gravity tidal acceleration for a pair of geodesic test particles

(This local frame-independent tensor equation can be calculated in either a LIF or a coincident LNIF)

is properly compared to the electromagnetic Lorentz force tensor equation

d^2x^u/ds^2 = (e/m)F^uvdx^v/ds for a single test particle with charge to mass ratio e/m.

In the language of Cartan's differential forms for space-time gravity for zero torsion limit

R = DW

R = curvature 2-form

W = spin-connection 1-form

D = d + W/\ = covariant exterior derivative

/\ = exterior multiplication

For general internal symmetry Yang-Mills fields (EM as trivial Abelian case)

D = d + A/
A = 4-potential connection 1-form

F = DA = Yang-Mills curvature 2-form ---> Yang-Mills 'force field" analog to Lorentz force eq.


DR = D^2W = 0 (Gravity) and DF = D^2A = 0 (Yang Mills) analog of Faraday's law & no magnetic monopoles


D*W = *J source eq. D*F = *J (Yang-Mills) source eq.

Here the analogy breaks down from the equivalence principle.

The Gravity source equation is a 4-form equation because W is a 1-form and *W is a 3-form. So *J is a 4-form.
J is 0-form local scalar.

With tetrad indices this is

D*Wab = *Jab

that maps to

Guv = kTuv

Local conservation of matter source stress-energy current density is from

D^2*Wab = D*Jab = 0

i.e. Tuv(Matter)^;v = 0

In contrast for Yang-Mills fields

D*F = *J is a 3-form eq. analog of Ampere's law and Gauss's law.

D*J = 0 local conservation of electrical current densities.

The equivalence principle is profound and many-faceted. There are deep similarities but also profound differences between the internal symmetry (extra dimensions of space) gauge theories and the space-time gauge theory of GR - the key difference is the equivalence principle.

The geodesic in Newton's theory is a straight line in Euclidean 3D space decoupled from 1D time at constant speed, no acceleration. Global inertial frames cover the entire space.

A test object falls freely to Earth. In Newton's theory the surface of earth is approximated as a GLOBAL INERTIAL FRAME. The parabolic trajectory of the test object is explained as a real gravity force in the Earth as an inertial frame. Einstein turns this upside down - PARADIGM SHIFT! The Earth is now a LOCAL NON-INERTIAL FRAME i.e. no global frames and the geodesic is completely different. The Einsteinian geodesic is in curved 4D space-time it can be a precessing ellipse in 3D space with accelerations relative to Earth frame at a focus. There is no pure gravity force on the test object because by definition real forces push test objects off geodesics.

Of course in both Newton's and Einstein's theories an observer on the freely-falling test object is weightless. However, the explanation for this invariant fact under morphing between the theories is very different.

Your error Paul is that you are trying to force Newton's explanation on Einstein's theory. It doesn't work! Apples and Oranges.

Any g-force is 100% an inertial reaction force in the local rest non-inertial frame of the test object generated by a non-gravity force (for all practical purposes an electromagnetic force).

Let's take some examples.

In globally flat empty space-time the metric is that of special relativity c = 1

ds^2 = dt^2 - dx^2 - dy^2 - dz^2

This formula is an INERTIAL REPRESENTATION of invariant ds^2 for tiny inertial observer-detectors on timelike geodesics. In this case the timelike geodesics are Newtonian to a good approximation.

Obviously the Levi-Civita connection field (whose special combination is g-force) is globally zero in this inertial representation because the first partial derivatives of the constant metric components vanish. Obviously the geodesic deviation curvature tensor components vanish also. Note two kinds of geodesic deviation - Ricci components for a cloud of GEODESIC TEST OBJECTS and WEYL CONFORMAL TIDAL STRETCH-SQUEEZE (e.g. gravity waves). The pure gravity vacuum field without dark energy and torsion is 100% Weyl tensor. See Penrose "Road to Reality" for more details on that distinction.

