Expression for Momentum Density

It is useful to describe the equation for momentum $\rho\mbox{\boldmath${v}$}$. Eulerian derivative of momentum density $\rho\mbox{\boldmath${v}$}$ is rewritten as

$$\frac{\partial \rho\mbox{\boldmath${v}$}}{\partial t} = \frac{\partial \rho}{\partial t}\mbox{\boldmath${v}$}+\rho\frac{\partial \mbox{\boldmath${v}$}}{\partial t},$$

$$= -\nabla(\rho\mbox{\boldmath${v}$})\mbox{\boldmath${v}$}-\rho\mbox... ...dmath${v\cdot\nabla v}$}-\mbox{\boldmath${\nabla}$}p+\rho\mbox{\boldmath${g}$},$$ (A.12)

where we used equations (A.7) and (A.11). In the Cartesian cordinate $(x,y,z)=(x_i,x_2,x_3)$, this becomes

$$\frac{\partial }{\partial t} \left(\begin{array}{c} \rho v_... ...\begin{array}{c} g_{x}\\ g_{y}\\ g_{z}\end{array}\right).$$ (A.13)

Using the momentum stress tensor

$$\pi_{i j}=\rho v_i v_j +\delta_{i j}p,$$ (A.14)

the above equation is written as

$$\frac{\partial \rho v_i}{\partial t}+\frac{\partial \pi_{i j}}{\partial x_j}=\rho g_i,$$ (A.15)

where $\delta_{i j}$ represents the Kronecker's delta as $\delta_{i j}=0$ for $i \ne j$ and $\delta_{i j}=1$ for $i = j$.

If there is no external force, rhs of equation (A.15) is equal to zero. If the momentum density $\rho\mbox{\boldmath${v}$}$ were to obey the continuity equation, equation (A.15) would be

$$\frac{\partial \rho v_i}{\partial t}+\frac{\partial \rho v_i v_j}{\partial x_j}=0.$$ (A.16)

However, this is incorrect, because there exists the pressure force in the fluid and thus the momentum of the fluid element is not conserved.