Fluid Mechanics

Recall

Recall from Postulate 1 that we have

• $M$: a material space, equipped with the ${\rho }_M$ mass measure
• $W$: a world space, equipped with the metric $⟨\cdot ,\cdot {⟩}_W$
• $\varphi (t):M\to W$, the “flow map” to show how the particles are embedded in the world space $W$.
• $\mathbf{u}\in \Gamma (TW)$ defined by $\dot{\varphi }=\mathbf{u}\circ \varphi$, the velocity field

We also have the Continuity Equation for mass conservation. If $q_W$ denotes the mass density, then

$${\dot{q}}_W+\mathbf{u}\cdot \nabla q_W+(\nabla \cdot \mathbf{u})q_W=0$$

or

$${\dot{q}}_W+\nabla \cdot (q_W\mathbf{u})=0$$

Derivation of Fluid Momentum Equation

We will derive the fluid momentum equation by examining the forces on the system. Recall that

$$\frac{D}{Dt}\mathbf{u}=\frac{\partial \mathbf{u}}{\partial t}+\mathbf{u}\cdot \nabla \mathbf{u}$$

is the world-space acceleration of a flowing particle. The $D/Dt$ operator is the material derivative of the velocity field $\mathbf{u}$ (so obviously, this gives us the acceleration of a particle). By Newton’s second law, we have

$$q_W\frac{D}{Dt}\mathbf{u}=:\text{(force density)}$$

Force density is the amount of force per unit volume.

$"\\usepackage{tikz}\n\n\\begin{document}\n\\begin{tikzpicture}[>=stealth, scale=1.3, line join=round, line cap=round]\n\n$

source code

The force is given by the pressure on the parcel of volume. Suppose $p:W\to \mathbb{R}$ is the scalar pressure field and the center of the cube is at $(x,y,z)$ with the above dimensions. Using Taylor expansion, the pressure at the right wall is $p(x,y,z)+\frac{\partial p}{\partial x}\frac{\Delta x}{2}$. Since forice is pressure times area, the force at the right wall pointing in the ${\vec{e}}_x$ direction is

$$-\left[\left(p(x,y,z)+\frac{\partial p}{\partial x}\frac{\Delta x}{2}\right)\Delta y\Delta z\right]\cdot {\vec{e}}_x$$

Similarly, at the hidden left wall, the force is

$$+\left[\left(p(x,y,z)-\frac{\partial P}{\partial x}\frac{\Delta x}{2}\right)\Delta y\Delta z\right]\cdot {\vec{e}}_x$$

The net force from all six sides is

$${\mathbf{F}}_{\text{net}}=-\underset{\nabla p}{\underbrace{\left(\frac{\partial p}{\partial x}{\vec{e}}_x+\frac{\partial p}{\partial y}{\vec{e}}_y+\frac{\partial p}{\partial z}{\vec{e}}_z\right)}}\cdot \Delta x\Delta y\Delta z$$

And so the force density is $-\nabla p$. Therefore, the equations of motion for a (compressible) fluid are

$$\{\begin{array}{rl}{\dot{q}}_W+\nabla \cdot (q_W\mathbf{u})& =0\\ q_W\left(\frac{D\mathbf{u}}{Dt}\right)& =-\nabla p\\ \boxed{\text{still need equation for p}}& \end{array}⟺\{\begin{array}{rl}{\dot{q}}_W+\nabla \cdot (q_W\mathbf{u})& =0\\ \frac{\partial }{\partial t}(q_W\mathbf{u})+\text{div}(q_W\mathbf{u}\otimes \mathbf{u}+p{\mathbf{I}}_{3\times 3})& =0\\ & ⋮\end{array}$$

Here, $\text{div}(p{\mathbf{I}}_{3\times 3})=\nabla p$. We can show equivalence of the second equation as follows.

$$\begin{array}{rl}\frac{\partial }{\partial t}(q_w\mathbf{u})+\text{div}(q_w\mathbf{u}\otimes \mathbf{u})+\text{div}(p{\mathbf{I}}_{3\times 3})& =\left(\mathbf{u}\frac{\partial q_w}{\partial t}+q_w\frac{\partial \mathbf{u}}{\partial t}\right)\\ & +(\mathbf{u}(\nabla \cdot (q_w\mathbf{u}))+q_w(\mathbf{u}\cdot \nabla )\mathbf{u})\\ & +\text{div}(p{\mathbf{I}}_{3\times 3})\\ 0& =\mathbf{u}\underset{\text{Mass Continuity}}{\underbrace{\left[\frac{\partial q_w}{\partial t}+\nabla \cdot (q_w\mathbf{u})\right]}}+q_w\underset{\text{Material Derivative}}{\underbrace{\left[\frac{\partial \mathbf{u}}{\partial t}+(\mathbf{u}\cdot \nabla )\mathbf{u}\right]}}+\underset{\nabla p}{\underbrace{\text{div}(pI)}}\\ 0& =q_w\frac{D\mathbf{u}}{Dt}+\nabla p\\ q_w\frac{D\mathbf{u}}{Dt}& =-\nabla p\end{array}$$

