What Creates Gravity?

Thanks for the suggestion but there is a quote popularly attributed to Einstein.
"If facts dont fit the theory change the facts."
It is anyones guess what will happen if I challenge the fundamentals of the theory.
However isnt it obvious that any Local frame must be capable of defining its measurables... I.e it must be able to define length , volume , time and several distributed properties like pressure , temperature.An infinitely small local frame can not define those concepts.
Local frame is different from a point mass or point velocity ...
Local frame is allows the local observer to define and make observations!!

We don't - at least those of us who know what we are talking about...But if you are ever to go beyond Einstein's Theory, you first have to understand Einstein's Theory. This is the hurdle you are falling over. Obviously not, given those of us who have carried on working on space-time physics following his death Errr, I think we are well aware of this - ever tried to model an electron with the Kerr-Newman metric? No? Remember the hurdle I mentioned? Who cares about your point of view? We are talking about mathematical physics. What do you call finite? As big as you like? No, it's not. Wasn't this clear from my previous message? If it wasn't clear, you need to brush up on your reading skills. So??? Yes, and again, so??? Given your utter ignorance of the theory of General Relativity, guess how much Your Opinion matters?The world lines of point particles trace out geodesics through space-time. The world-volumes of larger bodies will be complicated by the inter particle forces. Care to explain why this spells the downfall of GR???Edited by cavediver, : No reason given.

If the local volume is not zero and the gravitational force is not uniform then it is possible to distinguish between accelerating frame and gravitationally accelerating frame.
That was the whole point of this discussion.In other words a gravitational free fall can be distinguished from a induced free fall in zero gravity because both these forces act differently on a finite volume which is the metric of GR.Another example involves a spherical shell of mass...
In a spherical shell with uniformly distributed mass there is no net force on anything inside the sperical shell.
However in this case zero acceleration is not physically equivalent to a spherical shell.In other words observations made in the local frame of reference does not translate into gravitational frame of reference.
Had it been so a mass at rest would never know whether there is speherical shell encapsulating its space?When we study the free fall we make the local frames equivalent to gravitational frames.. in other words geometry of the free fall translates into gravity..quote from wikipedia Which means that the inertial frame of reference exists locally and is definable such that it is indistinguishable from any normal accelerating frame of reference.
It is important that by the definition of local frame of reference the physical laws should be definable and observable...
However in differential manifold such observations can not be made at all because the defintions of length,volume , velocity ,time can not be described by living inside the metric.
A local frame is definable if and only if laws of physics can be described and studied in the differential volume independent of the external observer.

If the local volume is not zero, and the gravitational field is not uniform over that volume, then you do not have a frame... So? You might want to think about re-writing these statements because at the moment they make little sense. A combination of poor English and weak understanding does not make for an edifying exposition of relativity... an inertial frame follows geodesics. An accelerating frame does not... I'm not sure what you are trying to say here.Are you still claiming that GR is invalid because the tangent space is not equivalent to its base space?Edited by cavediver, : No reason given.

To make the point more clear.
1.Let us assume that differential local frame of reference is definable.
i.e dV=dX.dY.dZ is definable as a frame of reference. (mathematically dV tends to zero)
and that the equivalence principle holds.
2.Since we have defined it as our frame of reference therefore it should be possible to measure dV , dX,dY and dZ .... since these are arbitarily small quantities therefore we must propose the existence of arbitarily sensitive instruments because dV is the local frame of reference and quantities in local frame of reference can be measured independent of external observers.
3.Given the arbitarily sensitive and arbitarily miniaturizable intruments it should be possible to define and measure dX/2,dY/2 and dZ/2.
4.Given a gravitational force in dZ direction ,the dV must experience an arbitarily small spatial difference in acceleration too(for example at dZ and dZ/2) which should be measurable by the instruments.
By defintion our instruments are capable of measuring differential change.
5.Therefore it should be possible to differentiate between gravtitationally accelerating frame and non-gravitationally accelerating frame.
6.This is in contradiction with the Equivalence Principle.
7.If local frame of reference can detect gravity then it is not possible to equate the local geometry with gravitational presence because they are differentiated by the differential measuring instruments.

Yes I am still claiming that the GR is invalid because the local equivalence principle can not hold.
In the previous example the differential acceleration is dA=|GM/R^3|dZ
Which our instruments can easily detect because local frame of reference is definable.

The only point you make clear is that you have no clue. Yes, a measurably large frame will be able to detect differences in curvature across that frame. Great. Wonderful. So what??? Perhaps you could follow through the mathematics to show me how this is a problem for GR. Forget about the equivalence principle. Show me the algebra... and then explain where GR would fail and what experimental observations would be affected.

I'm sorry, but under my doctor's recommendation, I'm only allowed to repsond to a certain quota of stupid each day, and I'm way over limits now.

Algebra can not be created because the basic principle of equivalence is wrong.
I fail to understand when you ask so what ?
Are you asking ,So what if the equivalence principle is wrong ?
In that case the theory of GR is wrong.I thought that the thread is dedicated to gravity and physics that is why I posted here.
Anyways here is my last attempt.
Equivalence principle states that locally accelerating frame can not be distinguished from gravitationally accelerating frame.
Locally accelerating frames can be defined if and only if the physics can be defined inside those frames.
A metric which is used to represent local frame of reference is arbitarily small.
What properties must be definable inside any inertial frame?
Inside a local inertial frame the observer should be able to measure length , height , width , change in length , change in time , change in velocity...etcWhat happens when we discuss a free falling BOX?
Suppose we replace that metric with a freely falling BOX then the BOX will experience an internal stress due to the dimension dependent gravitational force... the force of gravity at the top of the box will be different from the force of grvaity at the bottom.
Here we are not talking about the geodesic of the BOX in space but the force distribution on the BOX due to external gravity which is the basis of equivalence principle.
Therefore IN THE CASE OF A BOX we can easily find out whether an external force of gravity is acting on it or not by following different points on the BOX thus violating the equivalence principle.
That is why we say that GR is local theory because it must use a metric. It can not use any arbitary frame of reference.What happens when we discuss from the point of view of a Metric?
A metric is infinitely small inertial frame of reference.
Metric holds the equivalence principle.Everything is fine so far. BUT what is an inertial reference frame?
An inertial reference frames share common set of physical laws and it is not possible to define Absolute Value of velocity. The velocities of inertial frames are relative and equally true.
BUT a more important property which is assumed while defining these frames is that measurements in these frames can be made independently of an external observer.
The metric which is local and infinitely small must be also capable of measuring its infinitely small physical values independently of the external observer because it is an inertial frame.
Therefore in the local frame of GR an infinitely sensitive instrument must also exist so that it can independently define and verify its physical laws.
There is nothing in principle which stops this from happening. It is just matter of scaling.
ANd if infinitely small measurements are possible in infinitely small metric then the metric can be replaced with the BOX!!Because BOX is made up of several points in a space therefore metric must also be definable in a similar fashion otherwise it can not define lines , volumes ,velocity and acceleration independent of external observer.
As I said the complete laws of motion must remain definable inside the infinitely small metric.
Which means the differential force of gravity in the local reference frame(which is now equivalent to our poor Box) can be measured violating the equivalence principle.Hope that makes things little bit clear but I am convinced that GR doesnt hold its promise of a conceptually consitent theory.

The metric is the entirety of space-time. I would not call that arbitrarily small. No, it most certainly is not. It is symmetric tensor field spanning all of existence. Well, geodesics described by the metric give rise to the equivalence principle. This is simply nonsense. And this makes even less sense if that is possible. You have some very misguided ideas about GR. It is blatently obvious that having a curved manifold means that the equivalence principle will only be exact at a point, and any extended laboratory will reveal the local curvature. This is not a problem for GR, it is a f'ing prediction. If it is predicted by GR, how can it possibly be a problem for GR? How could the EP possibly suggest that an extended frame would not be able to reflect the local curvature? Your thinking is simply bizarre.

The Equivalence Principle holds only for very special Gravitational fields(homogenous).
However in actual scenario the "true" gravitational field can be found as a non vanishing metrical curvature which are held responsible for the tidal gravitational forces.Therefore it is not a principle in a sense defined by the original statement...
It is at best an instance of Minkowski space under homogenous gravitational force... which was extended to include non-homogenous forces of gravity.

No, the EP holds for all gravitational fields if you consider it at a point. Sheesh...Repeating the same obvious point over and over again does not make the case for General Relativity any weaker. Perhaps it is time for you to move on to another topic?

I was not repeating myself. The non-vanishing metric curvature is the proof of gravitational field. In a constantly acceleration frame the metric curvature of vanishes.
Anyways the point is it is most ill understood concepts of GR.

The metric IS the gravitational field... No - in a freely-falling frame, the LOCAL metric can be regarded as flat. This is a consequence of the Pseudo-Riemannian geometry of space-time. Yes, by those that do not understand General Relativity, this is very ill-understood. But those of us who have spent a good portion of our lives working in this area, it is very well understood.

g ,the Metric of general relativity, represents the grvaitational field however some scientists have said that R represents the gravitational field. R=0(reimann tensor) for Minkowski space with a homogeneous gravitational field.
However R is not equal to 0 if there is a real gravitational field.