- #1
snoopies622
- 840
- 28
The covariant derivative of any metric tensor is zero. Is this an axiom or something that is derived from other axioms?
Derived.snoopies622 said:The covariant derivative of any metric tensor is zero. 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.snoopies622 said:Is this an axiom or something that is derived from other axioms?
pmb_phy said:Derived.
Pete
Le me rephrase this. It can be derived.DavidWhitbeck said:Um no Halls of Ivy had it dead on.
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.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.
pmb_phy said:The vanishing of the covariant derivative of the metric tensor is neither a defining property of the metric or the covariant derivative. ^
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 said: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
No. I had just explaine why it isn't an axiom. I derived it for you.nrqed said:So it is an axiom, right?
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 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.
How so? I've never seen that before myself.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.
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.Hurkyl said:Sheesh, does nobody listen to a word I say?
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.nrqed said: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.
pmb_phy said: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.
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.Hurkyl said: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.
"Self-evident" is a terrible notion.snoopies622 said: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”?
There are two easy trivial ways to answer your challenge:pmb_phy said: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
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.Hurkyl said:First, there's nothing stopping me from studying a theory by taking as my axioms every single one 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.(Clearly, each of my axioms is true in the theory, and from my axioms, I can (trivially) deduce any of its theorems)
Okay. Then address the current problem that I just explained above regarding the unesccesary connection of affine geometries and metric geometries.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 said:DavidWhitbeck - what page in wald?
If you can take a look at it tommorow and note the page number I would be grateful. Thanks David!DavidWhitbeck said: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 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.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.
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 302In 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, ...
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.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.
See above. Does that address your question?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?
Ah! I see I should have looked more closely at Ohanian! since the authors wrotepmb_phy said:I believe that it is the vanishing of the covariant derivative of the metric tensor that establishes the connection between the two geometries.
from which the authors show that the covariant derivative of the metric tensor is zero.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 ...
The metric tensor is a mathematical object that describes the geometry of a space. It is important in understanding the covariant derivative because it allows us to define a measure of distance and angles in a curved space, which is necessary for calculating the covariant derivative.
The metric tensor is used to define the inner product of two vectors, which is necessary for calculating the parallel transport of a vector from one point to another in a curved space. It also determines the change in the vector's components due to the curvature of the space.
The covariant derivative is a generalization of the ordinary derivative to curved spaces. It takes into account the curvature of the space and how it affects the change in a vector field. The ordinary derivative, on the other hand, only considers the change in a vector field in flat, Euclidean space.
In general relativity, the covariant derivative is used to define the geodesic equation, which describes the motion of particles in curved spacetime. It is also used to calculate the curvature tensor, which is essential for understanding the gravitational field and the behavior of matter in the presence of gravity.
Yes, understanding the metric tensor and the covariant derivative is crucial in many areas of physics, including general relativity, cosmology, and quantum field theory. It is also used in engineering and computer science for applications such as image processing and computer graphics.