CITE THIS NOTEBOOK: The angular speed distribution of randomly moving-particle group by Tao Guo. Wolfram Community OCT 26 2022.
ORIGINAL ARTICLE: Ting Yang, Tao Guo, The angular speed distribution of randomly moving-particle group. AIP Advances 12, 045005 (2022). https://doi.org/10.1063/5.0082251
Article abstract: Rotation is a common phenomenon in nature, although the detailed statistical principle behind it is seldom discussed. This article aims to inspire people to think about such problems. In this article, probability theory and the corresponding modules of Mathematica software are used to study the distribution of angular speeds generated by a population of equal-speed particles moving randomly in space. The probability density expression of this distribution is derived successfully, and its accuracy is then verified using simulation data. Finally, it is confirmed that the angular speed follows a Maxwell distribution with a specified scale parameter. This research may be helpful for understanding why most celestial bodies or microparticles in the universe are in rotation.
NOTE: If there is no special case, the Mathematica code starts with gray "
In[]:=
" and is bold by default according to Mathematica's rules.

Part 1. The Deduce Process of Angular Speed Distribution

Given a number of particles (assume they uniformly distribute in spherical space) with the same speed (assume it is c) and randomly moving in 3-dimensional space, there must be a corresponding movement component to produce a rotation effect on the overall centroid at a certain moment in time.
Under the assumption that the random variables
N
1
NormalDistribution[],
N
2
NormalDistribution[] and
N
3
NormalDistribution[], are independent of each other, the probability density of one coordinate X that is equivalent to the three coordinates of R, which is the unit random vector, can be written as [G. Marsaglia, “Choosing a point from the surface of a sphere,” The Annals of Mathematical Statistics 43, 645--46 (1972).][M. E. Muller, “A note on a method for generating points uniformly on n-dimensional spheres,” Communications of the ACM 2, 19--20 (1959).]:
Clear["Global`*"];​​X=TransformedDistribution
N
1
2
N
1
+
2
N
2
+
2
N
3
,{
N
1
,
N
2
,
N
3
}ProductDistribution[{NormalDistribution[],3}];​​PDF[X,x]
Out[]=
1
2
-1≤x≤1
0
True
The expression of
Ω'
in the main text is calculated as follows.
In[]:=
X=cSin[ArcCos[Θ]]Cos[Η];​​Y=cSin[ArcCos[Θ]]Sin[Η];​​Z=0;​​FullSimplify
2
X
+
2
Y
+
2
Z
,Assumptionsc>0
Out[]=
c
1-
2
Θ
NOTE:
ω'
in the main text is substituted by
ω'
as the probability expression here.
In[]:=
ω'=TransformedDistributionc
1-
2
Θ
,ΘUniformDistribution[{-1,1}];​​FullSimplify[PDF[ω',x],Assumptionsc>0]
Out[]=
x
c
2
c
-
2
x
x>0∧c>x
0
True
NOTE:
ω
S,X
(x)
in the main text is substituted by ωsx here. Then, the probability density of X, which represents one of the three equivalent coordinates of the angular velocity
Ω
S
contributed by the random vector
V
S
, can be obtained:
In[]:=
ωsx=TransformedDistributionF1F2,F1ProbabilityDistribution
1
2
,{x,-1,1},F2ProbabilityDistribution
x
c
2
c
-
2
x
,{x,0,c};​​FullSimplify[PDF[ωsx,x],Assumptionsc>0]
Out[]=
-1
cos

x
c

2c
x0∨(x>0∧c>x)
2
-1
sin

x
c
+π
4c
c+x>0∧x<0
NOTE:
ω
B,X
(x,r) in the main text is substituted by ωbx here. Then, the probability density of the norm r of the radius r at which the starting point of the vector
V
B
is located within the ball can be obtained:
ωbx=TransformedDistribution
1
r
x,xProbabilityDistribution
ArcCos
x
c

2c
0≤x<c
2ArcSin
x
c
+π
4c
-c<x<0
,{x,-c,c},Assumptionsr>0∧c>0;​​PDF[ωbx,x]
Out[]=
r
-1
cos

rx
c

2c
c+rx>0∧c-rx>0∧rx≥0
r2
-1
sin

rx
c
+π
4c
c+rx>0∧c-rx>0∧rx<0
The distribution function of the contribution of
V
B
to the equivalent coordinate
Ω
B,X
of
Ω
B
is calculated as follows:
In[]:=
CDFProbabilityDistribution
rArcCos
rx
c

