- Paul,I continue to disagree with you. You are using circular reasoning in my view. I believe my analogy still stands. Differential geometry does not abrogate the laws of arithmetic.Nor have I ever seen any differential geometry or mathematical physics treatise that I can recall that makes the distinction in LC between "geometric" and "curved coordinate" corrections as you do below. I am not at all convinced of the validity of that distinction. The Weyl Decomposition Theorem does not in of itself require this.I have read the shorter Poltorak Paper, and such a proof is not in there. It does reference the earlier, longer Poltorak Paper which I have not had the time to read yet. Perhaps he proves it there, but I have not seen it yet. Even if it were so proven, that does not abrogate arithmetic.To use a concept from systems engineering, you seemingly wish to "allocate" to LC a tensor piece. But the standard definition of LC in differential geometry does not allocate that piece to LC in the Weyl Decomposition Theorem.Now I leave it to Jack, Profs. Kiehn or Rodriguez to weigh in on this mathematical allocation issue with greater knowledge than I likely have. I will not comment further on this topic in the near-term. The IRS would likely not be interested in the intricacies of the Weyl Decomposition Theorem <g>.Take care,Robert E. Becker
wrote:*Paul Zielinski <iksnileiz@...>*Hi Robert,

ROBERT BECKER wrote:

OK, I think we now all agree on that much.Paul,No, Paul, my counterargument does not fail.The Weyl Decomposition Theorem does is valid and so is your first equation.

Tensor algebra.By the simplest of algebra, your second equation below is also correct.

OK.

Yes. I understand this to mean that you can generate a torsion free non-metric affine affine connection A from the LC connectionWhat you fail to recall is that the reason these equations "work" is because the non-LC connection A itself contains a tensor term, namely, what you call Q.

G by adding a certain (1,2) tensor quantity, called by Weyl the "non-metricity" Q. In fact you can generate an infinite class of such

quantities.

What the Weyl-Poltorak LC decomposition says is that this operation is reversible: you can recover the LC connection

from an arbitrary torsion-free non-metric connection A by subtracting a tensor quantity.

It also says that this reverse operation is the same as *adding* to the LC connection *another* (1,2) tensor quantity -Q.

If one thinks about this in terms of matrix representations [Q^i_jk] and [-Q^i_jk], then the truth of this seems almost trivial to me.

NO, because quite generally if you add a non-tensor X to a tensor Y, you get a *non-tensor* Z:The result is that the*net*RHS of your second equation does not contain a tensor, which is why the LHS of your second equation, namely LC, does not contain "within it", a net non-zero tensor.

Z = X + Y

and then the existence of a tensor "inside" the non-tensor Z is guaranteed *by definition*.

This would seem to constitute a counterexample refuting your argument.

This is exactly how I understand the LC decomposition: the LC connection is composite of a tensor geometric correction

term and a non-tensor curved-coordinate correction term.

On a globally flat manifold, the tensor geometric term in LC "zeros out".

Except that in the case of the LC connection, we have a theory of parallel transport that allows a distinction to be madeYour assertion is completely analogous to asserting that in the following complex number equation, the LHS has an imaginary part:4 = C - 2iwhere C = 4 + 2i.

between the coordinate and geometric contributions to the LC connection. If the LC connection is a composite, then it makes

mathematical sense to decompose it independently of the formal algebra.

I think that is what your argument neglects. We are dealing here with an underlying geometric model, namely that of parallel

transport along a Riemannian manifold.

So the LC decomposition can be made independently meaningful with respect to the geometric model, which allows us to

think of the LC connection as being a composite of two mathematically distinct parts, a geometric part, and a curved-

coordinate part, where the latter is entirely responsible for the non-tensor behavior of the LC connection as a whole.

So you have the operation reversed: in fact we build the LC connection from a completely non-metric (torsion free) affine

connection A (which latter is by definition completely insensitive to the intrinsic geometry of the manifold) by adding a

"metricity" tensor (=-Q) to the affine connection A which takes account of the effect of the intrinsic geometry on the

derivatives of tensor field quantities defined along the manifold.

That makes

G = A - Q

just like

Z = X + Y

in my hypothetical example above.

Z.The more important point that I made, and that you should be focusing on, is in my 2nd paragraph, where I said that the ultimate resolution of the dispute over how mass distributions influence nature in GR will prove (has always been) the heart of the issue. Resolve that, all the rest of this is ancillary.Now, I must leave this and get back to "real" math, namely taxes <g>.Take care,As I understand him what Robert is trying to say here is that in the well-known Weyl decomposition

A = G + Q

of an arbitrary torsion-free non-metric connection A, while the non-metricity Q is clearly a tensor, the

quantity -Q in the corollary LC decomposition

G = A - Q

is not a tensor.

