In physics and engineering, the distances among the particles is
usually an important factor for analysis of the subject especially when we are
studying Newton’s law of universal gravitation in mechanic, Coulomb’s law in
electricity, Maxwell’s equations in electromagnetic theory and so on. But, do you think that the application of the theorems in related to the distances among the particles is limited to above laws in classic physics? Definitely no. The most crucial thing is to predict a flexible system which is moving. For instance, suppose you have a bowl filled by water in your hand. When you move, how can you predict the motion of water system into the bowl? Of course, when we are speaking about a rigid system, the prediction and analysis of this system is easy just like when you are moving by a car (an approximately rigid system).
In the reference with article of “A Specific Permutation And Applications”, we can derive many conjectures (or theorems) or properties in
related to the distances among the particles in the space. Consequently, this
could be conducted as a big project in which I will not be able to handle this
project alone.
In this series, I will post
little by little these new properties accompanied by examples only related to Newton's law of universal gravitation in mechanic. Since I am using
the big data analysis to extract these properties, I have not any pure
mathematics proof for them. Therefore, I will mark up them as the Conjectures
instead of the Theorems. As the matter of fact, this is the Big Data Analysis with Microsoft excel to create the value from data.
Conjecture (1):
If above article (A Specific Permutation…) has been applied for
only three choices from many events (sample space), we will have the concentric
spheres or layered spheres just like below figure:
In previous article, I told you a specific permutation with n = 6
and r =3, gives us 48 points on a sphere, If we increase the sample space (n)
to 8 and r = 3, we will have 4 concentric spheres which gives us 48 points on
each sphere and for n = 10, we will have 10 concentric spheres which gives us
48 points on each sphere and for n = 12, we will have 20 concentric spheres
which gives us 48 points on each sphere and etc..
In fact, the number of the layered spheres can be calculated by
using previous equation divided to
48.
Note: I will again comeback to this conjecture as soon as possible.
Conjecture (2):
The distance of each point on a sphere is equal to at least five
pair points on the same sphere.
Suppose points P1, P2, P3, P4, P5, P6, P7, P8, P9, P10, P11 have
been located on a sphere. This conjecture says to us:
d (P1, P2) = d (P1, P3) = A
d (P1, P4) = d (P1, P5) = B
d (P1, P6) = d (P1, P7) = C
d (P1, P8) = d (P1, P9) = D
d (P1, P10) = d (P1, P11) =
E
Where:
d (P1, P2) = the distance between point P1 to P2
The amounts of A, B, C, D, E are the members of real number
Note: I will again comeback to this conjecture as soon as possible.
Conjecture (3):
The distances among the points P1(x, y, z), P2 (z, x, y) and P3(y,
z, x) are always the same.
d (P1, P2) = d (P1, P3) = d (P2, P3)
Example:
Suppose point P1 has below coordination:
P1 (3, -5, 11)
Therefore, the coordination P1 and P2 will be:
P2 (11, 3, -5)
P3 (-5, 11, 3)
d (P1, P2) = d (P1, P3) = d (P2, P3) = 19.59592
Let me tell you an example about Newton’s law of universal
gravitation as follows:
The Mapping A System of Three Particles for Given Gravity Potential
Energy And The Distance
Suppose
we have three particles P1, P2 and P3 with the distances among them in
infinity which have the masses of m1,
m2 and m3. If an external force brings all these particles in new location just
like below figure, how can we calculate total work done on them?
It is clear; the work done on particles is equal to total gravity
potential energy of system but in opposite direction in which we can write
below equation for total gravity potential energy of system:
Where:
G = Gravitational constant = 6.672E-11
(N.m2/kg2)
U (r) = total
gravity potential energy of system (J)
r = the distance among the particles (m)
Question (1):
If we have:
U (r) = -0.04143312
J
r = 0.0002 m
How can we map these particles in the space?
