Defining the Tensor Product of Gradients for Different Coordinate Systems

[member 428835]

Does anyone know where I can find the definition of $\nabla \otimes \nabla f$? I tried googling this but nothing comes up. I know it will change depending on the coordinate system, so does anyone know the general definition OR a table for rectangular, spherical, cylindrical coordinates?

Thanks so much.

 

[Orodruin]

It does not change with the coordinate system. That is the entire point. However, its components in a particular coordinate system may be different.

It is the tensor you obtain by taking the gradient twice.

 

[member 428835]

It does not change with the coordinate system. That is the entire point. However, its components in a particular coordinate system may be different.

It is the tensor you obtain by taking the gradient twice.

Okay, so in cylindrical coordinates, for example, $\nabla f = \langle f_r , f_\theta r^{-1}, f_z\rangle$. So does this imply $$\nabla \otimes \nabla f =
\begin{bmatrix}
f_r^2 & f_r f_\theta r^{-1} & f_r f_z\\
f_\theta r^{-1} f_r & f_\theta^2 r^{-2} & f_\theta r^{-1} f_z\\
f_rf_z & f_z f_\theta r^{-1} & f_z^2
\end{bmatrix}$$

 

[Orodruin]

No, you are missing all of the terms involving Christoffel symbols that you get when taking the gradient of a vector

 

[member 428835]

Ohhhhh yeaaaa, because the unit vectors change with position. Is there a table anywhere with this information? I'd prefer not to derive it all from scratch if I can help it.

 

[Chestermiller]

All you need to do is look up the equations for the gradient of a vector for your particular coordinate system, since $\nabla f$ is a vector.