2
$\begingroup$

The gradient in spherical coordinates is given by:

$\nabla f = \left(\frac{\partial f}{\partial r}, \frac{1}{r} \frac{\partial f}{\partial \theta}, \frac{1}{r \sin \theta} \frac{\partial f}{\partial \phi} \right)$

on the other hand the gradient is supposed to give us:

$\nabla f \cdot d\vec r = df$

where $d\vec r$ is the displacement vector

if I write $\left(\frac{\partial f}{\partial r}, \frac{1}{r} \frac{\partial f}{\partial \theta}, \frac{1}{r \sin \theta} \frac{\partial f}{\partial \phi} \right) \cdot (dr,d\theta,d\phi)$

it will be wrong because:

$df = \frac{\partial f}{\partial r}dr + \frac{\partial f}{\partial \theta}d\theta + \frac{\partial f}{\partial \phi} d\phi$

I realize my displacement vector is wrong and is not $(dr,d\theta,d\phi)$, but on the other hand isn't the displacement vector by definition just composed of the small changes in each coordinate?

$\endgroup$
  • $\begingroup$ The metric in spherical coordinates is $dr^2 + r^2d\theta^2 + (r\sin{\theta})^2d\phi^2$. $\endgroup$ – Neal Jun 26 '12 at 17:17
1
$\begingroup$

$$ \left(\frac{\partial f}{\partial r}, \frac{1}{r} \frac{\partial f}{\partial \theta}, \frac{1}{r \sin \theta} \frac{\partial f}{\partial \phi} \right) \cdot (dr, r d\theta , r \sin \theta d\phi ) = \frac{\partial f}{\partial r} dr + \frac{\partial f}{\partial \theta} d \theta + \frac{\partial f}{\partial \phi} d \phi = d f $$

$\endgroup$

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.