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Maxwell-Boltzman distribution

  1. Mar 16, 2016 #1
    1. The problem statement, all variables and given/known data
    1. According to the Maxwell-Boltzman law of theoretical physics, the probability density of a gas molecule is ##f(v) = kv^2e^{-\beta v^2}## if ##v > 0## and ##f(v) = 0## otherwise, where ##\beta## depends on its mass and absolute temperature, and ##k## is an appropriate constant. Show that the p.d.f. of the kinetic energy ##E##, whose values are related to those of ##V## by means of the equation ##E = \frac{1}{2}mV^2## has a Gamma distribution.

    2. Repeat the derivation when ##V \sim N(0,\sigma^2)##.

    2. Relevant equations


    3. The attempt at a solution
    For 1. I'm not too sure what exactly they are asking. Are they asking for pdf of E and how does that relate to the pdf ##f(v)##???
    For 2. I'm sure it just follows on from 1.
     
  2. jcsd
  3. Mar 16, 2016 #2

    Ray Vickson

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    You ask: "Are they asking for pdf of E and how does that relate to the pdf ##f(v)##???" Yes, that is exactly what the question says!
     
  4. Mar 17, 2016 #3
    So we apply the change of variable technique here then???
     
  5. Mar 17, 2016 #4

    Charles Link

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    This one I think is best done by using probability distribution functions which are integrals of the density functions from minus infinity to the value v or E. The derivative of the distribution function is the density function. The G(E)=F(v(E)) where the capital letters are distribution functions. (Note v=v(E) i.e. since E=E(v), we can also write v=v(E) ) Using the calculus chain rule, you should be able to get g(E) from F(v) and f(v), etc. (f(v)=density function for speed v). Question 2 isn't completely clear to me. It seems to be using a standard normal distribution with mean 0 and standard deviation ## \sigma ##, but will need to take another look.
     
    Last edited: Mar 17, 2016
  6. Mar 17, 2016 #5

    andrewkirk

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    @squenshl There is a standard formula in probability theory (which can be found in most texts but also can be readily derived from first principles) for the distribution of ##Y\equiv g(X)## where ##g## is a monotonic function of a random variable ##X## that has a known distribution.

    You know the distribution of ##V##.

    Is the function ##f:v\mapsto \tfrac{1}{2}mv^2## monotonic on the domain you have specified for ##V##? If so then you can use that probability formula to find the distribution of ##E=f(V)##.
     
  7. Mar 17, 2016 #6

    Ray Vickson

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    Well, you ARE changing variables, so what do you think?
     
  8. Mar 17, 2016 #7
    Right then.
    I have ##f(v) = kv^2e^{-\beta v^2}## and we have ##E(V) = \frac{1}{2}mV^2## which is monotonically increasing for ##V >0##. Hence, we can apply the change of variable formula ##f(v(E))\left|\frac{dV}{dE}\right|##. Firstly, ##V(E) = \sqrt{\frac{2E}{m}}## and ##\frac{dV}{dE} = \frac{1}{m}\left(\frac{2E}{m}\right)^{\frac{1}{2}}##. So the new probability function is given by $$k\left(\frac{2E}{m}\right)e^{-\beta\left(\frac{2E}{m}\right)} \times \frac{1}{m}\left(\frac{2E}{m}\right)^{\frac{1}{2}} = \frac{k}{m}\left(\frac{2E}{m}\right)^{\frac{1}{2}}e^{-\frac{2\beta E}{m}}$$.

    Am I on the right track by comparing this to the Gamma distribution then just finding out the appropriate parameters to make this statement true???
     
  9. Mar 18, 2016 #8
    I'm sure I'm doing it wrong as I'm getting the same probability function
     
  10. Mar 19, 2016 #9

    Charles Link

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    It does look like you got the correct result other than you are missing a minus sign in the exponent 1/2 when you did dV/dE in post #7. Perhaps you are also missing the concept that a probability distribution function is the integral of the probability density function, thereby giving you the gamma function. The probability density function is often loosely called a distribution function, but the distribution function is more formally the integral of the density function. The probability distribution function , e.g. F(x) gives the probability that a variable X is less than or equal to x. It's derivative ## F'(x)=f(x) ## is the probability density function.
     
    Last edited: Mar 19, 2016
  11. Mar 19, 2016 #10

    Ray Vickson

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    I get exactly the same answer as you.

    I don't see your problem; you DO have a gamma distribution with easily-determined parameters!
     
  12. Mar 19, 2016 #11
    We have ##g(E) = \frac{k}{m}\left(\frac{2E}{m}\right)^{\frac{1}{2}}e^{-\frac{2\beta E}{m}} = \sqrt{2}k\left(\frac{1}{m}\right)^{\frac{3}{2}}E^{\frac{1}{2}}e^{-\frac{2\beta E}{m}}##.

    The PDF of the Gamma distribution is ##g(E) = \frac{1}{a^p\Gamma(p)}E^{p-1}e^{-\frac{E}{a}}## where ##\Gamma(p) = (p-1)!##.

    Comparing the 2 we get ##p = \frac{3}{2}## and ##a = \frac{m}{2\beta}## as our parameters but clearly this ain't the case so I'm a little stuck???
     
    Last edited: Mar 19, 2016
  13. Mar 19, 2016 #12

    Charles Link

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    Try again. Don't you get p=3/2 ?
     
  14. Mar 19, 2016 #13
    Right I stuffed it up a bit.
     
  15. Mar 19, 2016 #14
    So ##p = \frac{3}{2}## but I think my ##a## is wrong. I actually think ##p## is wrong too because the 2 distributions are just not comparing right.
     
  16. Mar 19, 2016 #15
    I get ##p = \frac{3}{2}## and ##a = \frac{m}{2\beta}##.
     
  17. Mar 22, 2016 #16
    I guess that ##p = \frac{3}{2}## and ##a = \frac{m}{2\beta}## are the correct parameters.
     
  18. Mar 22, 2016 #17

    Ray Vickson

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    Or, maybe, ##a = 2 \beta/m##, depending on which definition of the Gamma distribution you are using. (I have seen it done both ways in different sources.)
     
  19. Mar 22, 2016 #18
    Yes right thanks!!! I'm still not sure how these parameters give me a correct Gamma distribution vod plugging them into the Gamma dist doesn't give ##G(E)## for me
     
    Last edited: Mar 22, 2016
  20. Mar 23, 2016 #19

    Ray Vickson

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    Yes, it does, but you need to use the actual value of ##k##, which you were not given in the original question. How would you determine ##k##?
     
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