# Maxwell-Boltzmann distribution for transport equations

1. May 18, 2013

### Denver Dang

I have to calculate the transport coefficients for the Maxwell-Boltzmann distribution. But I'm not sure what distribution I have to use.
As far as I know it should not be the MB distribution for $v$-space (Velocity) or $E$-axis (Energy), since that will get me the wrong dimensions in the end. I have to use the distribution per state.

But I'm not sure how this looks. The integral I have to solve, for me getting the electrical conductivity (1st transport coefficient) I need, is given by:

$${{\mathcal{L}}^{\,\left( 0 \right)}}={{\left( \frac{2m}{{{\hbar }^{2}}} \right)}^{3/2}}\frac{{{e}^{2}}\tau }{{{\pi }^{2}}m}\int{\left( -\frac{\partial {{f}_{MB}}}{\partial \varepsilon } \right)}\,{{\varepsilon }^{3/2}}d\varepsilon,$$

at least, again, when trying to calculate the electrical conductivity, which in the end should end up being Drudes formula $\sigma =\frac{n{{e}^{2}}\tau }{m}$.

So basically, not hard. But I have to get the distribution function right.

As far as I know the MB-distribution is given by:

$${{f}_{MB}}\left( \varepsilon \right)=C{{e}^{-\varepsilon /{{k}_{B}}T}},$$

where $C$ is what I need to figure out, since that will determine the dimensions of my coefficients.

According to my book the normalized MB distribution function is:

$$\bar{n}=\frac{{\bar{N}}}{{{Z}_{1}}\left( T,V \right)}{{e}^{-\varepsilon /{{k}_{B}}T}},$$

where:

$$\frac{{{Z}_{1}}\left( T,V \right)}{{\bar{N}}}=\frac{V}{{\bar{N}}}\left( \frac{2\pi m{{k}_{B}}T}{{{h}^{2}}} \right){{Z}_{\operatorname{int}}}\left( T \right),$$

and ${{Z}_{\operatorname{int}}}\left( T \right) = 1$ in my case.

But I'm not quite sure how to about this? As far as I can see, it's not just inserting the reversed term of this in $C$ - at least not from what I can see. Maybe it's the $V/N$ I'm not sure about.

So, anyone who can give me a clue, or...?

2. May 19, 2013

### Thaakisfox

Your normalization factor should be :

$$\frac{{{Z}_{1}}\left( T,V \right)}{{\bar{N}}}=\frac{V}{{\bar{N}}}\left( \frac{2\pi m{{k}_{B}}T}{{{h}^{2}}} \right)^{3/2}$$

The n in the drude law is the number density. i.e. the number of electrons per unit volume. In your normalization constant what is \bar{n}, \bar{N} ?

3. May 19, 2013

### Denver Dang

The bar over $n$ and $N$ means the "mean" of whatever it is...

But do I know what this is ?
Or at least V/N, or...?