latentcorpse
Dec30-10, 03:25 PM
In the notes attached in this thread:
http://www.physicsforums.com/showthread.php?t=457123
How do we go about the exercise at the bottom of p67/top of p68?
And secondly, at the top of p69, he giveas the example and invites us to check that (\phi^* g)_{\mu \nu} = diag ( \sin^2{\theta})
However I find that
(\phi^* g)_{\mu \nu}= \left( \frac{ \partial y^\alpha }{ \partial \theta} \right) \left( \frac{\partial y^\beta}{ \partial \phi} \right) \delta_{\alpha \beta} = \left( \frac{ \partial y^\alpha }{ \partial \theta} \right) \left( \frac{\partial y^\alpha}{ \partial \phi} \right) = \begin{pmatrix} \cos{\theta} \cos{\phi} & \cos{\theta} \sin{\phi} & -\sin{\theta} \end{pmatrix} \begin{pmatrix} -\sin{\theta} \sin{\phi} \\ \sin{\theta}\cos{\phi} \\ 0 \end{pmatrix}
=-\cos{\theta} \cos{\phi} \sin{\theta} \sin{\phi}+\cos{\theta} \cos{\phi} \sin{\theta} \sin{\phi}=0
Where have I gone wrong?
Thanks
http://www.physicsforums.com/showthread.php?t=457123
How do we go about the exercise at the bottom of p67/top of p68?
And secondly, at the top of p69, he giveas the example and invites us to check that (\phi^* g)_{\mu \nu} = diag ( \sin^2{\theta})
However I find that
(\phi^* g)_{\mu \nu}= \left( \frac{ \partial y^\alpha }{ \partial \theta} \right) \left( \frac{\partial y^\beta}{ \partial \phi} \right) \delta_{\alpha \beta} = \left( \frac{ \partial y^\alpha }{ \partial \theta} \right) \left( \frac{\partial y^\alpha}{ \partial \phi} \right) = \begin{pmatrix} \cos{\theta} \cos{\phi} & \cos{\theta} \sin{\phi} & -\sin{\theta} \end{pmatrix} \begin{pmatrix} -\sin{\theta} \sin{\phi} \\ \sin{\theta}\cos{\phi} \\ 0 \end{pmatrix}
=-\cos{\theta} \cos{\phi} \sin{\theta} \sin{\phi}+\cos{\theta} \cos{\phi} \sin{\theta} \sin{\phi}=0
Where have I gone wrong?
Thanks