These equations are not complete. Our equations govern mass continuity and momentum conservation but we still have some unknowns. Indeed, we have three unknowns: $q_W$, $\mathbf{u}$, and $p$. We need an additional constraint on the fluids. One can take some assumptions about the pressure of the fluid. For example,

”$p$ is only a function of the spatial mass density $q_W$” (barotropic fluid)

to complete the equation. Or

”$p$ is a function of $q_W$ and a hidden variable $s$ with $\frac{Ds}{Dt}=0$”

Then $s$ is called the “entropy” of the fluid, and the fluid is called “isentropic”.

Barotropic Fluids

A barotropic fluid is a fluid whose density is a function of pressure only. Suppose $p:=p(q)$ is very “stiff”, i.e. the fluid is hard to compress.

$"\\usepackage{amsmath}\n\\usepackage{tikz}\n\n\\begin{document}\n\\begin{tikzpicture}[>=stealth, scale=1.5, line width=1pt]\n\n$

source code

then the fluid would want to stay at the same density $q_0$ and the fluid is effectively incompressible. The slope $dp/dq$ means that we can try to exert extreme pressure to change the density, but the fluid will resist it.

In fluid mechanics, the slope $dp/dq$ is the square of the speed of sound in that fluid’s material. The speed of sound is really how fast vibrations travel.

Linearized Fluid Equation

Let $q=q_0$ (some base density) and $\mathbf{u}=0$ (the fluid is not moving) be static solutions. Let

$$q=q_0+\epsilon \tilde{q}\text{and}\mathbf{u}=0+\epsilon \tilde{\mathbf{u}}\text{with ε<<1}$$

to $O({\epsilon }^2)$. Visually, this means we applied some perturbation to the system at the initial state $(q_0,0)$ and we want to see how the system evolves. We ignore $O({\epsilon }^2)$ terms because we are only interested in the first-order behavior of the system.

Our mass continuity equation is

$$\begin{array}{rl}\frac{\partial q}{\partial t}+\nabla \cdot (q\mathbf{u})=0⟶\frac{\partial }{\partial t}(q_0+\epsilon \tilde{q})+\nabla \cdot \left((q_0+\epsilon \tilde{q})(\epsilon \tilde{\mathbf{u}})\right)& =0\\ 0+\epsilon \dot{\tilde{q}}+\nabla \cdot \left(q_0\epsilon \tilde{\mathbf{u}}+{\epsilon }^2\tilde{q}\tilde{\mathbf{u}}\right)& =0\\ \epsilon \dot{\tilde{q}}+\nabla \cdot (q_0\epsilon \tilde{\mathbf{u}})& =0\end{array}$$

The momenentum equation is

$$q\frac{D\mathbf{u}}{Dt}=-\nabla p⟶(q_0+\epsilon \tilde{q})\left(\frac{\partial }{\partial t}(\epsilon \tilde{\mathbf{u}})+(\epsilon \tilde{\mathbf{u}}\cdot \nabla )(\epsilon \tilde{\mathbf{u}})\right)=-\nabla p(q_0+\epsilon \tilde{q})$$

The $(\epsilon \tilde{\mathbf{u}}\cdot \nabla )(\epsilon \tilde{\mathbf{u}})$ term generates a ${\epsilon }^2$ term which we can ignore. Likewise, the $\epsilon \tilde{q}$ term in the left side generates a ${\epsilon }^2$ term. Therefore, we have $q_0\epsilon \dot{\tilde{\mathbf{u}}}$. Now, we can perform Taylor expansion of the pressure field:

$$\begin{array}{rl}-\nabla p(q_0+\epsilon \tilde{q})& \approx -\nabla \left[p_0+\left(\frac{dp}{dq}{|}_{q_0}\right)\epsilon \tilde{q}\right]\\ & =-\nabla \left[p_0+c^2\epsilon \tilde{q}\right]\\ & =-c^2\epsilon \nabla \tilde{q}\end{array}$$

where $c^2$ is the speed of sound. Therefore, the linearized fluid equations are

$$\{\begin{array}{rl}\epsilon \dot{\tilde{q}}+\nabla \cdot \left(\epsilon q_0\tilde{\mathbf{u}}\right)& =0\\ q_0\epsilon \dot{\tilde{\mathbf{u}}}& =-c^2\epsilon \nabla \tilde{q}\end{array}$$

This implies

$$\ddot{\tilde{q}}=c^2\underset{\Delta (\text{Laplacian})}{\underbrace{\nabla \cdot \nabla }}\tilde{q}$$

This is the acoustic wave equation. The acoustic wave equation describes how sound (or vibrations) propagate through a medium (fluid). The speed of sound $c$ determines how fast these waves travel.

