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 Post subject: ParallelPosted: Mon Oct 24, 2016 12:20 am

Joined: Mon Oct 17, 2016 9:33 pm
Posts: 10
Hi, i´ve working hard on this problem but i don´t get the solution. It is the exercise 2.12 of this notes
http://www.maths.ed.ac.uk/~aar/papers/dupontnotes.pdf

I´ve computed Christoffel´ symbols of the induced conection, and the metric´s matrix, but i don´t know how to prove the final result, and why $V$ and $W$ are orthogonal ( using the induced metric, in cartesians is trivial).

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 Post subject: Re: ParallelPosted: Mon Oct 24, 2016 4:43 am
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Joined: Mon Nov 09, 2015 1:36 am
Posts: 460
Location: Ioannina, Greece
Assuming that you are referring to part a) of the exercise 2.12, did you try this?
\begin{align*}
\frac{D Z}{dt}(t)&=\frac{D}{dt}\big(\cos(\theta_0-\beta t)\,X(t)+\sin(\cos(\theta_0-\beta t)\,Y(t)\big)\\
&=\frac{D}{dt}\big(\cos(\theta_0-\beta t)\,X(t)\big)+\frac{D}{dt}\big(\sin(\cos(\theta_0-\beta t)\,Y(t)\big)\\
&=\frac{d}{dt}\cos(\theta_0-\beta t)\,X(t)+\cos(\theta_0-\beta t)\,\frac{DX}{dt}(t)+\frac{d}{dt}\sin(\theta_0-\beta t)\,Y(t)+\cos(\theta_0-\beta t)\,\frac{DY}{dt}(t)\\
\end{align*}
and then calculate $\frac{DX}{dt}(t)$, $\frac{DY}{dt}(t)$ separately.

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Grigorios Kostakos

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 Post subject: Re: ParallelPosted: Mon Oct 24, 2016 1:36 pm

Joined: Mon Oct 17, 2016 9:33 pm
Posts: 10
I came up to these point but I don´t know how derivate these fields because they´re in cartesians. On the other hand, Is there any geometric argument to prove the statement without doing any operation? Using only the compatibility of the connection with the metric, and the orthogonality of both fields in cartesians?

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 Post subject: Re: ParallelPosted: Mon Oct 24, 2016 6:08 pm
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Joined: Mon Nov 09, 2015 1:36 am
Posts: 460
Location: Ioannina, Greece
PJPu17 wrote:
...Is there any geometric argument to prove the statement without doing any operation? Using only the compatibility of the connection with the metric, and the orthogonality of both fields in cartesians?

Maybe you're right! But I have no answer about this. My suggestion is: write $X(t)=\mathop{\sum}\limits_{k=1}^3v^k(t)\,\partial_k\big|_{\gamma(t)}$ and then apply the formula $\frac{dX}{dt}(t)=\mathop{\sum}\limits_{k=1}^3\bigg(\frac{dv^i}{dt}+\mathop{\sum}\limits_{ij=1}^3\frac{d\gamma^j}{dt}\,\Gamma_{ji}^{k}\,v^i\bigg)\,\partial_k\big|_{\gamma(t)}\,.$
Do the same for $Y(t)$. And don't forget the restrictions for $\alpha$ and $\beta$.

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Grigorios Kostakos

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 Post subject: Re: ParallelPosted: Mon Oct 24, 2016 11:04 pm

Joined: Mon Oct 17, 2016 9:33 pm
Posts: 10
I can´t do this, because i have the metric induce by the euclidean ( the first fundamental form on the sphere), and these fields are in cartesians. :S

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 Post subject: Re: ParallelPosted: Tue Oct 25, 2016 3:51 am
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Joined: Mon Nov 09, 2015 1:36 am
Posts: 460
Location: Ioannina, Greece
PJPu17 wrote:
I can´t do this, because i have the metric induce by the euclidean ( the first fundamental form on the sphere), and these fields are in cartesians. :S

Which are the charts for the sphere that you are considering?

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Grigorios Kostakos

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