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Normalization of 1D velocity boltzmann distribution

  1. May 9, 2015 #1
    Suppose the pdf is A*exp(-mv^2/2kT) , where A is the normalization constant.

    To obtain A I would integrate the pdf over the all possible values of v. The question is, should the limits be (-infinity,infinity) or [0,infinity) ? It seems that only by choosing the former can I get the correct normalization, but if in essence this is derived from an energy distribution , why doesn't it run from 0 to infinity?

    If I try to work out <v> or <v^2> of this distribution, should I use (-infinity,infinity) or [0,infinity) ? because I always use the latter when working with the standard 3D maxwell boltzmann distribution (normalization, <v>, <v^2>)
     
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  3. May 9, 2015 #2

    Simon Bridge

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    The pdf should be integrated over all possible values of v.
    It is important to understand the processes you use, not just remember to use them.
     
  4. May 10, 2015 #3
    as far as i understand it, the function is derived from a distribution of speed, which starts from 0. But if i am to use it to find the distribution of velocity, I presume it is reasonable to start from -infinity.

    What I would then be confused about is the following: In the derivation of the 3d speed distribution , the same 1d velocity normalization constant (from above) is raised to the power 3. Why is a velocity normalization constant used in a speed distribution ?
     
  5. May 10, 2015 #4

    Simon Bridge

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    Consider:
    In each linear degree of freedom, the velocity distribution function would be even - since there is an equal chance of going either way.

    What is the difference when you try the derivation from velocity instead of speed?
     
  6. May 10, 2015 #5

    Simon Bridge

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    Consider:
    In each linear degree of freedom, the velocity distribution function would be even - since there is an equal chance of going either way.

    What is the difference when you try the derivation from velocity instead of speed?
     
  7. May 12, 2015 #6
    If I start from linear speed distributions from the start, I would get a normalization constant that is twice as large as that for a velocity distribution. Thus the 3d speed distribution would have an extra factor of 8 compared to the standard one, which is derived from linear velocity distributions
     
  8. May 14, 2015 #7

    Simon Bridge

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    OK - so that basically means that the area in ##(-\infty,\infty)## for the un-normalized speed-derived function is twice that for the un-normalized velocity distribution.
    Does that makes sense considering what you are doing?

    After normalization, the main difference will be that the probability of ##0<v<\infty## is 0.5 for one and 1 for the other.
     
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