# Continuum Mechanics: Rigid Heat Conductor

1. Dec 9, 2015

### LightofAether

1. The problem statement, all variables and given/known data
Determine the thermodynamic restrictions for a rigid heat conductor defined by the constitutive equations:
$$\DeclareMathOperator{\grad}{grad}\psi = \hat{\psi}\left(\theta,\grad \theta, \grad \grad \theta\right) \\ \eta = \hat{\eta}\left(\theta,\grad \theta, \grad \grad \theta\right)\\ \textbf{q} = \hat{\textbf{q}}\left(\theta,\grad \theta, \grad \grad \theta\right)$$
2. Relevant equations

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

3. 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?

Last edited: Dec 9, 2015
2. Dec 10, 2015

### 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.