Step (1):
To find masses of particles, we should apply the
method mentioned in article of “Solving a Nonlinear Equation with Many Independent Variables”
I found 6 answers for mass of the particles:
m1
|
m2
|
m3
|
6
|
300
|
400
|
6
|
400
|
300
|
15
|
380
|
300
|
22
|
100
|
1000
|
40
|
130
|
700
|
40
|
330
|
300
|
Step (2):
Since the distance is equal to 0.0002 m, we can find the
coordination of particle P1 by using the model presented in article of “A Model to Track the Location of a Particle in the Space”
I found 408 answers for coordination of particle P1
as follows:
Here I have posted some of the results:
Models
|
x
|
y
|
z
|
1
|
-0.0005
|
-0.00047
|
-0.00035
|
2
|
-0.0005
|
-0.00037
|
-0.00035
|
3
|
-0.0005
|
-0.00035
|
-0.00047
|
4
|
-0.0005
|
-0.00035
|
-0.00037
|
5
|
-0.00047
|
-0.0005
|
-0.00035
|
6
|
-0.00047
|
-0.00045
|
-0.00032
|
7
|
-0.00047
|
-0.00035
|
-0.0005
|
8
|
-0.00047
|
-0.00035
|
-0.00032
|
9
|
-0.00047
|
-0.00032
|
-0.00045
|
10
|
-0.00047
|
-0.00032
|
-0.00035
|
11
|
-0.00045
|
-0.00047
|
-0.00032
|
12
|
-0.00045
|
-0.00042
|
-0.00029
|
13
|
-0.00045
|
-0.00032
|
-0.00047
|
14
|
-0.00045
|
-0.00032
|
-0.00029
|
15
|
-0.00045
|
-0.00029
|
-0.00042
|
16
|
-0.00045
|
-0.00029
|
-0.00032
|
17
|
-0.00042
|
-0.00045
|
-0.00029
|
18
|
-0.00042
|
-0.0004
|
-0.00027
|
19
|
-0.00042
|
-0.00029
|
-0.00045
|
20
|
-0.00042
|
-0.00029
|
-0.00027
|
21
|
-0.00042
|
-0.00027
|
-0.0004
|
22
|
-0.00042
|
-0.00027
|
-0.00029
|
23
|
-0.0004
|
-0.00042
|
-0.00027
|
24
|
-0.0004
|
-0.00037
|
-0.00024
|
25
|
-0.0004
|
-0.00027
|
-0.00042
|
26
|
-0.0004
|
-0.00027
|
-0.00024
|
27
|
-0.0004
|
-0.00024
|
-0.00037
|
28
|
-0.0004
|
-0.00024
|
-0.00027
|
29
|
-0.00037
|
-0.0005
|
-0.00035
|
30
|
-0.00037
|
-0.0004
|
-0.00024
|
31
|
-0.00037
|
-0.00035
|
-0.0005
|
32
|
-0.00037
|
-0.00035
|
-0.00022
|
33
|
-0.00037
|
-0.00024
|
-0.0004
|
34
|
-0.00037
|
-0.00024
|
-0.00022
|
35
|
-0.00037
|
-0.00022
|
-0.00035
|
36
|
-0.00037
|
-0.00022
|
-0.00024
|
37
|
-0.00035
|
-0.0005
|
-0.00047
|
38
|
-0.00035
|
-0.0005
|
-0.00037
|
39
|
-0.00035
|
-0.00047
|
-0.0005
|
40
|
-0.00035
|
-0.00047
|
-0.00032
|
41
|
-0.00035
|
-0.00037
|
-0.0005
|
42
|
-0.00035
|
-0.00037
|
-0.00022
|
43
|
-0.00035
|
-0.00032
|
-0.00047
|
44
|
-0.00035
|
-0.00032
|
-0.00019
|
45
|
-0.00035
|
-0.00022
|
-0.00037
|
46
|
-0.00035
|
-0.00022
|
-0.00019
|
47
|
-0.00035
|
-0.00019
|
-0.00032
|
48
|
-0.00035
|
-0.00019
|
-0.00022
|
49
|
-0.00032
|
-0.00047
|
-0.00045
|
50
|
-0.00032
|
-0.00047
|
-0.00035
|
51
|
-0.00032
|
-0.00045
|
-0.00047
|
52
|
-0.00032
|
-0.00045
|
-0.00029
|
53
|
-0.00032
|
-0.00035
|
-0.00047
|
54
|
-0.00032
|
-0.00035
|
-0.00019
|
55
|
-0.00032
|
-0.00029
|
-0.00045
|
56
|
-0.00032
|
-0.00029
|
-0.00017
|
57
|
-0.00032
|
-0.00019
|
-0.00035
|
58
|
-0.00032
|
-0.00019
|
-0.00017
|
Please
be informed that each answer gives us a model of three particles P1, P2 and P3.
