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Amplitude of the velocity gradient

  1. Jun 4, 2010 #1
    Dear all,

    Someone could help me to understand how is mathermatically expressed the amplitude of the velocity gradient?

    For example if vector of velocity is V(Ux,Vy,Wz)

    The amplitude of the velocity gradient is? :

    grad(V)= d/dx(Ux) +d/dy(Uy) + d/dz(Uz)

    Is it fine?

    Thanks in advance,

    Indira
     
  2. jcsd
  3. Jun 4, 2010 #2

    arildno

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    No.

    The velocity gradient is a 3*3 matrix, and if you are to express the magnitude of that, you'll need to use som sort of norm.

    I can understand the usefulness of such a measure, for example to estimate the maximal effect of viscous forces. But I don't know what is conventional procedure here.
     
  4. Jun 4, 2010 #3

    HallsofIvy

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    What you have is NOT "grad V", it is "div V", [itex]\nabla\cdot V[/itex]
     
  5. Jun 5, 2010 #4
    Thanks a lot for your reply. If I understood well, then the gradient of the velocity vector is going to be:

    Be V the velocity vector V=(Ux,Vy,Wz)

    Grad (V) = (d/dx Ux + d/dx Uy + d/dx Uz
    d/dy Ux + d/dy Uy + d/dy Uz
    d/dz Ux + d/dz Uy + d/dz Uz)

    After this how can I get the magnitude of that?

    Thanks for further input...

    CI
     
  6. Jun 5, 2010 #5

    arildno

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    No, it is not that either.

    Letting i and j be indices running from 1 to three,
    [tex]u_{1}=u, u_{2}=v, u_{3}=w, x_{1}=x,x_{2}=y, x_{3}=z[/tex]

    The gradient of the velocity vector is the matrix [itex]a_{ij}[/itex], with elements:
    [tex]\frac{\partial{u}_{i}}{\partial{x}_{j}}[/tex]
     
  7. Jun 6, 2010 #6
    Dear Arildno,

    Thanks a lot for your reply.

    Is the following fine:

    Grad (u)= [d/dx1 U1 + d/dx2 U1 + d/dx3 U1

    d/dx1 U2 + d/dx2 U2 + d/dx3 U2

    d/dx1 U3 + d/dx2 U3 + d/dx3 U3]


    After this step, I need to calculate the magnitude and finally to multiply the result by the symbol of kronecker. For the symbol of Kronecker I suppose that I should only have the trace of the matrix 3x3 written above, but for the magnitude I don't have any idea....

    Thanks and best regards,

    CI
     
  8. Jun 6, 2010 #7

    arildno

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    No, it is not.
    One of the most important reasons for that is that any NORM should have as basic property that if the norm of a vector/matrix difference equals zero, then this implies that the vectors/matrices in question are identical.

    A simple sum of terms cannot achieve this.

    You may look at the following page to look upon a few examples of norms commonly used for matrices:
    http://en.wikipedia.org/wiki/Matrix_norm
     
  9. Jun 6, 2010 #8
    I think that for the gradient of U, I should put each term of the matrix without sum as follows (I don't know how to put in LateX the brackets to the matrix):

    Grad(U) = [d/dx1 U1 d/dx2 U1 d/dx3 U1

    d/dx1 U2 d/dx2 U2 d/dx3 U2

    d/dx1 U3 d/dx2 U3 d/dx3 U3]

    I guess this part is Ok. Could you please confirm it ?

    After that I should look the web page to see how can I get the NORM of that matrix.

    Thanks a lot,

    CI

    Thanks a lot,

    CI
     
  10. Jun 6, 2010 #9

    arildno

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    Yes, that is how you should construct your matrix.
     
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