Equations of fluid dynamics 1 Introduction 3 Newtonian compressible fluid

Chapter 2 General compressible fluid

2.1 Balance equations

2.1.1 Mass equation

The mass inside a local fluid element $M=\int_{V}\rho(r_{i})dV$ is conserved. Therefore we get the balance equation for the mass from the generalised continuity equation setting $\alpha=\rho$ in the form

\displaystyle\frac{\partial}{\partial t}\rho+\frac{\partial}{\partial r_{j}}(v% _{j}\rho)=0.

(2.1)

2.1.2 Momentum equation

The momentum $P_{i}$ inside a local fluid element is $P_{i}=\int_{V}\rho(r_{i})v_{i}dV$ . The change of the momentum with time $\frac{d}{dt}P_{i}$ is equal to the sum of the forces on the fluid element

\displaystyle\frac{d}{dt}P_{i}=F_{i}=F_{p,i}+F_{visc,i}+F_{g,i}.

(2.2)

Here we have restricted ourself to a viscous, selfgravitating fluid, so the sum of forces on each fluid element consists of

•

the thermodynamic pressure on the surface $A$ of the fluid element

$\displaystyle F_{p,i}=-\oint_{A}pn_{i}dA=-\int_{V}\frac{\partial}{\partial r_{% i}}pdV,$ (2.3)
•

the viscous force meaning the irreversible transfer of momentum due to friction between the surfaces of the fluid elements

$\displaystyle F_{visc,i}=\oint_{A}\sigma^{\prime}_{ij}n_{j}dA=\int_{V}\frac{% \partial}{\partial r_{j}}\sigma^{\prime}_{ij}dV,$ (2.4)
•

the gravitational force, where for a selfgravitationg fluid $g_{i}$ is generated by the fluid itself (not only by the local fluid element, because gravity is a long-range force)

$\displaystyle F_{g,i}=\int_{V}\rho g_{i}dV.$ (2.5)

Using the generalized continuity equation (1.9) we can express the temporal change of each component of the momentum of a fluid element setting $\alpha=\rho v_{i}$ in the form

\displaystyle\frac{\partial}{\partial t}(\rho v_{i})+\frac{\partial}{\partial r% _{j}}(v_{j}\rho v_{i})=-\frac{\partial}{\partial r_{i}}p+\frac{\partial}{% \partial r_{j}}\sigma^{\prime}_{ij}+\rho g_{i}.

(2.6)

With the help of the continuity equation (2.1) we can write the momentum equation in the often used form called Euler equation

\displaystyle\frac{\partial}{\partial t}(v_{i})+v_{j}\frac{\partial}{\partial r% _{j}}(v_{i})=-\frac{1}{\rho}\frac{\partial}{\partial r_{i}}p+\frac{1}{\rho}% \frac{\partial}{\partial r_{j}}\sigma^{\prime}_{ij}+g_{i}.

(2.7)

2.1.3 Kinetic energy equation

If we multiply equation (2.7) with the velocity $v_{i}$ , use $\frac{1}{2}\frac{dx^{2}}{dt}=x\frac{dx}{dt}$ and the continuity equation we get an equation for the kinetic energy of a local fluid element in conservation form

\displaystyle\frac{\partial}{\partial t}(\frac{1}{2}\rho v^{2})+\frac{\partial% }{\partial r_{j}}(v_{j}\frac{1}{2}\rho v^{2})=-v_{i}\frac{\partial}{\partial r% _{i}}p+v_{i}\frac{\partial}{\partial r_{j}}\sigma^{\prime}_{ij}+v_{i}\rho g_{i}

(2.8)

This equation shows us that locally the kinetic energy is not conserved (otherwise the right-hand side of the equation should be zero).

2.1.4 Internal energy equation

If we assume that each fluid element is in thermal equilibrium the first law of thermodynamics does hold locally and we can write for the internal energy of a fluid element⁵⁵Why is there no gravitational effect on the internal energy? Can self-gravity be understood similar to Van-der-Waals forces? This would imply that the pressure of an ideal gas has to be reduced by some factor $p=\frac{NkT}{V}-\frac{a}{V^{2}},a\sim G$ .

\displaystyle E_{int}=\int dE_{int}=\int TdS-\int pdV=\int\rho TsdV-\int pdV

(2.9)

with $s=\frac{dS}{dm}$ and $\rho=\frac{dm}{dV}$ . Because $T$ and $p$ are understood as the average temperature and pressure in the fluid element, they can be moved out of the integral so that

\displaystyle E_{int}=\int\rho e_{int}dV=T\int\rho sdV-p\int dV.

