- On Feb 28, 2009, at 8:51 PM, Paul Zielinski wrote:JACK SARFATTI wrote:

You've already been definitively refuted Jack. Now I'm just having some fun.:-)

On Feb 28, 2009, at 2:49 PM, Paul Zielinski wrote:"...it is not so easy to free oneself from the idea that coordinates must have a direct metric significance."

- A. Einstein, "Autobiographical Notes"

Back in San Franciscore:Paul, Polemics are the last refuge of the man who cannot make a rational relevant refutation. ;-)That is delusional Paul. You have not thought deeply about what you are saying. In any case try to get your idea published.I have given you a detailed argument that you have not refuted in a rational relevant way.

But your argument is based on your own confusion about local coordinates in relation to curved spacetime geometry.

Curving the 4D spacetime of 1916 GR deforms the *geometry* of the manifold (metric relations between spacetime

points), NOT the coordinate assignments.Actually you are wrong about that. The choice of static local non-inertial frames changes in the SSS case. The covariant acceleration of the static frame is GM/r^2 so when you write{LC(M')}^rtt - {LC(M)}^rtt = G(M' - M)/r^2this is a conceptual relationship between two physically different frames of reference that actually do not simultaneously exist in a real physical situation. Your thesis is not even wrong simply garbled fragmentary ungrounded thinking in my opinion. I use physical detectors as local frame of reference.The meaning of the classical general relativity principle is the computation of local invariants based upon local measurements of two locally coincident detectors in arbitrary relative instantaneous motion measuring the same process up to quantum uncertainty limits.You can leave the local coordinate charts (MAPS from R^4 to the points

of the spacetime manifold) undisturbed (FIXED, CONSTANT, UNCHANGING, INVARIANT) while you deform

the manifold in any local spacetime region of interest.

In other words, you can arrange things such that under the gravitational deformation of the metric around any given spacetime

pointyou end up with the same local chart around*x*as you started with. The change in the intrinsic geometry doesn't force*x*

ANY change in the local coordinate assignments around.*x*The very concept of "gravitational deformation" is purely theoretical and cannot actually be measured unless e.g. the Earth loses chunks of matter in a collision with a large asteroid for example. It is not a valid measure of the actual local gravitational field i.e. tetrad 1-form e^I (Rovelli's definition)

That is something you still don't seem to understand. Hence the Einstein quote. You have not been able to fully separate the

concept of coordinate intervals from the concept of metric relations between points. I think this is the root of your problem,

and is the reason why you are having so much trouble understanding Levi-Civita.

Wrong.Synopsis, since there are interesting foundational issues here ignored even in some texts, though not in Wheeler's.1) Every time you write a particular metric field guv(x), and its associated symmetric torsion-free Levi-Civita connection field for parallel transport of tensors along world lines in 4D spacetime you have chosen a representation in terms of a network of local detectors (AKA local frames of reference).

Irrational response.All GCT coordinate representations of a given metric g_uv(x) are representations of a *single* metric field.Red Herring. No one says otherwise.

You seem to be saying that there is a different "metric field" for every GCT that changes the components of the metric tensor.I have said nothing of the kind. You simply do not get the point I am making because your mind is closed on this silly idea you have. It's completely silly - embarrassing in fact to my mind at least.The metric fields guv & g'uv induced by distinct sources Tuv & T'uv are not related by a GCT. They are different intrinsic gravity field configurations and they have different sets of static observers (outside event horizons) and, therefore, their formal difference has no direct physical meaning, i.e. no measurement procedures exist. They cannot simultaneously exist physically.

If that is what you are saying, then you are missing the whole point of the metric being a GCT tensor.Of course that is NOT what I have been saying.You are also missing the

point of defining affine connections on curved manifolds that correct the partial derivatives of tensors for coordinate artifacts.

Changing the coordinate representation of the metric doesn't affect the intrinsic spacetime geometry.

You are seriously confused about this. Your Einsteinian optics are preventing you from understanding the actual math.The simplest "hydrogen atom" (as it were) example in Einstein's 1915 GR is the SSS vacuum solution, written in the texts asg00 = gtt = - 1/g11 = - 1/grr = 1 - rs/rrs/r > 1rs = 2GM(source)/c^2Area of concentric sphere is A = 4pir^2usual spherical polar & azimuthal coordinate line element added.You can use any local coordinate chart around any given (well-behaved) point of interest, and hold that local chart invariant

under a gravitational deformation of the manifold.You are wrong. Physically you need to specify a set (fiber bundle) of local tangent frames that in principle are implementable by actual detectors. Here the choice of static frames (where they exist) shows my point very clearly. This completely eludes you. It is the fundamental error you have made. You are thinking like a mathematician and completely miss the physics. Different source configurations have physically different bundles of e.g. static local frames that are the basis for the contingent representation chosen for pragmatic convenience only. In any case I stop reading here and end the discussion. Think what you like but1) No journal will accept your idea and rightly so.2) No one pays any attention to it (except me here) and rightly so.That's my opinion.

You are systematically confusing actual geometric changes in the first partial derivatives g_ij, k with coordinate change.

The primary purpose of constructing an affine connection is to correct the derivatives of the metric for coordinate artifacts.

So the premise here is that a distinction must be made between coordinate contributions and actual geometric contributions

to the partial derivatives of tensors. This also applies to the metric tensor g_uv!

Nonsense. You cannot invalidate a mathematical theorem by attaching a physical interpretation to the math. To suggest that youNow Paul you look at the above as a purely mathematical problem.

can is completely wrong headed IMO.

