snoopies622
The covariant derivative of any metric tensor is zero. Is this an axiom or something that is derived from other axioms?

Homework Helper
Gold Member
This is called the "metric compatibility" condition.

Homework Helper
It really comes from the definition of "covariant derivative". The difference between the "covariant" derivative and the "naive" derivative is that the covariant discounts changes in a function that are due solely to "non-flatness" in the space. Since the metric tensor just gives that "non-flatness", it is essentially a "constant" and have (covariant) derivative 0.

pmb_phy
The covariant derivative of any metric tensor is zero. Is this an axiom or something that is derived from other axioms?
Derived.

Pete

Staff Emeritus
Gold Member
Is this an axiom or something that is derived from other axioms?
Your question is irrelevant. The only difference between 'something that is derived' and 'an axiom' is the manner in which one decides to present the theory.

DavidWhitbeck
Derived.

Pete

Um no Halls of Ivy had it dead on. It's just part of the definition of a covariant derivative, it's not a derivable property. The covariant derivative just one of a few derivative operators that are useful to use.

snoopies622
Thanks.

Is entry #3 the definition of covariant derivative, or only a description of one of its features? If the latter, how exactly is it defined?

pmb_phy
Le me rephrase this. It can be derived.
It's just part of the definition of a covariant derivative, it's not a derivable property. The covariant derivative just one of a few derivative operators that are useful to use.
Since when?? I've never seen a treatment of general relativity or tensor analysis when that was taken as a defining property. The reason I said it is derived is that (1) I've derived that relationship myself and (2) most treatements I've seen contain this derivation.

The vanishing of the covariant derivative establishes a relationship between the affine geometry of a manifold and its metric geometry. The derivation proceeds as follows: One requires that the magnitude of the scalar product of two 4-vectors remains unchanged when the 4-vectors are parallel transported. i.e.

$A^\mu B^\nu g{\mu\nu})_{,\beta} = 0[/tex] Since [itex]A^\mu B^\nu g{\mu\nu})_{\beta}$ is a scalar, we can replace the ordinary derivative with the covariant derivative

$A^\mu B^\nu g{\mu\nu})_{;\beta} = 0[/tex] When this is written out there will appear two terms which are the covariant derivatives of the vectors. Since the only changes contributed are those from the parallel transport then it follows that those terms vanish. One the ends up with [itex]A^\mu\nu g{\mu\nu}_{;\beta} = 0[/tex] Since the two vector is arbitrary if follows that [itex]g{\mu\nu}_{;\beta} = 0[/tex] Hence this relationship can be derived. Pete Last edited: Science Advisor Homework Helper Gold Member The vanishing of the covariant derivative of the metric tensor is neither a defining property of the metric or the covariant derivative. ^ So it is an axiom, right? I mean one must assume that there is no torsion. In that case the conenction is given by the Chrristoffel symbols. But as far as I know, the fact that there is no torsion is an axiom. The reason I said it is derived is that (1) I've derived that relationship myself and (2) most treatements I've seen contain this derivation. Pete I am a bit confused. If one assumes that the connection si given the the Christoffel symbols then it can be derived. But again, this follows from making the assumption that there is no torsion. pmb_phy Sorry nut that 1/2 hour is not enough to get an edit wrinkle free. I had to repost the derivation. So I'll repost the entire post instead - The vanishing of the covariant derivative establishes a relationship between the affine geometry of a manifold and its metric geometry. The derivation proceeds as follows: One requires that the magnitude of the scalar product of two 4-vectors remains unchanged when the 4-vectors are parallel transported. i.e. [itex]A^\mu B^\nu g_{\mu\nu},\beta} = 0[/tex] Since [itex]A^\mu B^\nu g_{\mu\nu;\beta}$ is a scalar, we can replace the ordinary derivative with the covariant derivative

