, with the vector for
the point in question, r.The surface of a hypersphere in 5 dimensions can be described by the equation:
| x2 + y2 + z2 + u2 + w2 | = | R2 | (1) |
where x,y,z,u,w are the 5 spatial coordinates, and the origin of the coordinate system lies at the centre of the hypersphere.
Suppose this hypersphere is rotating as a rigid body. In general (in any
number of dimensions) the velocity v of any point of a rotating body is given
by the product of the body's angular velocity matrix, , with the vector for
the point in question, r.
| v | = | r |
(2) |
The angular velocity matrix must be antisymmetric: ij = –ji. To see this, note that is the derivative with respect to time
of the linear transformation that takes any point from its original position at
t = 0 to its rotated position; let's call the matrix for this transformation
M(t). Since the motion is rigid, if we take two basis vectors,
ei and ej, and rotate them with M(t), the dot product of the resulting
vectors (which measures the angle between them) must remain constant, and so the
rate of change of this with respect to time must be zero.
| d/dt [(M(t) ei) dot (M(t) ej)] | = | 0 | (3a) |
| d/dt [(M(t)kiek) dot (M(t)rjer)] | = | 0 | (3b) |
| d/dt [M(t)kiM(t)kj] | = | 0 | (3c) |
| M(t)ki(t)kj + (t)kiM(t)kj |
= | 0 | (3d) |
| (0)ij + (0)ji |
= | 0 | (3e) |
| ij |
= | –ji |
(3f) |
Here we've used the Einstein summation convention of summing over all values of
repeated indices, such as k and r. To get from (3d) to (3e), note that M(0) is
just the identity matrix. In writing (3f) we've dropped any dependence on time
from the angular velocity matrix, since we're assuming that the body isn't
subject to external forces, leaving constant.
In 5 dimensions, a completely general 5x5 antisymmetric matrix will have 10 independent parameters, but it's always possible to choose a
basis that reduces it to the “canonical form”:
|
= |
|
(4) |
Here, the x and y coordinates have been chosen to lie in one plane of rotation,
the z and u coordinates in the other, and the w coordinate lies perpendicular to
both planes. To see why it's always possible to choose a basis that puts the
matrix in this form, first note that the determinant of any antisymmetric NxN
matrix where N is odd must be zero. This is because det = det T, and the determinant is a sum of products of N terms which all
change sign in the transpose for an antisymmetric matrix, yielding det = –(det ) for N odd. This means that there must be at least one non-zero
vector in the null space of . Choosing this as the direction for the w
coordinate, the axis of rotation, makes the last row and the last column of all zeros, and reduces the problem to 4 dimensions. The 4-dimensional case is
described in detail below.
Multiplying the vector for a general point, r = (x,y,z,u,w), by this
canonical matrix for yields a velocity vector of:
| v | = | ( 1y, –1x, 2u, –2z, 0) |
(5) |
So, for any point with x = y = z = u = 0, the velocity is zero. The set of points that meet this condition is just the w axis, the body's axis of rotation. This intersects the surface of the hypersphere at w = ±R, giving two poles.
In the physically possible (but cosmologically unlikely) case that 2 = 0, merely setting x = y = 0 is enough to make the velocity zero, and
since the other three coordinates are free to take any values, they trace out a
3-dimensional volume. This volume intersects the hypersphere to form a single
connected “pole”, the 2-sphere
z2 + u2 + w2 = R2.
The two equatorial circles are {z = u = w = 0;
x2 + y2 = R2}, where the speed at which the surface is moving is 1R, and {x = y = w = 0;
z2 + u2 = R2}, where the speed is 2R.
In 4 dimensions, a completely general angular velocity matrix is described by 6 parameters:
| A | = |
|
(6) |
It's always possible to re-orient the coordinate system in such a way that A is
converted to the canonical form. One method of doing this is to find the
eigenvectors of AA, the matrix product of A with itself. This is a real
symmetric matrix, and hence it must have 4 orthogonal eigenvectors; it turns out
that they come in two pairs, with eigenvalues –12 and –22. This makes sense geometrically: applying A to any vector
that lies in one of the planes of rotation simply rotates that vector by 90° and
multiples it by the appropriate , so applying A twice reverses the vector's
direction and multiplies it by 2. So each of these pairs of eigenvectors
spans one of the planes of rotation.
Another way to find the planes of rotation involves a linear operator called the Hodge dual; the Hodge dual of a matrix M is usually written as *M. In general, this operator takes an algebraic description of a geometrical object, such as a plane, and produces the corresponding description of the perpendicular object; in the context of 4-dimensional Euclidean geometry, it maps planes to other planes. For example, if the 4 coordinates we're using are called x, y, z and u, the Hodge dual of the x-y plane is the z-u plane, and similarly the dual of the plane spanned by any two coordinates is the plane spanned by the other two. With the small added complication that you need to stick to a consistent orientation scheme (to decide between the two possible directions of rotation within each plane) this is enough for us to write the Hodge dual of the matrix A. We just treat A as a sum of rotations in all 6 coordinate planes, and take the duals of each of them; this amounts to swapping the x-y coordinate of A with the z-u coordinate, etc., and changing a few signs to keep the orientation consistent.
| *A | = |
|
(7) |
Now, what we want to do is write A as a sum of rotations in just two planes:
one plane whose matrix we'll call S, and the plane perpendicular to it, whose
matrix will be *S. In other words, we want to find S, 1, and 2 such that:
| A | = | 1S + 2*S |
(8) |
Taking the dual of this, and noting that applying the dual twice to anything just gives you back the original matrix, yields:
| *A | = | 1* S + 2S |
(9) |
Multiplying Eqn (8) by 1 and Eqn (9) by 2 and taking the difference
allows us to find S in terms of A, 1 and 2:
| S | = | (1A – 2*A) / (12 – 22) |
(10) |
To find the values of 1 and 2, note that if S is applied to a vector
perpendicular to its plane, the result must be zero, and this is only possible
if the determinant of S is zero. Writing out S in full, and then computing its
determinant, gives:
| S | = |
|
(11) |
| det S | = | [2/(12 – 22)]4 [D(1/2)2 + |A|2(1/2) + D]2 |
(12) |
where
|A|2 = (a2+b2+c2+d2+e2+f2), the square of the magnitude of A, and
D = (be–cd–af) is the square root of the determinant of A. Eqn (12) can be
solved to find the value of 1/2 that makes det S equal to zero; call this
value r.
| r | = | [± (|A|4 – 4D2) – |A|2] / 2D |
(13a) |
| 1+r2 | = | |A|2 [|A|2 – (|A|4 – 4D2)] / 2D2 |
(13b) |
Then if S is normalised by requiring that
|S|2 = |*S|2 = 1, 1 and 2 can be found individually. There's a nice
Pythagorean relationship between the squares of the magnitudes of the matrices
involved, once A is split into dual parts, which makes it easy to find the
individual rates of rotation.
| |A|2 | = | 12|S|2 + 22|*S|2 |
|
| = | 12 + 22 |
(14a) | |
| 12 |
= | |A|2 r2 / (1+r2) | |
| = | [|A|2 – (|A|4 – 4D2)] / 2 |
(14b) | |
| 22 |
= | |A|2 / (1+r2) | |
| = | 2D2 / [|A|2 – (|A|4 – 4D2)] |
||
| = | [|A|2 + (|A|4 – 4D2)] / 2 |
(14c) |
Choosing new coordinates that put A into canonical form then requires picking one pair of orthogonal vectors spanning the plane defined by S, and then another pair spanning the plane defined by *S. There are standard linear algebra techniques for doing this, starting with a pair of row or column vectors from each matrix.
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