Expansion of a local dissipation function

[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 physics 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.

 

[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?

 

[trabo]

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

 

[Chestermiller]

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

 

[PJC01]

I can provide some clarification and explanation for the mathematical relation that you are struggling with. The dissipation function D is a measure of the energy lost due to irreversible processes in a material. It is defined as the difference between the work done by external forces and the change in internal energy. In the context of finite strain theory, the dissipation function is related to the free energy of the material, which is a function of both the strain and an internal variable.

The first term in the expression for D, \Big (\tau-2\dfrac{\partial \Psi}{\partial b_e}b_e \Big) : d, represents the work done by external forces on the material, while the second term, 2\dfrac{\partial \Psi}{\partial b_e}b_e : \Big ( -\dfrac{1}{2} L_v(b_e) b_e^{-1} \Big ), represents the change in internal energy due to the strain rate and the Lie derivative of the left Cauchy-Green tensor b_e. The last term, -\dfrac{\partial \Psi }{\partial \xi} \dfrac{d\xi}{dt}, represents the change in internal energy due to the evolution of the internal variable \xi.

Now, the key to understanding the equality in question lies in the definition of the Lie derivative. The Lie derivative of a tensor b_e with respect to a vector field v is defined as the rate of change of b_e in the direction of v. In this case, L_v(b_e) represents the rate of change of b_e in the direction of itself, which is equivalent to the symmetric part of the spatial velocity gradient L. Therefore, the two expressions given for D are equal if and only if the two terms involving the Lie derivative are equal.

To show this equality, we can use the fact that the derivative of a tensor with respect to itself is symmetric, i.e. \dfrac{\partial b_e}{\partial b_e}=\dfrac{1}{2}(L+b_eL^t). Substituting this into the two terms and using the definition of the Lie derivative, we get:

\dfrac{\partial \Psi}{\partial b_e}:b_e L^t=\dfrac{\partial \Psi}{\partial b_e }:L^t b_e+\dfrac{\partial \Psi}{\partial b_e}b_e