But the negative of any matrix representation of Q in which the indexed components Q^i_jk transform

as a (1,2) tensor must itself be the matrix representation of another (1,2) tensor.

Thus Robert's counterargument fails.

Z.

Jack Sarfatti wrote:Last term on RHS of 2.81 is a tensor, but maybe a part of the first term cancels it?Even so it does not change the operational physics in creon's, Woodward's and my arguments that geodesics have zero physical acceleration and that locally you cannot distinguish an accelerating frame from a stationary frame in a non tidal gravity field. Creon's gedankenexperiment shows that. Next step is to calculate each term in 2.81 explicitly for static lnif observer in SSS vacuum. I am traveling next 10 days so may not be able to right away.

Sent from my iPhoneJack,The equations below are not getting through on e-mail, but I looked at them in Rovelli. I am not nearly as fluent in tetrads as you are, and don't have time in the midst of doing taxes <g> to look deeply into this now, but I still do not think there is a tensor as part of LC (in the simple non-torsion case where there is also no distinction between connections for the tangent and co-tangent spaces).His assertion that there is such a tensor arose from a simple arithmetic and algebraic error where the (valid) connection decomposition theorem was used to misattribute the tensor term in that theorem to the LC term, rather than to the non-LC connection that is also a term in that theorem. It is analogous to the simple complex number equation example of the same kind of misattribution that I gave in an earlier e-mail.Yes indeed, z's argument here is bogus. Rovelli's 2.81 is plain vanilla Einstein gr only unique torsion-free connection, no non-metric shadow connection needed.I also believe the recent discussion and debate between the two of you (and Prof. Woodward) on how the influence of source mass distributions manifests itself in nature and in GR gets far more to the heart of the origin of the reasons for the disagreement between the two sides on the proper interpretation of GR. The LC tensor term issue seems to have arisen as an attempt by one side of that debate to implement mathematically their viewpoint. The physics discussion I believe is more at the core. Resolve that, and the math should follow (but then, I am not a mathematician <g>).Take care,Robert E. Becker*Jack Sarfatti*

In fact, I show below what does appear to be a non-zero tensor in the

LC connection without using some bogus non-metric connection A as

Zielinski alleges. I mean the second spin-connection (omega) term on

RHS Rovelli's 2.81.

ï¿¼

Of course, Zielinski never made the detailed concrete math arguments

I give below. He never pointed to (2.81) and said "This is what I

mean." Instead he produced reams of obscure unintelligible jargon. In

theoretical physics it is not enough to be right about some alleged

statement of fact, one must also be right for the right reasons. It's

not like guessing the right answer on a dumbed-down multiple-choice

test.

Note also, Rovelli's (2.83)

ï¿¼

and more importantly Rovelli's 2.89

ï¿¼

Therefore, there seems to be a non-zero GCT tensor part to the LC

connection, not needing any non-metric connection, that tensor is

X^wuv = e^wIS^IuJe^Iv

because

ï¿¼

ï¿¼

On Mar 29, 2008, at 1:54 PM, Paul Zielinski wrote:

> Now you're reduced to picking up anything smelly that just happens

> to be close at hand to throw at me?

> Hoping at least some of it may stick?

Sure, why not? If it walks like a crackpot, if it talks like a

crackpot, then .... :-)

You and Potter share some common methods - excess verbiage,

obscurity, lack of real math argument. Massimo's distinction between

"culture" and actually doing theoretical physics applies to you! Your

"metatheoretics" is a mere veneer. It's fool's gold. It's the Gary

Zukav Effect! You are Mr. Malaprop!

http://www.fun-with-words.com/mala_malapropisms.html

In short, Sir, you are not to be taken seriously on any of your

obscurantics about Einstein's General Theory of Relativity about

which, as Professor Woodward has testified you lack essential

understanding, i.e.,

> "If one does not grasp this, one has no hope of understanding GRT."

On Mar 27, 2008, at 10:12 PM, james f woodward wrote:

>

> Well, since it is obvious that you are right and Z is wrong about

> accelerating reference frames and homogenous gravitational fields

> being

> distinguishable, I've not been following the details of the argument

> closely. The problem is that Z does not understand that

> "acceleration"

> in a uniform (no tidal stresses) gravitational field in GR is in fact

> inertial motion and therefore, in the sense he wants, not accelerated

> motion at all. "Accelerations" are DEVIATIONS from inertial motion

> (i.e., deviations from "free fall" in a uniform gravity field). If

> one

> does not grasp this, one has no hope of understanding GRT.