(2.10)

with $e_{int}=\frac{dE_{int}}{dm}$ . If we take the time derivative of the internal energy we get

\displaystyle\begin{split}\displaystyle\frac{d}{dt}\int\rho e_{int}dV&% \displaystyle=\frac{d}{dt}{\left(T\int\rho sdV\right)}-\frac{d}{dt}{\left(p% \int dV\right)}\\ &\displaystyle=T\frac{d}{dt}\int\rho sdV-p\frac{d}{dt}\int dV+\int\rho sdV% \frac{d}{dt}T-\int dV\frac{d}{dt}p\end{split}

(2.11)

Because we assume local thermodynamic equilibrium $\frac{d}{dt}T=0$ and $\frac{d}{dt}p=0$ and the last two terms on the right hand side vanish. The other terms can be computed by using the Reynolds transport theorem and we get the balance equation for the internal energy of a fluid element

\displaystyle\frac{\partial}{\partial t}\rho e_{int}+\frac{\partial}{\partial r% _{j}}v_{j}\rho e_{int}=T{\left(\frac{\partial}{\partial t}\rho s+\frac{% \partial}{\partial r_{j}}v_{j}\rho s\right)}-p\frac{\partial}{\partial r_{j}}v% _{j}

(2.12)

2.1.5 Global dissipation of kinetic energy

For investigating the conservation of the kinetic energy of the whole fluid we write using the Reynolds transport theorem (1.8)

\displaystyle\frac{d}{dt}E_{kin}=\frac{d}{dt}\int_{V}\frac{1}{2}\rho v^{2}dV=% \int_{V}\frac{\partial}{\partial t}{\left(\frac{1}{2}\rho v^{2}\right)}+\frac{% \partial}{\partial r_{j}}{\left(v_{j}\frac{1}{2}\rho v^{2}\right)}dV

(2.13)

The second term on the right hand side can be transformed with Gauss’s theorem to an integral over the surface of the whole fluid

\displaystyle\int_{V}\frac{\partial}{\partial r_{j}}{\left(v_{j}\frac{1}{2}% \rho v^{2}\right)}dV=\oint_{A}v_{j}\frac{1}{2}\rho v^{2}dA=0.

(2.14)

The surface integral is zero because the velocity on the boundary of the fluid $v_{j}=0$ . Therefore we can write

\displaystyle\frac{d}{dt}E_{kin}=\frac{\partial}{\partial t}E_{kin}=\int_{V}% \frac{\partial}{\partial t}{\left(\frac{1}{2}\rho v^{2}\right)}dV=\int_{V}\rho v% _{i}\frac{\partial}{\partial t}v_{i}+\frac{1}{2}v^{2}\frac{\partial}{\partial t% }\rho dV

(2.15)

Inserting the Euler equation (2.7) and using the continuity equation (2.1) and $g_{i}=\frac{\partial}{\partial r_{i}}\phi$ we can transform this to

\displaystyle\begin{split}\displaystyle\frac{\partial}{\partial t}E_{kin}=&% \displaystyle-\int_{V}\frac{\partial}{\partial r_{j}}{\left[v_{j}\rho{\left(% \frac{1}{2}v^{2}+\frac{p}{\rho}+\phi\right)}+v_{i}\sigma^{\prime}_{ij}\right]}% dV\\ &\displaystyle+\int_{V}p\frac{\partial}{\partial r_{j}}v_{j}dV-\int_{V}\sigma^% {\prime}_{ij}\frac{\partial}{\partial r_{j}}v_{i}dV-\int_{V}\phi\frac{\partial% }{\partial t}\rho dV.\end{split}

(2.16)

The first term on the right hand side can be transformed with Gauss’s theorem to a surface integral. Again this surface integral is zero because on the surface of the fluid the velocity $v_{i},v_{j}=0$ . So we are left with

\displaystyle\frac{\partial}{\partial t}E_{kin}=\int_{V}p\frac{\partial}{% \partial r_{j}}v_{j}dV-\int_{V}\sigma^{\prime}_{ij}\frac{\partial}{\partial r_% {j}}v_{i}dV-\int_{V}\phi\frac{\partial}{\partial t}\rho dV,

(2.17)

which is the generalization of equation (16.2) from Landau and Lifschitz (1991) for general compressible fluids. So we see that the total kinetic energy for an ideal, compressible fluid is not conserved.

