# Expansion of a local dissipation function

1. Jun 20, 2014

### trabo

Hello everyone,

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

Regards.

2. Jul 2, 2014

### Greg Bernhardt

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?

3. Jul 16, 2014

### trabo

Not yet I'm afraid. I have a book on line that states that, but I just can't figure it out yet !

4. Jul 16, 2014

### Staff: Mentor

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