Integrating Maxwell-Boltzmann Distribution

  • Thread starter ejensen6
  • Start date
  • #1
ejensen6
19
0
I don’t know how to integrate the Maxwell-Boltzmann distribution without approximation or help from Maple.

Given the Maxwell-Boltzmann distribution:

f(v) = 4*pi*[m/(2*pi*k*T)]^(3/2)*v^2*exp[(-m*v^2)/(2*k*T)]

Assuming a fixed temperature and mass, one can simplify this equation:

f(v) = a*v^2*exp[-bv^2]
a = 4*pi*[m/(2*pi*k*T)]^(3/2)
b = m/(2*k*T)

In order to calculate the fraction of particles between two speeds v1 and v2, one should evaluate the definite integral:

∫f(v)dv

Can this be done analytically?

Also, this function predicts a non-zero (albeit very small) probability for particles to have a speed greater than the speed of light. Is there a correction to this distribution that takes this into account?
 

Answers and Replies

  • #2
cesiumfrog
2,010
5
Simplest way is 1) find it in a standard table of integrands and 2) then differentiate the integral just to convince yourself.
 
  • #3
ejensen6
19
0
Table of Integrals

I haven't been able to find this integral in a table. I looked in Gradshteyn and Ryzhik's book and couldn't find it. Maybe I need to find a bigger book of integrals.
 
  • #4
StatMechGuy
223
2
There isn't one. Let's look at our good friend the Gaussian integral. If you're in a statistical mechanics course, you are going to become a master of these things by the time you're done.

Okay, so
[tex]I(\alpha) = \int_{-\infty}^\infty dx e^{- \alpha x^2}[/tex]
Let's square it. Then we get that
[tex]I(\alpha)^2 = \int_{-\infty}^\infty dx \int_{-\infty}^\infty dy e^{-\alpha(x^2 + y^2)}[/tex]
After making a change of coordinates to polar and integrating over the entire plane, we get that the integral is given by
[tex]I(\alpha) = \sqrt{\pi/\alpha}[/tex]
Okay, but what about terms like x^2? Well...
[tex] \int x^2 e^{-\alpha x^2} = \int - \frac{d}{d \alpha} e^{-\alpha x^2}[/tex]
Which then means that
[tex]\int x^2 e^{-\alpha x^2} = - \frac{\partial}{\partial \alpha} I(\alpha)[/tex]
This works for all even powers. This is a rather handy trick to know, and is pretty standard.
 
Last edited:
  • #5
ejensen6
19
0
Integral

Yes, the solution of that integral from -infinity to +infinity or zero to infinity is well known and can be used to show that the function normalizes. But what if you want to know the fraction of particles that are between two particular speeds? I did find some information at the following web site:

http://user.mc.net/~buckeroo/MXDF.html [Broken]

It states below equation 35 that the integral "cannot be solved in closed form". I also found a computational chemistry class that had an assignment to numerically integrate the distribution, which implies that this might be necessary:

http://www.glue.umd.edu/~nsw/ench250/boltzman.htm [Broken]
 
Last edited by a moderator:
  • #6
Dr Transport
Science Advisor
Insights Author
Gold Member
2,570
722
look for the function erf(c), ther error function and you should be able to integrate the Boltzmann function between two finite limits.
 
  • #7
ejensen6
19
0
Erf

I see how "erf" would work. Thanks much.

Does anybody have a reference for the relativistic correction to the speed distribution?
 
  • #8
seebabu@gmail
1
0
error function

Please describe what is mean by error function.Why we are introducing error function in Maxwell's function distribution?
 
  • #9
jdougherty
25
0
The error function is defined as
[itex]\textrm{erf}(x) = \frac{2}{\sqrt{\pi}}\int^{x}_{0} e^{-t^{2}}~\textrm{d}t[/itex]
and pops up a lot when one does integrals with exponentials, even though it doesn't always imply that there is some sort of "error" involved.


cf. http://en.wikipedia.org/wiki/Error_function
 

Suggested for: Integrating Maxwell-Boltzmann Distribution

Replies
0
Views
517
Replies
1
Views
406
  • Last Post
Replies
10
Views
1K
Replies
3
Views
496
Replies
1
Views
426
Replies
23
Views
839
Replies
7
Views
4K
Replies
17
Views
3K
  • Last Post
Replies
2
Views
2K
Top