Chapter 4 General incompressible fluid

When we talk about an incompressible fluid we mean that the density of the fluid is constant in time, i.e. $\frac{\partial }{\partial t}\rho =0$. In most cases it is also assumed that the fluid is not stratified, that means the density is also spatially constant, i.e. $\frac{\partial }{\partial r_j}\rho =0$. Therefore one could call an incompressible fluid also a constant density fluid.

With constant density the continuity equation (2.47) becomes

	${\displaystyle \frac{\partial \rho }{\partial t}}+{\displaystyle \frac{\partial }{\partial r_j}}(v_j\rho )$	$=0$		(4.1)
	$\iff {\displaystyle \frac{\partial \rho }{\partial t}}+v_j{\displaystyle \frac{\partial \rho }{\partial r_j}}+\rho {\displaystyle \frac{\partial v_j}{\partial r_j}}$	$=0$		(4.2)
	$\Rightarrow {\displaystyle \frac{\partial v_j}{\partial r_j}}$	$=0,\text{with }\rho \ne 0$		(4.3)

The momentum equation (2.48) and the energy equation (2.49) become

	$\rho \left[{\displaystyle \frac{\partial v_i}{\partial t}}+v_j{\displaystyle \frac{\partial v_i}{\partial r_j}}+v_i{\displaystyle \frac{\partial v_j}{\partial r_j}}\right]=$	$-{\displaystyle \frac{\partial }{\partial r_i}}p++\rho g_i+{\displaystyle \frac{\partial }{\partial r_j}}{\sigma }_{ij}^{\prime }$		(4.4)
	$\rho \left[{\displaystyle \frac{\partial e}{\partial t}}+v_j{\displaystyle \frac{\partial e}{\partial r_j}}+e{\displaystyle \frac{\partial v_j}{\partial r_j}}\right]=$	$-{\displaystyle \frac{\partial }{\partial r_j}}(v_{j}p)+v_i\rho g_i+{\displaystyle \frac{\partial }{\partial r_j}}(v_i{\sigma }_{ij}^{\prime })$		(4.5)

If we make use of the relation (4.3) in the momentum and energy equation we finally get as equations for an incompressible, selfgravitating, newtonian fluid

	${\displaystyle \frac{\partial v_j}{\partial r_j}}=$	$\mathrm{ }0$		(4.6)
	${\displaystyle \frac{\partial v_i}{\partial t}}+v_j{\displaystyle \frac{\partial v_i}{\partial r_j}}=$	$-{\displaystyle \frac{1}{\rho }}{\displaystyle \frac{\partial }{\partial r_i}}p+g_i+{\displaystyle \frac{1}{\rho }}{\displaystyle \frac{\partial }{\partial r_j}}{\sigma }_{ij}^{*}$		(4.7)
	${\displaystyle \frac{\partial e}{\partial t}}+v_j{\displaystyle \frac{\partial e}{\partial r_j}}=$	$-{\displaystyle \frac{1}{\rho }}{\displaystyle \frac{\partial }{\partial r_j}}(v_{j}p)+v_i{g}_i+{\displaystyle \frac{1}{\rho }}{\displaystyle \frac{\partial }{\partial r_j}}(v_i{\sigma }_{ij}^{*}).$		(4.8)

where ${\sigma }_{ij}^{*}$ is a divergence free stress tensor.

In the incompressible case we can set $\theta =0$ and $\frac{\partial }{\partial r_i}\rho =0$ in equation (2.52) and so we are left with the following equation for an general incompressible fluid

${\displaystyle \frac{{\partial }^2}{\partial r_i\partial r_j}}(v_i{v}_j)=-{\displaystyle \frac{1}{\rho }}{\displaystyle \frac{{\partial }^2}{\partial r_i^2}}p+{\displaystyle \frac{1}{\rho }}{\displaystyle \frac{{\partial }^2}{\partial r_i\partial r_j}}{\sigma }_{ij}^{*}-4\pi G\rho$

(4.9)

Solving for $p$ we get

${\displaystyle \frac{{\partial }^2}{\partial r_i^2}}p=-\rho {\displaystyle \frac{{\partial }^2}{\partial r_i\partial r_j}}(v_i{v}_j)+{\displaystyle \frac{{\partial }^2}{\partial r_i\partial r_j}}{\sigma }_{ij}^{*}-4\pi G{\rho }^2$

(4.10)

or in vector notation

$\mathrm{\Delta }p=-\rho \nabla \left[\nabla \cdot (v_i{v}_j)\right]+\nabla \left[\nabla \cdot {\sigma }_{ij}^{*}\right]-4\pi G{\rho }^2$

(4.11)

which can be interpreted as the equation of state for a general incompressible fluid.¹³¹³Why can’t we use $p=R_s\rho T$ as equation of state for an incompressible fluid?

For an incompressible fluid $\frac{\partial }{\partial r_h}\rho =0$ and inserting this into the vorticity equation (2.53) we get

${\displaystyle \frac{\partial }{\partial t}}{\omega }_g$

$-{ϵ}_{ghi}{\displaystyle \frac{\partial }{\partial r_h}}{ϵ}_{ijm}{v}_j{\omega }_m={\displaystyle \frac{1}{\rho }}{ϵ}_{ghi}{\displaystyle \frac{\partial }{\partial r_h}}{\displaystyle \frac{\partial }{\partial r_j}}{\sigma }_{ij}^{*}$

(4.12)

or in vector notation

${\displaystyle \frac{\partial }{\partial t}}𝝎-\nabla \times (𝒗\times 𝝎)={\displaystyle \frac{1}{\rho }}\left[\nabla \times (\nabla \cdot \widetilde{{\sigma }^{*}})\right]$

(4.13)