Fluids Lecture 03

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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 particles, specified by and , with a force interaction between them:

Instead of doing independent particle dynamics, we’ll do statistical mechanics in 6-D phase space, which is the space of all and for all particles. Let’s say we have a distribution function which describes the # density of particles in phase space. Thus, the # of particles with position between and and velocity between and is .

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

We can compute a general property as:

On such property is the average velocity , given by:

For any individual particle, we can compute its deviation from :

If , then:

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 . 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, must satisfy a continuity equation:

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and . Thus,

Expanding this equation, we have:

Using Hamiltonian dynamics where , we have that and , so:

Thus, the last two terms on right of our continuity equation sum to 0, giving us (in a collisionless system):

Thus, , where

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:

where the underlined portion is . Now is the Eulerian derivative, and 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 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,

where is the collision operator. is the net rate at which particles enter a phase space volume 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:

Momentum conservation means:

Energy conservation means:

So now we will try to get from the equation 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:

We will multiply this by and integrate over to give us an equation in terms of :

Then we multiply the Bolzmann equation by and integrating over to get:

We can find any of the macroscopic fluid equatiosn by taking moments of the Bolzmann equation. For some general moment , which may be , etc., we have:

We’ll work this out for each of above:

where in the last equation, we’ll assume is independent of . This is okay, even for the magnetic force, because is perpedicular to . Combining all of the above equations, we get:

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.