Maxwells velocity distribution

[rabbed]

On this page: http://galileo.phys.virginia.edu/classes/252/kinetic_theory.html
From the text beginning with "Let us now figure out the distribution of particles as a function of speed.",
whats the reasoning behind multiplying a radial distribution with a volume?
Is it the same thing as integrating the PDF radially, but only for one instantaneous value instead of carrying out the full integration?
Since we have a radial distribution and a formula relating spherical volume to the radius, shouldn't it be possible to use the PDF or CDF method to derive the volume distribution or is that not what we want?

 

[Simon Bridge]

To give you a simple example: to find the mass inside an arbitrary volume V, where the density $\rho$ is not uniform, you divide the volume into lots of infinitesimal small boxes with volume dV (called a volume element). The volume element at position (x,y,z), within V, has volume dV=dxdydz and mass dm=$\rho$(x,y,z)dxdydz
The total mass is found by adding up the masses of all the small boxes in the volume ... which is what "integrate" means: $m= \int_V dm$

How would you go about using a pdf or cdf method do get to the same information?

But note: the author does not want to do a volume integral. What the author wants is the number of particles with speed between v and v+dv.
What pdf or cdf method did you have in mind for that?

 

[rabbed]

But there is a coordinate transformation going on also?
We know that the fraction of particles at point (x,y,z) is N * X_PDF(x)*dx * Y_PDF(y)*dy * Z_PDF(z)*dz
The fraction of particles at radius r is N * R_PDF(r)*dr
To go between these steps, do we need to use the volume element for spherical coordinates, r^2*sin(a)?

 

[Simon Bridge]

The number (not fraction) of particles in the box (in velocity coordinates) is given by: $dv_xdv_ydv_z$ is $$dn = NA^3e^{-Bv^2}dv_xdv_ydv_z : v^2=v_x^2+v_y^2+v_z^2$$

The fraction of particles in that box is dn/N.

What the author wants is the number dN between v and v+dv. Thus dN = Nf(v)dv so f(v)dv is the fraction of particles with speeds between v and v+dv.
There is no need to change to spherical coordinates here - the author exploits the symmetry.

It's like if you have a sphere in x-y-z coords, you know the volume between $r=\sqrt{x^2+y^2+z^2}$ and $r+dr$ is going to be $dV=4\pi r^2\; dr$ without having to change to spherical coordinates and integrating over the angles. You can do that if you like: you get the same result.

The author is just deriving the formula in terms of speed rather than velocity. call that a change in coordinates if you like.

 

[rabbed]

Hm, could you say the following:
N * R_PDF(r)*dr is NOT the number of particles inside either one or all volume elements of the spherical shell at radius r.
What is needed is the number of particles inside all volume elements of the spherical shell at radius r.
Instead, N * R_PDF(r)*dr is the number of particles inside the cubic volume element at a point (x,y,z) where r = sqrt(x^2+y^2+z^2).
That is why we need to multiply N * R_PDF(r)*dr with the number of cubic volume elements contained in the spherical
shell at radius r - 4*pi*r^2 number of elements?

 

[Simon Bridge]

Not really following your notation.
"R_PDF(r)"?

But that sounds kinda right. You can do that if the particle number density is spherically symmetric.
Take care - the author isn't talking about a literal volume.

 

[rabbed]

Thanks, I think I understand.
Sorry, with R_PDF(r) I mean the radial/speed distribution.

Can you also say that the maxwell distribution gives a constant gas pressure no matter if you randomize the average speed radially or from independent cartesian coordinates, which is the reason to have such a radial distribution and uniform direction instead of distributing average velocity as uniform points on a sphere or as uniform direction and uniform volume or any other distribution?

It kind of blurs out the dimension boundaries and difference in cartesian and spherical spaces in a way.

There are so many layers of thought to this distribution!