Grad in polar coordinates

Question

As part of my lectures, it is noted that $\nabla \sin \theta \propto \frac{1}{r^2}$ and $\nabla \phi \propto \frac{1}{r^2}$ where we are working in spherical polar coordinates with $\theta$ as the polar angle and $\phi$ as the azimuthal (so the physics convention)...

Now I can’t seem to see why this is true. I’ve tried $$\nabla \sin \theta = \frac{\partial}{\partial r} (\sin \theta) + \frac{\partial}{\partial \theta} (\sin \theta) + \frac{\partial}{\partial \phi} (\sin \theta)$$ but I can’t see how a $\frac{1}{r^2}$ is going to come out of this.

I’ve also tried to work with grad in spherical polars but I still can’t seem to get the $\frac{1}{r^2}$, likewise for $\nabla \phi$

Help would be appreciated....

Answer 1

The $\mathrm{grad}$ operator is of the form $\partial_x\mathbf{i}+\partial_y\mathbf{j}+\partial_z\mathbf{k}$ in Catesian coordinates. In other coordinate systems, the operator has different forms. See this wikipedia article for a list of them.

You can easily see the reason for this when you write out the conversion formula from Cartesian coordinates to polar/spherical coordinates, and differentiate it. You get extra factors on the polar coordinates side, because the relationship is no longer linear. To systematically address this problem, we can introduce connections. Furthermore, this can be generalized to curved spaces.

Answer 2

The thing is that

$$ \nabla f = \frac{\partial f}{\partial u_1}\hat{\bf e}_1 + \frac{\partial f}{\partial u_2}\hat{\bf e}_2 + \frac{\partial f}{\partial u_3}\hat{\bf e}_3 $$

only works when $u$ are cartesian coordinates. In spherical coordinates is not as trivial,

$$ \nabla f = \frac{\partial f}{\partial r}\hat{\bf e}_r + \frac{1}{r}\frac{\partial f}{\partial \theta} \hat{\bf e}_\theta + \frac{1}{r\sin\theta}\frac{\partial f}{\partial \phi}\hat{\bf e}_\phi $$