Science > Physics > GRAVITY'S FORCE WITHOUT FORCE IN EINSTEIN'S GENERAL RELATIVITY
| Topic: |
Science > Physics |
| User: |
"Jack Sarfatti" |
| Date: |
11 Dec 2004 10:11:16 PM |
| Object: |
GRAVITY'S FORCE WITHOUT FORCE IN EINSTEIN'S GENERAL RELATIVITY |
Newton's "gravity force" is replaced by Einstein's "gravity force
without force".
On Dec 11, 2004, at 5:12 PM, wrote:
I am talking here about *Einstein* GR. I am not talking here about my
model, which is not *Einstein* GR
Your model is not even wrong. It's a huge Malapropism.
Do you think I ever denied that my model is inconsistent with Einstein
equivalence? Of course it is.
I am glad you admit that. Ipso facto you are off your rocker!
There is no "contradiction". (LC) can still be interpreted under an
Einstein equivalence model even if it is *mathematically* decomposable.
It is not so. That is false. I mean under the restricted rules of the
1916 theory itself. Only the LC connection. No rigid affine connection
allowed. The latter makes no sense physically. Metamathematically Alex
enlarges the formal system beyond Einstein's 1916 formal system. That is
not allowed in your problem.
"But if it is decomposable, it can *also* be interpreted under an
alternate non-Einsteinian model in which the equivalence hypothesis is
relaxed."
1. Your premise that LC is decomposable is false.
2. The equivalence principle is confirmed by experiments. You cannot
simply "relax" it. That is not physics. That is shmizzicks.
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!
It's not the simplest possible problem. The SHGF is.
I have repeatedly asked you to define "SHGF" you have not. If you cannot
solve the Schwarzschild problem within your own terms you have NOTHING
OF VALUE.
Also, that I have not yet done it doesn't mean it cannot be done, which
is all that logically matters.
Physics is more than your "logic" that Socrates called "sophistry".
This is Yilmaz's mistake. You do have a {LC} connection field in curved
spacetime.
Of course.
{LC} = 0 in any LIF at any point of the manifold.
{LC} = 0 in any LIF at *some* point on the manifold.
No - at ANY point in the manifold! You got that wrong.
{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
OK. I agree with all of this.
Here mc^2{LC}^i00 is the TRANSLATIONAL INERTIAL FORCE of gravity to use
Newton's term.
It is the translational *effect* of the permanent gravitational field,
to use my terms.
NO! That even works in Minkowski spacetime without ANY permanent
gravitational field at all! That is, the conformal curvature tensor
vanishes everywhere-when. In other words the curl of LC connection is
identically zero!
The latter is bad to do because it mixes the two paradigms causing your
confusion Paul.
I agree.
Good, you agree you are confused.
That is why I prefer to use "effect" instead.
I am not trying to modify the geometrodynamics.
In my model, also, the motion of freely falling test particle moving
solely under the influence of a gravitational field is in reality "force
free".
But NOT FOR THE RIGHT REASON!
Zielinski's FORCE = T + N = 0 in the LIF
where T =/= 0. That is what the argument is all about. I say T = 0 ALWAYS.
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.
Yes, exactly -- and this is *gravitational compensation* by a
non-gravity force, and not "inertial compensation" as you incorrectly
described it.
No, mc^2{LC}^r00 IS THE DEFINITION OF THE "GRAVITY" INERTIAL "FORCE
WITHOUT FORCE" IN THE NON-GEODESIC REST LNIF
It is the "local" cancellation of the permanent gravitational field by
a non-gravity
force.
What do you do in Minkowski spacetime where curl{LC} = 0 everywhere-when?
You do not understand what "annihilated" means. It means your T = 0 ALWAYS.
It means no such thing. It simply means that [LC] = [T] + [N] = 0, where
[ ] indicates a matrix representation relative to a specific CS.
That's what the fight is. I say [T] = 0 in Einstein's 1916 GR and any
attempt to change that is not allowed.
In the absence of further physical interpretation, this is a *purely
mathematical issue*.
False.
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!
OK.
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.
I have no idea why you are saying this.
That's exactly the problem!
In an LIF, locally at some spacetime point (LC) = 0 because [T] + [N] =
[0] in the LIF CS, as I have explained.
Nonsense because your assuming [T] =/= 0 ain't kosher!
You simply do not understand manifold theory and differential geometry.
Neither does Hal Puthoff!
?
My discussion with Matt Visser at GR 17.
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.
I'll take this as a "no". :-)
Obviously not.
OK.
I am saying you do not understand what those words mean PHYSICALLY!
Of course I do. Empirically and operationally, it means no *effective*
net translation g-field at some point in the LIF, and therefore no net
translational effects on the motion of a freely moving test particle *as
observed from that
reference frame*. That's true in either model.
No, you have side-stepped the [T] =/= 0.
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!
So you're appealing to *Copenhagenism* to defend Einstein's concept of
observer reference frames as "moving coordinate systems"?!
Yes.
Einstein must be rolling in his grave!
Not at all. Einstein had great regard for Bohr. Einstein did not think
that every idea Bohr had was wrong! There difference was that Einstein's
ontology was objective local and Bohr replaced ontology with an
epistemology that was subjective and nonlocal.
But you actually may be onto something here. There is indeed a
relationship in my view.
It's Wigner's view. Actually it's group representation theory in physics!
In quantum theory the basis (choice of coordinates) is in qubit "Hilbert
space".
