1. May 31, 2008

### snoopies622

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

2. May 31, 2008

### robphy

This is called the "metric compatibility" condition.

3. May 31, 2008

### HallsofIvy

Staff Emeritus
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.

4. May 31, 2008

### pmb_phy

Derived.

Pete

5. May 31, 2008

### Hurkyl

Staff Emeritus
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.

6. May 31, 2008

### DavidWhitbeck

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.

7. Jun 1, 2008

### 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?

8. Jun 1, 2008

### pmb_phy

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. 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 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. No. I had just explaine why it isn't an axiom. I derived it for you. 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. How so? I've never seen that before myself. Pete 11. Jun 1, 2008 ### Hurkyl Staff Emeritus Sheesh, does nobody listen to a word I say? 12. Jun 1, 2008 ### pmb_phy 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? 13. Jun 1, 2008 ### pmb_phy 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 14. Jun 1, 2008 ### Hurkyl Staff Emeritus 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. 15. Jun 1, 2008 ### DavidWhitbeck It's in Wald. 16. Jun 1, 2008 ### pmb_phy 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? 17. Jun 1, 2008 ### 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

Last edited: Jun 1, 2008
18. Jun 1, 2008

### 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

19. Jun 1, 2008

### 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

20. Jun 1, 2008

### 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

Last edited: Jun 1, 2008