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## Parallel

### Parallel

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" onclick="window.open(this.href);return false;

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).

Thank you for your time

http://www.maths.ed.ac.uk/~aar/papers/dupontnotes.pdf" onclick="window.open(this.href);return false;

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).

Thank you for your time

- Grigorios Kostakos
- Founder
**Articles:**0**Posts:**460**Joined:**Mon Nov 09, 2015 1:36 am**Location:**Ioannina, Greece

### Re: Parallel

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.

\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.

Grigorios Kostakos

### Re: Parallel

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?

- Grigorios Kostakos
- Founder
**Articles:**0**Posts:**460**Joined:**Mon Nov 09, 2015 1:36 am**Location:**Ioannina, Greece

### Re: Parallel

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 \[\frac{dX}{dt}(t)=\mathop{\sum}\limits_{k=1}^3\bigg(\frac{dv^i}{dt}+\mathop{\sum}\limits_{ij=1}^3\frac{d\gamma^j}{dt}\,\Gamma_{ji}^{k}\,v^i\bigg)\,\partial_k\big|_{\gamma(t)}\,.\]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?

Do the same for $Y(t)$. And don't forget the restrictions for $\alpha$ and $\beta$.

Grigorios Kostakos

### Re: Parallel

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

- Grigorios Kostakos
- Founder
**Articles:**0**Posts:**460**Joined:**Mon Nov 09, 2015 1:36 am**Location:**Ioannina, Greece

### Re: Parallel

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

Grigorios Kostakos