Gradient of vector field in spherical coordinates

Question

I need to calculate the Hessian matrix of a scalar in spherical coordinates. To do so, I tried to determine the gradient of the gradient. Hence, I want a gradient of a vector field. My question is: Can the formula for the Nabla operator simply be applied to each component of the vector field, or is it more tricky? I googled hard, but only found the formula for scalars...

Answer 1

I think I figured it out. This is my approach for polar coordinates, it should work likewise for sphericals. For a scalar function $f$, the gradient in polar coordinates $r$ and $\varphi$ is

$$\mathrm{grad}(f)=\dfrac{\partial f}{\partial r}\underline{e}_r+ \dfrac{1}{r}\dfrac{\partial f}{\partial\varphi}\underline{e}_\varphi,$$

where $\underline{e}_i$ are the unit basis vectors. Substitute $f$ by its own gradient

$$\mathrm{grad}(\mathrm{grad}(f))=\dfrac{\partial}{\partial r}\left(\dfrac{\partial f}{\partial r}\underline{e}_r+\dfrac{1}{r}\dfrac{\partial f}{\partial\varphi}\underline{e}_\varphi\right)\otimes\underline{e}_r+ \dfrac{1}{r}\dfrac{\partial}{\partial\varphi}\left(\dfrac{\partial f}{\partial r}\underline{e}_r+\dfrac{1}{r}\dfrac{\partial f}{\partial\varphi}\underline{e}_\varphi\right)\otimes\underline{e}_\varphi.$$

With

$$\dfrac{\partial\underline{e}_r}{\partial r}=0, \quad \dfrac{\partial\underline{e}_\varphi}{\partial r}=0, \quad \dfrac{\partial\underline{e}_r}{\partial\varphi}=\underline{e}_\varphi, \quad \dfrac{\partial\underline{e}_\varphi}{\partial\varphi}=-\underline{e}_r,$$

one gets

$$\mathrm{grad}(\mathrm{grad}(f))= \dfrac{\partial^2f}{\partial r^2}\underline{e}_r\otimes\underline{e}_r+ \dfrac{\partial}{\partial r}\left(\dfrac{1}{r}\dfrac{\partial f}{\partial\varphi}\right)\underline{e}_\varphi\otimes\underline{e}_r+$$ $$ \dfrac{1}{r}\dfrac{\partial^2f}{\partial r\partial\varphi}\underline{e}_r\otimes\underline{e}_\varphi+ \dfrac{1}{r}\dfrac{\partial f}{\partial r}\underline{e}_\varphi\otimes\underline{e}_\varphi+ \dfrac{1}{r^2}\dfrac{\partial^2f}{\partial\varphi^2}\underline{e}_\varphi\otimes\underline{e}_\varphi- \dfrac{1}{r^2}\dfrac{\partial f}{\partial\varphi}\underline{e}_r\otimes\underline{e}_\varphi,$$

or

$$\mathrm{grad}(\mathrm{grad}(f))=\begin{bmatrix} \dfrac{\partial^2f}{\partial r^2}&\dfrac{1}{r}\dfrac{\partial^2f}{\partial r\partial\varphi}-\dfrac{1}{r^2}\dfrac{\partial f}{\partial\varphi}\\ \dfrac{\partial}{\partial r}\left(\dfrac{1}{r}\dfrac{\partial f}{\partial\varphi}\right)&\dfrac{1}{r}\dfrac{\partial f}{\partial r}+\dfrac{1}{r^2}\dfrac{\partial^2f}{\partial\varphi^2} \end{bmatrix}$$ Does this make sense?

Answer 2

Laplacian is the trace of the Hessian or an inner product of the gradient of the gradient. Hessian is the outer product. So if nabla is a "column vector" of differential operators: $$\nabla = \left[\begin{array}{c}\frac{\partial }{\partial x}\\\frac{\partial}{\partial y}\end{array}\right]$$ we have: $${\bf L} = \nabla^T\nabla = \left[\begin{array}{cc}\frac{\partial}{\partial x} \frac{\partial }{\partial y}\end{array}\right]\left[\begin{array}{c}\frac{\partial }{\partial x}\\\frac{\partial}{\partial y}\end{array}\right] = \frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\\{\bf H} = \nabla \nabla^T = \left[\begin{array}{c}\frac{\partial }{\partial x}\\\frac{\partial}{\partial y}\end{array}\right]\left[\begin{array}{cc}\frac{\partial}{\partial x} \frac{\partial }{\partial y}\end{array}\right] = \left[\begin{array}{cc}\frac{\partial^2}{\partial x^2}& \frac{\partial^2 }{\partial xy}\\\frac{\partial^2 }{\partial yx}& \frac{\partial^2 }{\partial y^2}\end{array}\right]$$ Hessian $\bf H$ contains the terms of $\bf L $ in the diagonal, so ${\bf L} = \text{trace}( {\bf H} )$

Then the coordinate system determines the chain rule you will have to consider when carrying out the calculations. But that is on another level of abstraction.