Also for the record:

On Dec 30, 2007, at 2:00 PM, NASA AMES scientist Creon Levit wrote:

> Isn't it simply this: Say we erect a 100 meter tower on the

> surface of the earth and put the scale on top of it, at r'=r+100.

> Right next to the tower we have an identical scale, but with no

> tower to hold it up. To keep this second scale up at r'=r+100

> would require a rocket firing with constant thrust.

>

> My weight would read the same on both scales. There is no way to

> tell which scale I am on just by measuring the local spacetime

> curvature.

On Dec 31, 2007, at 2:07 PM, Creon Levit wrote:

>

> We went through this before. If you allow extended non-

> infinitesimal experiments then you can indeed tell whether you are

> accelerating to hover in a gravitational field or accelerating to

> speed up in flat space. Or some other combination.

>

> Consider the SSS (weak field) case: We have two observers. The

> first observer is "Jack". He is (of course) at at the center of

> the universe, at r=0. The second observer is "Paul". Paul is

> flying around in a rocket and neither he nor Jack know where Paul

> is, or where Paul is going.

>

> However, they can remedy this confusing situation by communicating

> with each other.

>

> They start to exchange messages (over the radio). They measure the

> round trip radio communication time. This is easy to do because

> they've each perfected the ability to reply immediately by quoting

> arbitrary parts of a previous message and mixing in words like

> "nonsense" or "crackpot".

>

> Paul has for the moment settled on a steady course of action. He

> is accelerating so that his accelerometer / therapy couch reads his

> weight as a steady 100 units.

>

> If the round trip message time Paul measures is constant, then he

> has just determined that he is expending all that expensive rocket

> fuel merely to overcome "gravitational inertial resistance" due to

> Jack's influence on his rocket, his body, and perhaps even his mind.

>

> Paul is accelerating merely to hover just out of Jack's reach, in a

> spacetime that is objectively curved, whose "gravitational tetrad

> components" he busily writes down, if only to prove to Jack and

> other doubters that he can.

>

> If, OTOH, Paul measures the round trip communication time

> increasing quadratically, then he knows that he is accelerating

> now, instead, to overcome "purely inertial resistance" which has

> nothing to do with Jack.

>

> Jack's influence has faded away along with the "objective

> gravitational components" of the tetrad field, to be replaced by

> (equal?) "purely inertial components". Paul is moving ever more

> quickly though a vacuum that is objectively flat, where his 100

> units of weight on the couch (unchanged) are now due instead to his

> body's "purely inertial resistance to acceleration".

>

> He has finally escaped Jack. The objective evidence (the only

> measuring device with a changed reading) is that the messages are

> arriving less frequently. Paul relaxes more fully onto the couch.

>

> He has just had a big dinner, so he adjusts the rocket's thrust to

> keep his couch reading weight=100.

>

> But he cannot relax for long. Perhaps Jack is only toying with

> him. Perhaps Jack's messages will start to come more frequently

> again. Perhaps Jack even has a way to extend his influence, to

> curve space at will, to fool Paul into thinking space is flat where

> it is not. Perhaps Jack has mastered the Dark Arts. Or perhaps

> Jack has figured a way to send superluminal messages to Creon.

>

> Now that Paul thinks about it, Creon has been transmitting messages

> more frequently. Perhaps Paul is not in objectively flat space at

> all, but instead hurtling towards the planet Creon! Paul feels a

> sudden disturbance in the tetrad field..... but is it gravitational

> or inertial....?

Then we have

> On Feb 2, 2008, at 1:47 PM, ROBERT BECKER wrote:

>>

>> Jack,

>>

>> Thanks loads!!

>>

>> It is going to take me some time to get back to that. (I have try

>> to help my significant other with an MBA HW assignment on interest

>> rate swaps first.

>>

>> But after reading his (Zielinski's) reply, I believe I can see

>> where some of his misconceptions arise from, and also some of our

>> confusion because of the way he has been presenting his theory in

>> poorly connected dribs and drabs.

>>

>> For instance, it was only after reading his reply that I realized

>> he is actually proposing that there are two connections - well,

>> he may not realize it yet, but an infinity of connections - in his

>> theory: One is the usual LC, which he does utilize for all vectors

>> and tensors except the metric tensor. The other connection is a

>> "special" connection, a non-LC connection, non-metric compatible

>> connection, which is the one he intends to correct for only

>> coordinate-dependent effects, "leaving in" GR only artifacts due

>> to intrinsic geometry. This non-LC connection is the one he

>> proposes to use only for the metric tensor. This is why you and I

>> thought he was discarding the LC: because he does in fact discard

>> LC for the metric tensor, and it has been the metric tensor that

>> has been the center of discussions about his theory in this forum.

