# What does this tensor symbol mean?

1. Jul 6, 2010

### rsq_a

In a paper, the authors write:

From what I know of the standard incompressible equations, it seems that

$$\rho \textbf{u} \otimes \textbf{u} = \rho (\textbf{u} \cdot \nabla)\textbf{u}$$,

but otherwise, I have no idea what the $$\otimes$$ symbol means.

2. Jul 6, 2010

3. Jul 6, 2010

### rsq_a

I thought about that, but that doesn't make sense. If we interpret the symbol as the outer product, then $(u,v)\otimes (u,v)$ is a 2x2 matrix. But as I pointed out, it seems that

$$\textbf{u} \otimes \textbf{u} = \textbf{u} \cdot \nabla \textbf{u}$$

which is simply a vector.

4. Jul 6, 2010

### Mute

The quote in the original post tells you for certain that $\textbf{u} \otimes \textbf{u}$ is a tensor, as the tensor $\tau_d$ is being subtracted from it (and you can't substract a 2x2 tensor from a vector). Your deduction that $\textbf{u} \otimes \textbf{u} = (\mathbf{u} \cdot \nabla) \mathbf{u}$ is incorrect. How did you arrive at this conclusion?

5. Jul 6, 2010

### rsq_a

Okay, maybe there's a simple way for you to explain it to me. Can you simply break down the original equation into vector form (2d is fine).

6. Jul 6, 2010

### zhermes

From

$$-\tau = \rho \textbf{u} \otimes \textbf{u} - \tau_d$$

it is clear that
$$\textbf{u} \otimes \textbf{u}$$
has to be a tensor; I don't know where or why you have,
$$\textbf{u} \otimes \textbf{u} = \textbf{u} \cdot \nabla \textbf{u}$$
but my guess is that it can't be correct.

7. Jul 6, 2010

### rsq_a

There was a typo in my 'guess'. It should have been

$$\nabla \cdot (\textbf{u} \otimes \textbf{u}) = \textbf{u} \cdot \nabla \textbf{u}$$

Checking it over, that's correct.

What a confusing way to write the Navier-Stokes equations. I have no idea why the authors didn't just write them in the typical way everybody else uses without the tensor notation.