Getting to the Boltzmann Equation from a Nonequilibrium Distribution Function

[Denver Dang]

Homework Statement

Hi.

I have a course where I am supposed to show how to get to the Boltzmann equation from a nonequilibrium distrubution function.
At the moment I'm kinda lost to how this is done, so hopefully a hint or two would be of some help :)

Homework Equations

The nonequilibrium distribution function:

[tex]g\left( \mathbf{k},t \right)={{g}^{0}}\left( \mathbf{k} \right)+\int_{-\infty }^{t}{dt'}{{e}^{-\left( t-t' \right)\,/\tau \left( \mathrm{ }\varepsilon\text{ }\left( \mathbf{k} \right) \right)}}\left( -\frac{\partial f}{\partial \mathrm{ }\varepsilon\text{ }} \right)\times \mathbf{v}\left( \mathbf{k}\left( t' \right) \right)\cdot \left[ -e\mathbf{E}\left( t' \right)-\nabla \mu \left( t' \right)-\frac{\mathrm{ }\varepsilon\text{ }\left( \mathbf{k} \right)-\mu }{T}\nabla T\left( t' \right) \right][/tex]
where g⁰ is the local equilibrium disibution function and f being the Fermi function.

The Boltzmann equation is given by:

[tex]\frac{\partial g}{\partial t}+\mathbf{v}\cdot \frac{\partial g}{\partial \mathbf{r}}+\mathbf{F}\cdot \frac{1}{\hbar }\frac{\partial g}{\partial \mathbf{k}}={{\left( \frac{\partial g}{\partial t} \right)}_{coll}}[/tex]

The Attempt at a Solution

As far as I have been told, I should be able to come to the Boltzmann equation from the first equation with the use of these equations:

[tex]{{\left( \frac{dg\left( \mathbf{k} \right)}{dt} \right)}_{\mathrm{coll}}}=-\int{\frac{d\mathbf{k}'}{{{\left( 2\mathrm{ }\pi\text{ } \right)}^{3}}}{{W}_{\mathrm{k},\mathrm{k}'}}\left[ g\left( \mathbf{k} \right)-g\left( \mathbf{k}' \right) \right]}[/tex]

and:

[tex]\int{\frac{d\mathbf{k}'}{{{\left( 2\mathrm{ }\pi\text{ } \right)}^{3}}}{{W}_{\mathrm{k},\mathrm{k}'}}\left[ g\left( \mathbf{k} \right)-g\left( \mathbf{k}' \right) \right]}=\frac{1}{\tau \left( \mathbf{k} \right)}\left[ g\left( \mathbf{k} \right)-{{g}^{0}}\left( \mathbf{k} \right) \right][/tex]

These last two equations seems to be somewhat related to the relaxtion-time-approximation.
But how I get from the first equation on the very top, to the Boltzmann equation, well, there I'm kinda lost tbh :/

So anyone with an idea?

It's from the book: "Solid state physics by Ashcroft and Mermin" if that helps.

 

[mark726]

Hi there,

I can give you some hints on how to derive the Boltzmann equation from the nonequilibrium distribution function. First, let's start with the definition of the nonequilibrium distribution function:

g\left( \mathbf{k},t \right)={{g}^{0}}\left( \mathbf{k} \right)+\int_{-\infty }^{t}{dt'}{{e}^{-\left( t-t' \right)\,/\tau \left( \mathrm{ }\varepsilon\text{ }\left( \mathbf{k} \right) \right)}}\left( -\frac{\partial f}{\partial \mathrm{ }\varepsilon\text{ }} \right)\times \mathbf{v}\left( \mathbf{k}\left( t' \right) \right)\cdot \left[ -e\mathbf{E}\left( t' \right)-\nabla \mu \left( t' \right)-\frac{\mathrm{ }\varepsilon\text{ }\left( \mathbf{k} \right)-\mu }{T}\nabla T\left( t' \right) \right]

This equation describes the evolution of the distribution function in time, where the first term on the right-hand side is the equilibrium distribution function and the second term accounts for deviations from equilibrium due to external forces, such as an electric field or a temperature gradient.

Now, let's consider the left-hand side of the Boltzmann equation:

\frac{\partial g}{\partial t}+\mathbf{v}\cdot \frac{\partial g}{\partial \mathbf{r}}+\mathbf{F}\cdot \frac{1}{\hbar }\frac{\partial g}{\partial \mathbf{k}}

The first term represents the time evolution of the distribution function, which is also described by the first term on the right-hand side of the nonequilibrium distribution function. The second term accounts for the spatial variation of the distribution function, and the third term accounts for the effect of external forces on the distribution function.

To derive the Boltzmann equation, we need to equate the two expressions for the time evolution of the distribution function. This means that we need to set the first term on the right-hand side of the nonequilibrium distribution function equal to the first term on the left-hand side of