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- Xref: sparky sci.physics:18998 sci.math:15074
- Newsgroups: sci.physics,sci.math
- Path: sparky!uunet!snorkelwacker.mit.edu!galois!riesz!jbaez
- From: jbaez@riesz.mit.edu (John C. Baez)
- Subject: Covariant vs. Lie Derivative in Gen. Rel.?
- Message-ID: <1992Nov16.221115.9273@galois.mit.edu>
- Sender: news@galois.mit.edu
- Nntp-Posting-Host: riesz
- Organization: MIT, Department of Mathematics
- Date: Mon, 16 Nov 92 22:11:15 GMT
- Lines: 60
-
- Ric Peregrino writes:
-
- >I've been struggling with the concept of covariant derivative for some
- time.
-
- >For a covariant tensor of 1st rank (order, whatever) A
- >
- > j k
- >A = dA /dx - G A
- > i;j i ij k
- >
- >as I'm sure you all know.
-
- Fine. Here G, usually written Gamma, is called the Christoffel symbol
- in the context of general relativity. The left hand side is called the
- covariant derivative of A.
-
- >The "connection" is an additional restraint that need not
- >be imposed to insure the tensor character of the resulting tensor.
-
- Huh? This makes no sense to me, which is probably related to
- the problem you are having. The "connection" is not a restraint, or
- constraint. There are many ways of thinking about connections but the
- most relevant here is that the connection simply IS the entity
-
- k
- G
- ij
-
- More precisely, the Christoffel symbol expresses the connection in a
- given coordinate system (or frame). It's obvious from this point of
- view that one needs a connection to define covariant derivatives.
-
- >>2) The covariant derivative only requires a tangent vector at one point
- >>of the manifold. The price you pay is this: to define it you need to
- >>choose a connection on the (tangent bundle of) the manifold. Of course,
- >>such a connection - the Levi-Civita connection - comes for free if your
- >>manifold has a Riemann metric on it.
-
- >Hmm.... If I look at the covariant derivative of the scalar I above, all
- >I need to know are...
-
- You misunderstood my point since I wasn't precise enough. To take the
- covariant derivative of a tensor A in a given direction v at a certain
- point x, all *v* needs to be is a tangent vector at x. I would write
- this covariant derivative as D_v A(x); in coordinates it's
-
- j
- v A (x)
-
- i;j
-
- Of course to figure this out we need to know A and all its first
- derivatives at x and we need to have a choice of connection, that is, we
- need to know the Christoffel symbol. My point was that we only need to
- have a tangent vector v at the point x, as opposed to what we need to
- compute a Lie derivative, which requires v to be a vector field in a
- neighborhood of the point x. (Well, one can get by with less to compute
- a Lie derivative, but please don't anyone point that out and comfuse
- things further.)
-