However, we can still do a GCT (General Coordinate Transformation) on our flat metric to get

ds^2 = guvdx^udx^v

Now we will get a non-vanishing connection field!

What does this mean?


Paul one of your errors is thinking only FORMALLY without thinking of the OPERATION MEANING of the GCT. What we have here is a NETWORK WEB of tiny NON-GEODESIC OBSERVERS firing their rocket engines arbitrarily. Not one of these observers could get on a non-geodesic were it not for the electromagnetic force. We can neglect consideration weak-strong forces here for all practical purposes since they are short range and we are at macro-range here.

Now compute the geodesic deviation - what is it? All components STILL ZERO. The curvature is zero!

How do the local non-geodesic observers do the experiment? -They can use tiny water drops if you like. The drop will not change shape. Better to use a dust cloud so as not to correct for surface tension.

Next in isotropic radial coordinate (that Hal Puthoff confounds with Schwarzschild radial coordinate in his wrong PV gravity theory in Eric Davis's USAF "Teleportation" study) in weak slow test particle speed limit, with UNIVERSAL Newtonian gravity potential energy V per unit test particle mass with a SSS (Static Spherically Symmetric) SOURCE:

ds^2 ~ (1 + 2V)dt^2 - (1 - 2V)[dx^2 + dy^2 + dz^2]

In Hal Puthoff's PV theory replace (1 + 2V) by e^2V

But this too is a representation in terms of HOVERING local non-inertial observers firing their rocket engines just so such that they all experience WEIGHT/MASS = UNIVERSAL "g-force" (loose terminology)

gi ~ - dV/dxi

ds^2 = [(1 - 2m/4r*)/(1 + 2m/4r*)]^2dt^2 - (1 + 2m/4r*)^4(dr*^2 + r*^2dAngle^2)

Note to get to usual Schwarzshild non-inertial representation use

r = r*(1 + 2m/4r*)^2

ds^2 = (1 - 2m/r)dt^2 - (1 - 2m/r)^-1dr^2 + r^2dAngle^2

really only for 2m/r < 1

Then V = - m/r

Area of concentric sphere about center of symmetry is 4pir^2

V is dimensionless in these units

g ~ -m/r^2 ~ 1/Length

However, now when they use tiny dust cloud of geodesic nanobots with transceivers to central computer they get in empty space the Weyl conformal curvature components C of the form (loosely speaking)

C ~ m/r^3 ~ 1/Area

Now here is the crucial point - in first case of globally flat space-time you can MOCK up the g-field to look just like the second case, but you will not find geodesic deviation i.e. you will find C = 0.


LOCALLY all g-force measurements are same in both cases. Only when you do a completely different kind of measurement can you tell that there is a source in second case and none in first case. In both cases the local observers are in NON-INERTIAL REPRESENTATIONS!

Another point Paul is that you cannot think in the Newtonian manner, as you do, that

g-force = objective intrinsic tensor g-force - compensating local frame-dependent g-force

such that

g-force = 0 in a Local Inertial Frame coincident with test object on geodesic.

Note that the geodesic equation for test object is in frame-invariant form

D^2x^u/ds^2 = 0

In the REST LIF of the test object this is simply

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

g-force (100% inertial) is only registered in the rest frame of the test object when it is OFF-GEODESIC i.e. a different equation

D^2x^u/ds^2 = F^u(non-gravity)/M(test object)

In the object's now local non-inertial frame

d^2x&u/ds^2 = 0 as before, but now

g-inertial force (AKA "weight") ~ Connection Field^i00 ~ F^i(non-gravity)/M(test object)

i = 1,2,3 space components.

You are confusing a coordinate transformation of a given geodesic trajectory with an actual change from
one geodesic trajectory to another. Even in standard GR there is at least a *mathematical* distinction between
the two cases.

No I am not.

Changes in observer motion alone cannot change the geodesic trajectory of any freely falling body;

Of course.

while a changing gravitational field can.

Of course.

