Christoffel symbol and covariant derivative

Question

I came across the Christoffel symbols via the geodesic equation, and I understand the extrinsic form and the intrinsic form and can prove that they are identical:

extrinsic form:

$$\Gamma^{j}_{~ik}=\frac{\partial^2 \vec{\mathbf{r}}}{\partial u^i\partial u^k}\cdot\frac{\partial\vec{\mathbf{r}}}{\partial u^m}\cdot g^{mj} $$

intrinsic form:

$$\Gamma^{j}_{~ik}= \frac{1}{2} \left(\frac{\partial g_{km}}{\partial u^i}+\frac{\partial g_{mi}}{\partial u^k} -\frac{\partial g_{ik}}{\partial u^m} \right)\cdot g^{mj}$$

Now, in context with the covariant derivative there is another version of Christoffel symbols.

I understand that in curvilinear coordinates, in order to get the derivatives of a vector, you have to differentiate the coefficients and the basis vectors

So, (Schutz chapter 5.3)

$$\frac{\partial V}{\partial x^\beta}= \frac{\partial V^\alpha}{\partial x^\beta}\vec{\mathbf{e}}_\alpha + V^\alpha\frac{\partial \vec{\mathbf{e}}_\alpha}{\partial x^\beta}$$

which I understand, but in the next line he writes

$$\frac{\partial \vec{\mathbf{e}}_\alpha}{\partial x^\beta} = \Gamma^{\mu}_{~\alpha\beta}\vec{\mathbf{e}}_\mu$$

which I do not understand. Why is that $\Gamma^{\mu}_{~\alpha\beta}$??

All textbooks I looked at, just define $\Gamma^{\mu}_{~\alpha\beta}$ that way, but what has it to do with the intrinsic and the extrinsic form?

Ideally, somebody could please show me how to calculate the intrinsic and/or the extrinsic form from the latter form (or vice versa)

Answer 1

A geodesic is a curve $\gamma$ which extremizes the path length functional $$S[\gamma]= \int d\lambda \ L(\gamma,\dot \gamma) =\int d\lambda \sqrt{g_{\mu\nu}(\gamma)\dot\gamma^\mu\dot\gamma^\nu}$$ The Euler-Lagrange equations are $$\frac{\partial L}{\partial \gamma^\alpha} = \frac{d}{d\lambda}\frac{\partial L}{\partial \dot\gamma^\alpha}$$ $$\implies (\partial_\alpha g_{\mu\nu})\dot \gamma^\mu\dot\gamma^\nu = \frac{d}{d\lambda} \left(2g_{\alpha\nu} \dot \gamma^\nu\right) = 2\partial_\mu g_{\alpha\nu} \dot \gamma^\mu \dot\gamma^\nu + 2g_{\alpha\nu} \ddot \gamma^\nu$$ $$\implies \ddot \gamma^\nu + \frac{1}{2}g^{\alpha\nu} (2\partial_\mu g_{\alpha\nu} - \partial_\alpha g_{\mu\nu} )\dot \gamma^\mu \dot \gamma^\nu = 0$$ which can be rewritten as

$$\ddot \gamma^\nu + \frac{1}{2} g^{\nu\alpha}( \partial_\nu g_{\alpha\mu} + \partial_\mu g_{\alpha \nu} - \partial_\alpha g_{\mu\nu})\dot \gamma^\mu \dot \gamma^\nu = 0$$

This is the geodesic equation. It depends only on the metric, and has precisely nothing to do with the connection.

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An autoparallel is a curve $\gamma$ whose tangent vector is parallel-transported along itself; in other words, the covariant derivative of $\dot \gamma$ along $\gamma$ is equal to zero:

$$\nabla_{\dot \gamma} \dot \gamma =0$$ $$\implies \ddot \gamma^\alpha + \Gamma^\alpha_{\ \ \mu \nu} \dot \gamma^\mu \dot \gamma^\mu = 0 $$

where $\Gamma^\alpha_{\ \ \mu\nu}$ are the connection coefficients defined by $\frac{\partial \hat e_\nu}{\partial x^\mu} = \Gamma^\alpha_{\ \ \mu\nu} \hat e_\alpha$. This has precisely nothing to do with the metric.

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Now we bring these things together by demanding that the autoparallels coincide with the geodesics - that is, that the shortest (or longest) lines are also the straightest lines. We must then have that

$$\Gamma^\alpha_{\ \ \mu\nu}\dot \gamma^\mu\dot \gamma^\nu = \frac{1}{2} g^{\nu\alpha}( \partial_\nu g_{\alpha\mu} + \partial_\mu g_{\alpha \nu} - \partial_\alpha g_{\mu\nu})\dot \gamma^\mu \dot \gamma^\nu$$

Since $\dot \gamma$ is arbitrary, one is tempted to simply drop the $\dot \gamma$'s from both sides, but this would be getting slightly ahead of ourselves. Note that $\Gamma^\alpha_{\ \ \mu\nu} = \Gamma^\alpha_{\ \ [\mu\nu]} + \Gamma^\alpha_{\ \ (\mu\nu)}$, and the antisymmetric part will be annihilated when contracted with $\dot \gamma^\mu\dot\gamma^\nu$ (which is symmetric). As a result, demanding that autoparallels = geodesics determines only the symmetric part of $\Gamma$.

However, if we also demand that $\Gamma^\alpha_{\ \ [\mu\nu]}=0$ - that is, that the connection be torsion-free - then our connection is uniquely determined by the metric. This is the Levi-Civita connection:

$$\Gamma^\alpha_{\ \ \mu\nu} = \frac{1}{2} g^{\nu\alpha}( \partial_\nu g_{\alpha\mu} + \partial_\mu g_{\alpha \nu} - \partial_\alpha g_{\mu\nu})$$

It must be emphasized that this is a particular choice of connection that we are making. The metric and the connection are, in principle, completely separate objects which we are choosing to be related to one another.

Answer 2

The components of the metric are $$g_{ab} = \vec{e}_a \cdot \vec{e}_b$$ in the notation of Schutz that you were using. Now you can compute $$\frac{\partial g_{ab}}{\partial x^c} = \frac{\partial}{\partial x^c} \left( \vec{e}_a \cdot \vec{e}_b \right)$$ which has terms as in the intrinsic form on the left-hand-side and terms like the expression that you questioned on the right-hand-side. You'll need to do some work to complete it, but you can use this to go in either of the directions that you wanted.

Answer 3

From what I see in the book, he also defines the Christoffel symbols in that way, it is later used in defining the covariant derivative. The calculation is on pages 133-134 in the same book. If I have misunderstood something in your question please elaborate.