Wednesday, May 18, 2005

The Meaning of General Relativity

On May 18, 2005, at 11:00 AM, Jack Sarfatti wrote:

Zielinski bends every word of Einstein to ludicrous extremes in a warped caricature of relativity.

On May 18, 2005, at 2:26 AM, wrote:

No, Jack, I'm afraid that Einstein was very serious about his "general relativity". In order to make mutually accelerating reference frames K and K' physically equivalent, he was led to posit an *identification* of gravitational and inertial fields, where one was no more or less physically real then the other -- which from my POV amounts to a conflation of *subjective appearances* with *objective physical realities*.

Zielinski bends every word of Einstein to ludicrous extremes in a warped caricature of relativity.

I showed you what that really means Paul. Einstein is completely correct.

Zielinski bends every word of Einstein to ludicrous extremes in a warped caricature of relativity.

GR is a mathematical theory.

What Einstein means Paul is in the math, i.e. in the REST LNIF of the test particle of total mass m

inertial "g-force" c^2(LC)^i00 = f^i(non-gravity force)/m

In particular when you arbitrarily choose HOVERING LIFS, then guv for that special choice in the SSS case for r > 2GM(source)/c^2 is LF invariant local Diff(4) scalar

ds^2 = guvdx^udx^v

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

The theory is completely consistent and beautiful.

Starting from there you can make Diff(4) GCT's Xu'^u LNIFu <-> LNIFu' or tetrads eu^a LNIFu <-> LIFa

This was later modified to *local intrinsic equivalence*, wherein Einstein still did not regard curvature as an essential property of the gravitational field.

Measuring the local geodesic deviation is a completely different experiment than measuring g-force. You never have to bring in curvature to discuss the gravity "field" in the sense of (LC)^i00. That's additional information about (LC)^uvw, l.

Gravity field g-force per test particle mass = c^2(LC)^i00

i = 1,2,3

The point is that you can have objective Einstein *geometrodynamics* (with objectively determined physical geodesics) without Einstein equivalence. All you really need is *weak* equivalence.

On May 18, 2005, at 2:13 AM, wrote:

Jack Sarfatti wrote:

Paul, the key to your confusion over "reality", "objectivity" and "existence" is your misunderstanding of what "zero" means for physical quantities.

So your position is that an object can be an object without being an object?

Zielinski bends every word of Einstein, Wheeler and Sarfatti to ludicrous extremes in a warped caricature of relativity's reality.

The point about a quantity that is not zero at a point, being reducible to zero at that point by a *coordinate transformation* is that it is not a *mathematical object* with respect to coordinate transformations on the space in which it is defined.

Who cares? It's physically real if it moves a pointer on a good detector - including "null measurements" of course! What you mean is that (LC)^uvw is not a local Diff(4) nonlinear tensor. It is global T4 linear "affine" tensor. Nothing physically unreal if a measured property is zero in a special frame and non-zero in another! x = 0 does not mean "x does not exist"! A physically real OBJECT does not need to be a tensor under ALL groups G. The laws of nature need to be tensor equations under all physical groups G, but their vacuum states need not be G-invariant - that's ODLRO.

You confuse "variable x having a value zero" with x not existing at all! I will find your quotes below on that.

Jack, show me a field quantity outside of Einsteinian physics that can be made zero at any given point by a coordinate transformation.

Magnetic field B from a test charge in a transformation to the rest frame of the test charge in special relativity. But that is Einsteinian physics.

In Galilean-Newtonian particle physics, "speed" of the particle is a physically real property that is zero in the rest frame of the particle.

Paul your criterion of reality is far removed from experimental physics.

A scalar is a tensor of rank zero. Therefore if a scalar quantity is not zero in one coordinate system, it is not zero in any other.

So what?

A variable that is not zero at a point in one coordinate system, but is zero at that point in another is not a scalar quantity.

Correct, relative to the group of coordinate transformations that you left unspecified so that your ill-posed remark was meaningless.

Am I really "confused"?

Yes! You said "speed" was a scalar! It's not a scalar under O(1,3) which is the relevant physical group there!

I am NOT saying that when (LC) -> 0 it is no longer *defined*; I'm saying that if it is not zero in one spacetime CS but vanishes in another, then it is not a *mathematical object* with respect to general spacetime coordinate changes.

So what? That has nothing to do with physics. (LC) is physically real. It is locally measured as g-force. Inertial frames by definition have zero g-force. Non-inertial frames by definition have non-zero g-force. To say that g-force is not physical real is a bad idea. Physically real properties are what can be measured by actual detectors in actual experiments. We infer the existence of an "object" from a collection of such measurements.

BTW with the LC field you can only make it 0 at a single event P in an LIF each time. You cannot make it zero globally when space-time is really variably curved!

True, but immaterial.

Only from your POV.

The point is that a non-trivial LC connection field can be made to vanish at any point by a spacetime coordinate transformation, so it does not represent a mathematical object in 4D spacetime.

We don't care if it's a "mathematical object". Experimental physicists are not interested in "mathematical objects". They are only interested in "physical objects".

One aspect of "objectivity" in physics is the search for spacetime *objects*. The LC connection is clearly not an object with respect to the full class of general spacetime coordinate transformations.

Of course it's not, but it's physically real!

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