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What does this tensor symbol mean?

  1. Jul 6, 2010 #1
    In a paper, the authors write:

    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.
     
  2. jcsd
  3. Jul 6, 2010 #2
  4. Jul 6, 2010 #3
    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.
     
  5. Jul 6, 2010 #4

    Mute

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    Homework Helper

    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?
     
  6. Jul 6, 2010 #5
    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).
     
  7. Jul 6, 2010 #6
    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.
     
  8. Jul 6, 2010 #7
    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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