Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
The metric in spherical coordinates is $dr^2 + r^2d\theta^2 + (r\sin{\theta})^2d\phi^2$. –  Neal Jun 26 '12 at 17:17
add comment

1 Answer

up vote 0 down vote accepted

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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