Ashcroft & Mermin, Solid State Physics, Chapter 16, Problem 2

[Denver Dang]

Solving Boltzmann equation

Homework Statement

Taken from Ashcroft & Mermin, Solid State Physics, Chapter 16, Problem 2:

"A metal is perturbed by a spatially uniform electric field and temperature gradient. Making the relaxation-time-approximation (16.9) (where g₀ is the local equilibrium distribution appropriate to the imposed temperature gradient), solve the Boltzmann equation (16.13) to linear order in the field and temperature gradient, and verify that the solution is identical to (13.43)."

Homework Equations

Eq. 16.9:
$${{\left( \frac{\partial g}{\partial t} \right)}_{coll}}\simeq -\frac{\left[ g\left( \mathbf{k} \right)-{{g}_{0}}\left( \mathbf{k} \right) \right]}{\tau \left( \mathbf{k} \right)}$$
Eq. 16.13:
$$\frac{\partial g}{\partial t}+\mathbf{v}\cdot \frac{\partial g}{\partial \mathbf{r}}+\mathbf{F}\cdot \frac{1}{\hbar }\frac{\partial g}{\partial \mathbf{k}}={{\left( \frac{\partial g}{\partial t} \right)}_{coll}}$$
Eq. 13.43:
$$g\left( \mathbf{k} \right)={{g}_{0}}\left( \mathbf{k} \right)+\tau \left( \mathrm{ }\!\!\varepsilon\!\!\text{ }\left( \mathbf{k} \right) \right)\left( -\frac{\partial f}{\partial \mathrm{ }\!\!\varepsilon\!\!\text{ }} \right)\mathbf{v}\left( \mathbf{k} \right)\cdot \left[ -e\mathbf{\varepsilon }+\frac{\mathrm{ }\!\!\varepsilon\!\!\text{ }\left( \mathbf{k} \right)-\mu }{T}\left( -\nabla T \right) \right]$$

The Attempt at a Solution

Well, I'm kinda lost. I put Eq. 16.9 into 16.13, and then what?
Can't really see how I get from the Boltzmann equation, with the relaxation-time-approximation, to 13.43 :/

So anyone who can help?

Thanks in advance.

 

[mark726]

Solving the Boltzmann equation can be a tricky task, but with the relaxation-time-approximation and some linear approximation, the solution can be simplified. Here are the steps to solve the problem:

1. Start by substituting Eq. 16.9 into Eq. 16.13 to get:

\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}}=-\frac{\left[ g\left( \mathbf{k} \right)-{{g}_{0}}\left( \mathbf{k} \right) \right]}{\tau \left( \mathbf{k} \right)}

2. Next, use the linear approximation for the electric field and temperature gradient, assuming they are small, to get:

\mathbf{F}\cdot \frac{1}{\hbar }\frac{\partial g}{\partial \mathbf{k}}\simeq -e\mathbf{\varepsilon }\cdot \left( -\frac{\partial f}{\partial \mathrm{ }\!\!\varepsilon\!\!\text{ }} \right)\mathbf{v}\left( \mathbf{k} \right)\cdot \left[ -e\mathbf{\varepsilon }+\frac{\mathrm{ }\!\!\varepsilon\!\!\text{ }\left( \mathbf{k} \right)-\mu }{T}\left( -\nabla T \right) \right]

3. Now, substitute g(k) from Eq. 13.43 into the above equation to get:

\mathbf{F}\cdot \frac{1}{\hbar }\frac{\partial g}{\partial \mathbf{k}}\simeq \tau \left( \mathrm{ }\!\!\varepsilon\!\!\text{ }\left( \mathbf{k} \right) \right)\left( -\frac{\partial f}{\partial \mathrm{ }\!\!\varepsilon\!\!\text{ }} \right)\mathbf{v}\left( \mathbf{k} \right)\cdot \left[ -e\mathbf{\varepsilon }+\frac{\