Let me add, one of my several objections to Zielinski's extraordinary

claim is that he is not proposing an alternative theory to Einstein's

1916 GR but that his theory is equivalent to it. Definitely not true.

At times Hal Puthoff seems to also make that claim for his PV model?

>>

>> Perhaps you can comment for the forum specifically on his belief

>> that the vanishing of the covariant derivative of the metric

>> tensor means that "the actual intrinsic first-order variation of

>> the metric along the manifold is always - intrinsically - zero".

Until Zielinski writes a relevant equation here I am not sure what he

means except the trite observation that the covariant derivative is

the actual intrinsic ... variation per unit infinitesimal

displacement, for example in a covariant version of Taylor series,

replacing ordinary derivatives by covariant derivatives along some

world line

e.g. Delta T ~ (DT/ds)delta s

where T is a tensor

DT/ds is also a tensor since D/ds is the directional covariant

derivative along some timelike world line labeled by its proper time s.

Of course Zielinski is incapable of making such a concrete example of

his high-falutin pompous meta-theoretical cryptic jargon.

>>

>> Actually, he says that this is what you believe, and that this is

>> what you have brainwashed me to believe. (I will deal with that

>> insult to me separately). Now, you never said this, and neither

>> did I. So, I can only assume that this is Paul's misconception,

>> not ours.

One of many!

>>

>> In any case, Paul does seem to believe that use of the vanishing

>> metric covariant derivative condition (i.e. adoption of LC) in GR

>> somehow excises or removes from GR the desirable intrinsic

>> geometric curved manifold effects (e.g. the curvature) that one

>> should be retaining in GR because they can be physically

>> attributed covariantly to a source mass field and gravitational

>> field. Conversely, he seems to believe that (what he says is the

>> incorrect) use of LC for the metric tensor retains in GR the

>> undesirable, purely coordinate system-dependent artifacts, which

>> is what he wants to excise from GR because he attributes the

>> physical origin of undesirable non-covariant, coordinate-dependent

>> accelerating frame artifacts in GR to these coordinate-dependent

>> artifacts in the LC. And it is to these undesirable accelerating

>> frame artifacts in GR that he attributes the necessity in GR for

>> the existence and use of an (undesirable and unnecessary) energy-

>> momentum pseudotensor.

>>

>> To summarize his view on this specific topic, I will repeat a

>> quote from him buried in his reply to me:

>>

>> "Notwithstanding the fact that the LC covariant derivative of the

>> metric is identically zero, the actual intrinsic first-order

>> variation of a curved Riemann metric is generally NOT zero. The

>> true intrinsic first-order variation of the metric along a Riemann

>> manifold is thus given NOT by the LC covariant derivative, but

>> rather by a covariant derivative associated with a certain non-

>> metric connection A." (And this A is the one non-LC connection

>> that corrects only for all coordinate-dependent effects.)

Of course, there is no physical motivation for claiming a new non-

metric connection A. Note that Zielinski never

1) defines "A" mathematically

2) defines it physically , operationally, how to detect if it's

really there.

Zielinski just uses "A" as an ill-posed symbol.

Also any theory which claims there are two connections (LC) and A is

not Einstein's 1916 GR as Zielinski also sometimes, at least, claims

- or seems to in his low signal-to-noise messages.

>>

>> So to me the points of confusion here seem to relate to the

>> geometric meaning of covariant deriviative. How that meaning

>> relates to that of a connection. And why the vanishing of the

>> metric covariant derivative does NOT mean that the intrinsic

>> curvature of the manifold (or the first derivatives of the metric)

>> must vanish identically.

>> Perhaps you can address this specific issue. I do not believe a

>> string of equations would be the best way to do that since the

>> confusion appears to be conceptual rather than purely "equation"-

>> driven.

The only way I know how to answer this is using the actual equations

of Einstein's theory. Zielinski relies on obscure excess verbal

rhetoric, which is a bad way to try to do theoretical physics. This

explains Feynman's contempt for the sort of philosophy of physics

that Zielinski does. Feynman called it "philofawzy."

Rovelli does this in Chap II of "Quantum Gravity" in a few lines

using Cartan's elegant notation.

