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For my analysis of Maxwell's derivation of the speed distribution see Velocity Distribution Derivation

For our 3D momentum case, the kinetic energy `E` can be written

`E=(p_x^2+p_y^2+p_z^2)/(2m)`

where `p_x`, `p_y` and `p_z` are the momentum components in the x, y and z directions and `m` is the mass of an atom which I have set to 1 for all atoms here.From the 3D Maxwell-Boltzmann distribution we can write for the energy distribution probability `P(E)`:

`P(EltE'ltE+deltaE)=sqrt(E/E_(avg))exp(-(3E)/(2E_(avg)))`

The peak value of `(dN)/(dE)` is obtained by taking its derivative:

`"set " r=E/E_(avg)`

`(d^2N)/(dE^2)=d/(dr)(sqrt(r)exp(-3/2r))=0`

`1/2(1/(sqrt(r)))-3/2sqrt(r)=0`

`1/(sqrt(r))-3sqrt(r)=0`

`3r=1`

`r=1/3`

`"thus "E_(peak)=1/3E_(avg)`

`(dN(p))/(dp)=(dN(E))/(dE)(dE)/(dp)=(dN(E))/(dE)(p/m)`

where `p=sqrt(p_x^2+p_y^2+p_z^2)`The signed values of `p_x`, `p_y`, and `p_z` are accumulated in bins of width `deltap=p_(max)/n_B` and these bins are plotted here on the range `-2p_(rms)->+2p_(rms)`.

Also I have plotted an algebraic curve.

First let:

`p_(avg)^2=2msum_(i=0)^(i=N)(E_i)`

Then if we let:`sigma=4/5sqrt(p_(avg)^2)`

Then the curve`f(p)=bbp/(|p|)exp(-p^2/(sigma^2))`

fits the bin data for either `p_x,p_y, or p_z` very well.To initiate the animation I choose all of the starting kinetic energies the same so you would see that the initial momenta form a circle in the `(p_x,p_y)` plane. The particles will usually collide with each other so that their energies get dispersed, some going higher and some going lower. In order to condense into the expected Maxwell-Boltzmann distribution, the lower energy particles will participate less in collisions since they are moving more slowly.

The learner has access to sliders with which to adjust the program parameters:

1. The Initial Kinetic Energy, `E_0`, per particle

2. Red Particle Radius

3. Blue Particle Radius

4. Total Number of Particles

The momentum bins data fits a gaussian of width `sigma` where for three dimensions (3D) `sigma` is

`sigma=4/5sqrt(1/(N)sum_(i=1)^(i=N)bbvecp_i*bbvecp_i`

`E_(avg)=1/(2m)1/(N)sum_(i=1)^(i=N)bbvecp_i*bbvecp_i`

`4E_(avg)`

`E_(avg)`

`0`

`sigma=4/5p_(rms)=4/5sqrt(1/(N)sum_(i=1)^(i=N)bbvecp_i*bbvecp_i`

`(dN)/(dbbvecp)=(bbvecp/|p|)exp(-(bbvecp*bbvecp)/sigma^2)`

`-sigma`

`+sigma`

`-2sigma`

`+2sigma`