If we substitute the first term on the right hand side with the balance equation for internal energy (2.12) and again makes use of Gauss’s theorem we get the following expression

\displaystyle\frac{\partial}{\partial t}E_{kin}=\frac{\partial}{\partial t}% \int_{V}\rho sdV-\frac{\partial}{\partial t}\int_{V}\rho e_{int}dV-\frac{% \partial}{\partial t}\int_{V}\rho\phi dV+\int_{V}\rho\frac{\partial}{\partial t% }\phi dV-\int_{V}\sigma^{\prime}_{ij}\frac{\partial}{\partial r_{j}}v_{i}dV

(2.18)

One might be tempted to the following conclusion, that

\displaystyle\frac{\partial}{\partial t}E_{tot}=\frac{\partial}{\partial t}E_{% kin}+\frac{\partial}{\partial t}E_{int}+\frac{\partial}{\partial t}E_{pot}=T% \frac{\partial}{\partial t}S-\int_{V}\sigma^{\prime}_{ij}\frac{\partial}{% \partial r_{j}}v_{i}dV+\int_{V}\rho\frac{\partial}{\partial t}\phi dV

(2.19)

If then one requires that the total energy should be constant, one would come to the conclusion

\displaystyle\frac{\partial}{\partial t}S=\frac{1}{T}\int_{V}\sigma^{\prime}_{% ij}\frac{\partial}{\partial r_{j}}v_{i}dV-\frac{1}{T}\int_{V}\rho\frac{% \partial}{\partial t}\phi dV

(2.20)

This would lead to the statement that the total entropy of an ideal fluid ( $\sigma^{\prime}_{ij}=0$ ) is not constant but dependent on the time derivative of the potential $\phi$ . But this is not true!

To get the right answer first notice that the potential energy of a selfgravitating system is not $\int\rho\phi dV$ but $\frac{1}{2}\int\rho\phi dV$ . So in the statement above the definition of the total energy was wrong. The potential of a certain density distribution is the solution of the poisson equation

\displaystyle\phi(x_{i},t)=-G\int_{V}\frac{\rho(x_{j},t)}{\left\lvert x_{i}-x_% {j}\right\rvert}dV_{j}

(2.21)

and therefore the potential energy of a selfgravitating system can be expressed as a double integral

\displaystyle E_{pot}=-\frac{G}{2}\iint\frac{\rho(x_{i},t)\rho(x_{j},t)}{\left% \lvert x_{i}-x_{j}\right\rvert}dV_{j}dV_{i}.

(2.22)

If we now take the time derivative of this expression for the potential energy of a selfgravitating system

\displaystyle\begin{split}\displaystyle\frac{\partial}{\partial t}E_{pot}&% \displaystyle=-\frac{G}{2}\iint\rho_{i}\frac{\frac{\partial}{\partial t}\rho_{% j}}{\left\lvert x_{i}-x_{j}\right\rvert}dV_{j}dV_{i}-\frac{G}{2}\iint\frac{% \partial}{\partial t}(\rho_{i})\frac{\rho_{j}}{\left\lvert x_{i}-x_{j}\right% \rvert}dV_{j}dV_{i}\\ &\displaystyle=-\frac{G}{2}\iint\rho_{i}\frac{\frac{\partial}{\partial t}\rho_% {j}}{\left\lvert x_{i}-x_{j}\right\rvert}dV_{j}dV_{i}-\frac{G}{2}\iint\rho_{j}% \frac{\frac{\partial}{\partial t}\rho_{i}}{\left\lvert x_{i}-x_{j}\right\rvert% }dV_{i}dV_{j}\\ &\displaystyle=-\frac{G}{2}\iint\rho_{i}\frac{\frac{\partial}{\partial t}\rho_% {j}}{\left\lvert x_{i}-x_{j}\right\rvert}dV_{j}dV_{i}-\frac{G}{2}\iint\rho_{i}% \frac{\frac{\partial}{\partial t}\rho_{j}}{\left\lvert x_{j}-x_{i}\right\rvert% }dV_{j}dV_{i}\\ &\displaystyle=-G\iint\rho_{i}\frac{\frac{\partial}{\partial t}\rho_{j}}{\left% \lvert x_{i}-x_{j}\right\rvert}dV_{j}dV_{i}\\ &\displaystyle=\int\rho\frac{\partial}{\partial t}\phi dV\end{split}

(2.23)

Here we used the abbreviation $\rho_{i}=\rho(x_{i},t)$ and $\rho_{j}=\rho(x_{j},t)$ . ⁶⁶It is also important to note, that even for a timedependent potential the Greenfunction $\frac{1}{\left\lvert x_{i}-x_{j}\right\rvert}$ is not timedependent!