S^I^J = spin connection 1-form

I,J are Lorentz group indices Rovelli's (ii) describing locally

coincident LIF <--> LIF' transformations

S^I^J = S^I^Judx^u

u are the Diff(4) i.e. local T4 indices i.e. Rovelli's (iii)

describing locally coincident LNIF <--> LNIF' transformations

the tetrad fields e^I themselves are simply (iv) the locally

coincident LIF <--> LNIF transformations

D = d + S/\

is the covariant exterior derivative

the curvature 2-form is

R^I^J = dS^I^J + S^IK/\S^IJ

a flat spacetime means

R^I^J = 0 over a 4D region of local coincidences

Note that

R^I^J = R^I^Juvdx^u/\dx^v

Ruvwl = ev^Iew^JR^I^J = Riemann-Christoffel 4th-rank intrinsic

curvature tensor

However, in Einstein's 1916 GR the torsion 2-form T^I is also zero,

which means

T^I = De^I = de^I + S^IJ/\e^J = T^Iuvdx^u/\dx^v = 0

where the GCT tensor part to be added to the LC curvature connection

is the contortion combination of permutations of the indices of 3rd

rank torsion tensor

T^wuv = e^wIT^Iuv

This implies Rovelli's 2.89 for the spin connection (S is omega in

Rovelli's notation) in terms of the tetrads

ï¿¼

note, however, that under a (iii) GCT transformation Rovelli's 2.61 that

ï¿¼

this is a first-rank GCT tensor homogeneous transformation.

Note also, Rovelli's (2.81)

ï¿¼

where also

>>

>> I will later take a final whack at all the other confusions. .

>>

>> Take care,

>>

>> Robert

>>

>

>

>

> The "elementary confusion" is all yours Jack, I can assure you.

You can assure me of nothing. You have negative credibility. James

Woodward is a professor of physics at Cal State and he teaches this

stuff. He said exactly what I said. Creon Levit of NASA gave a nice

gedankenexperiment. Robert Becker pointed out defects in your

"thesis" - you ignore everyone because you are a crackpot par

excellence! You are a wonderful case study for scholars of the crank

scientist mind - indeed a classic case.

> And your case is not helped by your blunderbuss

> attempts at assigning "guilt by analogy".

Not analogy - identity! Yes, Paul, identity. If the shoe fits, wear it.

>

>

> I'm looking forward to hearing your and Prof. Woodward's detailed

> responses to the specific questions

> I just asked about locally observable difference in the effects of

> objective gravitational acceleration vs. the effects

> of frame acceleration in a previous message.

Meaningless gibberish. What is the question? As usual you use some

"private language" in Wittgenstein's sense. The above sentence is

jargon like a Professor Irwin Corey monologue.

>

>

> That is, if you have any to offer.

An intelligent answer to a meaningless question is not possible.

Not all questions deserve answers.

But now to a substantive point. Rovelli's equation

>

>

>

>

> Jack Sarfatti wrote:

>>

>> Thanks Massimo for your dispassionate analysis on Mr Potter's UFO

>> engineering claims. I look forward to your more detailed

>> conclusions if and when you get the time. :-)

>>

>> Some of the things you say below also apply to Paul Zielinski's

>> theoretical claim based on an elementary confusion on how to use

>> the terms "acceleration" and "inertial" in Einstein's General

>> Relativity compared to Newton's theory of gravity. I have red-

>> lined some of those things below.

>>

>> On Mar 29, 2008, at 3:25 AM, Massimo Teodorani wrote:

>>> Dear Mr. Potter,

>>>

>>> I should say that the way in which you presented or posed

>>> yourself to this public doesn't certainly help the possibility

>>> that your work is considered, but I also see that the part of

>>> this public who intervened on the issue maybe dismissed your work

>>> too hastily. This is probably due to your exceedingly aggressive

>>> way of putting the entire thing in this mailing list, which is

>>> NOT the way a scientist should behave. I think there is a lot of

>>> many negative "human interactions" here, determined by your

>>> "vigorous" way of presenting your things and by the consequent

>>> (human) reactions to it.

>>>

>>> I have downloaded your book from your website (three parts: book,

>>> notes and biblio), around 550 pages, as I can evaluate at a first

>>> glance. It seems this book contains all Potter-thought in a

>>> compact and complete way. Now I have no time to study it, but I

>>> maybe would like to read it during my holidays. Anyway I gave a

>>> first glance at it (as maybe some of the others who intervened)

>>> and noticed the following things:

>>>

>>> 1) There is plenty of material and work (in terms of *ideas*),

>>> which perhaps should be examined in detail and by paying the

>>> necessary patience. I do not think that this large quantity of

>>> work can be dismissed so easily. There must be a reason why a

>>> person decides to embark himself in a so long opus on ufos and

>>> their alleged physics. Moreover I see that the bibliography cited

>>> is quite vast, therefore your book seems documented by several

>>> technical works that are cited therein. The illustrations are

>>> really nice, meticulous and many. Someone should wonder why a

>>> person should spend so much of his time to draw such very

>>> detailed sketches without a very good reason. They do not look

>>> only artwork. This is *work* that a scholar has done, anyway -

>>> his effort - independently from the criticism that his work may

>>> raise.