You have to distinguish mathematically between actual geometric changes and mere coordinate changes in the partial derivatives

of tensors, including those of the metric tensor.

That is what the affine deformation model I've been trying to explain to you does so nicely -- for the metric tensor g_uv and its

associated metric-compatible LC connection.

Look Jack, you just got the math wrong, that's all. Do you want to repair the situation, or not?That is a profound error of conceptualization that you as The Profundus Philofawzer of Physics Extraordinaire ;-) ought to be well aware of - the "informal language" (David Bohm) the interpretation of the naked mathematique.

As far as I'm concerned both the the mathematical issues *and* the issues of physical interpretation are now completely settled,

based on

(1) Einstein's 1917 interpretation of the LC connection field LC^i_jk() as representing the non-tidal field strength in GR;*x*

(2) Levi-Civita's authoritative proof that the difference between two LC connections (his 2 and 2') respectively compatible

with any two *geometrically* distinct metrics (his 1 and 1') on a single analytical manifold is a non-trivial 3-index GCT tensor

rho^i_jk.

At this point any quibbling on your side about how different coordinate representations of the Riemann metric are supposed to constitute

distinct "metric fields" (which is mathematical nonsense, since g_uv is a tensor object) is just whistling past the graveyard as far as

I'm concerned.

It's also a violation of the mathematics. Levi-Civita's proof shows that the frame acceleration field, the strength of which is represented byThe above representation is for static (i.e. constant r, theta, phi) detectors. When M =/= 0 , they are covariantly accelerating in order to stay still in curved spacetime. This is Alice in Wonderland. You are not in Kansas anymore. You are not in Flatland. This is a violation of the Victorian Age engineer's common sense.

a non-tensor coordinate correction term X^i_jk, is only accidentally "glued on" to the actual gravitational field, represented to first order

by the gravitational deformation tensor G^i_jk.

That is the whole point of the linear LC decomposition

LC^i_jk = G^i_jk + X^i_jk

where the gravitational deformation tensor G^i_jk is a particular instance of Levi-Civita's rho^i_jk.

Your insistence that different coordinate representations of g_uv constitute distinct metric fields is just a fallacy. It

ignores the tensor character of g_uv.

Splitting out the gravitational deformation tensor G^i_jk doesn't change any of that Jack. So why do you keep bringingNewton's First Law of Motion in Einsteinian 1916 classical geometrodynamical language generalizes toAll covariantly non-accelerating massive test particles follow timelike geodesics of the gravitational field.

this up?

All fully covariant and consistent with tensor G^i_jk. So what's your point?Newton's Second Law of Motion generalizes toD^2x^u/ds^2 = F^u/mwhere m is the invariant rest mass of the test particle (assumed here not to be ejecting mass like a rocket) and ds is the proper time of m along the worldline. F^u is the non-gravity applied force*pushing*the test particle off-geodesic inducing g-force on the test particle.

I'm sorry Jack but IMO this is complete and utter nonsense.More generally, Newton's 2nd Law in curved spacetime for massive test particle motion is(D/ds)(mdx^u/ds) = (Dm/ds)dx^u/ds + mD^2x^u/ds^2 = F^u(non-gravity)Newton's 3rd law is not valid for global total momenta because translation symmetry is broken - see Noether's theorem.Instead we have the local stress-energy current density conservation lawD^vTuv(non-gravity source fields curving spacetime) = 02) g-force and the Levi-Civita connection field {LC}Back to the simplest caseD^2x^u/ds^2 = F^u/mWe now do a Godel self-reference and let the test particle detect itself.Note, the connection field (LC Christoffel symbol )usually has nothing to do with the test particle, but is a property of the detector, except in the degenerate self-referential case.

The geodesics are covariantly defined in GR and have nothing to do with "detectors" or "observers" -- except as to mere

*appearances*. The LC connection field includes linearly independent contributions from the coordinates *and from the

intrinsic geometry*. Otherwise, how could it be a metric compatible connnection?

Of course a straight SR free-fall trajectory for example *looks* curved to a uniformly accelerating observer *who ignores

his own accelerating motion*! So what? That doesn't affect the actual geometry of the geodesic in the slightest. And neither

does curved geometry force any changes in local coordinate charts. You are completely and hopelessly confused here.

These are two completely different animals.

I'm not going to read any further.

