[Math] Gradient of vector field in spherical coordinates

vector analysis

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…

Best Answer

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?

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