About the metric tensor

  • #26
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... never mind. If the OP is happy then so am I. :smile:

Pete
 
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  • #27
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?
 
  • #28
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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! :smile:
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. :smile:
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?
See above. Does that address your question?

Pete
 
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  • #29
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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! :smile: 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. :biggrin:

Pete
 
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  • #30
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 [itex]t(f) = t^a\nabla_a f[/itex]
5. Torsion free-- [itex][\nabla_a,\nabla_b]f = 0[/itex]
 
  • #31
And the last one is optional.
 

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