# Christoffel symbol and covariant derivative

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)

• You labeled this special relativity but these symbols are typically introduced for GR. Dec 29 '20 at 18:34
• Never mind the names of the indices. Once I understood what I asked I can also formulate it coherently! Dec 29 '20 at 18:49
• @Brick, personally I think the tag SR fits, in the end Christoffel symbols aren't strictly about GR. Dec 29 '20 at 18:50
• The tag "Special Relativity" was added by the moderator, not myself. I had just tagged "Relativity". Is somebody thinking about my question, or just about tags? ;-) Dec 29 '20 at 18:57
• @Fuzzy I'm sure you're (mostly) joking, but as a side note, not everybody reads questions as they pop up on the front page. Many people sort questions by their favorite topics, so the tags are critical for getting your question in front of the people who are most able and willing to answer it. Dec 29 '20 at 20:57

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.

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.

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.

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.

• Brick, that was the kick in the direction I needed, thanks! (not supposed to say thanks in a comment? Huh?) I will work it out tomorrow, too tired now! Dec 29 '20 at 20:24
• You can't use this to identify the $\Gamma$'s from the connection with the $\Gamma$'s from the geodesic equation unless you also demand that $\Gamma$ be metric-compatible and torsion-free, i.e. that you're using the Levi-Civita connection. Dec 29 '20 at 20:24
• For sure @J.Murray is correct that at some point you must choose the metric-compatible connection and choose it to be torsion-free. That didn't seem to be the thrust of the question though, as I read it initially. Dec 29 '20 at 20:26
• I understood the question to be asking why the $\Gamma$'s - which are related to the metric and which show up in the geodesic equation - are equal to the connection coefficients which arise when differentiating basis vectors. Dec 29 '20 at 20:30
• Schutz, referenced by the OP, works it the other way around. He (a little bit on the sly via "physical arguments") asserts that the we need the metric-compatible, torsion-free connection, and then later shows that means that geodesics are autoparrallel. By the time he gets to the autoparallel issue though, he already has derived (under stated and unstated assumptions) the metric-component formula for the connections given by the OP and in your answer, @J.Murray. I like your argument better in some respects, but one has some choice as to what's asserted and what's derived. Dec 29 '20 at 21:02

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.

• Hi. snike, thanks, I see that, but I still don't see why the versions (intrinsic version and version in covariant derivative) should be identical, so he can use the same Gamma in the covariant derivative. In the covariant derivative there could be some other coeffients, but why are they Gamma? Dec 29 '20 at 18:43
• The whole point of defining the covariant derivative is that it is again a tensor. For example, if you just take a partial derivative of a tensor, then this derivative will not be a tensor. This problem is fixed with the covariant derivative and the help of the Christoffel symbols. In general, if we have a (p,q)-tensor, then its covariant derivative is a (p,q+1)-tensor. If you just take a partial derivative this will not be the case. Dec 29 '20 at 18:59
• Thanks, snike. I think I will have to dig into understanding what a Levy-Civita connection is. Dec 29 '20 at 19:39