# Maxwell-Boltzmann distribution for transport equations

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.

Well, anyone who can give me a clue?

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# Normalization Factor

Let us define a generalized Gaussian probability density function (PDF) as: $$f_{s}\left( x \right) = A_{o} \ e^{^{\displaystyle - \frac{ (x - x_{o})^{2} }{ 2 \sigma^{2} } }} \tag{0}$$ where $A_{o}$ is the normalization constant, $x$ is the argument, and $s$ denotes the set of distributions (e.g., particle species), $x_{o}$ is the offset of the peak from $x = 0$, and $\sigma$ is the variance of the distribution.

The normalization factor $A_{o}$ is found by using the following constraint: $$\int_{-\infty}^{+\infty} \ dx \ f_{s}\left( x \right) = 1 \tag{1}$$

The analytical solution to this integral can be found in any standard integral table or using something like Mathematica, where one finds: $$A_{o} = \frac{ 1 }{ \sqrt{2 \ \pi \ \sigma^{2}} } \tag{2}$$

# Maxwellian Velocity Distribution

To convert from the 1D Gaussian PDF in Equation 0 above to a Maxwell-Boltzmann distribution, or Maxwellian, we just convert the variables as follows:

• $x \rightarrow v$, where $v$ is the velocity argument of $f_{s}$ ranging from $-\infty$ to $+\infty$
• $x_{o} \rightarrow v_{o}$, where $v_{o}$ is drift velocity or bulk flow velocity
• $2 \ \sigma^{2} \rightarrow V_{Ts}$, where $V_{Ts}$ is the thermal speed (here the most probable speed)

Then we can see that the 1D Maxwellian is given by: $$f_{s}\left( v \right) = \frac{ 1 }{ \sqrt{\pi} \ V_{Ts} } \ e^{^{\displaystyle - \left( \frac{ v - v_{o} }{ V_{Ts} } \right)^{2} }} \tag{3}$$

The conversion to a full 3D distribution is simple enough so long as each velocity component, $v_{j}$, is not correlated with any other component. Then (dropping the subscript $s$ for brevity) we can define: \begin{align} f\left( v_{x}, v_{y}, v_{z} \right) & = f\left( v_{x} \right) \ f\left( v_{y} \right) \ f\left( v_{z} \right) \tag{4a} \\ & = \prod_{ k = x,y,z } \ A_{k} \ e^{^{\displaystyle - \left( \frac{ v_{k} - v_{ok} }{ V_{Tk} } \right)^{2} }} \tag{4b} \end{align} where the total normalization factor is given by: $$A_{x} \ A_{y} \ A_{z} \ = \frac{ 1 }{ \pi^{3/2} \ V_{Tx} \ V_{Ty} \ V_{Tz} } \tag{5}$$

# Conversion to Energy

To do this properly, one should use a momentum analog of the velocity-space version described above.

In the non-relativistic limit the conversion from momentum to energy is $E = \tfrac{p^{2}}{2 \ m}$ or $p = \sqrt{2 \ m \ E}$. Therefore, we can see that $dp/dE \propto E^{-1/2}$ or $\propto p^{-1}$. Since energy is a scalar, the limits of integration (e.g., when finding the normalization factor) change from $-\infty \leq v_{j} \leq +\infty$ to $0 \leq E \leq +\infty$.

Then the 3D version (e.g., Equation 4b above) becomes: $$f\left( E \right) = \frac{ 1 }{ Z } \ e^{^{\displaystyle - \left( \frac{ E }{ k_{B} \ T } \right) }}$$ where $Z$ is the partition function, $k_{B}$ is Boltzmann's constant, and $T$ is the temperature.

In the relativistic limit, the situation becomes incredibly complicated as discussed at What is the correct relativistic distribution function?.

# Transport Coefficients

One can calculate these through use of the moments of the distribution. I wrote a detailed answer for a non-relativistic velocity distribution at http://physics.stackexchange.com/a/218643/59023.

To make the probability distributions shown above consistent with the velocity moments discussed in my other answer, you can redefine $f_{s}$ such that when integrated over all velocity space one gets the number density of set $s$.

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