Metro in Russia has free wi-fi. So travelling back from work to hotel, Sanjay posted a riddle called Thomson Problem. With time and net available. This is what wikipedia said about it :
"The physical system embodied by the Thomson problem is a special case of one of eighteen unsolved mathematics problems proposed by the mathematician Steve Smale — "Distribution of points on the 2-sphere"
Below is my solution to Thomson Problem.
I am not sure because I have not investigated, but I am told in the FB post that its one of the unsolved riddles of this century with no solution which is universal in nature. Though I am loathe to do the mathematics (could be a big deal) but here is the solution to this riddle.
Variant of Thomson Problem:
How to locate 'n' equidistant points on the surface of a sphere?
(Original problem -Find location of equally similarly charged, repelling 'n' particles?)
Physical Solution:
Though, the only other cool thing it achieves (which I never knew how to do in geometry) is that if the neighbouring centres of spheres in the 'solution state' are joined with straight lines; then the resulting planar surfaces make 'n' faced uniform solid. Of course two balls make a line while three make a plane, 4 balls make a cube, 5, 6 and on are very interesting solutions because they are more difficult for mind to perceive. I find this pretty cool for a riddle-solution.
"The physical system embodied by the Thomson problem is a special case of one of eighteen unsolved mathematics problems proposed by the mathematician Steve Smale — "Distribution of points on the 2-sphere"
Below is my solution to Thomson Problem.
I am not sure because I have not investigated, but I am told in the FB post that its one of the unsolved riddles of this century with no solution which is universal in nature. Though I am loathe to do the mathematics (could be a big deal) but here is the solution to this riddle.
Variant of Thomson Problem:
How to locate 'n' equidistant points on the surface of a sphere?
(Original problem -Find location of equally similarly charged, repelling 'n' particles?)
Physical Solution:
If 'n' equidistant points are to be located, then take 'n' spheres of same radius (identical).
Imagine 'n' threads of same length 'l' each of which connects the centre (or may be even a hook on the surface of the sphere) to a fixed point (we refer to it 'centrum') anywhere (there is no problem in the centrum being anywhere vis-a-vis balls).
Now (in thought experiment) pull all the threads of exactly same length 'l' (connecting centre of each ball to the centrum) towards the centrum at exactly same velocity 'v'. The instant when all 'n' spheres touch each other and the thread cannot be anymore pulled into the centrum (which will be exactly in the centre, while 'n' balls will surround it) is the 'solution-state'.
The imaginary sphere joining the centres of each of these 'n' spheres will be the one which will have the centres of the 'n' spheres (in 'solution state') as equidistant points on its own surface.
Mathematical solution is achieved by solving the equations of motion of each of the spheres, add length of the radius to the 'n' solutions (points of contact of neighbouring spheres) & these will be the 'n' equidistant points on the sphere.
So, if one wants to find '2' points equidistant points. You will pull two balls together, apparently they will strike each other at a point on their surface beyond which the thread cannot be pulled. Draw a sphere using the centres of these spheres - two ends of the dia.
For '3' equidistant points, pull in three identical spheres - you get three points (for ease just hold three identical balls in your hand so that each touches other) on the sphere.
For '4' pull in 4 spheres you 4 points (basically points of contact of a cube that fits in a sphere)
For '5' you get '5' & so on.
There could be other ways to solve it also.
Imagine 'n' threads of same length 'l' each of which connects the centre (or may be even a hook on the surface of the sphere) to a fixed point (we refer to it 'centrum') anywhere (there is no problem in the centrum being anywhere vis-a-vis balls).
Now (in thought experiment) pull all the threads of exactly same length 'l' (connecting centre of each ball to the centrum) towards the centrum at exactly same velocity 'v'. The instant when all 'n' spheres touch each other and the thread cannot be anymore pulled into the centrum (which will be exactly in the centre, while 'n' balls will surround it) is the 'solution-state'.
The imaginary sphere joining the centres of each of these 'n' spheres will be the one which will have the centres of the 'n' spheres (in 'solution state') as equidistant points on its own surface.
Mathematical solution is achieved by solving the equations of motion of each of the spheres, add length of the radius to the 'n' solutions (points of contact of neighbouring spheres) & these will be the 'n' equidistant points on the sphere.
So, if one wants to find '2' points equidistant points. You will pull two balls together, apparently they will strike each other at a point on their surface beyond which the thread cannot be pulled. Draw a sphere using the centres of these spheres - two ends of the dia.
For '3' equidistant points, pull in three identical spheres - you get three points (for ease just hold three identical balls in your hand so that each touches other) on the sphere.
For '4' pull in 4 spheres you 4 points (basically points of contact of a cube that fits in a sphere)
For '5' you get '5' & so on.
There could be other ways to solve it also.
Though, the only other cool thing it achieves (which I never knew how to do in geometry) is that if the neighbouring centres of spheres in the 'solution state' are joined with straight lines; then the resulting planar surfaces make 'n' faced uniform solid. Of course two balls make a line while three make a plane, 4 balls make a cube, 5, 6 and on are very interesting solutions because they are more difficult for mind to perceive. I find this pretty cool for a riddle-solution.
Regds
DL
DL
No comments:
Post a Comment