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In this animation, we will examine the energy distributions of rigid arrays that interact with each other via potentials between the their elements and elements of other arrays. As such, these arrays will be subject to both repelling forces on their centers of mass (CM) as well as torques that tend to rotate the array about its CM. Therefore, for the 2 dimension case that we sill study here, we will have 3 degrees of freedom. These are position in the x and y direction as well as rotation about the CM. We should therefore expect that the energy distributions would be

`D(E)=sqrt(E)exp(-E/E_c)`

where `E_c` is some constant that is related to the average energy, E_(avg), and determined by the value of the ratio of the following integrals`E_(avg)=(int_0^ooED(E)dE)/(int_0^ooD(E)dE)`

It turns out the above ratio is `E_(avg)=(3/2)E_c` which means that `E_c=(2/3)E_(avg)` and then `D(E)` becomes

`D_p(E)=Csqrt(E)exp[-(3E)/(2E_(avg))]`

where `C` can be chosen so that D(E) represents either a probability (and integrates to 1.0) or, more conveniently for plotting, `C` can cause the peak value of `D(E)` to equal 1. To make `D(E)` equal 1 at its peak value, we differentiate the original expression with respect to `E` and find that the peak value of ` D(E)` is at `E_p=E_c/2`. Then the value of D(E) at its peak is`D_(max)(E_p)=sqrt(E_c/2)exp(-2E_c/E_c)`

Then we choose`C=exp(1/2)sqrt[2/E_c]`

wich causes`D_p(E_c/2)=exp(1/2)sqrt[E_c/(2E_c)]sqrt[2E_c/E_c]exp[-E_c/2E_c]=1`

as desired. So`D_p(E)=exp(1/2)sqrt(2/Ec)sqrt(E)exp(-2E/Ec)=exp(1/2)sqrt((3E)/E_(avg))exp(-3E/(2E_(avg)))`

The plots show the distributions of the total kinetic energy (black), the translational kinetic energy (green), the rotational kinetic energy (blue), and the expression above (black) which is a smooth curve whereas the others are stepped since they are histograms. Because it takes a long time to compute any single plot, I have chosen to present some Final Distribution Pictures which the learner is free to click on and view. In this case, I have had great difficulty with the usual `1/r^2` (where r is the distance between the elements) forces between elements. I have instead used the expression `F=q*qexp(-r/r_C)` where r_C is a constant chosen by the learner and q represents the charge on eah element. Over limited range of the variable parameters this force value is stable but cannot be expected to be stable for all parameters.
**If an unstable set of parameters is chosen, in the upper left corner of the canvas a message 'Maximum Energy Exceeded...'
will appear and you should change at less aggressiver parameter and press Start again.
**