Graduate Defining the Tensor Product of Gradients for Different Coordinate Systems

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The discussion centers on defining the tensor product of gradients, specifically ##\nabla \otimes \nabla f##, and clarifies that while the tensor itself does not change with the coordinate system, its components do. The tensor is derived by taking the gradient twice, and the example of cylindrical coordinates illustrates how the gradient is represented. Participants note the importance of including Christoffel symbols when calculating the gradient of a vector due to the changing unit vectors with position. A request for a reference table for different coordinate systems is made, emphasizing the desire to avoid deriving the information from scratch. Understanding the gradient's representation in various coordinates is essential for accurate calculations.
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.
 
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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.
 
Orodruin said:
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 $$\nabla \otimes \nabla f =
\begin{bmatrix}
f_r^2 & f_r f_\theta r^{-1} & f_r f_z\\
f_\theta r^{-1} f_r & f_\theta^2 r^{-2} & f_\theta r^{-1} f_z\\
f_rf_z & f_z f_\theta r^{-1} & f_z^2
\end{bmatrix}$$
 
No, you are missing all of the terms involving Christoffel symbols that you get when taking the gradient of a vector
 
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Likes 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.
 
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.
 

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