Z.The Baron Munchausen is weightless, no inertial g-force, in free-fall parabolic path (timelike geodesic in curved spacetime)He does feel g-force initially when cannon firesIn particular think of the above SSS case in which everything can be concretely computed. Z never bothers to compute an example. This is why his thesis here is not even wrong in my opinion.i = 1,2,3 = spacelike componentsi = 0 is the timelike componentLet's look only at the i = 1 radial component in the static LNIFD^2r/ds^2 = d^2r/ds^2 + {LC}^ruv(dx^u/ds)(dx^v/ds)However, in the self-referential "diagonal" casedx^i/ds = 0dx^o/ds = 1 (c = 1)d^2r/ds^2 = 0Therefore, the g-force per unit rest mass measured on the test particle itself is{LC}^r00 ~ +GM(source)/r^2This, however, is not an objective non-tidal non-zero tensor field. It is purely contingent based on an arbitrary, though pragmatically contingent, choice since we here on Earth closely approximate this situation.3) Z's Red Herring of the Gravity DeformationStart from M = 0 (flat space time) obviously{LC}^r00(M = 0) = 0here the static local non-inertial observer limits to the global inertial observer - a point overlooked by Z because he does not calculate a simple example.Now increase to &M{LC}^r00(&M) = G&M/r^2This is not a tensor obviously.The difference is not a tensor, i.e.,{LC}^r00(0) - LC^r00(&M) = G&M/r^2No tensors here!Note the natural assumption here is the adiabatic transformation of the global static inertial frame GIF to the local static non-inertial frame LNIF. The contingent choice of convenience is "static".Every time I hear a philofawzer say it's "pure math" or "your argument is mathematical nonsense" I reach for my delete button. ;-)"As far as the laws of mathematics refer to reality, they are not certain, as far as they are certain, they do not refer to reality."- Any intelligent fool can make things bigger, more complex, and more violent. It takes a touch of genius -- and a lot of courage -- to move in the opposite direction."
- "Imagination is more important than knowledge."
- "Reality is merely an illusion, albeit a very persistent one."
- "The only real valuable thing is intuition."
- "Anyone who has never made a mistake has never tried anything new."
- "Great spirits have often encountered violent opposition from weak minds."
- "God does not care about our mathematical difficulties. He integrates empirically."
- "Do not worry about your difficulties in Mathematics. I can assure you mine are still greater."
- "Not everything that counts can be counted, and not everything that can be counted counts." (Sign hanging in Einstein's office at Princeton)
- http://rescomp.stanford.edu/~cheshire/EinsteinQuotes.html

Jack Sarfatti wrote:What's fundamentally wrong in z' s argument is that he does not understand the breakdown of the global frame assumption of special relativity. You simply cannot use purely formal arguments. You must specify what detectors you mean. Physics is about detector responses.

Sent from my iPhoneStart with flat spacetime. The lc connection field is zero for the global inertial observer.Now introduce a small SSS Tuv matter source which induces an non zero lc field for the static non-inertial observer that adiabatically emerges if you impose a fixed r constraint.Since the difference is a tensorThat difference must be zero otherwise you have a contradiction.

Sent from my iPhoneNonsense.

The common non-tensor coordinate part just cancels out of the difference between the two LC connections,

which are themselves NOT GCT tensors due to the common coordinate parts that cancel.

You are just confused.

Z.

Jack Sarfatti wrote:Problem with the deformation approach is that it does not work. Start with flat spacetime and it implies lc is a tensor.Back in Sf

Sent from my iPhoneSo you are claiming that eq 2 is within a single network of observers e.g static Lnifs in the SSS case. We add a perturbation to tuv source inducing a metric perturbation. The take difference in lc connections and that is the 3rd rank tensor?Why did you not say that clearly to begin with?In any case it is not conceptually important to the foundations of relativity.Only the tetrad and spin connection cartan 1-forms are intrinsic nontidal measures of the gravity field ie scalar zero rank tensors under gct aka local t4. About to take off from jfk.

Sent from my iPhoneYou showed in detail that if you replace Levi-Civita's two different metrics 1 and 1' with the *same*

metric represented in two different coordinate charts, you get a very different quantity when you take

the difference of the corresponding compatible connections that obviously is not a tensor..

Which is just another way of saying that you don't understand -- or don't want to understand -- Levi-

Civita's proof.

I told you what Levi-Civita's rho^i_jk means mathematically: it represents the effect of a deformation

of the geometry of the manifold (from metric 1 to metric 1') on the Levi-Civita connection, free of

any non-tensor contribution from the coordinates.

In the context of GR, since in Einstein's model the net observed non-tidal gravitational field strength is

represented by the LC connection, clearly the tensor rho^i_jk represents the effect on the non-tidal field

strength of the change in the 4D spacetime geometry when you go from 1 to 1' -- free of any contribution

from observer frame acceleration.

If you take metric 1 to be geometrically Minkowski, then the difference G^i_jk between the two LC

connections obviously represents the effect on the gravity-free LC connection field of the change from

Minkowski geometry (n_uv) to gravitationally deformed geometry (g_uv).

Which means that the tensor field G^i_jk(x) represents the *actual* (i.e., objective, frame-independent)

non-tidal (first-order) gravitational field strength in GR, free of frame acceleration artifacts ("fictitious

forces").

Z.

Jack Sarfatti wrote:There is no physics at all in that snippet eq 2.

Sent from my iPhone at jfk boarding for SfoActually, we are talking about Levi-Civita's idea here. He's the one with two different metrics on a single

analytical manifold, with a common local coordinate chart around each point, under general transformations

of which his rho^i_jk (the difference between two metric-compatible connections) transforms as a 3-index

tensor.I showed in detail it is not a tensor if the usual physics is added to the naked math. It is a contingent relation between two networks of pairs of locally coincident detectors. - you are chasing your tail againthis is not interestingOn Mar 1, 2009, at 8:10 PM, Paul Zielinski wrote:JACK SARFATTI wrote:

I didn't say that Einstein ever said you could do this. After all, Einstein was no dummy.On Mar 1, 2009, at 6:12 PM, Paul Zielinski wrote:What this picture illustrates is exactly what I said it does: there is no actual physical equivalence of gravitation and frame acceleration

in a non-homogeneous field (which every physical gravitational field is, except locally).To me it shows exactly the opposite of what you say. You do not understand the meaning of the word "gravitational field" in Einstein's 1915 General Relativity.It shows that you cannot replicate such a field

globally with frame acceleration; and neither can you globally transform such a field away with frame acceleration.Red Herring. Einstein never said you could.