In relativity the basis (choice of coordinates) is in spacetime.
Yes, I agree that there is an interesting analogy here.
It's more than an analogy, it's a proportionality. This is what Feynman
said about Dirac's association of the classical action/hbar to the phase
of the quantum history amplitude.
When spacetime is flat you can use global coordinates.
Exactly. Then, in a global Cartesian CS, N = 0.
What do you mean by N?
In the global inertial coordinates in Minkowski spacetime, Cartesian or
not what you have is {LC} = 0 everywhere-when from which it trivially
follows that covariant curl {LC} = 0 everywhere-when.
When its curved you need local coordinate patches or charts weaved into
an Atlas with overlap transition functions that cover the manifold.
Yes, of course.
The overlaps are the GCTs.
What? GCTs can act locally on local charts.
Obviously. What do you think the local Jacobian matrix Xu^u'(P) means?
It's the overlap transition function between x^u local coordinates at P
and x^u' local coordinates at the same P, where now P is an objective
physical event - not a mere formal point in the manifold.
"Physics is simple when it is local." Wheeler. GR is a local classical
field theory that allows time travel GLOBALLY.
Are you confusing *transition functions*, that sew coordinate patches
together into atlases, with GCTs?
I am not confusing them. They are the same!
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!
I don't now about Hal, but I have no problem with this.
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)
Meaning R^u_vwl = 0?
First of all since there are NO material sources here you should know that
Riemann = Ricci + Conformal
where
Ricci = 0 when Tuv(matter) = 0
So you should say, in ordinary vacuum where-when
C^u_vwl = GCT COVARIANT CURL {LC} = 0 everywhere-when.
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)
This is THE LOCAL EQUIVALENCE PRINCIPLE in reverse.
Problem, is (2) a curved spacetime or not?
Solution.
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.
OK.
What do you mean OK? Then you have thrown in the towel? That is the
complete solution of your original question which was "How do you tell
the difference between a real gravity field from objective curvature of
space-time from an artificial gravity field in the starship and flying
saucer?
This property of the curl is INVARIANT under any further GCTs.
OK. Although of course the components are different in different CSs.
Obviously. Do not quibble over trivia.
This is the complete solution to the problem you raised.
Which problem?
"The Question is: What is The Question?" Wheeler
"To Be or Not to Be? That is The Question! Hamlet.
How long must I suffer The Slings and Arrows of Zielinski's Pointless
Dulled Points?
That the choice of coordinates is supposed in the Einstein model to
locally annihilate the gravitational field together with its physical
stress-energy?
Oh yes, that! Another time. How awkward!
Of course we know that the non-trivial curvature field does not vanish
in the LIF. But in the Einstein model the gravitational field is
supposedly still annihilated at a point in the LIF.
Yes, and you find that troublesome. I do not. That is exactly how it
should be.
In the contemporary interpretation, it is the curvature field that
represents the actual gravitational field, while the translation
"g-field" is all that is locally "annihilated".
Only in an LIF. If you are in an LNIF at a fixed distance from say the
center of Earth you feel the non-zero {LC} from the non-gravity force
holding you on the non-geodesic worldline keeping you at that fixed
distance!
Of course the effective net translation field does go to zero in the LIF
in either model; but in my model the geometric component [T] =/= 0 is
distinct and separate from the coordinate-originated inertial component
[N] =/= 0,
A false distinction. You cannot write this distinction explicitly in the
simplest Schwarzschild case. I say you can, but [T] = 0 in EVERY PROBLEM
in 1916 GR.
and it is only the former, together with Riemann curvature, that is a
manifestation of the "permanent" or "actual" gravitational field.
Excess ugly baggage. All you need do is calculate Curl{LC} is it zero or
not? That's all that matters.
I regard the contemporary model as incoherent at worst and strongly
counterintuitive at best, since it holds that the physical
gravitational field remains in existence while its local stress-energy
density is "zeroed out" in the LIF -- or else that there simply is no
such thing as local gravitational vacuum stress-energy.
I find this same situation quite coherent and satisfactory. I think your
discomfort is due to chronic acid reflux dyspepsia. I prescribe "Prevacid".
My model resolves this by separating the pure inertial contribution
I^u_vw from G^u_vw and R^u_vwl, and using the tensor quantities G^u_vw
and R^u_vwl together to describe the physical effects of the
distribution of gravitating matter. Then the effective g-field can
vanish in an LIF without annihilating thelocalized frame-independent
stress energy content of the "permanent" field.
How awkward.
Jack wrote: 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.
Z: Right.
Jack wrote: 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).
Z: OK.
Jack: What do you mean "OK" Jose? A mere puny "OK"? "My best conundrum
wasted!" Jack Point in Yeoman of the Guard. This is the beginning of a
great discovery!
.
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| User: "vonroach" |
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| Title: Re: GRAVITY'S FORCE WITHOUT FORCE IN EINSTEIN'S GENERAL RELATIVITY |
12 Dec 2004 06:10:58 PM |
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On Sun, 12 Dec 2004 04:11:16 GMT, Jack Sarfatti <sarfatti@pacbell.net>
wrote:
Humm.... Very interesting but could you elaborate a little on
`gravity's force without force'? `Gravity' can be reliable measured
by simple instruments regardless of how one choose to explain the
measurement (force?). It produces an acceleration of a mass which
implies a vector of force acting in the direction of the acceleration.
It a little curious when one explains this interaction of two masses
entirely on the basis of space distortion .
.
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