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The metric describing the surface of the unit sphere (with $x^1 = \theta$ and $x^2 = \phi$ ) is $$ [g_{ij}] = \begin{pmatrix} 1 & 0 \\ 0 & \sin^2{\theta} \end{pmatrix} $$ Find the covariant form of $ [A^i] = \begin{pmatrix} \pi \\ \pi / 4\end{pmatrix} $

Using $$ A_i = \sum_{j=1}^2 g_{ij}A^j $$ I get $ [A_i] = \begin{pmatrix} \pi \\ 0\end{pmatrix} $. Is this correct?

strider
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  • the metric tensor has never rank less than maximal. If that happens then at that point the coordinate system is no longer defined (as a coordinate system). You also don't want to use the components of the vector fields as values for the coordinates. – Thomas Sep 07 '17 at 11:04
  • @Thomas how do I use the metric tensor then to transform the vector to covariant? I really don't understand this bit. – strider Sep 07 '17 at 11:15
  • I answered your question. You have also a type in the indices in your formula for $A_i$ – Thomas Sep 07 '17 at 11:33

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You apply the formula you've written down. The problem with your answer is that your are evaluating the metric tensor in $\pi$. What you should calculate is $$A_1 = \sum_j g_{1j} A^j = g_{11} A^1 = 1 \star \pi $$ and $$A_2 = \sum_j g_{2j} A^j = g_{22} A^2 = \sin^2 \theta \frac{\pi}{4} $$

Thomas
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  • Thanks. What's confusing me is the $\sin^2{\theta}$. This comes from a multiple choice question, but none of the options include $\sin^2{\theta}$. Excuse me please, knowledge extremely limited. – strider Sep 07 '17 at 11:47
  • @strider it's just applying the formula. A point on the sphere is described by a pair $(\phi, \theta)$. At that point you are given (?) that the metric tensor has the form you wrote down. if you are given the tangent vector $(\pi, \pi/4)^T$ (independent of $(\phi, \theta)$) then $A_2$ is just $g_{12} A^1 + g_{22} A^2$. I just plugged in the values. If you get something else then maybe they fixed $(\phi, \theta)$ and you are supposed to evaluate at the given point? Maybe you just add the possible choices to your question or even better, write down the question as a whole. – Thomas Sep 07 '17 at 12:23
  • @strider (I may have mixed up $\phi$ and $\theta)$, but you will get such a $\sin$ term, with components interchanged then) – Thomas Sep 07 '17 at 12:24
  • thanks, yeah maybe the question got it mixed up.. – strider Sep 07 '17 at 12:45