0
$\begingroup$

For the transformation from spherical coordinates to cartesian

$$F(r,\theta,\phi) = (r\cos\theta\sin\phi,r\sin\theta\sin\phi,r\cos\phi)$$

Calculate the push forward of the vector field $V = \frac{\partial}{\partial \theta}$

Is this just computing the Jacobian of $F$ and multiplying it by $V$?

In that case, is the answer just

$$\begin{pmatrix} \cos\theta\sin\phi & -r\sin\theta\sin\phi & r\cos\theta\cos\phi \\ \sin\theta\sin\phi & r\cos\theta\sin\phi & r\sin\theta\cos\phi \\ \cos\phi & 0 & -r\sin\phi \end{pmatrix} \begin{pmatrix}0 \\ 1 \\ 0\end{pmatrix} = -r\sin\theta\sin\phi\partial_x + r\cos\theta\sin\phi \partial_y$$

The next part confuses me more regarding the directions of the transformation

Question 2:

Find the pushforward via $F^{-1}$ of $x\partial_y - y\partial_x$.

$\endgroup$
1
$\begingroup$

I would write the image just in terms of $x,y,z$ and $\partial_x, \partial_y, \partial_z$. That will help you do the second question. :)

$\endgroup$
  • $\begingroup$ My problem is mainly understanding what is exactly asked for, is the pushforward of $V$ its image under the Jacobian from spherical to cartesian coordinates? Also, I don't quite get what is the second question asking for. $\endgroup$ – The Bosco Apr 12 '18 at 3:43
  • $\begingroup$ Yes, your answer is correct. I'm suggesting that you write it in cartesian coordinates. Off a small set, $F$ is a diffeomorphism to its image, and $F^{-1}$ maps backward from cartesian to spherical. $\endgroup$ – Ted Shifrin Apr 12 '18 at 3:47
  • $\begingroup$ For question 2, I know that I have to do it through $J^{-1} v_{cartesian} = v_{spherical}$, but if I solve for $v_{cartesian}$ I end up with the exact same answer and procedure as just getting $F^*(v_{spherical}) = v_{cartesian}$ $\endgroup$ – The Bosco Apr 12 '18 at 4:04
  • $\begingroup$ In principle, you need the inverse matrix of the Jacobian in terms of $x,y,z$. But did you realize that your answer to question 1 was precisely $-y\partial_x + x\partial_y$? $\endgroup$ – Ted Shifrin Apr 12 '18 at 5:55

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.