1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Tensor derivates

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

    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

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


    User Avatar
    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:

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

    Here p is a constant and and u a vector.

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


    User Avatar
    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]?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted