I know that $\gamma $ is a geodesic if and only if $$\nabla _{\dot \gamma}\dot\gamma =0.$$ Using this, I'm trying to re find the equation $$\ddot x^k +\Gamma_{i\ell}^k \dot x^i\dot x^\ell=0,$$ but I don't get it.

Let $x^1,...,x^n$ local coordinate. Then $\gamma (t)=(x^1(t),...,x^n(t))$ and $\dot\gamma (t)=\dot x^i(t)\frac{\partial }{\partial x^i}$ (using Einstein convention). Then, $$0=\nabla _{\dot \gamma (t)}\dot \gamma (t)=\nabla _{\dot x^i \frac{\partial }{\partial x^i}}\dot x^\ell \frac{\partial }{\partial x^\ell}=\dot x^i\frac{\partial \dot x^\ell}{\partial x^i}\frac{\partial }{\partial x^\ell}+\dot x^i\dot x^\ell \underbrace{\nabla _{\frac{\partial }{\partial x^i}}\frac{\partial }{\partial x^\ell}}_{=\Gamma_{i\ell}^m\frac{\partial }{\partial x^m}}$$

but I don't get the right equation. Any idea ?

  • 3
    $\begingroup$ You lost a dot. Try looking under the couch. $\endgroup$ – amcalde Jun 9 '16 at 11:04
  • $\begingroup$ @amcalde: I corrected it, thanks. $\endgroup$ – MSE Jun 9 '16 at 12:08
  • $\begingroup$ You actually are very close. Just remember that you're on the curve $\gamma(t)$, so you need to interpret $\dfrac{\partial \dot x^\ell}{\partial x^i}$ by using the chain rule, and you'll get $\dfrac{\ddot x^\ell}{\dot x^i}$. Then just fiddle with indices and you have it. It's probably better to write everything explicitly with a composition and use the chain rule from the start. $\endgroup$ – Ted Shifrin Jun 9 '16 at 16:02

First of all, remember that the arguments of $\nabla$ must be tangent fields on the manifold, which $\dot \gamma$ is not, therefore one has to first give a rigorous meaning to the notation $\nabla _{\dot \gamma} \dot \gamma$. Once we've clarified this, the rest will come naturally, without any effort.

Let $\gamma : [a,b] \to M$ and let $t_0 \in [a,b]$. Let $p = \gamma (t_0)$. Let $U$ be some small neighbourhood of $p$ such that the portion of $\gamma$ that stays inside $U$ has no self-intersection, and let $X \in \mathcal X (U)$ be a local field that extends $\dot \gamma$, i.e. $X _{\gamma (t)} = \dot \gamma (t)$. Then $\left( \nabla _{\dot \gamma} \dot \gamma \right) (t_0)$ means $(\nabla _X X) _p$.

With this, things become easy because the condition $\nabla _{\dot \gamma} \dot \gamma = 0$ expands as:

$$0 = \nabla _X X = \nabla _{X^i \partial _i} (X^j \partial _j) = X^i \nabla _{\partial _i} (X^j \partial _j) = X^i \Big( (\nabla _{\partial _i} X^j) \partial _j + X^j (\nabla _{\partial _i} \partial _j) \Big) = \\ X^i \Big( (\partial _i X^j) \partial _j + X^j \Gamma _{ij} ^k \partial _k \Big) = (X^i \partial _i) (X^j) \partial _j + \Gamma _{ij} ^k X^i X^j \partial _k = \Big( X (X^k) + \Gamma _{ij} ^k X^i X^j \Big) \partial _k ,$$

which implies that

$$\tag {#} X (X^k) + \Gamma _{ij} ^k X^i X^j = 0 \; \forall k .$$

Notice now that for any smooth $f$,

$$X(f) (p) = \textrm d f (X) (p) = \textrm d _p f (X_p) = \textrm d _{\gamma (t_0)} f (\dot \gamma (t_0)) = \frac {\textrm d} {\textrm d t} \Bigg| _{t = t_0} f \circ \gamma ,$$


$$X(X^k) (p) = \frac {\textrm d} {\textrm d t} \Bigg| _{t = t_0} X^k \circ \gamma = \frac {\textrm d} {\textrm d t} \Bigg| _{t = t_0} \dot \gamma ^k = \ddot \gamma ^k (t_0) ,$$

hence, upon evaluation in $p$, $(\#)$ becomes

$$\ddot \gamma ^k (t_0) + \Gamma _{ij} ^k (\gamma (t_0)) \ \dot \gamma ^i (t_0) \dot \gamma ^j (t_0) = 0 ,$$

which upon removal of the arguments becomes the desired

$$\ddot \gamma ^k + (\Gamma _{ij} ^k \circ \gamma) \ \dot \gamma ^i \dot \gamma ^j = 0 .$$

All this pain that we have gone through was required by the fact that $\nabla _{\dot \gamma} {\dot \gamma}$ does not obey the usual algebraic manipulation rules as $\nabla _X Y$ does. (For instance, what would you make of $\nabla _{\partial _i} \dot \gamma ^j$, given that $\dfrac {\partial \dot \gamma ^j} {\partial x_i}$ has no meaning - how would you derive a thing that depends on $t$ with respect to the variables $x_1, \dots, x_n$?)

  • $\begingroup$ Gorgeous ! Thank you for the time you spent to write it. I'll stud this and make you a feedback if necessary. Thank you for everything :) $\endgroup$ – MSE Jun 9 '16 at 13:54
  • $\begingroup$ Actually, it's perfectly clear ! Nothing to say except : thank you very very much. $\endgroup$ – MSE Jun 9 '16 at 14:11

Reconsider your line $$ 0=\nabla _{\dot \gamma (t)}\dot \gamma (t)=\nabla _{\dot x^i \frac{\partial }{\partial x^i}}\dot x^\ell \frac{\partial }{\partial x^\ell}=\dot x^i\underbrace{\frac{\partial x^\ell}{\partial x^i}}_{=\delta_{i\ell}}\frac{\partial }{\partial x^\ell}+\dot x^i\dot x^\ell \underbrace{\nabla _{\frac{\partial }{\partial x^i}}\frac{\partial }{\partial x^\ell}}_{=\Gamma_{i\ell}^m\frac{\partial }{\partial x^m}}. $$ In particular I think you have lost a 'dot' in $$ \dot x^i\underbrace{\frac{\partial x^\ell}{\partial x^i}}_{=\delta_{i\ell}}\frac{\partial }{\partial x^\ell}, $$ which should be $$ \dot x^i\underbrace{\frac{\partial \dot x^\ell}{\partial x^i}}_{}\frac{\partial }{\partial x^\ell}. $$

  • $\begingroup$ You right, I corrected it. But I still cannot conclude :-) $\endgroup$ – MSE Jun 9 '16 at 11:46
  • $\begingroup$ Unfortunately, the formula $\nabla _X Y = \nabla _{X^i \partial _i} {Y^j \partial _j}$ is valid only for fields defined on some neighbourhood of $\gamma ([0,1])$, which $\dot \gamma$ is not. Things are more complicated than what you write, even though the basic idea is correct. The missing steps, though, are significant and should not be omitted. $\endgroup$ – Alex M. Jun 9 '16 at 12:23
  • $\begingroup$ @AlexM.: Could you please post a complete answer (if you can), because I really have problem with this exercise, and I have problem to see every identification and step. I think, to have one detailed example would help me a lot. Thank you. $\endgroup$ – MSE Jun 9 '16 at 13:37
  • $\begingroup$ @MSE: Complete answer posted. :) $\endgroup$ – Alex M. Jun 9 '16 at 13:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.