# Tensor Fields - Tensor Product of Two Gradient Operators

1. Feb 9, 2015

### barnflakes

I'm trying to re-derive a result in a paper that I'm struggling with. Here is the problem:

I wish to calculate $$(\nabla \otimes \nabla) h$$ where $\nabla$ is defined as $$\nabla = \frac{\partial}{\partial r} \hat{\mathbf{r}}+ \frac{1}{r} \frac{\partial}{\partial \psi} \hat{\boldsymbol{\psi}}$$ and $h$ is a scalar field.

I got something like this:

$$(\nabla \otimes \nabla) = \frac{\partial^2}{\partial r^2}\hat{\mathbf{r}} \otimes \hat{\mathbf{r}} + \frac{\partial}{\partial r} \left( \frac{1}{r} \frac{\partial}{\partial \psi} \right)\hat{\mathbf{r}} \otimes \hat{\boldsymbol{\psi}} + \frac{1}{r} \frac{\partial^2}{\partial \psi \partial r}\hat{\boldsymbol{\psi}} \otimes \hat{\mathbf{r}} + \frac{1}{r^2} \frac{\partial^2}{\partial \psi^2} \hat{\boldsymbol{\psi}} \otimes \hat{\boldsymbol{\psi}}$$

but I have a feeling this is wrong.

After that I wish to calculate $$\mathbf{r} \cdot (\nabla \otimes \nabla)$$ where $\hat{\mathbf{r}}$ is a position vector in cylindrical coordinates, so $\mathbf{r} = r \hat{\mathbf{r}}$. However, I'm now struggling with the fact that $\mathbf{r}$ can be written as either $\mathbf{r} \otimes \mathbf{1}$ or $\mathbf{1} \otimes \mathbf{r}$ and I'm not sure which one to choose.

Any insight would be much appreciated.

Last edited: Feb 9, 2015
2. Feb 9, 2015

### Ben Niehoff

This is wrong. However, your notation is ambiguous. Do you mean

$$(\nabla \otimes \nabla) h = \nabla h \otimes \nabla h \quad ?$$
or do you mean the Hessian

$$(\nabla \otimes \nabla) h = (\nabla_i \nabla_j h) \, e^i \otimes e^j \quad ?$$
In either case, your expression is wrong. In particular, if what you mean is the Hessian, then you will have to take into account that your unit vectors in polar coordinates are not constant.

Your tensor is symmetric, so it doesn't matter. But in general, since you're dotting $\mathbf{r}$ from the left, you should choose $\mathbf{r} \otimes \mathbf{1}$.

3. Feb 9, 2015

### barnflakes

Thanks for your reply. In response to your first question, I mean the latter. Do you have a reference as to what the correct expression would be?

4. Feb 9, 2015

### Ben Niehoff

No, but you can Google something like "Hessian in spherical coordinates".