# Fluids Lecture 03

### Kinetic Theory and Fluid Equations

We’ve derived many of the fluid equations looking at large-scale structure in a fluid. Now we are going to re-derive these equations looking at individual particles. This will help to tell us when the fluid equations are valid and how transport occurs.

We’ll start with a system of $N$ particles, specified by ${\vec {x}}$ and ${\vec {v}}$ , with a force interaction between them:

$m_{i}{d \over dt}{{\vec {v}}_{i}}={\vec {F}}_{i}\,\!$ Instead of doing independent particle dynamics, we’ll do statistical mechanics in 6-D phase space, which is the space of all ${\vec {x}}$ and ${\vec {v}}$ for all particles. Let’s say we have a distribution function $f({\vec {x}},{\vec {v}},t)$ which describes the # density of particles in phase space. Thus, the # of particles with position between ${\vec {x}}$ and ${\vec {x}}+d{\vec {x}}$ and velocity between ${\vec {v}}$ and ${\vec {v}}+d{\vec {v}}$ is $f({\vec {x}},{\vec {v}}.t)d^{3}{\vec {x}}d^{3}{\vec {v}}$ .

Now the macroscopic fluid variables which we’ve discussed so far are averages (velocity averages/moments) of $f$ . For example, $n$ , the number density of th fluid, is given by:

$n=\int {d^{3}{\vec {v}}f({\vec {x}},{\vec {v}},t)}\,\!$ We can compute a general property $Q({\vec {x}},t)$ as:

$\left\langle Q({\vec {x}},t)\right\rangle ={\frac {1}{n}}\int {d^{3}{\vec {v}}Qf({\vec {x}},{\vec {v}},t)}\,\!$ On such property is the average velocity ${\vec {u}}$ , given by:

${\vec {u}}={\frac {1}{n}}\int {d^{3}{\vec {v}}\cdot {\vec {v}}f({\vec {x}},{\vec {v}},t)}\,\!$ For any individual particle, we can compute its deviation from ${\vec {u}}$ :

${\vec {w}}={\vec {v}}-\left\langle {\vec {v}}\right\rangle ={\vec {v}}-{\vec {u}}\,\!$ If $\left\langle {\vec {w}}\right\rangle =0$ , then:

${\frac {1}{2}}m\left\langle w^{2}\right\rangle =avg\ random\ KE={3 \over 2}kT\,\!$ implying that we are in thermal equilibrium. We can use this technique to compute various macroscopic fluid variables, like viscous stress, pressure, and heat flux.

To understand these macroscopic variables, we need the find the evolution in 6-D phase space–that is, we need an equation for $f$ . To do this, we’ll first assume a collisionless system. Let’s take a 6-D phase space volume, which has a 5-D surface. In a collisionless system, there are no sources or sinks of particles in phase space (their distribution slowly, smoothly evolve with time). The result of this is that the # of particles in this volume only changes because particles move across the boundary. Thus, $f$ must satisfy a continuity equation:

$\displaystyle {\partial \over \partial t}{f}+“\nabla ”\cdot f“{\vec v}”=0 \,\!$

The quotes here are to acknowledge that $\displaystyle “{\vec v}”$ has 2 contributions:

${\dot {\vec {x}}}_{i}$ and ${\dot {\vec {v}}}_{i}$ . Thus,

${\partial \over \partial t}{f}={\partial \over \partial x_{i}}\underbrace {({\dot {x}}_{if})} _{flux\ through \atop spatial\ boundary}+{\partial \over \partial v_{i}}\underbrace {({\dot {v}}_{if})} _{velocity\ flux}=0\,\!$ Expanding this equation, we have:

${\partial \over \partial t}{f}+{\dot {x}}_{i}{\partial \over \partial x_{i}}{f}+{\dot {v}}_{i}{\partial \over \partial v_{i}}{f}+{\partial \over \partial x_{i}}{{\dot {x}}_{i}}+{\partial \over \partial v_{i}}{{\dot {x}}_{i}}=0\,\!$ Using Hamiltonian dynamics where $H=H({\vec {p}},{\vec {x}},t)$ , we have that ${\dot {x}}_{i}={\partial \over \partial p_{i}}{H}$ and ${\dot {p}}_{i}=-{\partial \over \partial x_{i}}{H}$ , so:

{\begin{aligned}{\partial \over \partial x_{i}}{{\dot {x}}_{i}}&={\partial H \over \partial x_{i}\partial p_{i}}\\{\partial \over \partial v_{i}}{{\dot {v}}_{i}}&={\partial \over \partial p_{i}}{{\vec {p}}_{i}}=-{\partial H \over \partial x_{i}\partial p_{i}}\\\end{aligned}}\,\! Thus, the last two terms on right of our continuity equation sum to 0, giving us (in a collisionless system):

${\partial \over \partial t}{f}+{\dot {x}}_{i}{\partial \over \partial x_{i}}{f}+{\dot {v}}_{i}{\partial \over \partial v_{i}}{f}=0\,\!$ Thus, ${Df \over Dt}=0$ , where

${D \over Dt}={\partial \over \partial t}+{\dot {x}}_{i}{d \over dx_{i}}+{\dot {v}}_{i}{\partial \over \partial v_{i}}={\partial \over \partial t}+v_{i}{\partial \over \partial x_{i}}+a_{i}{\partial \over \partial v_{i}}\,\!$ This is the collisionless Bolzmann equation.

