# Normalization of 1D velocity boltzmann distribution

1. May 9, 2015

### throneoo

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>)

2. May 9, 2015

### Simon Bridge

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. May 10, 2015

### throneoo

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. May 10, 2015

### Simon Bridge

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. May 10, 2015

### Simon Bridge

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 12, 2015

### throneoo

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. May 14, 2015

### Simon Bridge

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.