Definition (Mach Number)

The Mach number is a dimensionless quantity defined as

$$\frac{|\mathbf{u}|}{c}$$

When the speed of sound $u\ll c$, then we may assume $q=q_0$ for constant $q_0$ and has some perturbation $\epsilon \tilde{\mathbf{u}},\epsilon \tilde{q}$ satisfying the linear acoustic wave equation, the fluid has a low Mach number and is called quasi-static. In other words, it is incompressible. When the speed of sound $u$ is comparable to $c$, then the fluid is compressible and we need to use the full nonlinear equations.

Incompressible Fluids

An incompressible fluid is a fluid whose density $q$ is constant $q_0$ throughout the flow and the divergence of the velocity field $\mathbf{u}$ is zero ($\nabla \cdot \mathbf{u}=0$). We have two definitions of incompressibility.

The strong version is that the density $q$ is constant. The equations of motion are

$$\{\begin{array}{rl}q& =q_0\\ \nabla \cdot \mathbf{u}& =0\end{array}$$

The weak version only requires $\nabla \cdot \mathbf{u}=0$. This is an iff condition with $\dot{\mathbf{q}}+\mathbf{u}\cdot \nabla q=0$. So $q$ can be nonconstant, just that the more desnse part is advected (move or transport by bulk motion) by the fluid. Indeed, it is also volume preserving over time. For $\Omega \subset M$¹,

$$\dot{\mathbf{q}}+\mathbf{u}\cdot \nabla q=0⟺\nabla \cdot \mathbf{u}=0⟺\frac{d}{dt}{\int }_{{\varphi }_t(\Omega )}dV=0$$

Theorem (Hodge Decomposition)

Let $\chi$ be the space of vector fields $\Gamma (TW)$.

$"\\usepackage{amsmath}\n\\usepackage{amssymb}\n\\usepackage{amsfonts}\n\\usepackage{xcolor}\n\n\\begin{document}\n\\begin{tikzpicture}[>=stealth, scale=1.2, line join=round, line cap=round]\n\n$

source code

Define

$${\chi }_{\text{div}}=\{\mathbf{u}\in \Gamma (TW)|\nabla \cdot \mathbf{u}=0\wedge (\mathbf{u}\cdot \mathbf{n}){|}_{\partial W}=0\}$$

If $\chi$ is the space of all possible vector fields, then ${\chi }_{\mathrm{div}}$ is the space of all physically valid incompressible velocity fields. In other words

• $\nabla \cdot \mathbf{u}=0$. The fluid is divergence-free (perfectly incompressible).
• $(\mathbf{u}\cdot \mathbf{n}){|}_{\partial W}=0$. The fluid velocity normal $\mathbf{n}$ to the boundary $\partial W$ is zero. It can slide along the walls, but it cannot flow through the walls.

Define the inner product (fluid kinetic energy) on $\Gamma (TW)$ by

$$\Vert \mathbf{u}{\Vert }^2={\iiint }_W{q}_0\Vert \mathbf{u}{\Vert }^{2}dV$$

What does it mean to be completely orthogonal to ${\chi }_{\mathrm{div}}$? Define

$${\chi }_{\text{div}}^{\perp }=\left\{\frac{\nabla p}{q}|p:W\to \mathbb{R}\right\}$$

The theorem states that the orthogonal complement ${\chi }_{\text{div}}^{\perp }$ is the space of all pressure gradients $\nabla p$.

Proof: We need to show the two vector spaces are orthogonal. This is done by seelcting any arbitrary vector $\mathbf{u}\in {\chi }_{\mathrm{div}}$ and an arbitrary vector from the second space $\nabla p/q$ always has in inner product of $0$.

$$\begin{array}{rl}{\int }_{W}q\left(\frac{\nabla p}{q}\right)\cdot \mathbf{u}& ={\int }_W(\nabla p\cdot \mathbf{u})dV\\ & ={\oint }_{\partial W}p(\mathbf{u}\cdot \mathbf{n})dS-{\int }_{W}p(\nabla \cdot \mathbf{u})dV\\ & =0-0\\ & =0\\ ⟺& \nabla \cdot \mathbf{u}=0\wedge \frac{\partial \mathbf{u}}{\partial \mathbf{n}}=0\end{array}$$

The second equality is by Divergence Theorem.

---

The corollary is that $p$ is uniquely determined by orthogonal projection. The pressure force becomes the Lagrange multipler for the $\nabla \cdot \mathbf{u}=0$ constraint. Because these two spaces are orthogonal, any arbitrary vector field $w$ can be uniquely sliced into two independent pieces

$$\mathbf{w}={\mathbf{u}}_{\text{incompressible}}+\nabla p$$

Footnotes

1. See Postulate 2 for an explanation for this notation. ↩