In GR, frame acceleration simply changes the *coordinate representation* of a geodesic.

Correct, and that's consistent with what I do above.

Suppose the test object is charged in an electric field, then in a local LIF, NO LONGER A REST FRAME of the test object

d^2x^u/ds^2 = F^u/M =/= 0.

connection field zero in that momentary LIF.

In addition, it should be intuitively obvious that there are two separate components to the Einstein g-field, one
physical and frame-independent, and the other subjective and frame-dependent.

You are wrong. Your intuition is no good here. That is proved by the fact you have not been able to formulate your intuition here mathematically in many years and you will never be able to do so because it is impossible. You are forcing Newton's world picture into Einstein's incompatible world picture - never mind that they give same answers to a good approximation in many ordinary situations.

Let's suppose there is a material source
located at some point, generating a static gravitational field of some kind, and we are observing this in a frame of
reference that is stationary with respect to the source. Now transform into a frame of reference that is accelerating
away from the source. If, keeping that frame of reference fixed, we suddenly remove the source to infinity, both the
tidal and the physical part of the g-field vanish, leaving only the inertial field. This shows immediately the physical
link between tidal curvature and the objective physical part of the g-field. In GR, there is also a corresponding
mathematical link between the two properties (tidal and g-field) of the curved metric. This also shows that the
distinction between the inertial and gravitational components of the combined g-field is *physically meaningful*
(since one component causally depends on the matter distribution, while the other does not).

What I am doing here is treating the objective physical g-field just like, say, an electrostatic field.

Yes, I know you are and that is an error!

The problem is that we have a metric field guv ~ Newton's potential

Einstein's connection ~ gradients of Newton's potential

Einstein's curvature ~ gradients of gradients of Newton's potential

Your analogy to electrostatics is naive, superficial and completely wrong. It's what all amateurs do!

V(Newton) ~ -GM/r

V(Coulomb) ~ - Q/r

However, in Maxwell EM field theory

V(Coulomb) & Vector Potential A form the CONNECTION FIELD in internal S1 fiber space!

This is where your analogy profoundly breaks down and your entire thesis is not even wrong!

To take your analogy to it's absurd conclusion:

Einstein's metric would be a space-time connection since Newton's potential is part of it!

Now you can also see your error also in the Reissner-Nordstrom geometry with a charge on the SSS source where

(1 - 2m/r) is replaced by (1 - 2m/r + Q^2/r^2)

If we were to use your false analogy this should be, by your faulty intuition,

(1 + 2V(Gravity) +2 V(Coulomb) ) = (1 - 2m/r + Q/r)

The point is that in Maxwell's local gauge theory of the electromagnetic field, the Coulomb potential is the timelike part of a 4-potential connection field in internal symmetry space. Think fiber bundles.

But you cannot think of Newton's potential as part of the space-time connection field!

The difference is the equivalence principle. There is no equivalence principle for the EM force (not allowing Kaluza-Klein theory here).

Note that the gauge freedom of the connection fields is GCT in the case of the space-time connection.

Gauge transformations of EM/Yang-Mills connection fields are INHOMOGENEOUS just like space-time connection.

So in Einstein's GR gravity you have 3 levels

Metric (Newton's potential in weak field limit) tensor with local invariant ds^2

Connection (g-force) non-tensor NO local invariants! (barring tensor torsion of course)

Curvature tensor with local invariants.

There is also the 4 subspace level of tetrads where the metric is symmetric bilinear in the tetrads.

Curved tetrad is from local gauging of T4 - zero torsion.

Then in my emergent gravity theory the tetrads come from variations in the World Hologram Goldstone Phases of the Higgsian type Vacuum ODLRO inflation field. So this is a 5th level.

But in Maxwell & Yang-Mills local gaugings of internal U(1em), SU(2 weak) & SU(3 strong)


The Coulomb potential is part of the CONNECTION

The tensor FIELD CURVATURE gives the electrostatic force that you FALSELY compare with g-force.

