What does this tensor symbol mean?

[rsq_a]

In a paper, the authors write:

At the continuum level, the dynamics of incompressible flow has to obey conservation laws of mass and momentum:

\rho \partial_t \textbf{u} = \nabla \cdot \tau

and

\nabla \cdot \textbf{u} = 0

where the momentum flux -\tau = \rho \textbf{u} \otimes \textbf{u} - \tau_d. Here \rho is the density of the fluid which is assumed to be constant, \textbf{u} = (u,v) is the velocity field, and \tau_d is the stress tensor.

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.

 

[zhermes]

That's a 'tensor product,' or 'outer product.'
See http://en.wikipedia.org/wiki/Tensor_product

Maybe some one has a good conceptual description?

 

[rsq_a]

That's a 'tensor product,' or 'outer product.'
See http://en.wikipedia.org/wiki/Tensor_product

Maybe some one has a good conceptual description?

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.

 

[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?

 

[rsq_a]

The quote in the original post tells you for certain that \textbf{u} \otimes \textbf{u} is a tensor, as a tensor is being subtracted from it. Your deduction that \textbf{u} \otimes \textbf{u} = \mathbf{u} \cdot \nabla \mathbf{u} is incorrect. How did you arrive at this conclusion?

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).

 

[zhermes]

From

<br /> -\tau = \rho \textbf{u} \otimes \textbf{u} - \tau_d<br />

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

 

[rsq_a]

From

<br /> -\tau = \rho \textbf{u} \otimes \textbf{u} - \tau_d<br />

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

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.