>>>

>>> 2) The book seems to be totally speculative, and practically all

>>> the issues are presented in a purely "wordy" way, in spite of the

>>> technical way in which the various chapter titles are often

>>> written. I see (forgive me if I looked badly, in case) no

>>> equation (except for a few very simple ones here and there), no

>>> graph, no numerical order of magnitude, no numerical simulation,

>>> no prediction of the numerical values of possible *observables*

>>> such as expected luminosity, energy density, spectral energy

>>> distribution, spectral lines, pulsation rates and time

>>> variability, magnetic intensity, etc. For all of these reasons I

>>> might think or understand that this your book is just a book of

>>> popularization character. But you seem to present this as a

>>> "research book". It is not (at least for a physicist), unless you

>>> mention at least one of your own technical publications on the

>>> subject (also technical reports, congress communications, not

>>> only peer-reviewed papers) to which your book might be referred,

>>> but in the biblio part I see none of them. This, and only this

>>> probably, would make your book credible to scientific readers and

>>> peers, so that anyone would be invited to study it in detail as a

>>> leading guide to your technical and operational research.

>>>

>>> 3) Nikola Tesla, as many know, had a very strange approach to the

>>> scientific method. At the best of my knowledge he didn't use to

>>> write down equations (this doesn't mean that they didn't exist in

>>> his own mind in whatever form), but he seemed to pass directly to

>>> experimental and technological demonstrations (many of them), the

>>> most important of which (such as the induction engine for

>>> alternate current, or also the telecommanded vessel, and many

>>> others) worked very well indeed. So this means that a few

>>> scientists operate in a very strange way, that anyway seems to

=== message truncated ===

ated ===

- Kay zum Felde wrote:

Most of the formulas are textbook standard. If you have difficulties translatingHi Paul, in general I've difficulties putting your explanations in any kind of formulas,

my explanations into formulas, I can only conclude that like Jack and others

you do not have a good intuitive grasp of some these basic concepts in

differential geometry.

Sean Carroll is very good on this. You could look at his "Notes on GR", Ch. 3.

That might save a lot of unnecessary typing here. If you don't already have them,

they are available online at

http://xxx.lanl.gov/abs/gr-qc/9712019

Right.since sometimes I don't really understand it and sometimes I am not familiar with the terms,

Right. This is the point of "metricity" or "metric compatibility" of a connection: to preserveso I also have some questions: --- Paul Zielinski <iksnileiz@...> wrote:

Hello Kay, Kay zum Felde wrote:

Hi Paul, for the following see f.e. R.M. Wald, "General Relativity". Parallel transport is represented by a the

expression

t^a D_a, where t^a is some vector and D_a is some connection. Now Wald takes some inner product of

two

vectors, defined by g_ab v^a w^b (note this is a generalization of the infinitesimal metric

ds^2=g_ab

the inner product defined by the Riemann metric under the infinitesimal transport of vectors

along the manifold. This allows the covariant derivative of a "metric" connection to separate

the inherent physical variation of a vector field quantity (or, more generally, a tensor field

quantity) defined on the spacetime manifold of GR from the variation in the inner products

of all transported vectors that is purely due to the intrinsic geometry (differential variation

of the metric -- and thus inner products -- along the manifold).

The essential point here is that you can define connections ("non-metric connections") for

which the covariant derivatives do not preserve the inner product under infinitesimal transport

of vectors, but such derivatives are still "covariant derivatives" in that they correct for

curved-coordinate artifacts.

A good informal discussion of the mathematical "artifacts" arising from the use of curved

coordinates and their contribution to the Levi-Civita connection in the context of GR can

be found here:

http://www.mathpages.com/rr/s5-06/5-06.htm

Right. That is what I call the "metricity" condition. This picks out a unique torsion-freedx^a dx^b), and constructs the quantity t^a D_a(g_bc v^b w^c) and demands invariance under parallel transport,

affine connection on any Riemannian manifold, the familiar LC connection:

Ñ*g*= 0

But this condition has nothing to do with the *covariance* of a tensor derivative, which

has to do with the behavior of the derivatives of tensor field quantities under non-linear

coordinate transformations.

The easiest way to understand this is to look at how the LC covariant derivative acts on

a flat manifold, where there is no correction needed to the inner products of transported

tensors -- since there the intrinsic geometry (represented by the metric) is completely

uniform to all orders.

I don't claim any such thing.e.g.:

t^a D_a(g_bc v^b w^c)=0. It follows: t^a v^b w^c D_a g_bc=0. And this is only satisfied

in

general if D_a g_bc=0. Now, you claim that

parallel

transport is not conserved.

