Normalising a velocity spectrum

[natski]

For any given velocity distribution, you have a y-axis with probability and an x-axis of velocity. Without really thinking much about it, I had assumed the normalisation condition was that the area under the graph (the integral of the function w.r.t. velocity) would be equal to 1. Of course, the area under the graph has the dimensions of velocity and so it doesn't make sense to set the area to 1.

So my question is, what IS the normalisation condition for a velocity distribution??

 

[Meir Achuz]

The y-axis is the PROBABILITY, not the velocity, so normalize to 1.

 

[natski]

This is what I said? The y-axis is probability.

But I don't think it is right to set the mean velocity to a probability of 1 because now the total probability under the graph will be greater than 1.

 

[Meir Achuz]

\int Pdx=1, where P is the probability distribution. P is on the y axis.
\int vPdx=<v>.

 

[lalbatros]

natski,

The y-axis is the probability distribution (say f), not the probability (say P).
The probability distribution with respect to the velocity v is defined such that

dp = f(v) dv​

gives the probability that the velocity is in the interval [v,v+dv].

The probability that the velocity is between v1 and v2 may be obtained by integration:

[tex]P(v1,v2) = \int_{v1}^{v2} f(v') dv'[/tex]​

This last expression is also a probability, also called cumulated probability.

The dimensions of f(v) are the inverse of the dimensions of a velocity.
The dimensions of P(v1,v2) are those of a simple number, no dimension.

In practical applications, like particle size analysis in the industry or sales statistics or reliability data ..., I prefer to use the cumulated probability. The probability density is not convenient in these practical situations. For example the area under a curve is not really a visible data while the value of the cumulated probability along the y-axis is a clear information. And of course, there is this problem with "strange" units for f . In theoretical physics of course the probability density is more convenient.

Michel

 

[natski]

Just wanted to say thanks for your help on this. Problem now solved.