Editing Continuum mechanics (section)

===Clausius–Duhem inequality===
The [[Clausius–Duhem inequality]] can be used to express the second law of thermodynamics for elastic-plastic materials.  This inequality is a statement concerning the irreversibility of natural processes, especially when energy dissipation is involved.

Just like in the balance laws in the previous section, we assume that there is a flux of a quantity, a source of the quantity, and an internal density of the quantity per unit mass.  The quantity of interest in this case is the entropy.  Thus, we assume that there is an entropy flux, an entropy source, an internal mass density <math>\rho</math> and an internal specific entropy (i.e. entropy per unit mass) <math>\eta</math> in the region of interest.

Let <math>\Omega</math> be such a region and let <math>\partial \Omega </math> be its boundary.  Then the second law of thermodynamics states that the rate of increase of  <math>\eta</math> in this region is greater than or equal to the sum of that supplied to <math>\Omega</math> (as a flux or from internal sources) and the change of the internal entropy density <math>\rho\eta</math> due to material flowing in and out of the region.
  
Let <math>\partial \Omega </math> move with a flow velocity <math>u_n</math> and let particles inside <math>\Omega</math> have velocities <math>\mathbf{v}</math>.  Let <math>\mathbf{n}</math> be the unit outward normal to the surface <math>\partial \Omega </math>.  Let <math>\rho</math> be the density of matter in the region, <math>\bar{q}</math> be the entropy flux at the surface, and <math>r</math> be the entropy source per unit mass.  
Then the entropy inequality may be written as
:<math>
    \cfrac{d}{dt}\left(\int_{\Omega} \rho~\eta~\text{dV}\right) \ge
    \int_{\partial \Omega} \rho~\eta~(u_n - \mathbf{v}\cdot\mathbf{n}) ~\text{dA} + 
    \int_{\partial \Omega} \bar{q}~\text{dA} + \int_{\Omega} \rho~r~\text{dV}.
  </math>
The scalar entropy flux can be related to the vector flux at the surface by the relation <math>\bar{q} = -\boldsymbol{\psi}(\mathbf{x})\cdot\mathbf{n}</math>.  Under the assumption  of incrementally isothermal conditions, we have
:<math>
    \boldsymbol{\psi}(\mathbf{x}) = \cfrac{\mathbf{q}(\mathbf{x})}{T} ~;~~ r = \cfrac{s}{T}
  </math>
where <math>\mathbf{q}</math> is the heat flux vector, <math>s</math> is an energy source per unit mass, and <math>T</math> is the absolute temperature of a material point at <math>\mathbf{x}</math> at time <math>t</math>.

We then have the Clausius–Duhem inequality in integral form:
:<math>
    {
    \cfrac{d}{dt}\left(\int_{\Omega} \rho~\eta~\text{dV}\right) \ge
    \int_{\partial \Omega} \rho~\eta~(u_n - \mathbf{v}\cdot\mathbf{n}) ~\text{dA} - 
    \int_{\partial \Omega} \cfrac{\mathbf{q}\cdot\mathbf{n}}{T}~\text{dA} + \int_\Omega \cfrac{\rho~s}{T}~\text{dV}.
    }
  </math>
We can show that the entropy inequality may be written in differential form as
:<math>
    {
    \rho~\dot{\eta} \ge - \boldsymbol{\nabla} \cdot \left(\cfrac{\mathbf{q}}{T}\right)
       + \cfrac{\rho~s}{T}.
    }
  </math>
In terms of the Cauchy stress and the internal energy, the Clausius–Duhem inequality may be written as 
:<math>
    {
      \rho~(\dot{e} - T~\dot{\eta}) - \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} \le 
           - \cfrac{\mathbf{q}\cdot\boldsymbol{\nabla} T}{T}.
    }
  </math>