Complicated Maxwell Boltzman Distribution Integration


by TFM
Tags: boltzman, complicated, distribution, integration, maxwell
TFM
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#1
Apr25-08, 02:38 PM
P: 1,031
1. The problem statement, all variables and given/known data

Calculate the integral

[tex]\left v\langle\right\rangle = \int^\infty_0 v f(v) dv[/tex].

The function

[tex]f(v)[/tex]

describing the actual distribution of molecular speeds is called the Maxwell-Boltzmann distribution,

[tex]f(v) = 4\pi (\frac{m}{2\pi kT})^3^/^2 v^2 e^{-mv^2/2kt}[/tex]

(Hint: Make the change of variable

[tex] v^2 = x [/tex]

and use the tabulated integral

[tex]\int^\infty _0 x^ne^\alpha^x dx = \frac{n}{\alpha^n^+^1}[/tex]

where n is a positive integer and [tex] \alpha [/tex] is a positive constant.)
Express your answer in terms of the variables T, m, and appropriate constants.

2. Relevant equations



3. The attempt at a solution

I think I have got some way in, but I am not sure how to go from here:

firstly the:

[tex] 4\pi (\frac{m}{2\pi kT})^{3/2} [/tex]

is a constant, so can put in c for now

[tex]f(v) = (c) v^2 e^{-mv^2/2kt}[/tex]

ANd can remove from the integration

[tex]v = c \int^\infty_0 v (v^2 e^{-mv^2/2kt}) dv[/tex]

then, replace [tex]v^2[/tex] with x:

[tex]v = c \int^\infty_0 v (x e^{-mx/2kt}) dv[/tex]

and change the integration,

[tex]x = v^2, thus \frac{dx}{dv} = 2x thus dv = \frac{dx}{2v}[/tex]

Which gives:

[tex]v = c \int^\infty_0 v (x e^{-mx/2kt}) \frac{dx}{2v}[/tex]

and

[tex]v = c \int^\infty_0 (x e^{-mx/2kt}) \frac{dx}{2}[/tex]

Taking out the half:

[tex]v = c/2 \int^\infty_0 x e^{-mx/2kt} dx[/tex]

rearraniging for the tabulated integral,

[tex]\alpha = -\frac{m}{2kt}[/tex]

So:

[tex]v = c/2 \int^\infty_0 x e^{-\alpha x} dx[/tex]

Which can be integrated using tabulated given in question:

[tex]c/2 \left[\frac{1}{\alpha^2}\right][/tex]

Putting back [tex]\alpha[/tex]

[tex]c/2 \left[\frac{1}{(- \frac{m}{2kt})^2}\right][/tex]

And:

[tex]c/2 \left[\frac{1}{(- \frac{m^2}{4k^2t^2})}\right][/tex]

Which I believe can go around to:

[tex]c/2 (\frac{4k^2 t^2}{m^2})[/tex]

Putting back the c:

[tex](4\pi (\frac{m}{2\pi kT})^3^/^2)/2 (\frac{4k^2 t^2}{m^2})[/tex]

Which I think can be rarranged a bit more to give:

[tex](2\pi (\frac{m}{2\pi kT})^3^/^2) (\frac{4k^2 t^2}{m^2})[/tex]

But I am not quite sure where to go from here.

Any ideas? Does it look right?

TFM
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Tom Mattson
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#2
Apr25-08, 03:35 PM
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I am finding your solution difficult to read, partly because of errors on LaTeX and typographical errors, and partly because you are making this way too complicated. It really isn't that much work.

But here goes...

Quote Quote by TFM View Post
[tex]c/2 \left[\frac{1}{\alpha^2}\right][/tex]
This is correct.

Putting back [tex]/alpha[/tex]

[tex]c/2 \left[\frac{1}{(- \frac{m}{2kt})^2}\right][/tex]
Here's one of your typos: The lower case [itex]t[/itex] in the exponent should be a [itex]T[/itex], should it not?

[tex](2\pi (\frac{m}{2\pi kT})^3^/^2) (\frac{4k^2 t^2}{m^2})[/tex]
Replace [itex]t[/itex] with [itex]T[/itex] and this is correct, too.