What he did do is consider the case of an imaginary perfectly homogeneous gravitational field, comparing it to a uniformly

accelerating frame of reference.

Now, there is in fact no such thing as a matter-induced *globally* homogeneous gravitational field.

So according to you, I suppose, his famous thought experiment "has no operation meaning"? :-)

That's right, but the only reason for this is Einstein's version of the equivalence principle -- which as it turns out isYou seem confused on every major idea distinguishing Einstein's explanation of gravity from Newton's. In Einstein's theory there is no such thing as an objective local non-tidal gravity force field.

empirically false, and not only that, it has no actual support in the mathematical framework of GR!

Dump Einstein's equivalence idea (I mean the one that literally identifies the intrinsic physical nature of gravitation

and frame acceleration) and the whole mess clears up Everything is then fully covariant (described by the tensors G^i_jk

and R^i_jkl) and we recover the traditional fundamental distinction between fictitious frame acceleration fields, on the

one hand, and actual non-tidal gravitational fields, on the other.

Two very different animals. Even in GR.

Same in GR, once you put Einstein's "white rabbit" back into the magician's hat.There is that, of course, in Newton's theory of gravity.

I think this is going to change.

Then the Einsteinian die hards retreat to their next trench and say, "well, not globally, but *locally* you can do this. Therefore, a gravitational

field is *locally* equivalent to frame acceleration."This shows how crazy you are. You are the die hard here. You are the throw back. In any case, the "Einstein die hards" as you call them control physics as a profession.

The equivalence principle is empirically false -- you can locally distinguish non-tidal gravitational attraction from frame acceleration byYou don't have a chance with your crank thesis - and rightly so in my opinion.

empirical means. This is because you cannot produce non-tidal gravitational attraction of free test particle geodesics with respect to sources

with frame acceleration! You can't do it Jack. Apples and oranges! And this is fully local, empirical, and operational.

As the relativist J. L. Synge famously wrote, to the extent that Einstein's equivalence principle (even in its "local" version) is non-trivial,

it is empirically false!

No response to (2) from Jack. Probably because he doesn't have one?But this doesn't work either, because

(1) Local tidal effects cannot be replicated and cannot be transformed away by frame acceleration, either globally or locally, with

the sole exception of a globally perfectly homogeneous matter-induced gravitational field -- which doesn't exist;Another Red Herring, Straw Man.

(2) Measurements of non-tidal gravitational acceleration d^2r/dt^2 of test objects with respect to sources can be made completely

"local", and since such gravitational acceleration cannot be produced or eliminated by frame acceleration, we can use such measurements

to *locally* distinguish between a frame acceleration field and a real gravitational field.

So the "locality" condition fails to do the job, Einstein's equivalence idea thus goes up in smoke. RED HERRING.

A heuristically fruitful red herring, but a red herring none the less. Not to be taken seriously -- in the context of justification.

Z.

It should be obvious that you can *locally* measure the acceleration of free test objects with respect to a typical gravitational source as

they approach the source boundary -- and this is a frame-invariant empirically measurable quantity that cannot be produced or eliminated,

or augmented or diminshed, by accelerating any observer's frame of reference!

So, Einstein equivalence as written is *empirically* false! There is in fact no "local equivalence" of frame acceleration and actual gravitation.

The Emperor, as they say, has no clothes.

Z

JACK SARFATTI wrote:On Mar 1, 2009, at 3:46 PM, Paul Zielinski wrote:Nice illustration.

However, how is it relevant to what we have been talking about?Paul, your argument is so vague and shifting that I don't know what it is except to say that I do not buy your thesis that there is a 3rd rank non-tidal gravity field tensor in Einstein's 1915 GR or in any sane variation on it. The expression you suggest does not work because it has no local observable meaning in terms of possible operational definitions that an experimental physicist can do. Your formal arguments fall on deaf ears and you are banging your head against a brick wall. There is a non-tidal gravity field however, it is the set of four tetrad Cartan 1-forms e^I = e^Iue^u as explained by Rovelli in Ch 2 of his "Quantum Gravity". The set {e^I} is a set of zero-rank GCT tensors, i.e. invariant local scalar fields under iii local T4 (x) group and it is a Lorentz group first-rank tensor under ii.<mime-attachment.jpeg>

You are systematically confusing the objective geometrical changes in the geodesics with changes in the

"rest frames" of free test objects.No I am not. This shows you have not understood even one word of my argument. You are completely confused. The hovering static LNIF accelerating observers (rockets blasting to stand still in the gravity field) pictured by Hawking are not the rest frames of free test objects.<mime-attachment.jpeg>Furthermore the objective changes in the timelike geodesics are seen in the change in the geodesic deviation of neighboring unaccelerating detectors. Gravity deformation is not the proper way to define the gravity field which is there even if there is no such deformation. Your position is clueless to my mind.What you write below has nothing to do with what I have been professing here. Nothing at all. You are on a different planet in an alternate universe not even in the same ball park.This allows you to pretend that you can transmute statements about

objective spacetime geometry into statements about observer reference frames.

You are just playing verbal tricks on yourself. Don't you get it? This is just sophistry.

Look, everyone agrees that you cannot globally "transform away" a non-homogeneous gravitational field,

which is the whole point of Hawking's picture.

So what's *your* point Jack?

Z.

JACK SARFATTI wrote:PS Hawking's picture Fig 1.11 lower right should make my point obvious. Imagine changing the mass of Earth with a density, but keeping the hovering rockets at same distance from center of Earth.

<mime-attachment.jpeg>On Mar 1, 2009, at 2:03 PM, JACK SARFATTI wrote:Paul none of this stuff had been published. What has been published has no relationship to your interpretation of what it means. Your basic methodology is flawed, which is why I say your thesis is not even wrong. You do not understand the subtle role of local frames. When you deform the gravity field by changing the source matter field configuration Tuv(matter) ---> T'uv(matter'), you also adiabatically change the local reference frames! You cannot consistently keep those frames fixed for a given kind of constraint e.g. the static observer constraint. You can imagine it mathematically, but it is not possible physically, therefore, you cannot claim, as you do claim, that your purely formal "deformation" change is the consistent basis for an actual operational definition of an intrinsic objective real locally observable non-tidal gravity field that is a GCT 3rd rank tensor.As an example: deform M to M' in the SSS case for static locally accelerating detectors each at fixed r, theta, phig = GM/r^2g' = GM'/r^2r is same in both cases, but g =/= g' and these g's are the actual accelerations of the local static frames in each case. They are different frames!You cannot subtract them and claim you have a tensor in the same frame! It is physical nonsense operationally even though you can make an abstract formal process inside your demented mind! ;-) It does not exist in the laboratory, which is what physics is all about.On Mar 1, 2009, at 1:16 PM, Paul Zielinski wrote:JACK SARFATTI wrote:

Most of this stuff has already been published.On Feb 28, 2009, at 8:51 PM, Paul Zielinski wrote:JACK SARFATTI wrote:

You've already been definitively refuted Jack. Now I'm just having some fun.:-)

On Feb 28, 2009, at 2:49 PM, Paul Zielinski wrote:"...it is not so easy to free oneself from the idea that coordinates must have a direct metric significance."

- A. Einstein, "Autobiographical Notes"

Back in San Franciscore:Paul, Polemics are the last refuge of the man who cannot make a rational relevant refutation. ;-)That is delusional Paul. You have not thought deeply about what you are saying. In any case try to get your idea published.

If you think you can go up against Levi-Civita on the math, then I say it is you who is delusioal.

Really?I have given you a detailed argument that you have not refuted in a rational relevant way.

But your argument is based on your own confusion about local coordinates in relation to curved spacetime geometry.

Curving the 4D spacetime of 1916 GR deforms the *geometry* of the manifold (metric relations between spacetime

points), NOT the coordinate assignments.Actually you are wrong about that.

Because the geometry changes. Of course the change of geometry affects the geodesics. What is an "LIF"The choice of static local non-inertial frames changes in the SSS case.

in one geometry will not be an "LIF" in another. However, this has nothing essentially to do with coordinate

transformations. The changes in the geodesics are geometrical.

So you are still seriously confused. If you want to understand this, you should think about coordinate systems,

not observer frames.

The relationship we are really talking about here is between two sets of geodesics covariantly determined by differentThe covariant acceleration of the static frame is GM/r^2 so when you write{LC(M')}^rtt - {LC(M)}^rtt = G(M' - M)/r^2this is a conceptual relationship between two physically different frames of reference that actually do not simultaneously exist in a real physical situation.

metrics according to the textbook standard geodesic equation.

Jack, it's too late for this. You just got the math wrong, that's all.Your thesis is not even wrong simply garbled fragmentary ungrounded thinking in my opinion.

The change from n_ij to g_ij does not force *any* changes in local coordinate charts around any given point of

interest on the manifold. The geodesics will of course change with the geometry, but coordinate transformations

are simply not relevant here. It is the geometry that changes, not the coordinates.

Your belief to the contrary is what is truly delusional here. You cannot invalidate mathematical theorems with physical

interpretations.

We've already deconstructed this weasel term "local". In this context it means, among other things, that you can'tI use physical detectors as local frame of reference.The meaning of the classical general relativity principle is the computation of local invariants based upon local measurements of two locally coincident detectors in arbitrary relative instantaneous motion measuring the same process up to quantum uncertainty limits.

talk about the frame-invariant gravitational acceleration of test object geodesics with respect to world lines of sources

-- which is what I call Einstein's "make-believe physics". This is like putting blinkers on a cart horse.

The fact is that once you take account of the gravitational acceleration of test objects with respect to sources, Einstein's

equivalence model simply evaporates. You cannot produce gravitational acceleration by accelerating an observer's frame

of reference -- not in Newton's theory, and not in GR either. To suppose that you can is a form of insanity IMHO.

What do you think Einstein's field equationsYou can leave the local coordinate charts (MAPS from R^4 to the points

of the spacetime manifold) undisturbed (FIXED, CONSTANT, UNCHANGING, INVARIANT) while you deform

the manifold in any local spacetime region of interest.

In other words, you can arrange things such that under the gravitational deformation of the metric around any given spacetime

pointyou end up with the same local chart around*x*as you started with. The change in the intrinsic geometry doesn't force*x*

ANY change in the local coordinate assignments around.*x*The very concept of "gravitational deformation" is purely theoretical

G_uv = kT_uv

are all about Jack? They tell us how matter deforms the geometry of spacetime! This *is* Einstein's theory of gravity!

You have to be kidding, Given a single SSS source, all you have to do is measure g, the acceleration of test particle geodesics inand cannot actually be measured unless e.g. the Earth loses chunks of matter in a collision with a large asteroid for example. It is not a valid measure of the actual local gravitational field i.e. tetrad 1-form e^I (Rovelli's definition)

relation to the corresponding flat-spacetime geodesics, that is due to gravity. This is all determined in GR by the geodesic equation:

<moz-screenshot-43.jpg>

This is a generally covariant equation.

