Expansion of a local dissipation function

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 3K views
trabo
Messages
7
Reaction score
0
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 physics 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.
 
Physics news on Phys.org
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?
 
Not yet I'm afraid. I have a book on line that states that, but I just can't figure it out yet !
 
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