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Expansion of a local dissipation function

  1. Jun 20, 2014 #1
    Hello everyone,

    I'm studying the finite strain theory and have come across the maximum dissipation principle. It implies a dissipation function defined as
    [itex]D=\tau:d-\dfrac{d\Psi}{dt}[/itex]​
    [itex] \tau[/itex] denotes the Kirchhoff stress tensor, [itex]d[/itex] the eulerian deformation rate and [itex] \Psi=\Psi(b_e,\xi)[/itex] the free energy, [itex]b_e[/itex] the left Cauchy-Green tensor, and [itex]\xi[/itex] an internal variable.
    I quiet understood the physic but there is a mathematical relation that I don't understand. Given the above definition, we claim in a book that
    [itex]D=\Big (\tau-2\dfrac{\partial \Psi}{\partial b_e}b_e \Big) : d + 2\dfrac{\partial \Psi}{\partial b_e}b_e : \Big ( -\dfrac{1}{2} L_v(b_e) b_e^{-1} \Big )-\dfrac{\partial \Psi }{\partial \xi} \dfrac{d\xi}{dt}[/itex]​
    where [itex]d[/itex] is the symmetric part of the spatial velocity gradient [itex]L[/itex] and [itex]L_v(b_e)[/itex] denotes the Lie derivative of [itex]b_e[/itex]. We can show that
    [itex]\dfrac{d}{dt}b_e=Lb_e+b_eL^t+L_v(b_e)[/itex]​
    thus the both expressions given to [itex]D[/itex] are equal if and only if [itex]\dfrac{\partial \Psi}{\partial b_e}:b_e L^t=\dfrac{\partial \Psi}{\partial b_e }:L^t b_e[/itex], but how can this last equality be true :confused: ?
    I can't tell why, if you do please share it :wink:

    Regards.
     
  2. jcsd
  3. Jul 2, 2014 #2
    I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?
     
  4. Jul 16, 2014 #3
    Not yet I'm afraid. I have a book on line that states that, but I just can't figure it out yet !
     
  5. Jul 16, 2014 #4
    I wasn't able to show that last equality either. But, are you sure that one of those L transposes is not an L? Just a thought.

    Chet
     
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