Are you still trying to say that in GR there is no way of comparing the curved space geodesic trajectories with the corredponding flat

space trajectories? And that there is no way of observing such changes empirically?

That would make Einstein's theory completely useless as a theory of gravity. But of course this is nonsense.

You just said otherwise. What did you mean then by "a particular metric field guv(x)"?That is something you still don't seem to understand. Hence the Einstein quote. You have not been able to fully separate the

concept of coordinate intervals from the concept of metric relations between points. I think this is the root of your problem,

and is the reason why you are having so much trouble understanding Levi-Civita.

Wrong.Synopsis, since there are interesting foundational issues here ignored even in some texts, though not in Wheeler's.1) Every time you write a particular metric field guv(x), and its associated symmetric torsion-free Levi-Civita connection field for parallel transport of tensors along world lines in 4D spacetime you have chosen a representation in terms of a network of local detectors (AKA local frames of reference).

Irrational response.All GCT coordinate representations of a given metric g_uv(x) are representations of a *single* metric field.Red Herring. No one says otherwise.

It sounded like you were saying this.

You seem to be saying that there is a different "metric field" for every GCT that changes the components of the metric tensor.I have said nothing of the kind.

That changes in the geometry do not force changes in the local coordinate charts in GR? You think that's a silly idea?You simply do not get the point I am making because your mind is closed on this silly idea you have. It's completely silly - embarrassing in fact to my mind at least.

It is a mathematical fact with the consequence that the Einstein field, represented by LC^i_jk(x), is decomposable into an

actual gravitational field represented by a tensor G^i_jk, and a non-tensor X^i_jk which accounts for all non-tensor coordinate

contributions to the LC connection field:

LC^i_jk = G^i_jk + X^i_jk

Which means that in 1916 GR, frame acceleration fields are every bit as fictitious in relation to actual gravitational fields as

they are in Newton's theory in relation to actual Newtonian forces.

And you think this is a "silly idea"?

OK. Good.The metric fields guv & g'uv induced by distinct sources Tuv & T'uv are not related by a GCT. They are different intrinsic gravity field configurations and they have different sets of static observers (outside event horizons) and, therefore, their formal difference has no direct physical meaning, i.e. no measurement procedures exist. They cannot simultaneously exist physically.

If that is what you are saying, then you are missing the whole point of the metric being a GCT tensor.Of course that is NOT what I have been saying.

Then you should have no trouble understanding why the difference between an L:C connection compatible with a curved metric and

the LC connection compatible with a flat-level Minkowski metric is a tensor.

That is your delusion Jack.You are also missing the

point of defining affine connections on curved manifolds that correct the partial derivatives of tensors for coordinate artifacts.

Changing the coordinate representation of the metric doesn't affect the intrinsic spacetime geometry.

You are seriously confused about this. Your Einsteinian optics are preventing you from understanding the actual math.The simplest "hydrogen atom" (as it were) example in Einstein's 1915 GR is the SSS vacuum solution, written in the texts asg00 = gtt = - 1/g11 = - 1/grr = 1 - rs/rrs/r > 1rs = 2GM(source)/c^2Area of concentric sphere is A = 4pir^2usual spherical polar & azimuthal coordinate line element added.You can use any local coordinate chart around any given (well-behaved) point of interest, and hold that local chart invariant

under a gravitational deformation of the manifold.You are wrong.

In Einstein's coordinate frame model, which is a relic of 1905 SR, yes.Physically you need to specify a set (fiber bundle) of local tangent frames that in principle are implementable by actual detectors.

This is where you get really confused. Of course changes in the spacetime geometry will change the classification of trajectories,Here the choice of static frames (where they exist) shows my point very clearly.

since the geodesics chang -- but this happens regardless of whether you hold the local coordinate charts invariant under the

deformation of the geometry, or not.

Of course you can say that the *rest* frames of freely falling test objects will change, and that this is associated with a change in

the coordinate charts associated with the rest frames, but this is immaterial and irrelevant to the point I'm making. It is

a trivial observation. There is nothing that compels us to observe a freely falling test object from its own rest frame, regardless

of changes in the geometry.

You might as well observe that the rest frame of any object changes when you accelerate the object. So what?

What eludes me is why you think this matters. It's a tautology. Trivial.This completely eludes you.

If what you mean here by "physics" is the completely trivial observation that the *rest* frames of freely falling test objects change whenIt is the fundamental error you have made. You are thinking like a mathematician and completely miss the physics.

the geodesics change, then I think I am quite correct in ignoring it. It is of no consequence. It is immaterial to my argument.

It is trivial and irrelevant. It adds nothing of any substance to the fact that when the spacetime geometry changes, the geodesics change.

This is why I think it's better to think about mathematical coordinates instead of observer frames -- to avoid these kinds of pitfalls. I think

this is a good example of the kind of confusion that Einstein's coordinate frame model can generate in the mind of a physicist.

As far as I'm concerned, this is just another way of saying that the geodesics change wen the spacetime geometry changes.Different source configurations have physically different bundles of e.g. static local frames that are the basis for the contingent representation chosen for pragmatic convenience only.

You are simply substituting pseudo-relativistic talk about "LIFs" and "LNIFs" for objective talk about covariantly determined

gravitationally deformed geodesics. Your favorite card trick!

I think it's time to move on. Let's talk about gauge theory.In any case I stop reading here and end the discussion.

Most of this has *already* been published Jack. Some of it -- like Levi-Civita's theorem -- many years ago.Think what you like but1) No journal will accept your idea and rightly so.

Poltorak has already published his stuff, working with a metric-affine model.

A. A. Logunov has published reams on this. Like V. Fock, he says: "No such thing as general relativity".

No one serious in this field actually believes in classic Einstein equivalence anymore Jack. Just look at the most

recent text books on GR.2) No one pays any attention to it (except me here) and rightly so.

And you are perfectly entitled to your opinions. You are free to make all your own mistakes.That's my opinion.

Z.

You are systematically confusing actual geometric changes in the first partial derivatives g_ij, k with coordinate change.

The primary purpose of constructing an affine connection is to correct the derivatives of the metric for coordinate artifacts.

So the premise here is that a distinction must be made between coordinate contributions and actual geometric contributions

to the partial derivatives of tensors. This also applies to the metric tensor g_uv!

Nonsense. You cannot invalidate a mathematical theorem by attaching a physical interpretation to the math. To suggest that youNow Paul you look at the above as a purely mathematical problem.

can is completely wrong headed IMO.

You have to distinguish mathematically between actual geometric changes and mere coordinate changes in the partial derivatives

of tensors, including those of the metric tensor.

That is what the affine deformation model I've been trying to explain to you does so nicely -- for the metric tensor g_uv and its

associated metric-compatible LC connection.

Look Jack, you just got the math wrong, that's all. Do you want to repair the situation, or not?

As far as I'm concerned both the the mathematical issues *and* the issues of physical interpretation are now completely settled,

based on

(1) Einstein's 1917 interpretation of the LC connection field LC^i_jk() as representing the non-tidal field strength in GR;*x*

(2) Levi-Civita's authoritative proof that the difference between two LC connections (his 2 and 2') respectively compatible

with any two *geometrically* distinct metrics (his 1 and 1') on a single analytical manifold is a non-trivial 3-index GCT tensor

rho^i_jk.

At this point any quibbling on your side about how different coordinate representations of the Riemann metric are supposed to constitute

distinct "metric fields" (which is mathematical nonsense, since g_uv is a tensor object) is just whistling past the graveyard as far as

I'm concerned.

It's also a violation of the mathematics. Levi-Civita's proof shows that the frame acceleration field, the strength of which is represented by

a non-tensor coordinate correction term X^i_jk, is only accidentally "glued on" to the actual gravitational field, represented to first order

by the gravitational deformation tensor G^i_jk.

That is the whole point of the linear LC decomposition

LC^i_jk = G^i_jk + X^i_jk

where the gravitational deformation tensor G^i_jk is a particular instance of Levi-Civita's rho^i_jk.

Your insistence that different coordinate representations of g_uv constitute distinct metric fields is just a fallacy. It

ignores the tensor character of g_uv.

Splitting out the gravitational deformation tensor G^i_jk doesn't change any of that Jack. So why do you keep bringingNewton's First Law of Motion in Einsteinian 1916 classical geometrodynamical language generalizes toAll covariantly non-accelerating massive test particles follow timelike geodesics of the gravitational field.

this up?

All fully covariant and consistent with tensor G^i_jk. So what's your point?Newton's Second Law of Motion generalizes toD^2x^u/ds^2 = F^u/mwhere m is the invariant rest mass of the test particle (assumed here not to be ejecting mass like a rocket) and ds is the proper time of m along the worldline. F^u is the non-gravity applied force*pushing*the test particle off-geodesic inducing g-force on the test particle.

I'm sorry Jack but IMO this is complete and utter nonsense.More generally, Newton's 2nd Law in curved spacetime for massive test particle motion is(D/ds)(mdx^u/ds) = (Dm/ds)dx^u/ds + mD^2x^u/ds^2 = F^u(non-gravity)Newton's 3rd law is not valid for global total momenta because translation symmetry is broken - see Noether's theorem.Instead we have the local stress-energy current density conservation lawD^vTuv(non-gravity source fields curving spacetime) = 02) g-force and the Levi-Civita connection field {LC}Back to the simplest caseD^2x^u/ds^2 = F^u/mWe now do a Godel self-reference and let the test particle detect itself.Note, the connection field (LC Christoffel symbol )usually has nothing to do with the test particle, but is a property of the detector, except in the degenerate self-referential case.

The geodesics are covariantly defined in GR and have nothing to do with "detectors" or "observers" -- except as to mere

*appearances*. The LC connection field includes linearly independent contributions from the coordinates *and from the

intrinsic geometry*. Otherwise, how could it be a metric compatible connnection?

Of course a straight SR free-fall trajectory for example *looks* curved to a uniformly accelerating observer *who ignores

his own accelerating motion*! So what? That doesn't affect the actual geometry of the geodesic in the slightest. And neither

does curved geometry force any changes in local coordinate charts. You are completely and hopelessly confused here.

These are two completely different animals.

I'm not going to read any further.

