# Covariant derivative and geodesic

Let $f: U \subset \mathbb{R}^2 \rightarrow \mathbb{R}^3$ be a surface patch. Then if we have two vector fields $$X = \sum_i \xi^i \frac{\partial f}{\partial u^i}$$

and $$Y = \sum_i \eta^i \frac{\partial f}{\partial u^i}$$

the covariant derivative is given by $$\nabla_X Y:= \sum_{i,k} \xi^{i} \left( \frac{\partial \eta^k}{\partial u^i} + \sum_j \Gamma_{i,j}^k \eta^j\right)\frac{\partial f}{\partial u^k}$$

Now my question is: If I have a curve $\gamma: I \rightarrow U$ and $c:=f \circ \gamma,$

then my textbook claims that

$$\nabla_{c'}c' = \sum_{i,k} \dot{u}^{i}(t) \left( \frac{\partial \dot{u}^{k}(t)}{\partial u^i} + \sum_j \Gamma_{i,j}^k \dot{u}^{j}(t)\right)\frac{\partial f}{\partial u^k}$$

Although I don't know what this function $u$ should actually denote, this equation looks somehow wrong, as $\dot{u}$ is a function of $t$ but is differentiated with respect to $u^{i}$, which does not make sense to me. Could anybody explain this to me?

So I have essentially two questions:

1.) How is $u$ defined?

2.) How does this equation follow from the definition of the covariant derivative and why is the derivative of $\dot{u}$ with respect to $u^{i}$ not zero?

• have you considered using Einstein summation notation? lol Apr 11, 2019 at 10:57

Observe that through the map $$\left(\begin{array}{c} u^1\\ u^2 \end{array}\right) \stackrel{f}\to \left(\begin{array}{c} x^1\\ x^2\\ x^3 \end{array}\right),$$ you have a GPS for points in the surface. It is with this that you get $\frac{\partial f}{\partial u^1}$ and $\frac{\partial f}{\partial u^2}$, which are the two columns of the jacobian $Jf$, and serve to define the tangent space at any point on the surface.

So to travel along a curve in the surface you need to travel along a curve in the domain $U$. This implies that $u^1,u^2$ depends on a parameter, which we can denote with $t$.

Schematically $$t\stackrel{\gamma}\to \left(\begin{array}{c} u^1(t)\\ u^2(t) \end{array}\right) \stackrel{f}\to \left(\begin{array}{c} x^1(t)\\ x^2(t)\\ x^3(t) \end{array}\right),$$ will give the way to control the positions on curve over the surface.

Now $\dot{c}=\frac{d}{dt}(f\circ\gamma)=Jf\cdot\dot{\gamma}$ and since $\dot{\gamma}=\dot{u^1}e_1+\dot{u^2}e_2$ then $$\dot{c}=Jf(\dot{u^1}e_1+\dot{u^2}e_2)=\dot{u^1}Jf(e_1)+\dot{u^2}Jf(e_2),$$ so $$\dot{c}=\dot{u^1}\frac{\partial f}{\partial u^1}+\dot{u^2}\frac{\partial f}{\partial u^2}.$$

This way you can use it in $\nabla_{\dot{c}}\dot{c}$ to get $$\nabla_{\dot{c}}\dot{c}=\sum_s(\ddot{u^s}+\sum_{ij}\Gamma^s{}_{ij}\dot{u^i}\dot{u^j})\frac{\partial f}{\partial u_s}.$$

This last formula is get as follow: $$\nabla_{\dot{c}}\dot{c}= \nabla_{\sum_s\dot{u^s}\frac{\partial f}{\partial u_s}} \sum_{\sigma}\dot{u^{\sigma}}\frac{\partial f}{\partial u_{\sigma}},$$

$$\quad=\sum_{s{\sigma}}\dot{u^s}\nabla_{\frac{\partial f}{\partial u_s}} \dot{u^{\sigma}}\frac{\partial f}{\partial u_{\sigma}},$$

$$\qquad=\sum_{s{\sigma}}\dot{u^s} \frac{\partial \dot{u^{\sigma}}}{\partial u_s}\frac{\partial f}{\partial u_{\sigma}}+\sum_{s{\sigma}}\dot{u^s}\dot{u^{\sigma}}\nabla_{\frac{\partial f}{\partial u_s}}\frac{\partial f}{\partial u_{\sigma}},$$

$$\qquad=\sum_{s{\sigma}}\dot{u^s} \frac{\partial \dot{u^{\sigma}}}{\partial u_s}\frac{\partial f}{\partial u_{\sigma}}+\sum_{s{\sigma}r}\dot{u^s}\dot{u^{\sigma}}\Gamma^r{}_{s{\sigma}}\frac{\partial f}{\partial u_r},$$

$$\qquad=\sum_{s{\sigma}} \frac{\partial \dot{u^{\sigma}}}{\partial u_s}\frac{du^s}{dt}\frac{\partial f}{\partial u_{\sigma}}+\sum_{s{\sigma}r}\dot{u^s}\dot{u^{\sigma}}\Gamma^r{}_{s\sigma}\frac{\partial f}{\partial u_r},$$

$$\qquad=\sum_{{\sigma}} \frac{d^2u^s}{dt^2}\frac{\partial f}{\partial u_{\sigma}}+\sum_{s\sigma r}\dot{u^s}\dot{u^r}\Gamma^{\sigma}{}_{sr}\frac{\partial f}{\partial u_{\sigma}},$$

$$\qquad=\sum_{{\sigma}}\left( \frac{d^2u^{\sigma}}{dt^2}+\sum_{sr}\dot{u^s}\dot{u^r}\Gamma^{\sigma}{}_{sr}\right) \frac{\partial f}{\partial u_{\sigma}}.$$

• yes, could you precisely explain how you come to this last equation, this is where I have troubles? Cause I just don't see how this representation of $\dot{c}$ is reconcilable with the covariant derivative.
– user159356
Jan 3, 2015 at 9:40
• Done. @TobiasHurth: To get the expression for $\dot{c}$ one uses that $Jf$ is a matrix that is distributed over $\dot{\gamma}=\dot{u^1}e_1+\dot{u^2}e_2$. Jan 3, 2015 at 15:53
• okay thankyou, there is one step that I don't understand. It is the derivative $\frac{\partial \dot{u}^{\sigma}}{\partial u_s}$. I mean $\dot{u}^{\sigma}$ is a function of $t$, so I really don't see what this derivative should actually mean? Consider a circle $\gamma(t) = \cos(t) e_1 + \sin(t) e_2$, then $\dot{\gamma}(t) = - \sin(t) e_1 + \cos(t) e_2$. Now how would you calculate such a derivative?
– user159356
Jan 3, 2015 at 16:15
• you don't need to worry about, because $\sum_s\frac{\partial\dot{u^k}}{\partial u^s}\frac{du^s}{dt}=\frac{d^2u^k}{dt^2}$ by the Chain's Rule Jan 3, 2015 at 16:41
• for your example: $\dot{u^1}=-\sin(t)=-u^2$ and $\dot{u^2}=\cos(t)=u^1$ then $\frac{d\dot{u^1}}{du^1}=0$ and $\frac{d\dot{u^1}}{du^2}=-1$ as well $\frac{d\dot{u^2}}{du^1}=1$ and $\frac{d\dot{u^2}}{du^2}=0$. Jan 3, 2015 at 20:50