2c
0≤x<
c
r
r2ArcSin
rx
c
+π
4c
-
c
r
<x<0
,x,-
c
r
,
c
r
,Assumptionsr>0∧c>0,x
Out[]=
1
2
c
r
-x>0∧x0
1
c
r
-x<0∧x≥0
-
2
c
-
2
r
2
x
+rx
-1
cos

rx
c
+2c
2c
c
r
-x0∧x≥0∨
c
r
-x≥0∧x>0
2
2
c
-
2
r
2
x
+2rx
-1
sin

rx
c
+πrx
4c
x<0∧
c
r
-x>0∧
c
r
+x>0
We remove the meaningless parts of the above function and integrate the function in the whole unit ball (This code takes approximately 24 seconds):
In[]:=
FullSimplify
1
∫
0
4π
2
r
2
2
c
-
2
r
2
x
+2rxArcSin
rx
c
+πrx
4c
-
c
r
<x<0
-
2
c
-
2
r
2
x
+rxArcCos
rx
c
+2c
2c
0≤x<
c
r
r,Assumptionsc>0
Out[]=
-
2
π
3
c
32
3
x
c+x≤0
(128-3π)π
3
c
96
3
x
c<x
π3
2
c
x
2
c
-
2
x
-30
3
x
2
c
-
2
x
+3
4
c
3
-1
sin

x
c
+4log(c)-log
2
c
-
2
x
-x+24
4
x
-1
cos

x
c
+64c
3
x
48c
3
x
c≥x∧x≥0
π(8
4
x
-3
4
c
)
-1
sin

x
c
+x
2
c
-
2
x
(10
2
x
-
2
c
)+4
4
c
log(c)-log
2
c
-
2
x
+x+4π
4
x
16c
3
x
True
We find the first derivative of the above result with respect to x:
In[]:=
FullSimplifyD
-
2
π
3
c
32
3
x
x≤-c
(128-3π)π
3
c
96
3
x
x>c
π3
2
c
x
2
c
-
2
x
-30
3
x
2
c
-
2
x
+3
4
c
3
-1
sin
x
c
+4log(c)-log
2
c
-
2
x
-x+24
4
x
-1
cos
x
c
+64c
3
x
48c
3
x
0<x≤c
π(8
4
x
-3
4
c
)
-1
sin
x
c
+x
2
c
-
2
x
(10
2
x
-
2
c
)+4
4
c
log(c)-log
2
c
-
2
x
+x+4π
4
x
16c
3
x
-c<x<0
,x
Out[]=
3
2
π
3
c
32
4
x
c+x≤0
π(3π-128)
3
c
32
4
x
c<x
-
πx
2
c
-
2
x
(3
2
c
+2
2
x
)+3
4
c
3
-1
sin

x
c
+4log(c)-log
2
c
-
2
x
-x-8
4
x
-1
cos

x
c

16c
4
x
x>0∧c≥x
π(9
4
c
+8
4
x
)
-1
sin

x
c
+x
2
c
-
2
x
(3
2
c
+2
2
x
)-12
4
c
log(c)-log
2
c
-
2
x
+x+4π
4
x
16c
4
x
c+x>0∧x<0
We modify the above results to the form of a continuous function and integrate the function in the whole interval
(-∞,∞)
.
In[]:=
Integrate
3
2
π
3
c
32
4
x
c+x≤0
3
2
π
3
c
32
4
x
c<x
-
πx
2
c
-
2
x
(3
2
c
+2
2
x
)+3
4
c
3
-1
sin
x
c
+4log(c)-log
2
c
-
2
x
-x-8
4
x
-1
cos
x
c
16c
4
x
x>0∧c≥x
π(9
4
c
+8
4
x
)
-1
sin
x
c
+x
2
c
-
2
x
(3
2
c
+2
2
x
)-12
4
c
log(c)-log
2
c
-
2
x
+x+4π
4
x
16c
4
x
c+x>0∧x<0
,{x,-∞,∞},Assumptionsc>0
Out[]=
4π
3
The above function is normalized according to the integration results:
In[]:=
FullSimplify
1
4π
3
3
2
π
3
c
32
4
x
c+x≤0
3
2
π
3
c
32
4
x
c<x
-
πx
2
c
-
2
x
(3
2
c
+2
2
x
)+3
4
c
3
-1
sin

x
c
+4log(c)-log
2
c
-
2
x
-x-8
4
x
-1
cos

x
c

16c
4
x
x>0∧c≥x
π(9
4
c
+8
4
x
)
-1
sin

x
c
+x
2
c
-
2
x
(3
2
c
+2
2
x
)-12
4
c
log(c)-log
2
c
-
2
x
+x+4π
4
x
16c
4
x
c+x>0∧x<0

The standard deviation of ωx is:

Part 2. Figures Used in This Article

Figure 1

Figure 2

Figure 3