So we see that for a selfgravitating system

\displaystyle\int\rho\frac{\partial}{\partial t}\phi dV=\frac{\partial}{% \partial t}{\left(\frac{1}{2}\int\rho\phi dV\right)}.

(2.24)

If we use this expression in (2.18) and use $E_{pot}=\frac{1}{2}\int\rho\phi dV$ we get

\displaystyle\frac{\partial}{\partial t}E_{tot}=\frac{\partial}{\partial t}E_{% kin}+\frac{\partial}{\partial t}E_{int}+\frac{\partial}{\partial t}E_{pot}=T% \frac{\partial}{\partial t}S-\int_{V}\sigma^{\prime}_{ij}\frac{\partial}{% \partial r_{j}}v_{i}dV

(2.25)

Demanding that the total energy of the whole fluid should be conserved leads to the right expression for the time evolution of the total entropy

\displaystyle\frac{\partial}{\partial t}S=\frac{1}{T}\int_{V}\sigma^{\prime}_{% ij}\frac{\partial}{\partial r_{j}}v_{i}dV

(2.26)

So total entropy is conserved for an ideal fluid, even if we take selfgravity into account⁷⁷Nevertheless Penrose (Penrose, 1989) pointed out, that the entropy of a selfgravitating fluid should rise up to the point when all matter is collapsed to a black hole, see also Appendix Q.1..

2.1.6 Local dissipation of kinetic energy

For analyzing the dissipation of kinetic energy in a local fluid element we have to make the same calculations as we did to get equation (2.18), but without the assumption of $v=0$ on the boundary. Doing this we arrive at

\displaystyle\begin{split}\displaystyle\frac{d}{dt}E_{kin}=&\displaystyle\int_% {V}\left[\frac{\partial}{\partial r_{j}}{\left(v_{i}\sigma^{\prime}_{ij}-v_{j}% p\right)}+T{\left(\frac{\partial}{\partial t}\rho s+\frac{\partial}{\partial r% _{j}}v_{j}\rho s\right)}-{\left(\frac{\partial}{\partial t}\rho e_{int}+\frac{% \partial}{\partial r_{j}}v_{j}\rho e_{int}\right)}\right.\\ &\displaystyle\left.-{\left(\frac{\partial}{\partial t}\rho\phi+\frac{\partial% }{\partial r_{j}}v_{j}\rho\phi\right)}+\rho\frac{\partial}{\partial t}\phi-% \sigma^{\prime}_{ij}\frac{\partial}{\partial r_{j}}v_{i}\right]dV.\end{split}

(2.27)

If we identify the second, third and fourth term on the right hand side with the total time derivative of the entropy, the internal energy and the potential energy respectively we get

\displaystyle\frac{d}{dt}E_{kin}+\frac{d}{dt}E_{int}+\frac{d}{dt}E_{pot}=\int_% {V}\frac{\partial}{\partial r_{j}}{\left(v_{i}\sigma^{\prime}_{ij}-v_{j}p% \right)}+T\frac{d}{dt}S+\rho\frac{\partial}{\partial t}\phi-\sigma^{\prime}_{% ij}\frac{\partial}{\partial r_{j}}v_{i}dV.

(2.28)

If we additionally assume that locally the same entropy equation holds as globally

\displaystyle\frac{d}{dt}S=\int_{V}\sigma^{\prime}_{ij}\frac{\partial}{% \partial r_{j}}v_{i}dV

(2.29)

or in differential form

\displaystyle\frac{\partial}{\partial t}\rho s+\frac{\partial}{\partial r_{j}}% v_{j}\rho s=\sigma^{\prime}_{ij}\frac{\partial}{\partial r_{j}}v_{i},

(2.30)

we are led to the following balance equation for the total energy $e_{tot}=e_{k}+e_{int}+\phi$ for a local fluid element

\displaystyle\frac{\partial}{\partial t}\rho e_{tot}+\frac{\partial}{\partial r% _{j}}v_{j}\rho e_{tot}=-\frac{\partial}{\partial r_{j}}{\left(v_{j}p\right)}+% \frac{\partial}{\partial r_{j}}{\left(v_{i}\sigma^{\prime}_{ij}\right)}+\rho% \frac{\partial}{\partial t}\phi.

