| Topic: |
Science > Physics |
| User: |
"Mike Fontenot" |
| Date: |
06 Nov 2007 03:53:05 PM |
| Object: |
Covariant Derivatives, and Invariance of Vectors |
I posted the following question about a week ago on
sci.physics.foundations, and I got several responses from
knowledgeable people. But I didn't get any responses from
anyone familiar with Simmonds' book. I'd like to hear
some comments from someone who is knowledgeable about
Simmonds' book, and who can address my misunderstanding in
the context of that book. Here's the posting:
________________________________________________________________
I've recently worked my way through Simmonds' "A Brief on Tensor
Analysis", and one result he gives has made me realize that
I apparently don't really understand what is meant by "invariance",
in particular, the invariance of vectors.
I THOUGHT that the "invariance of vectors" referred to the fact
that vectors have an existence that is independent of which
co-ordinate system is used to describe them (i.e., they have
a "direction" in space (or spacetime), and a length, that are
the same in ALL co-ordinate systems).
But one of Simmonds' results contradicts my understanding. When
he introduces the covariant derivative, del_sub_i, on page 79
(of the 2nd edition), he states two properties that the
covariant derivative must possess, the first of which is
(a) "The covariant derivative of an invariant, e.g., a scalar,
a vector, or a 2nd order tensor, coincides with its partial
derivative".
He then shows that the covariant derivative of each of the
base vectors g_sub_i (as well as of each of the reciprocal
base vectors g_sup_i) is identically zero, i.e., that
del_sub_j (g_sub_i) = 0.
But in arriving at that result, it is clear that he is NOT
assuming that the partial of g_sub_i, wrt the co-ordinate
u_sup_j, is zero. And it IS clear that the partial derivatives
of each of the base vectors aren't identically zero, since each
base vector can in general have different lengths and directions
at different points in the space.
So property (a) above is NOT true for each base vector, yet
he is requiring that it be true for any "invariant" vector.
So the base vectors apparently aren't "invariant vectors",
but I don't understand why not. A given base vector, at
some given point in the space, certainly has a specific
length and direction, independent of all co-ordinate systems.
(Of course, it's not a BASE vector in those other co-ordinate
systems, but it IS a vector with a specific length and
direction).
And clearly Simmonds isn't using the term "invariant" to
just mean a vector which has the same length and direction
at each point in the space...he applies property (a) to
general vector fields, which vary at different points in the
space, and which clearly have non-zero partials.
Does anyone on this newsgroup know what my misunderstanding
is here? I'd appreciate very much any words of wisdom that
anyone can supply.
Mike Fontenot
.
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| User: "Rock Brentwood" |
|
| Title: Re: Covariant Derivatives, and Invariance of Vectors |
06 Nov 2007 04:09:52 PM |
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On Nov 6, 3:53 pm, Mike Fontenot <mlf...@comcast.net> wrote:
(a) "The covariant derivative of an invariant, e.g., a scalar,
a vector, or a 2nd order tensor, coincides with its partial
derivative".
What you stated for (a) is wrong. If Simmons also stated it, then it's
(still) wrong. Covariant derivatives coincide with partial derivatives
for the scalar functions, only. There is no meaningful partial
derivatives for geometric objects on a manifold, since Calculus is
done on numeric objects, not geometric objects (and more generally, if
one is to get technical, Calculus is done in Banach spaces, but that
gets aside from the issue here).
Different ways to define derivatives are therefore introduced for the
various geometric objects on manifolds. The Lie Derivative is the
closest thing you have to a partial derivative, and it's defined for
tensors. Thus, Lie_(d/dx^i) (T) for a tensor T has, as its components,
the partials with respect to x^i of the respective components of T.
The Lie Derivative is a structure native to the manifold and does not
require any additional structure atop the manifold.
The covariant derivative requires the definition of a connection (and
serves, also, to define a connection). There is no a' priori
definition for the covariant derivative of the basis vector e_i = d/
dx^i. It's arbitrary and, in fact, serves to define the components of
the connection via the formula
del_i (e_j) = sum Gamma^k_{ij} e_k.
Different choices for the set of components (Gamma^k_{ij}) give you
different definitions of different connections and covariant
derivatives. There is no criterion at the outset for further
restricting what these values ought to be. In particular, condition
(a) does NOT apply to the coordinate basis vectors (e_i = d/dx^i), nor
to any other basis.
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| User: "Eric Gisse" |
|
| Title: Re: Covariant Derivatives, and Invariance of Vectors |
06 Nov 2007 04:40:20 PM |
|
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On Tue, 06 Nov 2007 14:53:05 -0700, Mike Fontenot <mlfasf@comcast.net>
wrote:
[...]
(a) "The covariant derivative of an invariant, e.g., a scalar,
a vector, or a 2nd order tensor, coincides with its partial
derivative".
The phrase "in flat space" needs to be added to the end.
[...]
.
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