$A^\mu B^\nu g_{\mu\nu;\beta} = 0[/tex] When this is written out there will appear two terms which are the covariant derivatives of the vectors. Since the only changes contributed are those from the parallel transport then it follows that those terms vanish. One the ends up with [itex]A^\mu B\nu g_{\mu\nu;\beta} = 0[/tex] Since the two vector is arbitrary if follows that [itex]g{\mu\nu;\beta} = 0[/tex] Hence this relationship can be derived. So it is an axiom, right? No. I had just explaine why it isn't an axiom. I derived it for you. I mean one must assume that there is no torsion. In that case the conenction is given by the Chrristoffel symbols. But as far as I know, the fact that there is no torsion is an axiom. That is usually assumed when one is working with the affine connection, but yes. Its an assumption that the given that the manifold is torsion free. I am a bit confused. If one assumes that the connection si given the the Christoffel symbols then it can be derived. But again, this follows from making the assumption that there is no torsion. How so? I've never seen that before myself. Pete Staff Emeritus Science Advisor Gold Member Sheesh, does nobody listen to a word I say? pmb_phy Sheesh, does nobody listen to a word I say? Yes. Sometimes we agree with what you post and sometimes it is dismissed as irrelevant. In this case a derivation of the relation starting with basic properties was quite useful, especially since there is no reason to assume that anyone has ever chosen different axioms where the vanishing of the covariant derivative of the metric is zero. Especially since this derivation is found in such GR texts as that of Ohanian and Ruffini. Does that bother you for some reason? pmb_phy I mean one must assume that there is no torsion. In that case the conenction is given by the Chrristoffel symbols. But as far as I know, the fact that there is no torsion is an axiom. Torsion is not an axiom, it is a property of a manifold. In retrospect I don't see any place in the above derivation which assumes zero torsion. Pete Staff Emeritus Science Advisor Gold Member Showing that one set of properties is equivalent to another set of properties: useful. Quibbling about which set of properties is "right" to use as axioms: not useful. Acting as if 'axiomness' is a quality inherent to a set of properties: wrong. DavidWhitbeck Le me rephrase this. It can be derived. Since when?? I've never seen a treatment of general relativity or tensor analysis when that was taken as a defining property. It's in Wald. pmb_phy Showing that one set of properties is equivalent to another set of properties: useful. Quibbling about which set of properties is "right" to use as axioms: not useful. Acting as if 'axiomness' is a quality inherent to a set of properties: wrong. None of which I or the OP seem to care about. A derivation was given where only the typical definitions were used. If that bothers you then ignore it. I never said nor implied that you were wrong so I don't see why you're so concerned about this. In any case I'm not even intersted at this point. DavidWhitbeck - what page in wald? pmb_phy re - [itex]g(\nabla_v(X), Y) = 0$ - Where did this come from? I was referring to the fact that the first two terms in

$$A^\mu_{;\beta}B^\nu g_{\mu\nu} + A^\mu_{;\beta} B^\nu + A^\mu B^\nu g_{\mu\nu;\beta} = 0$$

vanish. I.e. since the only changes in the vectors $A^\mu$ and $B^\mu$ are those contributed by parallel transport then the covariant derivatives of those vectors vanish and the above equation reduces to

$$A^\mu B^\nu g_{\mu\nu;\beta} = 0$$

Pete

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pmb_phy
Note: In the derivation I posed above there were supposed to be parentheses around the expression so that it'd be apparent that it was the derivative of the scalar product that the derivative was being taken of. I didn't notice that they were missing until just now.

Due to the recent change in software which allows us only 30 minutes to edit a post you can expect this to continue into the future. As such I'd like to take this opportunity to say that I highly object to such a short limit on the time allowed to edit posts. I use the submit button to view the posts and as such it often takes more than 30 minutes to get a post just right. I stopped using the Preview Post function a very long time ago when I realized that there was a problem with it (I forget exactly what the problem was though).

Pete

pmb_phy
Hurkyl - So let me get this straight. You fully believe that, in all possible cases, any theorem that exists, or can exist, in a mathemtical theory can be rephrased as an axiom with appropriate changes in definions and basic axiom set. Is that correct? If so then I find that hard to believe, especially if you expect us to accept it without proof. Do you have or know of a proof of this claim/assertion?

Pete

pmb_phy
Let me give you an example which makes this idea of yours hard to believe. If one defines either the metric or the covariant derivative such that the covariant derivative of the metric vanishes then I'd say that your assertion is not possible. This is because such a definition would require all metric spaces to have an affine connection defined on them and all affine spaces to have a metric defined on them. That would require a change in the definition of both metric geometry and affine geometry since as it is now an affine space can exist without an affine connection on it and an affine space can exist without a metric defined on it.

Pete

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snoopies622
Thanks, Pete. I appreciate your efforts.

Hurkyl’s point is well taken, but I think it is always better in math to start with something self-evident and build up from there, rather than simply to declare that it’s all equivalent so it doesn’t matter where one begins. Euclid could have made the Pythagorean Theorem an axiom and used it to prove the parallel postulate, but that would’ve been kind of strange, in my opinion.

Oh, and isn’t the last word in entry #14 supposed to be “priceless”?

Staff Emeritus
Gold Member
Thanks, Pete. I appreciate your efforts.