(2.31)

which is basically the sum of the three balance equations

$\displaystyle\frac{\partial}{\partial t}\rho\phi+\frac{\partial}{\partial r_{j% }}v_{j}\rho\phi$	$\displaystyle=+\rho\frac{\partial}{\partial t}\phi$	(2.32)
$\displaystyle\frac{\partial}{\partial t}\rho e_{k}+\frac{\partial}{\partial r_% {j}}v_{j}\rho e_{k}$	$\displaystyle=-v_{j}\frac{\partial}{\partial r_{j}}p+v_{i}\frac{\partial}{% \partial r_{j}}\sigma^{\prime}_{ij}$	(2.33)
$\displaystyle\frac{\partial}{\partial t}\rho e_{int}+\frac{\partial}{\partial r% _{j}}v_{j}\rho e_{int}$	$\displaystyle=-p\frac{\partial}{\partial r_{j}}v_{j}+\sigma^{\prime}_{ij}\frac% {\partial}{\partial r_{j}}v_{i}$	(2.34)

If we do not include potential energy in the total energy but use instead $e=e_{k}+e_{int}$ we can write

\displaystyle\frac{\partial}{\partial t}\rho e+\frac{\partial}{\partial r_{j}}% v_{j}\rho e=-\frac{\partial}{\partial r_{j}}{\left(v_{j}p\right)}+\frac{% \partial}{\partial r_{j}}{\left(v_{i}\sigma^{\prime}_{ij}\right)}-v_{i}\rho% \frac{\partial}{\partial r_{i}}\phi.

(2.35)

which can be split into two balance equations

	$\displaystyle\frac{\partial}{\partial t}\rho e_{k}+\frac{\partial}{\partial r_% {j}}v_{j}\rho e_{k}$	$\displaystyle=-v_{j}\frac{\partial}{\partial r_{j}}p+v_{i}\frac{\partial}{% \partial r_{j}}\sigma^{\prime}_{ij}-v_{i}\rho\frac{\partial}{\partial r_{i}}\phi$		(2.36)
	$\displaystyle\frac{\partial}{\partial t}\rho e_{int}+\frac{\partial}{\partial r% _{j}}v_{j}\rho e_{int}$	$\displaystyle=-p\frac{\partial}{\partial r_{j}}v_{j}+\sigma^{\prime}_{ij}\frac% {\partial}{\partial r_{j}}v_{i}$		(2.37)

where gravity is now just treated as a source in the balance equation of the kinetic energy.

One might recognize, that we didn’t mention selfgravity. In fact for the derivation of the energy equation (2.31) we assumed, that the potential energy of the local fluid element is not due to selfgravity but due to some external potential generated by the rest of the fluid. This assumption is valid in case the local fluid element does not contribute much to the global potential $\phi$ . This means, that

\displaystyle\phi_{local}=-G\int_{V_{local}}\frac{\rho{\left(x_{j}\right)}}{% \left\lvert x_{i}-x_{j}\right\rvert}dV_{j}\ll\phi_{global}=-G\int_{V_{global}}% \frac{\rho{\left(x_{j}\right)}}{\left\lvert x_{i}-x_{j}\right\rvert}dV_{j}

(2.38)

Because we know from the last chapter that the term $-v_{i}\rho\frac{\partial}{\partial r_{i}}\phi$ transforms into $-\frac{1}{2}\frac{\partial}{\partial t}\rho\phi$ in the selfgravity case we could implement a correction in (2.35), which accounts for a rising ”selfgravitiness” of a local fluid element like

\displaystyle\begin{split}\displaystyle\frac{\partial}{\partial t}\rho e+\frac% {\partial}{\partial r_{j}}v_{j}\rho e=&\displaystyle-\frac{\partial}{\partial r% _{j}}{\left(v_{j}p\right)}+\frac{\partial}{\partial r_{j}}{\left(v_{i}\sigma^{% \prime}_{ij}\right)}-v_{i}\rho\frac{\partial}{\partial r_{i}}\phi\\ &\displaystyle+\frac{\phi_{local}}{\phi_{global}}{\left(\frac{1}{2}\frac{% \partial}{\partial t}\rho\phi-\frac{\partial}{\partial r_{j}}v_{j}\rho\phi-% \rho\frac{\partial}{\partial t}\phi\right)}\end{split}

(2.39)

Nevertheless a more practical solution for numerical simulations might be to refine the grid in such a way that the condition $\phi_{local}\ll\phi_{global}$ holds everywhere in the computational domain. ⁸⁸Equivalent to the criterion described in this paragraph might be the so called Truelove criterion (Truelove et al., 1997).

