- #1

RedX

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Is this the right way to think about the covariant derivative, and if not, what improvements would you suggest to visualize the meaning of the covariant derivative?

[tex](1)\mbox{ }\vec{e}_i(x')=\vec{e}_i(x)+\frac{\partial \vec{e}_i(x)}{\partial x^j}dx^j

=\vec{e}_i(x)+\Gamma^{k}_{ij}\vec{e}_k(x)dx^j[/tex]

so that [tex]\vec{e}_i(x)=\vec{e}_i(x')-\Gamma^{k}_{ij}\vec{e}_k(x)dx^j [/tex].

This means that the vector V at x, when written in the x' basis, is given by contracting both sides with V

[tex](2) \mbox{ }\vec{V}(x)=V^i(x) [\vec{e}_i(x')-\Gamma^{k}_{ij}\vec{e}_k(x)dx^j]=

V^i(x) [\vec{e}_i(x')-\Gamma^{k}_{ij}\vec{e}_k(x')dx^j]

[/tex]

(in the last step I replaced the basis in x with the basis in x' since the difference is a product of two differentials which is small)

Now, suppose that V(x) changes to V(x') as you move from x to x'. V(x') can be written in the x' basis as:

[tex]\vec{V}(x')=[V^i(x)+\frac{\partial V^i(x)}{\partial x^j}dx^j]\vec{e}_i(x') [/tex]

Subtracting V(x') from V(x):

[tex] \left[\frac{\partial V^i(x)}{\partial x^j}+V^k\Gamma^{i}_{kj}\right]\vec{e}_i(x')dx^j=\left[\frac{\partial V^i(x)}{\partial x^j}+V^k\Gamma^{i}_{kj}\right]\vec{e}_i(x)dx^j[/tex]

I also have a question about equation (1). To relate the basis vectors at two different points, do you have to use an embedding space? How is it possible to write the basis at a point as a linear combination of the basis at different point without an embedding space?

Also, on something that I think is related to my 1st question, for equation (2), shouldn't this equality be exact? But if (2) represents a parallel transport of V(x) to x' (as it should since the covariant derivative is the difference between the new vector and the parallel transported old vector, so (2) has to represent the parallel transported old vector), then the vectors aren't equal. A parallel transported vector is different from the initial vector in the embedding space.

edit:

I was doing some reading on Wikipedia, and I ran across the article on an "affine connection". It would make sense to me if the Christoffel symbols "lifted" the basis vectors from one tangent space to another. However, the Christoffel symbols do not behave this way, since they are just an array of ordinary numbers! In other words, a linear combination of the basis vectors in the tangent space still exists in the tangent space, so how can the Christoffel symbols form an affine connection to a different tangent space (I'm not sure if I'm using the terminology right)?

[tex](1)\mbox{ }\vec{e}_i(x')=\vec{e}_i(x)+\frac{\partial \vec{e}_i(x)}{\partial x^j}dx^j

=\vec{e}_i(x)+\Gamma^{k}_{ij}\vec{e}_k(x)dx^j[/tex]

so that [tex]\vec{e}_i(x)=\vec{e}_i(x')-\Gamma^{k}_{ij}\vec{e}_k(x)dx^j [/tex].

This means that the vector V at x, when written in the x' basis, is given by contracting both sides with V

^{i}[tex](2) \mbox{ }\vec{V}(x)=V^i(x) [\vec{e}_i(x')-\Gamma^{k}_{ij}\vec{e}_k(x)dx^j]=

V^i(x) [\vec{e}_i(x')-\Gamma^{k}_{ij}\vec{e}_k(x')dx^j]

[/tex]

(in the last step I replaced the basis in x with the basis in x' since the difference is a product of two differentials which is small)

Now, suppose that V(x) changes to V(x') as you move from x to x'. V(x') can be written in the x' basis as:

[tex]\vec{V}(x')=[V^i(x)+\frac{\partial V^i(x)}{\partial x^j}dx^j]\vec{e}_i(x') [/tex]

Subtracting V(x') from V(x):

[tex] \left[\frac{\partial V^i(x)}{\partial x^j}+V^k\Gamma^{i}_{kj}\right]\vec{e}_i(x')dx^j=\left[\frac{\partial V^i(x)}{\partial x^j}+V^k\Gamma^{i}_{kj}\right]\vec{e}_i(x)dx^j[/tex]

I also have a question about equation (1). To relate the basis vectors at two different points, do you have to use an embedding space? How is it possible to write the basis at a point as a linear combination of the basis at different point without an embedding space?

Also, on something that I think is related to my 1st question, for equation (2), shouldn't this equality be exact? But if (2) represents a parallel transport of V(x) to x' (as it should since the covariant derivative is the difference between the new vector and the parallel transported old vector, so (2) has to represent the parallel transported old vector), then the vectors aren't equal. A parallel transported vector is different from the initial vector in the embedding space.

edit:

I was doing some reading on Wikipedia, and I ran across the article on an "affine connection". It would make sense to me if the Christoffel symbols "lifted" the basis vectors from one tangent space to another. However, the Christoffel symbols do not behave this way, since they are just an array of ordinary numbers! In other words, a linear combination of the basis vectors in the tangent space still exists in the tangent space, so how can the Christoffel symbols form an affine connection to a different tangent space (I'm not sure if I'm using the terminology right)?

Last edited: