Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question

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
  • $2 \ \sigma^{2} \rightarrow V_{Ts}^{2}$, 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 $s$.

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.