2.2 Divergence equation

Using

\displaystyle v_{j}\frac{\partial}{\partial r_{j}}(v_{i})=\frac{\partial}{% \partial r_{j}}(v_{i}v_{j})-v_{i}\frac{\partial}{\partial r_{j}}v_{j}

we can express the euler equation (2.7) like

\displaystyle\frac{\partial}{\partial t}(v_{i})+\frac{\partial}{\partial r_{j}% }(v_{i}v_{j})-v_{i}\frac{\partial}{\partial r_{j}}v_{j}=-\frac{1}{\rho}\frac{% \partial}{\partial r_{i}}p+\frac{1}{\rho}\frac{\partial}{\partial r_{j}}\sigma% ^{\prime}_{ij}+g_{i}.

Taking the divergence of this equation yields

	$\displaystyle\frac{\partial}{\partial t}{\left(\frac{\partial v_{i}}{\partial r% _{i}}\right)}$	$\displaystyle+\frac{\partial^{2}}{\partial r_{i}\partial r_{j}}(v_{i}v_{j})-{% \left(\frac{\partial v_{i}}{\partial r_{i}}\right)}{\left(\frac{\partial v_{j}% }{\partial r_{j}}\right)}-v_{i}\frac{\partial}{\partial r_{i}}{\left(\frac{% \partial v_{j}}{\partial r_{j}}\right)}=$
		$\displaystyle\frac{1}{\rho^{2}}{\left(\frac{\partial\rho}{\partial r_{i}}% \right)}{\left(\frac{\partial p}{\partial r_{i}}\right)}-\frac{1}{\rho}\frac{% \partial^{2}}{\partial r_{i}^{2}}p-\frac{1}{\rho^{2}}{\left(\frac{\partial\rho% }{\partial r_{i}}\right)}{\left(\frac{\partial}{\partial r_{j}}\sigma^{\prime}% _{ij}\right)}+\frac{1}{\rho}\frac{\partial^{2}}{\partial r_{i}\partial r_{j}}% \sigma^{\prime}_{ij}+\frac{\partial}{\partial r_{i}}g_{i}$

With the poisson equation (Q.2) and using the notation from Truesdell for the divergence $\theta=\frac{\partial v_{i}}{\partial r_{i}}$ we get an equation for the divergence of an arbitrary fluid

\displaystyle\begin{split}\displaystyle\frac{\partial}{\partial t}\theta-% \theta^{2}-v_{i}\frac{\partial}{\partial r_{i}}\theta+\frac{\partial^{2}}{% \partial r_{i}\partial r_{j}}(v_{i}v_{j})=&\displaystyle\frac{1}{\rho^{2}}{% \left(\frac{\partial\rho}{\partial r_{i}}\right)}{\left[\frac{\partial p}{% \partial r_{i}}-\frac{\partial}{\partial r_{j}}\sigma^{\prime}_{ij}\right]}\\ &\displaystyle-\frac{1}{\rho}{\left[\frac{\partial^{2}}{\partial r_{i}^{2}}p-% \frac{\partial^{2}}{\partial r_{i}\partial r_{j}}\sigma^{\prime}_{ij}\right]}-% 4\pi G\rho\end{split}

(2.40)

or in vector notation

\displaystyle\begin{split}\displaystyle\frac{\partial}{\partial t}\theta-% \theta^{2}-(\boldsymbol{v}\cdot\nabla)\theta+\nabla{\left[\nabla\cdot(v_{i}v_{% j})\right]}=&\displaystyle\frac{1}{\rho^{2}}\nabla\rho\cdot{\left[\nabla p-% \nabla\cdot\sigma^{\prime}_{ij}\right]}\\ &\displaystyle-\frac{1}{\rho}{\left[\Delta p-\nabla(\nabla\cdot\sigma^{\prime}% _{ij})\right]}-4\pi G\rho\end{split}

(2.41)

2.3 Vorticity equation

Using the vector identity (M.2) we can express the euler equation (2.7) like

\displaystyle\frac{\partial}{\partial t}(v_{i})+\frac{\partial}{\partial r_{i}% }{\left(\frac{1}{2}v_{j}v_{j}\right)}-\epsilon_{ijm}v_{j}\epsilon_{mkl}\frac{% \partial}{\partial r_{k}}v_{l}=-\frac{1}{\rho}\frac{\partial}{\partial r_{i}}p% +\frac{1}{\rho}\frac{\partial}{\partial r_{j}}\sigma^{\prime}_{ij}+g_{i}.