For instance, look at Model (1) in above table. By applying the conjecture (3),
you can find below system of articles:
x
|
y
|
z
|
Distance
|
|
P1
|
-0.0005
|
-0.00047
|
-0.00035
|
0.00020
|
P2
|
-0.00035
|
-0.0005
|
-0.00047
|
0.00020
|
P3
|
-0.00047
|
-0.00035
|
-0.0005
|
0.00020
|
As you can see, the distance among P1, P2 and P3 is equal to 0.0002
m.
Now, please comeback to step (1), we have 6 answers for the masses
and for each answer of masses, we have 408 models. It means that we found 2448
systems of three particles which have the same total gravity potential energy.
Question (2):
If we have:
U (r) = -0.01748064
J
r = 0.0002 m
How can we map these particles in the space?
It is clear, the number of models is 408 but the answers of masses
are as follows:
m1
|
m2
|
m3
|
4
|
100
|
500
|
10
|
240
|
200
|
20
|
20
|
1300
|
20
|
220
|
200
|
30
|
380
|
100
|
31
|
200
|
200
|
40
|
60
|
500
|
It means that we found 2856 systems of three particles which have
the same total gravity potential energy.
Question (3):
If we have:
U (r) = -2.27021E-06 J
r = 1.54 m
How can we map these particles in the space?
The answers for masses are just like to question (2):
m1
|
m2
|
m3
|
4
|
100
|
500
|
10
|
240
|
200
|
20
|
20
|
1300
|
20
|
220
|
200
|
30
|
380
|
100
|
31
|
200
|
200
|
40
|
60
|
500
|
But the answers for step (2) are as follows:
Models
|
x
|
y
|
z
|
1
|
-0.31
|
0.09
|
0.923333
|
2
|
-0.31
|
0.523333
|
0.923333
|
3
|
-0.31
|
0.923333
|
0.09
|
4
|
-0.31
|
0.923333
|
0.523333
|
5
|
-0.27667
|
0.123333
|
0.956667
|
6
|
-0.27667
|
0.556667
|
0.956667
|
7
|
-0.27667
|
0.956667
|
0.123333
|
8
|
-0.27667
|
0.956667
|
0.556667
|
9
|
-0.24333
|
0.156667
|
0.99
|
10
|
-0.24333
|
0.59
|
0.99
|
11
|
-0.24333
|
0.99
|
0.156667
|
12
|
-0.24333
|
0.99
|
0.59
|
13
|
0.09
|
-0.31
|
0.923333
|
14
|
0.09
|
0.923333
|
-0.31
|
15
|
0.123333
|
-0.27667
|
0.956667
|
16
|
0.123333
|
0.956667
|
-0.27667
|
17
|
0.156667
|
-0.24333
|
0.99
|
18
|
0.156667
|
0.99
|
-0.24333
|
19
|
0.523333
|
-0.31
|
0.923333
|
20
|
0.523333
|
0.923333
|
-0.31
|
21
|
0.556667
|
-0.27667
|
0.956667
|
22
|
0.556667
|
0.956667
|
-0.27667
|
23
|
0.59
|
-0.24333
|
0.99
|
24
|
0.59
|
0.99
|
-0.24333
|
25
|
0.923333
|
-0.31
|
0.09
|
26
|
0.923333
|
-0.31
|
0.523333
|
27
|
0.923333
|
0.09
|
-0.31
|
28
|
0.923333
|
0.523333
|
-0.31
|
29
|
0.956667
|
-0.27667
|
0.123333
|
30
|
0.956667
|
-0.27667
|
0.556667
|
31
|
0.956667
|
0.123333
|
-0.27667
|
32
|
0.956667
|
0.556667
|
-0.27667
|
33
|
0.99
|
-0.24333
|
0.156667
|
34
|
0.99
|
-0.24333
|
0.59
|
35
|
0.99
|
0.156667
|
-0.24333
|
36
|
0.99
|
0.59
|
-0.24333
|
In fact, we can find 252 systems of three particles which have the
same total gravity potential energy.
As you can see, when the distances increase, the work done on
systems will decrease.
