- #1

snoopies622

- 846

- 28

- TL;DR Summary
- I'm wondering under what circumstances deriving Christoffel symbols in two different ways will produce the same results.

I've noticed that for both the surface of a sphere and a paraboloid, one arrives at the same Christoffel symbols whether using

[tex] \Gamma^i_{kl} = \frac {1}{2} g^{im} ( \frac {\partial g_{mk} }{\partial x^l} + \frac {\partial g_{ml}}{\partial x^k} - \frac {\partial g_{kl}} {\partial x^m} )

[/tex] which assumes [itex]

\nabla g=0 [/itex], or by embedding them in (flat) three dimensional space and taking the inner product of the partial derivatives of the basis vectors and the contravariant forms of those basis vectors, as in

[tex] \Gamma ^k_{ij} = \frac { \partial \bf{e}_i}{ \partial x^j} \cdot { \bf {e} ^k} [/tex]

and I'm wondering why this is the case, and if it is always so. Is it because I'm tacitly assuming [itex]

\nabla g=0 [/itex] in the three-space as well?

[tex] \Gamma^i_{kl} = \frac {1}{2} g^{im} ( \frac {\partial g_{mk} }{\partial x^l} + \frac {\partial g_{ml}}{\partial x^k} - \frac {\partial g_{kl}} {\partial x^m} )

[/tex] which assumes [itex]

\nabla g=0 [/itex], or by embedding them in (flat) three dimensional space and taking the inner product of the partial derivatives of the basis vectors and the contravariant forms of those basis vectors, as in

[tex] \Gamma ^k_{ij} = \frac { \partial \bf{e}_i}{ \partial x^j} \cdot { \bf {e} ^k} [/tex]

and I'm wondering why this is the case, and if it is always so. Is it because I'm tacitly assuming [itex]

\nabla g=0 [/itex] in the three-space as well?