(2.42)

We can obtain an equation for the vorticity of the flow field by taking the curl of this form of the euler equation

\displaystyle\begin{split}\displaystyle\epsilon_{ghi}\frac{\partial}{\partial r% _{h}}\frac{\partial}{\partial t}(v_{i})+\epsilon_{ghi}\frac{\partial}{\partial r% _{h}}\frac{\partial}{\partial r_{i}}{\left(\frac{1}{2}v_{j}v_{j}\right)}-% \epsilon_{ghi}\frac{\partial}{\partial r_{h}}\epsilon_{ijm}v_{j}\epsilon_{mkl}% \frac{\partial}{\partial r_{k}}v_{l}=\\ \displaystyle-\epsilon_{ghi}\frac{\partial}{\partial r_{h}}{\left(\frac{1}{% \rho}\frac{\partial}{\partial r_{i}}p\right)}+\epsilon_{ghi}\frac{\partial}{% \partial r_{h}}{\left(\frac{1}{\rho}\frac{\partial}{\partial r_{j}}\sigma^{% \prime}_{ij}\right)}+\epsilon_{ghi}\frac{\partial}{\partial r_{h}}g_{i}.\end{split}

(2.43)

If the time and space derivative of the velocity field commute, if the spatial derivatives of the velocity field commute⁹⁹This is true if the velocity field is continuously differentiable twice and if the gravitational field can be expressed by $g_{i}=-\frac{\partial}{\partial r_{i}}\phi$ (means as the gradient of a potential) we see that the second term on the left hand side and also the term due to gravity are zero. This is so because these terms are equivalent to the curl of a gradient of a scalar field which is a zero vector. So we get

\displaystyle\begin{split}\displaystyle\frac{\partial}{\partial t}\epsilon_{% ghi}\frac{\partial}{\partial r_{h}}v_{i}-\epsilon_{ghi}\frac{\partial}{% \partial r_{h}}\epsilon_{ijm}v_{j}\epsilon_{mkl}\frac{\partial}{\partial r_{k}% }v_{l}=&\displaystyle-\epsilon_{ghi}{\left[\frac{1}{\rho}\frac{\partial}{% \partial r_{h}}{\left(\frac{\partial}{\partial r_{i}}p\right)}+{\left(\frac{% \partial}{\partial r_{i}}p\right)}{\left(\frac{\partial}{\partial r_{h}}\frac{% 1}{\rho}\right)}\right]}\\ &\displaystyle+\epsilon_{ghi}{\left[\frac{1}{\rho}\frac{\partial}{\partial r_{% h}}{\left(\frac{\partial}{\partial r_{j}}\sigma^{\prime}_{ij}\right)}+{\left(% \frac{\partial}{\partial r_{j}}\sigma^{\prime}_{ij}\right)}{\left(\frac{% \partial}{\partial r_{h}}\frac{1}{\rho}\right)}\right]}\end{split}

(2.44)

The first term on the right hand side is zero again, because it is the curl of a gradient field and so we get as the vorticity equation ¹⁰¹⁰An equivalent equation would be an equation for the rotation tensor, because the rotation tensor is dual to the vorticity vector (also see appendix I). The advantage of formulating an equation for the rotation tensor would be, that the rotation tensor can be consistently defined in other dimensions than three and also in curved space. for some arbitrary fluid with $\omega_{g}=\epsilon_{ghi}\frac{\partial}{\partial r_{h}}v_{i}$

\displaystyle\begin{split}\displaystyle\frac{\partial}{\partial t}\omega_{g}-% \epsilon_{ghi}\frac{\partial}{\partial r_{h}}\epsilon_{ijm}v_{j}\omega_{m}=&% \displaystyle-\frac{1}{\rho^{2}}{\left[\epsilon_{ghi}{\left(\frac{\partial}{% \partial r_{h}}\rho\right)}{\left(\frac{\partial}{\partial r_{i}}p\right)}-% \epsilon_{ghi}{\left(\frac{\partial}{\partial r_{h}}\rho\right)}{\left(\frac{% \partial}{\partial r_{j}}\sigma^{\prime}_{ij}\right)}\right]}\\ &\displaystyle+\frac{1}{\rho}\epsilon_{ghi}\frac{\partial}{\partial r_{h}}% \frac{\partial}{\partial r_{j}}\sigma^{\prime}_{ij}\end{split}

