Complicated Maxwell Boltzman Distribution Integration

In summary, the Maxwell Boltzmann distribution is a probability distribution that describes the speeds of particles in a gas at a given temperature. It is important because it allows us to understand the behavior of gases and predict their properties, such as pressure and temperature. The term "complicated" refers to the fact that the integration of the distribution can be mathematically complex and challenging. It is integrated using calculus methods and can be affected by factors such as temperature, mass of particles, and nature of collisions. The integration of the distribution has significance in scientific research as it allows for analysis and understanding of gas behavior and is used in fields such as thermodynamics, statistical mechanics, and kinetic theory.
  • #1
TFM
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Homework Statement



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.

Homework Equations





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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  • #2
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...

TFM said:
[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].
 
  • #3
Tom Mattson said:
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
 
  • #4
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
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
 
  • #6
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]
 
Last edited:
  • #7
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 :smile:

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
 
  • #8
TFM said:
[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 canceled down to:

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

Does this look right?

TFM
 
  • #9
TFM said:
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 canceled 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/hbase/kinetic/kintem.html#c3

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

or <v2>.
 
Last edited:
  • #10
Thaks for all the Help,

TFM
 

1. What is the Maxwell Boltzmann distribution and why is it important?

The Maxwell Boltzmann distribution is a probability distribution that describes the speeds of particles in a gas at a given temperature. It is important because it allows us to understand the behavior of gases and predict their properties, such as pressure and temperature.

2. What does "complicated" refer to in the term "Complicated Maxwell Boltzmann distribution integration"?

The term "complicated" refers to the fact that the integration of the Maxwell Boltzmann distribution can be mathematically complex and challenging, especially in cases where the distribution is not a simple function.

3. How is the Maxwell Boltzmann distribution integrated?

The Maxwell Boltzmann distribution is integrated by using calculus methods, such as the substitution or integration by parts. In some cases, it can also be approximated using numerical methods.

4. What factors affect the integration of the Maxwell Boltzmann distribution?

The integration of the Maxwell Boltzmann distribution can be affected by factors such as the temperature of the gas, the mass of the particles, and the nature of the collisions between particles. In addition, the shape of the distribution and the limits of the integration also play a role.

5. What is the significance of the integration of the Maxwell Boltzmann distribution in scientific research?

The integration of the Maxwell Boltzmann distribution is significant in scientific research as it allows us to analyze and understand the behavior of gases and their properties. It is used in fields such as thermodynamics, statistical mechanics, and kinetic theory to make predictions and develop models for various phenomena in nature.

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