$\mathbf g=\nabla \varphi$, so that we can re-write this as follows:
@f{eqnarray*}
-\nabla \cdot (2 \eta \varepsilon ({\mathbf u})) + \nabla p &=&
- -\beta\; T\; \mathbf{g} +
+ -\rho_{\text{ref}} \; beta\; T\; \mathbf{g} +
\rho_{\text{ref}} [1+\beta T_{\text{ref}}] \nabla\varphi.
@f}
The second term on the right is time independent, and so we could
with which the Stokes equations would read:
@f{eqnarray*}
-\nabla \cdot (2 \eta \varepsilon ({\mathbf u})) + \nabla p_{\text{dyn}} &=&
- -\rho \; \beta \; T \mathbf{g},
+ -\rho_{\text{ref}} \; \beta \; T \; \mathbf{g},
\\
\nabla \cdot {\mathbf u} &=& 0.
@f}
that considers the total pressure instead:
@f{eqnarray*}
-\nabla \cdot (2 \eta \varepsilon ({\mathbf u})) + \nabla p &=&
- \rho_{\text{ref}} [1-\beta(T-T_{\text{ref}})] \mathbf{g},
+ \rho(T)\; \mathbf{g},
\\
\nabla \cdot {\mathbf u} &=& 0.
@f}
numerically about the same. So, we now get this for the Stokes system:
@f{eqnarray*}
-\nabla \cdot (2 \eta \varepsilon ({\mathbf u})) + \nabla p &=&
- -\rho \; \beta \; T \mathbf{g},
+ \rho(T) \; \mathbf{g},
\\
\frac{\eta}{L} \nabla \cdot {\mathbf u} &=& 0.
@f}
@f{eqnarray*}
-\nabla \cdot (2 \eta \varepsilon ({\mathbf u})) +
\nabla \left(\frac{\eta}{L} \hat p\right) &=&
- -\rho \; \beta \; T \mathbf{g},
+ \rho(T) \; \mathbf{g},
\\
\frac{\eta}{L} \nabla \cdot {\mathbf u} &=& 0.
@f}