(2.45)

or in vector notation with $\boldsymbol{\omega}=\nabla\times\boldsymbol{v}$

\displaystyle\frac{\partial}{\partial t}\boldsymbol{\omega}-\nabla\times(% \boldsymbol{v}\times\boldsymbol{\omega})=-\frac{1}{\rho^{2}}{\left[(\nabla\rho% )\times(\nabla p)-(\nabla\rho)\times(\nabla\cdot\tilde{\sigma})\right]}+\frac{% 1}{\rho}{\left[\nabla\times(\nabla\cdot\tilde{\sigma})\right]}

(2.46)

2.4 Summary

Balance equations

$\displaystyle\frac{\partial}{\partial t}\rho+\frac{\partial}{\partial r_{j}}(v% _{j}\rho)$	$\displaystyle=0$	(2.47)
$\displaystyle\frac{\partial}{\partial t}(\rho v_{i})+\frac{\partial}{\partial r% _{j}}(v_{j}\rho v_{i})$	$\displaystyle=-\frac{\partial}{\partial r_{i}}p+\frac{\partial}{\partial r_{j}% }\sigma^{\prime}_{ij}+\rho g_{i}$	(2.48)
$\displaystyle\frac{\partial}{\partial t}(\rho e)+\frac{\partial}{\partial r_{j% }}(v_{j}\rho e)$	$\displaystyle=-\frac{\partial}{\partial r_{j}}(v_{j}p)+\frac{\partial}{% \partial r_{j}}(v_{i}\sigma^{\prime}_{ij})+v_{i}\rho g_{i}$	(2.49)

with Newtonian gravity (Poisson Equation):

\displaystyle\frac{\partial}{\partial r_{j}}g_{j}=4\pi G\rho

(2.50)

and an equation of state dependent on the material of the fluid. ¹¹¹¹See Appendix A.

Global dissipation of kinetic energy $\mathcal{E}$

\displaystyle\mathcal{E}=\frac{1}{V}\frac{\partial}{\partial t}E_{kin}=\frac{1% }{V}\int_{V}p\frac{\partial}{\partial r_{j}}v_{j}dV-\frac{1}{V}\int_{V}\sigma^% {\prime}_{ij}\frac{\partial}{\partial r_{j}}v_{i}dV-\frac{1}{V}\int_{V}\phi% \frac{\partial}{\partial t}\rho dV

(2.51)

Divergence equation

\displaystyle\begin{split}\displaystyle\frac{\partial}{\partial t}\theta-% \theta^{2}-v_{i}\frac{\partial}{\partial r_{i}}\theta+\frac{\partial^{2}}{% \partial r_{i}\partial r_{j}}(v_{i}v_{j})=&\displaystyle\frac{1}{\rho^{2}}{% \left(\frac{\partial\rho}{\partial r_{i}}\right)}{\left[\frac{\partial p}{% \partial r_{i}}-\frac{\partial}{\partial r_{j}}\sigma^{\prime}_{ij}\right]}\\ &\displaystyle-\frac{1}{\rho}{\left[\frac{\partial^{2}}{\partial r_{i}^{2}}p-% \frac{\partial^{2}}{\partial r_{i}\partial r_{j}}\sigma^{\prime}_{ij}\right]}-% 4\pi G\rho\end{split}

(2.52)

Vorticity equation

\displaystyle\begin{split}\displaystyle\frac{\partial}{\partial t}\omega_{g}-% \epsilon_{ghi}\frac{\partial}{\partial r_{h}}\epsilon_{ijm}v_{j}\omega_{m}=&% \displaystyle-\frac{1}{\rho^{2}}{\left[\epsilon_{ghi}{\left(\frac{\partial}{% \partial r_{h}}\rho\right)}{\left(\frac{\partial}{\partial r_{i}}p\right)}-% \epsilon_{ghi}{\left(\frac{\partial}{\partial r_{h}}\rho\right)}{\left(\frac{% \partial}{\partial r_{j}}\sigma^{\prime}_{ij}\right)}\right]}\\ &\displaystyle+\frac{1}{\rho}\epsilon_{ghi}\frac{\partial}{\partial r_{h}}% \frac{\partial}{\partial r_{j}}\sigma^{\prime}_{ij}\end{split}

(2.53)