I got a question (in a Question paper) as follows:-
A three-sphere is like a two-sphere. It consists of all points equidistant from a fixed point (the origin) in four dimensional space. Consider a particle free to move on a three sphere. How many conserved quantities does this system possess?
The answer say's 6 conserved quantities are there, but how is it possible? Can anyone kindly explain.
The dynamics of a particle moving on any-fixed-manifold is given by the Lagrangian
L = (1/2)gIJ(x)(dxI/dt)(dxJ/dt)
where xI(t) are the coordinates in the ambient space and gIJ(x) is the metric of the manifold, in the present case, a 3-sphere in 4-dimensional Euclidian space.
Spherical symmetry means that angular momentum is conserved-all of its components.
The components of angular momentum are those of the tensor MIJ=xIpJ-xJpI, where pI=gIJ(x)(dxJ/dt) and pI=gIJ(x)pJ. Here gIJ(x) is the inverse metric:
gIJgjK=δI K.
(Therefore pI=dxI/dt; whereas pJ≠dxJ/dt, except for a flat manifold.)
Said in an equivalent way: If gIJ(x) is the metric of a sphere, it's a standard exercise that these quantities are all conserved for the xI(t) that solve the equations of motion, i.e. the Euler-Lagrange equations.
In d dimensions (of ambient space) there are d(d-1)/2 non-zero components of MIJ; for d=4 this gives (4x3)/2 = 6.
Stam Nicolis Thank you very much Sir for helping me understanding this problem.
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