I'm saying that the LC covariant derivative is the correct covariant derivative for all physical tensor field

quantities defined on any Riemannian manifold, EXCEPT for the metric tensor itself (the latter itself being

considered, as it is in GR, as a physical field quantity).

Quite generally, by "general tensor field quantity" I mean any physical tensor quantity defined as a "tensor field" overI don't know if this mathematically a problem, but what does physics

have

to say about it ?

I claim no such thing. For general vector field quantities,

what are 'general vector field quanties' ?

the 4D spacetime manifold of GR, *other* than the metric itself (which in GR is itself treated as a physical field).

For all field quantities on the spacetime manifold -- vector or higher-rank tensor --the LC

connection is still the correct connection for defining the covariant derivative.

I hope I understand this automatically, when I got your definition above.

with the *sole exception* of the metric, the LC connection is still the appropriate

connection for defining the covariant derivative in my model for GR.

I'm saying that the appropriate connection for defining a covariant derivative of the

metric on *any* Riemannian manifold is not the LC connection, but is a different

*non-metric* affine connection for which the associated covariant derivative of the

metric is *not* identically zero (unlike the LC covariant derivative, which is

identically zero).

For purely mathematical reasons.

Everyone I've talked to about this appears to be very seriously confused about

what the "metricity" condition satisfied by the LC connection actually means,

both mathematically, and physically in the context of GR.

Yes. The metric is also regarded as a physical tensor field on the spacetime manifoldI am saying that it is not the right connection for defining a covariant derivative for the metric itself,

A covariant derivative for the metric itself ???

in GR.

I mean a certain *non-metric* covariant derivative of the metric. The metric-compatible LC covariantEither I completely don't understand anything in this sentence or you mean D_a in: D_a t^b=partial_a t^b-Gamma^b_ac t^b

derivative is not useful when applied to the metric itself, because it is always zero by definition,

regardless of the intrinsic geometry. That is obviously not the true value of the first order differential

variation of the metric on an arbitrary curved manifold, but only in certain special cases such as "flat"

manifolds.

I mean that the LC covariant derivative is expressly designed to compensate the partial derivatives ofsince the LC covariant

derivative automatically self-compensates the actual first-order variation of the metric along the manifold to zero (Ñ/g/ = 0).

automatically self-compensating ? What do you mean by this ?

tensors for the effects of the first-order variation of the metric along the manifold on their *inner products*.

That is what is meant in differential geometry by a "metric" connection!

Are you familiar with this term? Among other things, the "metric compatible" LC connection exactly

compensates parallel-transported vectors for changes in their inner products that are induced on curved

Riemannian manifolds by the metric (which actually is an inner product). LC parallel transport "preserves"

the inner product defined by the metric at any given point under infinitesimal transport to neighboring points

on the manifold. That's how a metric-compatible connection separates the inherent physical first-order

differential variation of a vector field quantity from the gravitational part that is purely due to the intrinsic

geometry of the manifold.

Why would we wish to compensate the first partial derivatives of the metric itself for the intrinsic

first-order differential variation of the metric?!

Of course you always get zero if you do that. Useless.

The reason for that may be that you don't know what I mean by a "non-metric connection".I am saying that there is a *completely non-metric affine connection*

I don't know what you mean by 'completely non-metric affine connection' ?

Are you familiar with this concept?

Yes, curved geometry has an effect on the inner products of infinitesimally parallel-transportedWald says that an affine parametrization preserves t^aD_a t^b=0, where t^a is the tangent vector of a geodesic. You mean this ?

vectors (and higher tensors) that has nothing to do with the physical variation of the field quantity

in and of itself, but rather with the intrinsic variation of the metric (inner product) along the manifold

-- so in Riemannian geometry we use a *metric compatible* connection (namely, the unique LC

connection) to compensate for the effects of the metric on the derivatives of physical field

quantities on curved manifolds.

That is the whole point of "metricity" AKA "metric compatibility" (of a Riemannian connection).

That is not to be confused with the issue of "covariance", which has to do with correcting

for *curved-coordinate artifacts* (even on globally flat manifolds!), and not with the effects

of "curved" intrinsic geometry on displaced vectors.

Very few physicists I've discussed this topic with seem to understand this distinction.

I don't understand the question. Geodesics are determined by the actual covariant intrinsic first-orderIf you mean this, what do you put into instead of a geodesic ?

variation of the metric along the manifold, not by the choice of spacetime coordinates. The objective

gravitational distortion of the geodesics in GR is a function of the *covariant* values of the first derivatives

of the metric field. Yet the coordinate values of the derivatives of vector and tensor field quantities depend

in a non-covariant manner on the choice of coordinates, even on flat manifolds. That's the problem that

a "covariant derivative" is constructed to solve -- by definition.

