Kinematics problem on Tetrahedron

[Orodruin]

I would say the problem is that in 2D the symmetry group was simply a cyclic group, which results in ant1->ant2->ant3->...->antN->ant1 respecting the full symmetry group. Three dimensional objects tend to have more complicated non-abelian symmetry groups with several generators.

 

[Satvik Pandey]

Suppose there are three particles at the vertex of an equilateral triangle.Suppose 1[itex]^{st}[/itex] moves towards 2[itex]^{nd}[/itex] and it moves towards 3[itex]^{rd}[/itex].
In this they will meet at the centroid of triangle.
They will always on the vertex of an equilateral which rotates and get small.
What role does symmetry play here.And which symmetry you all are talking about.
This is a basic concept and I have problem in understanding it.

 

[ehild]

The velocities are along the sides of the equilateral triangle. All ants move with the same speed, so they cover the same ds=vdt distance in dt time. All sides of the triangle shortens with the same amount, so it stays an equilateral triangle.
Draw the radial and tangential components of the velocities: All radial components are vcos(30°) and all tangential components are vsin(30°). The radius of the triangle gets shorter by vcos(30)dt , and it turns by the angle dψ = (vsin(30) dt )/r .
The same happens during the second dt time: The triangle shrinks and turns... and so on.
In case of the tetrahedron, the ants moved along four edges, and they stayed equal but the other two edges behaved differently.

ehild

 

[Orodruin]

Another way of stating this would be the following: If you rotate the triangle into itself (1/3 of a full turn), then you will have exactly the same problem again (up to a renaming of the ants). This means that rotations by this amount is a symmetry of the problem and that also the solution of the problem will have the very same symmetry. This results in all of the sides keeping the same length.

In the case of the tetrahedron, not all rotations that leave the tetrahedron in the same position as it was originally (up to reordering of the edges) are symmetries of the problem. For example, you may rotate one edge that was being walked along into one that was not and still have the same tetrahedron. Thus this would be a symmetry of the tetrahedron, but not of the problem. This means that the solution to the problem will typically not respect the tetrahedral symmetry. Now, there *is* a symmetry of the full problem that is being respected, but it is only a smaller part of the tetrahedral symmetry.

 

[Satvik Pandey]

The radius of the triangle gets shorter by vcos(30)dt , and it turns by the angle dψ = (vsin(30) dt )/r .

ehild

In very small time dt angle by which rotates is dψ.(assuming that in dt r does not get shorter where r is the distance between centroid of triangle and its vertex)

rdψ=vsinθdt
dψ=vsin(30)dt/r
Is that all what you wanted to say?

 

[Orodruin]

No, r also gets smaller by an infinitesimal amount. It is the corresponding equations in three dimensions that need to be solved to find the solution to your original problem. The difference is that the shape of the original object will change in 3D.

 

[Satvik Pandey]

No, r also gets smaller by an infinitesimal amount. It is the corresponding equations in three dimensions that need to be solved to find the solution to your original problem. The difference is that the shape of the original object will change in 3D.

So how ehild found dψ = (vsin(30) dt )/r in his post#38.

 

[Orodruin]

By projecting the velocity onto the tangential direction.

 

[Satvik Pandey]

In tangential direction component of the velocity will be vsinθ.
I do not understand from which reference line does ehild has measured dψ.

 

[Orodruin]

Yes, the velocity is v sinθ. So the distance (in the tangential direction) traveled in a short time dt is going to be v sinθ dt. However, this distance can also be written r dψ as this is the distance you move if you change the polar coordinate ψ by dψ. Thus

r dψ = v sinθ dt
dψ = v sinθ dt/r

This is what you wrote earlier, but it does not imply that r is constant.

 

[Satvik Pandey]

Yes, the velocity is v sinθ. So the distance (in the tangential direction) traveled in a short time dt is going to be v sinθ dt. However, this distance can also be written r dψ as this is the distance you move if you change the polar coordinate ψ by dψ. Thus

r dψ = v sinθ dt
dψ = v sinθ dt/r

This is what you wrote earlier, but it does not imply that r is constant.

If r is not constant then how can we treat it as a constant.
In time dt r will be decreased by vcos(30)dt. Doesn't it matters?

 

[Orodruin]

Why do you think we are treating it as constant? The equation states the relation between an infinitesimal change in time and an infinitesimal change in ψ. In a similar way you can write down the corresponding equation for dr and you will end up with a set of two coupled differential equations.

 

[Satvik Pandey]

Why do you think we are treating it as constant? The equation states the relation between an infinitesimal change in time and an infinitesimal change in ψ. In a similar way you can write down the corresponding equation for dr and you will end up with a set of two coupled differential equations.

Thank You Orodruin.I didn't have knowledge about coupled differential equation that's why I had problem in understanding it but I understood it now.

 

[Satvik Pandey]

I need a help from you Orodruin.Could you please tell me which things should I know before attempting to solve that problem on tetrahadron(in post#1)?

 

[Orodruin]

Mathematically, it would be possible to understand the solution with some knowledge about differential equations and integration as well as basic vector analysis if one simply accepts that the configuration will be a (flattened) tetrahedron at any given point in time. After that it was a matter of selecting a good coordinate system to work with.

To really show that the configuration will be a flattened tetrahedron at any time, some group theory would be necessary. The alternative is to reason along more hand-waving lines.

 

[Satvik Pandey]

Thank You Orodruin.I am not very familiar with the concepts on differential equations.I will try to solve it later.
I hope that you will help me again.