Normalization of 1D velocity boltzmann distribution

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  • #1
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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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  • #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.
 
  • #3
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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.
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 ?
 
  • #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?
 
  • #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?
 
  • #6
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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?
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
 
  • #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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