Hurkyl’s point is well taken, but I think it is always better in math to start with something self-evident and build up from there, rather than simply to declare that it’s all equivalent so it doesn’t matter where one begins. Euclid could have made the Pythagorean Theorem an axiom and used it to prove the parallel postulate, but that would’ve been kind of strange, in my opinion.

Oh, and isn’t the last word in entry #14 supposed to be “priceless”?
"Self-evident" is a terrible notion.

Anyways, I wasn't trying to say the axiomatic method was a bad idea -- I'm trying to warn against trying to ascribe any meaning to the axioms beyond "I'm taking this as a starting point". Pedagogically, it's quite useful to start from a core set of assumptions and see what can be derived from them. And formally, axioms are a convenient way to specify theories and work with them, much in the same way a basis provides a convenient way to work with a vector space. (In fact, axioms for a theory and a spanning set for a vector space are both examples of the more general notion of a 'generating set')

And generally speaking, it's a good idea to work from many different starting points and see how they relate. For a relevant example, consider Spivak's Differential Geometry, one of the primary references on the subject. He gives no less than three different definitions of "tangent vector" (by the curve it's tangent to, by the differential operator it defines, and by its components relative to all coordinate charts), and then later gives a fourth abstract axiomatic definition of "tangent bundle", showing that it's unique up to isomorphism and all three of the original definitions really are examples of tangent bundles. The format of http://www.mathpop.com/bookhtms/dg2.htm]Spivak's second book [Broken] is similar; through several of its chapters, he rederives differential geometry from several different starting points.

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Staff Emeritus
Gold Member
Hurkyl - So let me get this straight. You fully believe that, in all possible cases, any theorem that exists, or can exist, in a mathemtical theory can be rephrased as an axiom with appropriate changes in definions and basic axiom set. Is that correct? If so then I find that hard to believe, especially if you expect us to accept it without proof. Do you have or know of a proof of this claim/assertion?

Pete

First, there's nothing stopping me from studying a theory by taking as my axioms every single one of its theorems. (Clearly, each of my axioms is true in the theory, and from my axioms, I can (trivially) deduce any of its theorems)

A less over-the-top response would be for me to take as my set of axioms the theorem you selected together with the axioms you were using.

Finally, just reading many math textbooks will provide numerous examples. Look for any theorem that states several different conditions are equivalent; that will yield several different ways to axiomatically define whatever concept they are describing. Also look for theorems describing "necessary and sufficient conditions".

pmb_phy
First, there's nothing stopping me from studying a theory by taking as my axioms every single one of its theorems.
That is far from clear. In mathematics one needs to be able to prove assertions like the one you make and not make assumptions such as this one. Although it may appear obvious to you it doesn't mean that its true. Math is too tricky for that.
(Clearly, each of my axioms is true in the theory, and from my axioms, I can (trivially) deduce any of its theorems)
Theorems often have restrictions to which it applies so what one takes as axioms isn't as easy as you make it seem to appear.
Finally, just reading many math textbooks will provide numerous examples. Look for any theorem that states several different conditions are equivalent; that will yield several different ways to axiomatically define whatever concept they are describing. Also look for theorems describing "necessary and sufficient conditions".
Okay. Then address the current problem that I just explained above regarding the unesccesary connection of affine geometries and metric geometries.

Pete

Staff Emeritus
Gold Member
So as not to derail the thread, I will continue the discussion of formal logic via private message.

pmb_phy
... never mind. If the OP is happy then so am I.

Pete

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DavidWhitbeck
DavidWhitbeck - what page in wald?

Arg, Pete if I saw this earlier while I was still in the office I could have told you. It is in the chapter on derivative operators and curvature which should be chapter three.

I would have to study it carefully, but I think that you might be stipulating additional properties that Wald does not assume. If I could see axiomatically how you arrived at the covariant derivatives of vectors vanished then maybe I could understand your p.o.v.

