How to find the reasons for change of linear momentum?

Question

I would like to have an equation that says "The spatial change of the linear momentum vector is equal to the sum of these terms...", or written as an equation: $$\mathrm{grad}\left(\rho \mathbf{u} \right) = \mathbf{A} + \mathbf{B} + \mathbf{C} + \dots$$ where $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$ (and so on) are terms that depend on density $\rho$, fluid velocity $\mathbf{u}$, pressure $p$ and/or pressure gradient $\mathrm{grad}\left(p\right)$, and the stress tensor $\mathbf{\tau}$, its gradient or divergence...

I have all the quantities, they are a result of a computational fluid dynamics simulation. But I want to show that, for example, $\partial\,\rho u_i / \partial\,x_j$ (i.e. one component of the linear momentum vector gradient) is the result of (for example) the increase in pressure (positive pressure gradient) or something similar.

This is how far I have come while trying to solve this problem:

Start with the total differential of the linear momentum vector $\rho \mathbf{u}$ (material derivative): $$\frac{\mathrm{d}\,\rho \mathbf{u}}{\mathrm{d}\,t} = \underbrace{\frac{\partial \, \rho \mathbf{u}}{\partial \, t}}_{=0} + \mathrm{grad}\left(\rho \mathbf{u} \right) \cdot \mathbf{u}$$ The total differential $\mathrm{d}\,\rho \mathbf{u}/\mathrm{d}\,t$ is the infinitesimal change in linear momentum as the fluid particle travels an infinitesimal distance in the direction of its velocity vector $\mathbf{u}$.

Next, look at the local form of the (steady state) linear momentum balance in the Navier-Stokes-Equations: $$\mathrm{div}\left(\rho \mathbf{u} \otimes \mathbf{u} \right) = \mathrm{grad}\left(\rho \mathbf{u} \right) \cdot \mathbf{u} + \rho \mathbf{u} \, \mathrm{div}\left(\mathbf{u}\right) = -\mathrm{grad}\left(p\right) + \mathrm{div}\left(\tau\right)$$ Solve for $\mathrm{grad}\left(\rho \mathbf{u} \right) \cdot \mathbf{u}$ and thus $$\frac{\mathrm{d}\,\rho \mathbf{u}}{\mathrm{d}\,t} = \mathrm{grad}\left(\rho \mathbf{u} \right) \cdot \mathbf{u} = -\rho \mathbf{u}\, \mathrm{div}\left(\mathbf{u}\right) -\mathrm{grad}\left(p\right) + \mathrm{div}\left(\tau\right)$$

This explains the change of the linear momentum vector along its direction of travel, but not along a chosen direction (i.e. $x_1$, $x_2$ or $x_3$ ...). These are three equations, but I need 9: The linear momentum gradient has 9 components and is not necessarily symmetric.

Does someone know how to solve this?

Answer 1

Nevermind, i figured it out myself. It was just an error in reasoning.

Taking the linear momentum in the $x_1$ direction as an example: $$\frac{\partial\,\rho u_1}{\partial\,x_1} u_1 = - \frac{\partial\,\rho u_1}{\partial\,x_2} u_2 - \frac{\partial\,\rho u_1}{\partial\,x_3} u_3 - \rho u_1 \mathrm{div}\left(\mathbf{u}\right) - \frac{\partial\,p}{\partial\,x_1} + \sum_{i=1}^3 \frac{\partial\,\tau_{1i}}{\partial\,x_i}$$ Linear momentum in $x_1$ direction can be

1. converted/redirected into linear momentum in another direction ($u_2 \partial\,\rho u_1/\partial\,x_2 \neq 0$ and/or $u_3 \partial\,\rho u_1/\partial\,x_3\neq 0$)
2. "stored" or "released" when the divergence of the velocity field is positive or negative ($\rho u_1 \mathrm{div}\left(\mathbf{u}\right)\neq 0$)
3. increased or decreased by a pressure gradient ($\frac{\partial\,p}{\partial\,x_1} \neq 0$)
4. increased or decreased by stresses ($\sum_{i=1}^3 \frac{\partial\,\tau_{1i}}{\partial\,x_i}\neq0$)

Thus, there are only three equations that can be rearranged in 9 ways to give the desired information.