Maxwell speed distribution


by Je m'appelle
Tags: distribution, maxwell, speed
Je m'appelle
Je m'appelle is offline
#1
Apr22-11, 10:23 AM
P: 109
1. The problem statement, all variables and given/known data
Show that the number [tex]N(0,v)[/tex], of molecules of an ideal gas with speeds between 0 and v is given by

[tex]N(0,v) = N \left[ erf(\xi) - \frac{2}{\sqrt{\pi}} \xi e^{-\xi^2} \right][/tex]

Where,

[tex]erf(\xi) = \frac{2}{\sqrt{\pi}} \int_{0}^{\xi} e^{-x^2} dx [/tex]

And,

[tex]\xi^2 = \left(\frac{mv^2}{2kT} \right)[/tex]

2. Relevant equations

[tex]\frac{dN_v}{N} = \sqrt{\frac{2}{\pi}} \left( \frac{m}{kT}\right)^{\frac{3}{2}} v^2 e^{\frac{-mv^2}{2kT}} \ dv [/tex]

3. The attempt at a solution

Alright, so I managed to get to the following

[tex]\frac{dN_v}{N} = \frac{4}{\sqrt{\pi}} x^2 e^{-x^2} dx [/tex]

Where,

[tex]\alpha = \frac{m}{2kT}, \ x = \sqrt{\alpha}v [/tex]

=========================\\===========================

So far so good, but now when checking the solution on the textbook it claims the following algebraic manipulation which I can't follow

[tex]\frac{2N}{\sqrt{\pi}}\int_{0}^{\xi} x (2xe^{-x^2} dx) = \frac{2N}{\sqrt{\pi}} x e^{-x^2} |_{\xi}^{0} \ + \ N \frac{2}{\sqrt{\pi}} \int_{0}^{\xi} e^{-x^2} dx [/tex]

What was done on the integral above, how can you "split" it like that?
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L-x
L-x is offline
#2
Apr22-11, 10:49 AM
P: 61
Integration by parts
Dick
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#3
Apr22-11, 11:00 AM
Sci Advisor
HW Helper
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P: 25,165
It's integration by parts. udv=d(uv)-vdu with u=x and dv=(2x)e^(-x^2)dx.

Je m'appelle
Je m'appelle is offline
#4
Apr22-11, 11:10 AM
P: 109

Maxwell speed distribution


Quote Quote by L-x View Post
Integration by parts
But of course! Agh, how didn't I see that, I feel very stupid right now.

Thanks for the highlight, L-x!
L-x
L-x is offline
#5
Apr22-11, 11:15 AM
P: 61
No worries. It's not immidiately obvious, although the integral is written in a way which gives you a clue: x(2x) instead of 2x^2
Dick
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#6
Apr22-11, 11:29 AM
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P: 25,165
Don't double post, Je m'appelle.

http://www.physicsforums.com/showthread.php?t=492505


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