In a previous post you said something to the effect of one can't define a covariant derivative operator without implicitly assuming something about the existence of an affine connection, I don't really get that remark and I'm probably not reading it correctly, I just never studied at that level, but it seems to me that you can only run into a problem with a definition if it's
(a) ill-defined, i.e. lacking in logical consistency
(b) well defined, but doesn't actually exist

I'm hoping that it's not (a). Are you saying that you can't prove the existence of such an operator without making additional assumptions?

pmb_phy
Arg, Pete if I saw this earlier while I was still in the office I could have told you. It is in the chapter on derivative operators and curvature which should be chapter three.
If you can take a look at it tommorow and note the page number I would be grateful. Thanks David!
I would have to study it carefully, but I think that you might be stipulating additional properties that Wald does not assume. If I could see axiomatically how you arrived at the covariant derivatives of vectors vanished then maybe I could understand your p.o.v.
I recall reading the proof in more than one places but at the moment I can only recall the one in Gravitation and Spacetime - Second Edition, by Ohanian and Ruffini page 326-327. When I posted the derivation here I was following that text (after all its not like I've memorized all these derivations. My memory stinks!). Basically the authors parallel transport two vectors A and B defined from a point P on a manifold to another point Q. Since the vectors must remain unchanged by the parallel transport, by definition, then the only thing that can change are its components in a curvilinear system. For this reason the covariant derivative of each vector must vanish. If you'd like I can scan the text in of this proof and upload it unto my website and Pm the link to you.
In a previous post you said something to the effect of one can't define a covariant derivative operator without implicitly assuming something about the existence of an affine connection, I don't really get that remark and I'm probably not reading it correctly, ...
It was implied that one could take the vanishing of the covariant derivative of the metric tensor as an axiom. That was the comment that I was talking about. Regarding the metric tensor and the affine connection - Since Ohanian and Ruffini say this so beautifully I'll simply quote that text. From page 302
Mathematically, a Riemannian space is a differentiable manifold endowed with a topological structure and a geometrical structure. In the discussion of the geometric structure of a curved space we can make a distinction between the affine geometry and the metric geometry. These two kinds of geometries correspond to two different ways in which we can detect the curvature of a space. One way is by examining the behavior of of parallel line segments, or parallel vectors. For example, on the surface of a sphere, we can readily detect the curvature by transporting a vector around a close path, always keeping the vector parallel to itself as possible. [fig not shown here] shows what happens if we parallel transport a vector around a triangular path on the sphere. The final vector differs in the direction from the initial vector, whereas on a flat surface the final vector would not differ. Such changes in a vector produced by parallel transport characterize the affine geometry (the word affine means connected, and refers to how parallels at different places are connected, or related). ... Another way in which we can detect curvature is by measurements of lengths and areas. For example, we can draw a circle on the sphere, and check the radius vs. circumference, or the radius vs. area. Both the circumference and the area of such a circle are smaller than for a circle on a flat surface [figure not shown here]. Such deviations from what we expect on a flat surface characterize the metric geometry.
One can have an affine space with no metric defined on it and one can have a metric space with no affine connection defined on it. If one were to take the covariant derivative of the metric equals zero as an axiom then I don't see how we can keep those geometries seperate. I believe that it is the vanishing of the covariant derivative of the metric tensor that establishes the connection between the two geometries. I guess the problem I have is that stating the covariant derivative of the metric tensor = 0 as an axiom seems to me to border on a definition of the metric instead. Too high-guru math for me to see it as being obvious or not.
I just never studied at that level, but it seems to me that you can only run into a problem with a definition if it's
(a) ill-defined, i.e. lacking in logical consistency
(b) well defined, but doesn't actually exist

I'm hoping that it's not (a). Are you saying that you can't prove the existence of such an operator without making additional assumptions?

Pete

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pmb_phy
I believe that it is the vanishing of the covariant derivative of the metric tensor that establishes the connection between the two geometries.
Ah! I see I should have looked more closely at Ohanian! since the authors wrote
Next, we establish the relation between the affine geometry and the metric geometry, that is, the relationship between the Christoffel symbols (...) and the metric tensor ...
from which the authors show that the covariant derivative of the metric tensor is zero.

Its nice to see that my instincts still work ... sometimes.

Pete

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DavidWhitbeck
Alright Pete it looks like you are dealing with defining a derivative operator such that it parallel transports vectors, and then you prove that the metric is constant with respect to that operator.

Wald defines derivative operators by five rules on page 31, and then on page 35 he proves that if you additionally require that it keeps the metric constant, then there exists a unique derivative operator that satisfies those six properties, and he calls it the covariant derivative.

So you go: parallel transport along geodesics => metric covariantly constant
Wald goes: metric covariantly constant=> parallel transport along geodesics

It looks like the person who is really right on this thread is Hurkyl for pointing out that you can do it either way.

So these are the five properties that a derivative operator must satisfy:

1. Linearity.
2. Leibnitz Rule.
3. Commutativity with contraction.
4. Tangent vectors satisfy $t(f) = t^a\nabla_a f$
5. Torsion free-- $[\nabla_a,\nabla_b]f = 0$

DavidWhitbeck
And the last one is optional.