Z.The Baron Munchausen is weightless, no inertial g-force, in free-fall parabolic path (timelike geodesic in curved spacetime)He does feel g-force initially when cannon firesIn particular think of the above SSS case in which everything can be concretely computed. Z never bothers to compute an example. This is why his thesis here is not even wrong in my opinion.i = 1,2,3 = spacelike componentsi = 0 is the timelike componentLet's look only at the i = 1 radial component in the static LNIFD^2r/ds^2 = d^2r/ds^2 + {LC}^ruv(dx^u/ds)(dx^v/ds)However, in the self-referential "diagonal" casedx^i/ds = 0dx^o/ds = 1 (c = 1)d^2r/ds^2 = 0Therefore, the g-force per unit rest mass measured on the test particle itself is{LC}^r00 ~ +GM(source)/r^2This, however, is not an objective non-tidal non-zero tensor field. It is purely contingent based on an arbitrary, though pragmatically contingent, choice since we here on Earth closely approximate this situation.3) Z's Red Herring of the Gravity DeformationStart from M = 0 (flat space time) obviously{LC}^r00(M = 0) = 0here the static local non-inertial observer limits to the global inertial observer - a point overlooked by Z because he does not calculate a simple example.Now increase to &M{LC}^r00(&M) = G&M/r^2This is not a tensor obviously.The difference is not a tensor, i.e.,{LC}^r00(0) - LC^r00(&M) = G&M/r^2No tensors here!Note the natural assumption here is the adiabatic transformation of the global static inertial frame GIF to the local static non-inertial frame LNIF. The contingent choice of convenience is "static".Every time I hear a philofawzer say it's "pure math" or "your argument is mathematical nonsense" I reach for my delete button. ;-)"As far as the laws of mathematics refer to reality, they are not certain, as far as they are certain, they do not refer to reality."- Any intelligent fool can make things bigger, more complex, and more violent. It takes a touch of genius -- and a lot of courage -- to move in the opposite direction."
- "Imagination is more important than knowledge."
- "Reality is merely an illusion, albeit a very persistent one."
- "The only real valuable thing is intuition."
- "Anyone who has never made a mistake has never tried anything new."
- "Great spirits have often encountered violent opposition from weak minds."
- "God does not care about our mathematical difficulties. He integrates empirically."
- "Do not worry about your difficulties in Mathematics. I can assure you mine are still greater."
- "Not everything that counts can be counted, and not everything that can be counted counts." (Sign hanging in Einstein's office at Princeton)
- http://rescomp.stanford.edu/~cheshire/EinsteinQuotes.html

Jack Sarfatti wrote:What's fundamentally wrong in z' s argument is that he does not understand the breakdown of the global frame assumption of special relativity. You simply cannot use purely formal arguments. You must specify what detectors you mean. Physics is about detector responses.

Sent from my iPhoneStart with flat spacetime. The lc connection field is zero for the global inertial observer.Now introduce a small SSS Tuv matter source which induces an non zero lc field for the static non-inertial observer that adiabatically emerges if you impose a fixed r constraint.Since the difference is a tensorThat difference must be zero otherwise you have a contradiction.

Sent from my iPhoneNonsense.

The common non-tensor coordinate part just cancels out of the difference between the two LC connections,

which are themselves NOT GCT tensors due to the common coordinate parts that cancel.

You are just confused.

Z.

Jack Sarfatti wrote:Problem with the deformation approach is that it does not work. Start with flat spacetime and it implies lc is a tensor.Back in Sf

Sent from my iPhoneSo you are claiming that eq 2 is within a single network of observers e.g static Lnifs in the SSS case. We add a perturbation to tuv source inducing a metric perturbation. The take difference in lc connections and that is the 3rd rank tensor?Why did you not say that clearly to begin with?In any case it is not conceptually important to the foundations of relativity.Only the tetrad and spin connection cartan 1-forms are intrinsic nontidal measures of the gravity field ie scalar zero rank tensors under gct aka local t4. About to take off from jfk.

Sent from my iPhoneYou showed in detail that if you replace Levi-Civita's two different metrics 1 and 1' with the *same*

metric represented in two different coordinate charts, you get a very different quantity when you take

the difference of the corresponding compatible connections that obviously is not a tensor..

Which is just another way of saying that you don't understand -- or don't want to understand -- Levi-

Civita's proof.

I told you what Levi-Civita's rho^i_jk means mathematically: it represents the effect of a deformation

of the geometry of the manifold (from metric 1 to metric 1') on the Levi-Civita connection, free of

any non-tensor contribution from the coordinates.

In the context of GR, since in Einstein's model the net observed non-tidal gravitational field strength is

represented by the LC connection, clearly the tensor rho^i_jk represents the effect on the non-tidal field

strength of the change in the 4D spacetime geometry when you go from 1 to 1' -- free of any contribution

from observer frame acceleration.

If you take metric 1 to be geometrically Minkowski, then the difference G^i_jk between the two LC

connections obviously represents the effect on the gravity-free LC connection field of the change from

Minkowski geometry (n_uv) to gravitationally deformed geometry (g_uv).

Which means that the tensor field G^i_jk(x) represents the *actual* (i.e., objective, frame-independent)

non-tidal (first-order) gravitational field strength in GR, free of frame acceleration artifacts ("fictitious

forces").

Z.

Jack Sarfatti wrote:There is no physics at all in that snippet eq 2.

Sent from my iPhone at jfk boarding for SfoActually, we are talking about Levi-Civita's idea here. He's the one with two different metrics on a single

analytical manifold, with a common local coordinate chart around each point, under general transformations

of which his rho^i_jk (the difference between two metric-compatible connections) transforms as a 3-index

tensor.I showed in detail it is not a tensor if the usual physics is added to the naked math. It is a contingent relation between two networks of pairs of locally coincident detectors.