0
$\begingroup$

Given a surface $S \subset \mathbb R^3$ and a curve $\alpha \subset S$. Let $V$ be a vector field defined along $\alpha $.

Write $$\begin{aligned}\alpha &=(u^1(t), u^2(t)) ,\\ \alpha '(t) &= \frac{du^1}{dt} \partial_1 + \frac{du^2}{dt} \partial_2 ,\\ V(t) & = v^1(t) \partial_1 + v^2(t) \partial_2 \end{aligned}$$

Then $$\begin{aligned}\nabla_{\alpha '} V & = \frac{du^i}{dt}(\partial_i v^j)\partial_j + \frac{du^i}{dt} v^j \Gamma^k_{ij} \partial _k \\ & = (\frac{d}{dt} v^k + \frac{du^i}{dt} \Gamma^k_{ij} v^j ) \partial_k \end{aligned}$$

My question is: how to get $$\frac{du^i}{dt}(\partial_i v^j)\partial_j = \frac{d}{dt} v^k \partial_k$$

Thank you!

$\endgroup$
  • 1
    $\begingroup$ Its by chain rule. $\endgroup$ – Sou Nov 13 '17 at 19:58
  • $\begingroup$ @Sou燈馬想 I'm quite confused with these symbols, can you explain a little bit more? Thank you! $\endgroup$ – mathshungry Nov 14 '17 at 3:35
  • 1
    $\begingroup$ $\partial_i v^j = \frac{\partial v^j}{\partial u^i}$. So $$ \frac{du^i}{dt}\partial_i v^j = \frac{du^i}{dt} \frac{\partial v^j}{\partial u^i} = \frac{dv^j}{dt}$$ $\endgroup$ – Sou Nov 14 '17 at 7:25
  • $\begingroup$ @Sou燈馬想 Please, have a look at math.stackexchange.com/questions/2552333/… I couldn't reach you, my whisperer doesn't always your name. $\endgroup$ – user122424 Dec 5 '17 at 15:31

Your Answer

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

Browse other questions tagged or ask your own question.