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Tensor derivates

  1. Dec 3, 2013 #1
    1. The problem statement, all variables and given/known data
    Hi

    I am reading about some fluid mechanics, when suddenly I read saw that someone took the derivate of a tensor. It is in this thesis, on page 26 eq. (70). It is the final equality I can't understand.

    So the author is taking the derivate [itex]\partial_{x_{\alpha}} P_{\alpha\beta}[/itex] of the momentum flux tensor. How on earth does this end up giving [itex]
    \rho u_{\alpha}\partial_{x_\alpha}u_{\beta} + \partial_{x_\alpha}p
    [/itex]?


    Thanks in advance for hints/help.
     
  2. jcsd
  3. Dec 3, 2013 #2

    ShayanJ

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    Gold Member

    You give us the definition of [itex] P_{\alpha\beta} [/itex] and we will answer your question!!! Deal?
     
  4. Dec 4, 2013 #3
    Sorry, here it is:

    [tex]
    P_{\alpha\beta} = p\delta_{\alpha\beta} + (u_1^2, u_1u_2; u_2u_1, u_2^2)
    [/tex]

    Here p is a constant and and u a vector.

    Deal! :redface:
     
    Last edited: Dec 4, 2013
  5. Dec 4, 2013 #4

    ShayanJ

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    Gold Member

    Your tensor can also be written as [itex] P_{\alpha\beta}=p \delta_{\alpha\beta}+u_{\alpha}u_{\beta} [/itex].

    Let [itex] \partial_{\alpha}=\partial_{x_{\alpha}} [/itex].

    Then we have [itex] \partial_{\alpha}P_{\alpha\beta}=\partial_{\alpha} p \delta_{\alpha\beta}+u_{\beta}\partial_{\alpha}u_{\alpha}+ u_{\alpha} \partial_{\alpha} u_{\beta} [/itex]

    This is all that can be said without adding other assumptions.Only that [itex] \partial_{\alpha} p \delta_{\alpha\beta}=0 [/itex] because p is a constant!

    So,you should see whether there are other assumptions too or not.For example [itex] \partial_{\alpha}u_{\alpha} [/itex] is the divergence of the vector u.It may be zero so we will have [itex] \partial_{\alpha}P_{\alpha\beta}=u_{\alpha} \partial_{\alpha} u_{\beta}(+ \partial_{\alpha} p \delta_{\alpha\beta}=0) [/itex] which is near to what you want.But I don't know where that [itex] \rho [/itex] comes from.Can you give the definition of u and also other equations involving them?
     
  6. Dec 4, 2013 #5
    I'll check it out, but it seems [itex]p\propto \rho[/itex] (from the thesis). It doesn't say anything about the gradient of u though.

    Due to Einstein summation [itex]\partial_{\alpha}u_\alpha[/itex] is the gradient of u, but what is [itex]u_{\alpha\partial_\alpha u\beta}[/itex]?
     
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