# Complicated Maxwell Boltzman Distribution Integration

1. Apr 25, 2008

### TFM

1. The problem statement, all variables and given/known data

Calculate the integral

$$\left v\langle\right\rangle = \int^\infty_0 v f(v) dv$$.

The function

$$f(v)$$

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

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

(Hint: Make the change of variable

$$v^2 = x$$

and use the tabulated integral

$$\int^\infty _0 x^ne^\alpha^x dx = \frac{n}{\alpha^n^+^1}$$

where n is a positive integer and $$\alpha$$ 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:

$$4\pi (\frac{m}{2\pi kT})^{3/2}$$

is a constant, so can put in c for now

$$f(v) = (c) v^2 e^{-mv^2/2kt}$$

ANd can remove from the integration

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

then, replace $$v^2$$ with x:

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

and change the integration,

$$x = v^2, thus \frac{dx}{dv} = 2x thus dv = \frac{dx}{2v}$$

Which gives:

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

and

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

Taking out the half:

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

rearraniging for the tabulated integral,

$$\alpha = -\frac{m}{2kt}$$

So:

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

Which can be integrated using tabulated given in question:

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

Putting back $$\alpha$$

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

And:

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

Which I believe can go around to:

$$c/2 (\frac{4k^2 t^2}{m^2})$$

Putting back the c:

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

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

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

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

Any ideas? Does it look right?

TFM

2. Apr 25, 2008

### Tom Mattson

Staff Emeritus
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...

This is correct.

Here's one of your typos: The lower case $t$ in the exponent should be a $T$, should it not?

Replace $t$ with $T$ and this is correct, too.

Start cancelling things. For instance, $\frac{(2kT)^2}{(2kT)^{3/2}}=(2kT)^{1/2}$.

3. Apr 25, 2008

### TFM

where did the $$(2kt)^2$$ come from, and how have you removed the m and$$\pi$$ from the

$$(\frac{m}{2\pi kT})^3^/^2$$ ?

TFM

4. Apr 25, 2008

### Tom Mattson

Staff Emeritus
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.

5. Apr 26, 2008

### TFM

Looking at it , does this look right:

$$2\pi(\frac{m^3^/^2}{(2\pi kt)^3^/^2})(\frac{(2kt)^2}{m^2})$$

Taking out the pi from the first fraction:

$$(\frac{2 \pi}{\pi^3^/^2})(\frac{m^3^/^2}{(2kt)^3^/^2})(\frac{(2kt)^2}{m^2})$$

Does this look okay?

TFM

6. Apr 26, 2008

### Astronuc

Staff Emeritus
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

$$e^{-mv^2/2kT}$$

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

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

Then $$\frac{1}{\alpha^2} = \frac{4k^2T^2}{m^2}$$

Last edited: Apr 26, 2008
7. Apr 26, 2008

### TFM

Thanks for this ^ , I never knew this before

Does this mean I got the calculations in my last post wrong?

TFM

8. Apr 26, 2008

### TFM

From this, I have since got:

$$(\frac{2 \pi}{\pi ^{3/2}})(\frac{m^{3/2}}{m^2})(\frac{(2kT)^2}{(2kT)^{3/2}})$$

Which I cancelled down to:

$$(\frac{2 \pi}{\pi ^{3/2}})(\frac{1}{\sqrt{m}})(\sqrt{}2kT)$$

Does this look right?

TFM

9. Apr 27, 2008

### Astronuc

Staff Emeritus
Correct!

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

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

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

$$2 \sqrt{\frac{2kT}{\pi m}}$$

which is the same as

$$\sqrt{\frac{8kT}{\pi m}}$$ which is correct.

and now that you've gone through this exercise

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

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

or <v2>.

Last edited: Apr 27, 2008
10. Apr 27, 2008

### TFM

Thaks for all the Help,

TFM