What does this tensor symbol mean?

  • Thread starter rsq_a
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  • #1
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Main Question or Discussion Point

In a paper, the authors write:

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

[tex]\rho \partial_t \textbf{u} = \nabla \cdot \tau[/tex]

and

[tex]\nabla \cdot \textbf{u} = 0[/tex]

where the momentum flux [tex]-\tau = \rho \textbf{u} \otimes \textbf{u} - \tau_d[/tex]. Here [tex]\rho[/tex] is the density of the fluid which is assumed to be constant, [tex]\textbf{u} = (u,v)[/tex] is the velocity field, and [tex]\tau_d[/tex] is the stress tensor.
From what I know of the standard incompressible equations, it seems that

[tex]\rho \textbf{u} \otimes \textbf{u} = \rho (\textbf{u} \cdot \nabla)\textbf{u}[/tex],

but otherwise, I have no idea what the [tex]\otimes[/tex] symbol means.
 

Answers and Replies

  • #2
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  • #3
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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 [itex](u,v)\otimes (u,v)[/itex] is a 2x2 matrix. But as I pointed out, it seems that

[tex]\textbf{u} \otimes \textbf{u} = \textbf{u} \cdot \nabla \textbf{u}[/tex]

which is simply a vector.
 
  • #4
Mute
Homework Helper
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The quote in the original post tells you for certain that [itex]\textbf{u} \otimes \textbf{u}[/itex] is a tensor, as the tensor [itex]\tau_d[/itex] is being subtracted from it (and you can't substract a 2x2 tensor from a vector). Your deduction that [itex]\textbf{u} \otimes \textbf{u} = (\mathbf{u} \cdot \nabla) \mathbf{u}[/itex] is incorrect. How did you arrive at this conclusion?
 
  • #5
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The quote in the original post tells you for certain that [itex]\textbf{u} \otimes \textbf{u}[/itex] is a tensor, as a tensor is being subtracted from it. Your deduction that [itex]\textbf{u} \otimes \textbf{u} = \mathbf{u} \cdot \nabla \mathbf{u}[/itex] 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).
 
  • #6
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From

[tex]
-\tau = \rho \textbf{u} \otimes \textbf{u} - \tau_d
[/tex]

it is clear that
[tex]
\textbf{u} \otimes \textbf{u}
[/tex]
has to be a tensor; I don't know where or why you have,
[tex]
\textbf{u} \otimes \textbf{u} = \textbf{u} \cdot \nabla \textbf{u}
[/tex]
but my guess is that it can't be correct.
 
  • #7
107
1
From

[tex]
-\tau = \rho \textbf{u} \otimes \textbf{u} - \tau_d
[/tex]

it is clear that
[tex]
\textbf{u} \otimes \textbf{u}
[/tex]
has to be a tensor; I don't know where or why you have,
[tex]
\textbf{u} \otimes \textbf{u} = \textbf{u} \cdot \nabla \textbf{u}
[/tex]
but my guess is that it can't be correct.
There was a typo in my 'guess'. It should have been

[tex]\nabla \cdot (\textbf{u} \otimes \textbf{u}) = \textbf{u} \cdot \nabla \textbf{u}[/tex]

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.
 

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