But I am not quite sure where to go from here.
Start cancelling things. For instance, [itex]\frac{(2kT)^2}{(2kT)^{3/2}}=(2kT)^{1/2}[/itex].
TFM
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#3
Apr25-08, 04:22 PM
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Quote Quote by Tom Mattson View Post
Start cancelling things. For instance, [itex]\frac{(2kT)^2}{(2kT)^{3/2}}=(2kT)^{1/2}[/itex].
where did the [tex](2kt)^2[/tex] come from, and how have you removed the m and[tex]\pi[/tex] from the

[tex](\frac{m}{2\pi kT})^3^/^2[/tex] ?

TFM

Tom Mattson
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#4
Apr25-08, 05:42 PM
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Complicated Maxwell Boltzman Distribution Integration


No, those are still there. I just showed you how to do one of the cancellations. Remember you are allowed to move factors around and group them together.
TFM
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#5
Apr26-08, 04:07 AM
P: 1,031
Looking at it , does this look right:

[tex]2\pi(\frac{m^3^/^2}{(2\pi kt)^3^/^2})(\frac{(2kt)^2}{m^2})[/tex]

Taking out the pi from the first fraction:

[tex](\frac{2 \pi}{\pi^3^/^2})(\frac{m^3^/^2}{(2kt)^3^/^2})(\frac{(2kt)^2}{m^2})[/tex]

Does this look okay?

TFM
Astronuc
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#6
Apr26-08, 07:27 AM
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When doing an exponential which involves an expression, put the entire expression after the ^ in parentheses { }. This groups the entire expression, e.g. e^{-mv^2/2kT} yields

[tex]e^{-mv^2/2kT}[/tex]

It's also best to follow convention and use t for time, and T for temperature.

Also, in the OP, [tex]\alpha = \frac{m}{2kT}[/tex], so that the definite integral is finite.

Then [tex]\frac{1}{\alpha^2} = \frac{4k^2T^2}{m^2}[/tex]
TFM
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#7
Apr26-08, 12:42 PM
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When doing an exponential which involves an expression, put the entire expression after the ^ in parentheses { }. This groups the entire expression, e.g. e^{-mv^2/2kT} yields
Thanks for this ^ , I never knew this before

Also, in the OP, [tex]\alpha = \frac{m}{2kT}[/tex], so that the definite integral is finite.

Then [tex]\frac{1}{\alpha^2} = \frac{4k^2T^2}{m^2}[/tex]
Does this mean I got the calculations in my last post wrong?

TFM
TFM
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#8
Apr26-08, 03:02 PM
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Quote Quote by TFM View Post
[tex]2\pi(\frac{m^{3/2}}{(2\pi kt)^{3/2}})(\frac{(2kt)^2}{m^2})[/tex]

[tex](\frac{2 \pi}{\pi^3^/^2})(\frac{m^3^/^2}{(2kt)^3^/^2})(\frac{(2kt)^2}{m^2})[/tex]
From this, I have since got:

[tex] (\frac{2 \pi}{\pi ^{3/2}})(\frac{m^{3/2}}{m^2})(\frac{(2kT)^2}{(2kT)^{3/2}}) [/tex]

Which I cancelled down to:

[tex] (\frac{2 \pi}{\pi ^{3/2}})(\frac{1}{\sqrt{m}})(\sqrt{}2kT) [/tex]

Does this look right?

TFM
Astronuc
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#9
Apr27-08, 08:32 AM
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Quote Quote by TFM View Post
From this, I have since got:

[tex] (\frac{2 \pi}{\pi ^{3/2}})(\frac{m^{3/2}}{m^2})(\frac{(2kT)^2}{(2kT)^{3/2}}) [/tex]

Which I cancelled down to:

[tex] (\frac{2 \pi}{\pi ^{3/2}})(\frac{1}{\sqrt{m}})(\sqrt{2kT}) [/tex]

Does this look right?

TFM
Correct!

Well continuing the cancellation of pi in the last expression, one obtains

[tex] \left(\frac{2}{\pi^{1/2}}\right)\left(\frac{1}{\sqrt{m}}\right)(\sqrt{2kT}) [/tex],

Then one can bring pi and m under the square root as the denominator under the numerator 2kT,

[tex] 2 \sqrt{\frac{2kT}{\pi m}} [/tex]

which is the same as

[tex] \sqrt{\frac{8kT}{\pi m}} [/tex] which is correct.

and now that you've gone through this exercise

http://hyperphysics.phy-astr.gsu.edu...kintem.html#c3

One should also try the relationsip between mean kinetic energy and gas temperature,

or <v2>.
TFM
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#10
Apr27-08, 08:49 AM
P: 1,031
Thaks for all the Help,

TFM


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