Continuum Mechanics: Rigid Heat Conductor

[LightofAether]

Homework Statement

Determine the thermodynamic restrictions for a rigid heat conductor defined by the constitutive equations:
\DeclareMathOperator{\grad}{grad}\psi = \hat{\psi}\left(\theta,\grad \theta, \grad \grad \theta\right) \\<br /> \eta = \hat{\eta}\left(\theta,\grad \theta, \grad \grad \theta\right)\\<br /> \textbf{q} = \hat{\textbf{q}}\left(\theta,\grad \theta, \grad \grad \theta\right)

Homework Equations

[/B]
\rho \left( \dot{\psi}+\dot{\theta} \eta \right)-\textbf{T}:\textbf{D}+\frac{\textbf{q}}{\theta}\cdot \grad\theta\leq0

The Attempt at a Solution

I have already found the thermodynamic restrictions for \psi because it's straightforward (take the material derivative and apply the chain rule), but I don't know where to start for \eta or \textbf{q}. We're using the Coleman-Noll approach and I understand the procedure once I have \psi = \hat{\psi}\left(\theta,\grad \theta, \grad \grad \theta\right)=something, but I'm struggling with finding a good starting place for \eta. From what my professor has said, it seems like they can be arbitrary as long as they contain \theta,\grad \theta, \grad \grad \theta. That doesn't seem like a very good way to go about this, though. A result of plugging in \dot{\psi} into the relevant equation above (with \textbf{T}:\textbf{D}=0 because it's rigid) is \hat{\eta}=-\frac{\partial \hat{\psi(\theta)}}{\partial \theta}. Can I just plug that into the relevant equation above while keeping \dot{\psi} as \dot{\psi} to find the thermodynamic restrictions for \eta? The equation for \textbf{q} is probably \textbf{q}=-\textbf{K}\textbf{g}.

What do you think? Am I on the right track?

 

[LightofAether]

I talked with my professor today and it turns out that I already finished the problem! Finding \dot{\psi}, plugging it in, and "solving" for the thermodynamic restrictions was the whole thing. Please consider this thread solved.