I have a difficulty in the following derivation of the geodesic equation for a curve. Let $\gamma:R\rightarrow M$ be a curve on the manifold $M$. We have a connection and we are trying to find the geodesic curves for that connection. We define the tangent vector of $\gamma$ in the coordinate system $ \{x^i\} $by : $X_{\gamma} = (x^i \circ \gamma)'\frac{\partial}{\partial x^i} $

\begin{align} \nabla_{X_\gamma} X_\gamma &= 0 \\ &=\nabla_{(x^i \circ \gamma)'\frac{\partial}{\partial x^i}} (x^j \circ \gamma)'\frac{\partial}{\partial x^j} \\ &= \dot{\gamma}^i\frac{\partial \dot{\gamma}^j}{\partial x^i}\frac{\partial}{\partial x^j} + \dot{\gamma}^i\dot{\gamma}^j\Gamma_{ji}^q\frac{\partial}{\partial x^q}\\ &=_? \ddot{\gamma}^i \frac{\partial}{\partial x^j} + \dot{\gamma}^i\dot{\gamma}^j\Gamma_{ji}^q\frac{\partial}{\partial x^q} \end{align}

I simply do not understand the derivation implied in the double derivative. When I expand the expression it looks like :

\begin{align} (x^i \circ \gamma)'\frac{\partial (x^j \circ \gamma)'}{\partial x^i} &= \frac{d(x^i \circ \gamma)}{dt}\frac{\partial \frac{d(x^i \circ \gamma)}{dt}}{\partial x^i} \\ &= \frac{d(x^i \circ \gamma)}{dt} \partial_i(\frac{d(x^i \circ \gamma)}{dt} \circ x^{-1}) \end{align}

And I just can't seem to fall back to the result.

I am following this course : https://youtu.be/2eVWUdcI2ho?t=34m27s And the professor says " One can do this cleaner, otherwise I have an other 2 blackboards " which implies the derivation is not as straightforward as I imagined.


1 Answer 1


This isn't really a derivation because some of the expressions are not even defined! Usually, the covariant derivative is defined first as an operator that allows one to differentiate vector fields on (an open subset of a manifold) $M$. However, the velocity $\dot{\gamma} = X_{\gamma}$ of $\gamma$ is not a vector field on (an open subset of) $M$ but a vector field along $\gamma$. That is, for each time $t$ we get a tangent vector $\dot{\gamma}(t) \in T_{\gamma(t)}(M)$ so one must understand first how to covariantly differentiate vector fields along $\gamma$. This is done using a mechanism called the induced covariant derivative and then the derivation looks like this:

$$ \frac{D \dot{\gamma}(t)}{dt} = \frac{D \left( \dot{\gamma}^i(t) \frac{\partial}{\partial x^i}|_{\gamma(t)} \right)}{dt} = \left( \frac{d}{dt} \dot{\gamma}^i(t) \right) \frac{\partial}{\partial x^i}|_{\gamma(t)} + \dot{\gamma}^i(t) \frac{D \left( \frac{\partial}{\partial x^i}|_{\gamma(t)} \right) }{dt} \\= \ddot{\gamma}^i(t) \frac{\partial}{\partial x^i}|_{\gamma(t)} + \dot{\gamma}^i(t) \dot{\gamma}^j(t) \Gamma_{ij}^k(\gamma(t)) \frac{\partial}{\partial x^k}|_{\gamma(t)}. $$

What you are trying to understand doesn't even compile because the expression $\frac{\partial \dot{\gamma}^j}{\partial x^i}$ doesn't make sense. You can only differentiate a curve $\gamma$ in the direction of the curve itself, not in the directions $x^i$. However, using the induced covariant derivative, you only need to differentiate the vector field $\frac{\partial}{\partial x^i}$ along $\gamma$ and this makes sense because this vector field is the restriction of a local honest to god vector field on $M$. For more details, see any good book on differential geometry (such as Do Carmo or Lee).

  • 1
    $\begingroup$ Is schuller wrong here then? $\endgroup$
    – Babu
    Commented May 11, 2021 at 17:52
  • $\begingroup$ @Buraian: Not wrong, just sweeping a lot of details under the rug. If you listen to the lecture, at 35:35 he admits that "strictly speaking this (expression) is not defined" and asks from the audience to "trust me that it's (the final conclusion) right". $\endgroup$
    – levap
    Commented May 11, 2021 at 18:39
  • $\begingroup$ thx bro (or sis), this answer and comment was really helpful @levap $\endgroup$
    – Babu
    Commented May 11, 2021 at 18:50

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