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.Is this an axiom or something that is derived from other axioms?
Le me rephrase this. It can be derived.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.
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 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.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.
No. I had just explaine why it isn't an axiom. I derived it for you.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.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.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.
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.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.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: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?
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.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".