"The derivative of a tensor is not a tensor."

The basic purpose of a "covariant derivative" is to give us objective coordinate-invariant values for thethat can be derived from the LC connection which gives you the appropriate covariant derivative of the metric field, in that it gives you the true covariant value of

what is a 'true covariant value' ?

derivatives of a tensor field quantity, *free of coordinate artifacts*. The components of the derivatives of

tensor field quantities should transform according to tensor rules. The problem is that they don't,

unless you artificially construct a *covariant* derivative.

A covariant derivative may or may not be "metric compatible". The question here is, *which* covariant

derivatives should be used for *which* tensor field quantities?

If you don't already know about all this then I think you will need to review some basic differential

geometry before we go much further. You need to understand the purpose of connections and their

associated covariant derivatives on curved manifolds -- which is to give "true" values of the derivatives

of tensor field quantities on curved manifolds -- and even on flat manifolds when curved coordinate

systems (such as polar coordinates) are admitted.

Some covariant derivatives ("metric-compatible" connections) correct for intrinsic geometry, while other

covariant derivatives ("non-metric" connections) correct for intrinsic geometry in varying degrees, or not

at all. There are infinitely many covariant derivatives on any Riemannian manifold. One -- the unique

Levi-Civita connection -- is "metric-compatible" in the sense that the LC covariant derivative of the metric

vanishes identically,

Ñ*g*= 0,

on any Riemannian manifold, regardless of the intrinsic geometry.

It is the actual coordinate-invariant first-order variation of the metric that is due to the intrinsic geometrythe geometric first- order variation of the metric along the manifold,

what is 'the geometric first-order variation of the metric along the manifold' ?

of the manifold.

The raw partial derivatives of the metric on the other hand contain coordinate artifacts.

This is elementary differential geometry Kay. You need to do some homework. You and Jack!

Read up on connections and covariant derivatives. I'm not making this stuff up. It's in all the textbooks.

Look at Sean Carroll's Notes on General Relativity, or Matt Visser's Notes on Differential Geometry -- both

available in PDF online. See above.

OK, I can see that you need some basic tutoring on this subject.free of curved-coordinate artifacts.

what are 'curved-coordinate artifacts' ?

I would suggest you read Sean Carroll on connections and covariant derivatives. He explains all this

quite nicely. Read Visser.

Let's not waste time on elementary concepts and theorems that are not disputed by any mathematician

who is competent in this area.

OK, now it's obvious that you don't understand the concept of the "metric compatibility" ofThe metric is a special case. We don't want to correct the differential variation of the metric along the manifold for the differential variation of the metric along the manifold.

What ???

a connection. You may not even properly understand the concept of a covariant derivative!

Read Carroll.

What are the true values of the first partial derivatives of the metric on a completely flat

manifold? Obviously zero. But in polar coordinates, for example, the partial derivatives

of the metric are *not* zero.

Does that help? See the problem?

What is the relationship between the geodesics and LC parallel transport, as youIt means to me that you start at some point x in spacetime and moving along one direction on some trajectory and somehow you are able to return back (still following the initial direction) to x in

any

case. What does this mean ? It simply means that

the

universe (gravity) behaves at least according Einstein's basic 'special principle of

relativity'.

What I meant here was, that isotropy is based on the SR-principle and that GR (as Jack puts it GR-1916) is based on geodesics, e.g., oint ds=0 and this is what parallel transport wants to express, as far as I understood yet. I am, yet, not familiar with torsion.

understand it?

The LC connection contains information about curvature. That's why you can deriveNo, because in SR, R^u_vwl = 0. Where there is curvature, yo get a different result. That's why on curved manifolds the coordinate derivatives of general vectors and tensors have to be corrected using affine connections.

To me curvature does not appear in the formula t^a v^b w^c D_a g_bc=0, since this is simply not its purpose or better since curvature appears (is defined) in the following formula: (D_aD_b-D_bD_a)t_c=R_abcd^d t_d Thus it is a quantity of second order in D.

an expression for the Riemann tensor in terms of the LC connection and its derivatives:

Right, but it is well known that you can derive the Riemann curvature tensor fromThe parallel transport is a quantity first order in D

first-order quantities, based on the path-dependence of the first-order effects of

transport. I think that's called "holonomy".

I suggest that you read Carroll's Notes Ch 3.

Best wishes,

Paul ZielinskiKind regards Kay ____________________________________________________________________________________ You rock. That's why Blockbuster's offering you one month of Blockbuster Total Access, No Cost. http://tc.deals.yahoo.com/tc/blockbuster/text5.com