It should now be obvious why your entire thesis is PROFOUNDLY NOT-EVEN-WRONG.

Paul you have made a very common naive error made by many amateurs trying to make FREE ENERGY MACHINES and explain UFO saucer propulsion.

I am not reading the stuff below based on your false analogy of gravity to electrostatics.

Another way to see it is that

EM, Weak, Strong are renormalizable SPIN 1 theories.

Gravity is unrenormalizable SPIN 2 - but that gets into quantum field theory complications

The apparent net
effect of the electrostatic field on the motion of a charged particle depends on the observer's frame of reference,
because in certain observer frames the effects of the electrostatic field are compensated by the effects of frame
acceleration. Of course that doesn't mean that the electrostatic field itself is actually changing -- it is simply being
offset, as viewed from a given frame of reference, by an inertial field (of fictitious forces). Clearly in this case
the electrostatic and inertial fields are "apples and oranges", even in 100% orthodox textbook GR.

The situation is exactly the same for gravitational fields in GR, in the model I am proposing. The electrostatic example
clearly shows that such an "inertial compensation" interpretation of GR is feasible as a physical model.

This is why quantum fluctuations are in creative tension with classical equivalence since they are non-geodesic from the Bohm quantum potential which mimics electrical forces in a sense. The intrinsic gravity field is the relationship among the geodesics - hence curvature as "geodesic deviation". That you can write the connection field as a smooth field is relative to a set of detectors like hovering detections in standard REP is deceptive causing your confusion. But it has no scalar invariants i.e. is not a tensor field.

Jack, it is you who is confused here. The "forces" that are felt when you push a freely falling body off a geodesic
have nothing to do with *fictitious forces*. In GR this is simply a form of inertial resistance, which *appears* in
certain frames as inertia, while it *appears* in certain other frames as weight. This equally apparent character of
inertia and weight is what Einstein hoped would give him his "general relativity" and explain the equality of
gravitational and inertial mass. But when you look at his final 1916 theory, you immediately see that geodesics are
defined in a full covariant and completely objective manner, and frame transformations have the effect of changing
only the coordinate representation of geodesics, and not the actual geodesics themselves. So the most natural
interpretation of the 1916 formalism is the inertial compensation model I am proposing, which is however inconsistent
with Einstein's original concept of "general relativity".

It's just that in my model, the combined gravitational-inertial field of Einstein has a separate and distinct objective
component that is causally tied to the source(s) of the field. Eliminate the sources -- in a fixed frame of reference --
and you automatically eliminate both the tidal effects *and* the effects of the physical observer-independent part of the
g-field. Thus both the tidal and physical g-fields are causally related to the source of the actual physical
gravitational field. What is left after you remove the source while keeping the reference frame fixed is the pure inertial
part of Einstein's field.

This is like

F(x,y) = X(x) + Y(y)

So these two components really are apples and oranges, even in GR. In this respect GR is actually not significantly
different from Newtonian theory, despite all the hype. Strange, but true.

This is tricky and deep.

It is tricky.


On Aug 1, 2006, at 12:07 AM, Paul Zielinski wrote:

Jack Sarfatti wrote:

> Z: One aspect of this field is the action of so-called "g- forces" that act on
> test bodies;

WRONG. The g-force is NOT a property of the the gravitational field!

OK, then here is the nub of our disagreement.

I'm saying that there is an irreducible frame-independent component of the combined
g-field that is directly linked to curvature. I am saying that if you remove the sources
of the gravitational field, in a fixed observer frame, then this physical g-field
component goes to zero, along with the tidal forces, while the inertial g-field
component is unaltered.

The g-force is a property of non-gravity forces that create non- geodesic paths!

I say this only applies to the frame-dependent part (inertial part) of the combined

Pure gravity is all geodesic paths.
Curvature is a property of geodesics.
There is no such thing as pure gravity force in GR.

Yes there is. That's the whole point.

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