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Maxwell Boltzmann Distribution Question

  1. Jan 3, 2009 #1
    Hello everyone

    1. The problem statement, all variables and given/known data
    The equivalent of the Maxwell-Boltzman distribution for a two-dimensional
    gas is
    [itex]P(v) = Cv e^-\frac {mv^2}{kt}[/itex]

    Determine [itex]C[/itex] so that

    [itex]\int_0^\infty P(v)dv = N[/itex]

    2. Relevant equations
    Not really sure

    3. The attempt at a solution
    I wasn't really sure how to tackle this question so I figured i'd integrate [itex]P(v)[/itex] since the question says that'll equal N.

    [itex]\int_0^\infty P(v)dv[/itex]
    [itex]\int_0^\infty Cv e^-\frac {mv^2}{kt} dv[/itex]
    [itex]C\int_0^\infty v e^-\frac {mv^2}{kt} dv[/itex]

    [itex] u = \frac {mv^2}{kt}[/itex]

    [itex]\frac {du}{dv} = \frac {2mv}{kt}[/itex]

    [itex]dv = \frac {du kt}{2mv}[/itex]

    [itex]C\int_0^\infty v e^{-u} \frac {du kt}{2mv}[/itex]

    [itex]C\int_0^\infty e^{-u} \frac {du kt}{2m}[/itex]

    [itex]\frac {Ckt}{2m} \int_0^\infty e^{-u} du[/itex]

    [itex] = \frac {Ckt}{2m} \bigg[{-e^{-u}\bigg]_0^\infty[/itex]

    [itex] = \frac {Ckt}{2m} \bigg[{-e^{-\frac {mv^2}{kt}}\bigg]_0^\infty[/itex]

    I'm not really sure where to go from here. How would I evaluate this between infinity and zero?

  2. jcsd
  3. Jan 3, 2009 #2


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    HINT: You said the entire integral has to equal N, correct? Well, your last line is equal to the integral. So if A = B and B=C ... ?
  4. Jan 3, 2009 #3
    ok, so you're saying

    [itex] N = \frac {Ckt}{2m} \bigg[{-e^{-\frac {mv^2}{kt}}\bigg]_0^\infty[/itex]

    which yeah, makes sense. But do you want me to rearrange it to make C the subject while not evaluating the integral?
  5. Jan 3, 2009 #4


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    No. Evaluate the integral and then solve for C. That should be your answer.
  6. Jan 3, 2009 #5
    I thought so. This might seem stupid, but I really don't know how to evaluate the integral when one of the limits is [itex]\infty[/itex]. Could you shed some light on that please?
  7. Jan 3, 2009 #6


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    To evaluate your integral, evaluate the following instead:

    [tex]N = \frac {Ckt}{2m} \bigg[{-e^{-\frac {mv^2}{kt}}\bigg]_0^t[/tex]

    Now take that whole resulting expression and take the limit as t[itex]\rightarrow\infty[/itex].

    Now can you solve for C?
  8. Jan 3, 2009 #7
    Ok, so [itex]\bigg[{-e^{-\frac {mv^2}{kT}}\bigg]_0^t[/itex]

    will give me

    [itex]\lim_{t \to \infty}({-e^{-\frac {mt^2}{kT}} + 1)[/itex]


    Sorry if this is annoying, I've never actually done something like this which seems a bit strange considering a question requires it. I appreciate the help.
  9. Jan 12, 2009 #8
    I used a site that provides a nice integral and limit calculator:
    From that, I get:

    [tex]\[\lim_{v\to0}\left(-{{k\,T\,e^ {- {{m\,v^2}\over{k\,T}} }}\over{2\,m}}\right) = -{{k\,T}\over{2\,m}}\][/tex]
    [tex]\[C \times \left[ \frac{kTe^{\frac{-mv^2}{kT}}}{2m}\right]^{\infty}_0=C\times \frac{kT}{2m}\][/tex]

    Have I missed something obvious?
    Last edited: Jan 12, 2009
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