Gradient in spherical coordinates.

Question

I have looked at other posts about deriving the gradient in spherical coordinates and understand the concept, but now am looking at a task which doesn't make sense to me. I am being asked to determine the nabla-operator for the coordinate-tripel $\{x_1,x_2,x_3\}\rightarrow\{r,\rho,\vartheta\}$ in the general form $$\sum_{i=1}^3 ê_{x_i}|\frac{\partial \vec{r}}{\partial x_i}|^{-1}\frac{\partial}{\partial x_i}$$ with position-vector $\vec{r}(r,\rho, \vartheta)$. My issue with this is, how would I take the partial of $\vec{r}$ with respect to $x$ if $x$ does not appear in it?

Answer 1

Suppose you start with cartesian coordinates ${x_i}$ and you have a scalar function $f: R^{n}\mapsto R$. The Nabla or Gradient operator applied to $f$ in cartesian coordinates is given by

$$\vec{\nabla}f=\Sigma_{i}^{n}ê_i\dfrac{\partial f}{\partial x_i}$$ $$\vec{\nabla}=\Sigma_{i}^{n}ê_i\dfrac{\partial }{\partial x_i}$$

where $\{ê_i\}$ is some cartesian basis in $R^{n}$. If you want to change from coordinates, you should change your basis as now vectors are seen from another perspective. The change of basis to $ê'_i$ follows a similar rule as the gradient:

$$ê_i=\Sigma_{j}^{n}ê'_j\dfrac{\partial x_i}{\partial \theta_j}$$ where the $\theta_j$ are the new coordinates expressed as a function of the cartesian ones $x_i$; therefore $\vec{\nabla}$ changes to:

$$\vec{\nabla}=\Sigma_{i}^{n}ê_i\dfrac{\partial }{\partial x_i}=\Sigma_{i}^{n}\Sigma_{j}^{n}ê'_j\dfrac{\partial x_i}{\partial \theta_j}\dfrac{\partial }{\partial x_i}=\Sigma_{j}^{n}ê'_j\left(\Sigma_{j}^{n}\dfrac{\partial x_i}{\partial \theta_j}\dfrac{\partial }{\partial x_i}\right)=\Sigma_{j}^{n}ê'_j\dfrac{\partial }{\partial \theta_j}.$$

So you see that the expression of $\vec{\nabla}$ doesn't change when you change from coordinates, what it changes is the basis.

To answer your question, for example, in $R^3$ from cartesian to spherical coordinates you have: $x=r cos\phi \sin\theta$, $y=r sin\phi \sin\theta$ and $z=r cos\theta$, and the inverse $r=\sqrt{x²+y²+z²}$, $\phi=arctan(y/x)$ and $\theta=arccos(z/\sqrt{x²+y²+z²}).$ Indeed your vector $\vec{r}(r,\phi,\theta)$ does depend on $x,y,z$, and you can compute the partial detivatives.