- #1
Denver Dang
- 148
- 1
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 [itex]v[/itex]-space (Velocity) or [itex]E[/itex]-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:
[tex]{{\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,[/tex]
at least, again, when trying to calculate the electrical conductivity, which in the end should end up being Drudes formula [itex]\sigma =\frac{n{{e}^{2}}\tau }{m}[/itex].
So basically, not hard. But I have to get the distribution function right.
As far as I know the MB-distribution is given by:
[tex]{{f}_{MB}}\left( \varepsilon \right)=C{{e}^{-\varepsilon /{{k}_{B}}T}},[/tex]
where [itex]C[/itex] 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:
[tex]\bar{n}=\frac{{\bar{N}}}{{{Z}_{1}}\left( T,V \right)}{{e}^{-\varepsilon /{{k}_{B}}T}},[/tex]
where:
[tex]\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),[/tex]
and [itex]{{Z}_{\operatorname{int}}}\left( T \right) = 1[/itex] 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 [itex]C[/itex] - at least not from what I can see. Maybe it's the [itex]V/N[/itex] I'm not sure about.
So, anyone who can give me a clue, or...?
As far as I know it should not be the MB distribution for [itex]v[/itex]-space (Velocity) or [itex]E[/itex]-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:
[tex]{{\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,[/tex]
at least, again, when trying to calculate the electrical conductivity, which in the end should end up being Drudes formula [itex]\sigma =\frac{n{{e}^{2}}\tau }{m}[/itex].
So basically, not hard. But I have to get the distribution function right.
As far as I know the MB-distribution is given by:
[tex]{{f}_{MB}}\left( \varepsilon \right)=C{{e}^{-\varepsilon /{{k}_{B}}T}},[/tex]
where [itex]C[/itex] 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:
[tex]\bar{n}=\frac{{\bar{N}}}{{{Z}_{1}}\left( T,V \right)}{{e}^{-\varepsilon /{{k}_{B}}T}},[/tex]
where:
[tex]\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),[/tex]
and [itex]{{Z}_{\operatorname{int}}}\left( T \right) = 1[/itex] 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 [itex]C[/itex] - at least not from what I can see. Maybe it's the [itex]V/N[/itex] I'm not sure about.
So, anyone who can give me a clue, or...?