It is important to realize that there are two different kinds of derivatives we can use: Lagrangian and Eulerian:

{\begin{aligned}{\partial \over \partial t}{\rho }+\nabla \cdot \rho {\vec {v}}&=0\\\underbrace {{\partial \over \partial t}{\rho }+{\vec {v}}\cdot \nabla \rho } +\rho \nabla {\vec {v}}&=0\\\end{aligned}}\,\! where the underlined portion is ${d \over dt}{\rho }=({\partial \over \partial t}+{\vec {v}}\nabla )p$ . Now ${\partial \over \partial t}$ is the Eulerian derivative, and $({\partial \over \partial t}+{\vec {v}}\nabla )$ is the Lagrangian derivataive, and is comoving. When we say that a fluid is in equilibrium, we mean that the Lagrangian derivative is 0.

### Collisions

Collisions are instantaneous: the time between collisions is much greater than the duration of the collision. We will treat collisions as the instantaneous change of velocities of particles (not positions). Collisions can be treated as a source/sink term in the momentum/velocity component of our phase space equation. Thus,

${D \over Dt}{f}={\left(f\right)_{c}}\,\!$ where ${\left(\right)_{c}}$ is the collision operator. ${\left(f\right)_{c}}$ is the net rate at which particles enter a phase space volume $d^{3}{\vec {x}}d^{3}{\vec {v}}$ because collisions have changed the velocity of particles.

In general t hese equatsions are hard to solve, so to make this problem more tractable, we’ll examine the elastic collisions of point particles having no internal structure, so that we have a monotonic ideal gas, and collisions conserve particle energy and momentum. Now, to describe what the collision function looks like, we’ll use that collions conserve mass, momentum, and energy. Mass conservation states that:

$\int {d^{3}{\vec {v}}{\left({\partial \over \partial t}{f}\right)_{c}}}=0\,\!$ Momentum conservation means:

$\int {d^{3}{\vec {v}}{\left({\partial \over \partial t}{f}\right)_{c}}{\vec {v}}}=0\,\!$ Energy conservation means:

$\int {d^{3}{\vec {v}}{\left({\partial \over \partial t}{f}\right)_{c}}v^{2}}=0\,\!$ So now we will try to get from the equation ${D \over Dt}{f}={\left({\partial \over \partial t}{f}\right)_{c}}$ to the fluid equations we derived earlier. This means we need to make a transition from the 6-D phase space equatsions to the 3-D fluid equations. We start with:

${\partial \over \partial t}{f}+v_{i}{\partial \over \partial z_{i}}{f}+a_{i}{\partial \over \partial v_{i}}{f}={\left({\partial \over \partial t}{f}\right)_{c}}\,\!$ We will multiply this by $m$ and integrate over $d^{3}{\vec {v}}$ to give us an equation in terms of $\rho =m\int {fd^{3}{\vec {v}}}$ :

${\partial \over \partial t}{\rho }=mass\ conservation\ for\ fluid\ mechanics\,\!$ Then we multiply the Bolzmann equation by $m{\vec {v}}$ and integrating over $d^{3}{\vec {v}}$ to get:

${\partial \over \partial t}{\rho {\vec {u}}}=momentum\ equation\,\!$ We can find any of the macroscopic fluid equatiosn by taking moments of the Bolzmann equation. For some general moment $\chi ({\vec {v}})$ , which may be $m,m{\vec {v}},{\frac {1}{2}}mv^{2}$ , etc., we have:

{\begin{aligned}\int {d^{3}{\vec {v}}\chi {\left(f\right)_{c}}}&=0\\\int {d^{\vec {v}}\chi \left[\overbrace {{\partial \over \partial t}{f}} ^{1}+\overbrace {v_{i}{\partial \over \partial x_{i}}{f}} ^{2}+\overbrace {a_{i}{\partial \over \partial v_{i}}{f}} ^{3}\right]}&=0\\\end{aligned}}\,\! We’ll work this out for each of $1,2,3$ above:

{\begin{aligned}\int {d^{3}{\vec {v}}\chi {\partial \over \partial t}{f}}&={\partial \over \partial t}{}\int {d^{3}{\vec {v}}\chi f}={\partial \over \partial t}{}n\left\langle \chi \right\rangle \\\int {d^{3}{\vec {v}}\chi v_{i}{\partial \over \partial x_{i}}{f}}&={\partial \over \partial x_{i}}{}\int {d^{3}{\vec {v}}\chi v_{i}f}={\partial \over \partial x_{i}}{}(n\left\langle \chi v_{i}\right\rangle \\\int {d^{3}{\vec {v}}\chi a_{i}{\partial \over \partial v_{i}}{f}}&=\int {d^{3}{\vec {v}}{\partial \over \partial v_{i}}{}(\chi a_{i}f)}-\int {d^{3}{\vec {v}}a_{i}f{\partial \over \partial v_{i}}{\chi }}\\&=-\int {d^{3}{\vec {v}}a_{i}{\partial \over \partial v_{i}}{\chi }f}=-n\left\langle a_{i}{\partial \over \partial v_{i}}{\chi }\right\rangle \\\end{aligned}}\,\! where in the last equation, we’ll assume $a_{i}$ is independent of $v_{i}$ . This is okay, even for the magnetic force, because $a_{i}$ is perpedicular to ${\vec {v}}\times {\vec {B}}$ . Combining all of the above equations, we get:

${{\partial \over \partial t}{}(n\left\langle x\right\rangle )+{\partial \over \partial x_{i}}{}(n\left\langle \chi v_{i}\right\rangle )-n\left\langle a_{i}{\partial \over \partial v_{i}}{\chi }\right\rangle =0}\,\!$ This is the general moment equation. In general, we can’t just compute this because each moment requires the computation of a higher-order moment.