[Maxima-discuss] More on quaternion simplification

[Henry B. <hb...@pi...>]

Quaternion simplification is essentially
the expanding of quaternion expressions by
multiplying and distributing everything out. 
This converts quaternion products and vector
cross products into linear combinations of
"atomic" vectors such as A, B, AxB, etc.,
where the coefficients are scalar expressions
involving scalar parts of quaternions, norms
of quaternions, etc.

For example, if A,B,C are vectors (scalar
part =0), the quaternion product of all 3 is

(%i1) A.B.C;
(%o1)     - dot(A, B) C - A dot(B, C) + dot(A, C) B - triple(A, B, C)
(%i2)

The pattern changes for A.B.C.D, which uses a
linear combination of A,AxB,AxC,AxD:

(%i16) ABCD:A.B.C.D$
(%i17) S(ABCD);
(%o17)  dot(A, B) dot(C, D) - dot(A, C) dot(B, D) + dot(A, D) dot(B, C)
(%i18) ratcoef(ABCD,A);
(%o18)                         - triple(B, C, D)
(%i19) ratcoef(ABCD,cross(A,B));
(%o19)                            - dot(C, D)
(%i20) ratcoef(ABCD,cross(A,C));
(%o20)                             dot(B, D)
(%i21) ratcoef(ABCD,cross(A,D));
(%o21)                            - dot(B, C)
(%i22) 

The final step in quaternion simplification is
choosing an orthogonal basis for representing
the expression.  Thus, if a quaternion expression
looks like s+a*A+b*B+c*C+d*D..., where A,B,C,D,...
are vectors and s,a,b,c,d are scalar expressions,
we know that we can represent this same expression
as s+a'*A'+b'*B'+c'*C', where A',B',C' are exactly
3 mutually orthogonal vectors, and a',b',c' are
scalar expressions, one or more of which might be
zero.  We do NOT require that A',B',C' be unit
vectors, so |A'| is not necessarily 1; ditto for
|B'| and |C'|.

One obvious choice of a basis would be the unit
quaternions I,J,K:

Q = S(Q)+(Q.I)I+(Q.J)J+(Q.K)K

However, in most calculations, I,J,K will not
be the most convenient or perspicuous basis; it
will be better to choose a basis that matches
one or more of the vectors found in the expression
itself.

Given vectors A,B, we can quickly construct an
orthogonal (but not orthonormal) basis using the
basis vectors A, AxBxA, AxB.  (We note that the
expression AxBxA is not ambiguous, because although
the vector cross product "x" is not in general
associative, Ax(BxA)=(AxB)xA=(A.A)B-(A.B)A)

Given 3 mutually orthogonal vectors U,V,W, we
can now represent quaternion Q as

Q = S(Q)+((Q.U)/(U.U))U+((Q.V)/(V.V))V+((Q.W)/(W.W))W

Given any 2 vectors A,B such that AxB/=0, we
can represent any Q in terms of A,AxBxA, AxB.

Our routine express2(A.B.C,A,B) will express
the product A.B.C in terms of the vectors A,
B, AxB; it is a bit messy, so we pull it apart
for you:

(%i1) ABC:express2(A.B.C, A, B),expand$
(%i2) S(ABC);
(%o2)                          - triple(A, B, C)
(%i3) VABC:V(ABC),factor$
(%i4) denom(VABC);
                                           2
(%o4)                       N(A) N(B) - dot (A, B)
(%i5) NVABC:VABC*%,factor$
(%i6) ratcoef(NVABC,A),expand;
                                   2
(%o6) - N(A) N(B) dot(B, C) + 2 dot (A, B) dot(B, C) - dot(A, B) dot(A, C) N(B)
(%i7) ratcoef(NVABC,B),expand;
(%o7)           N(A) dot(A, C) N(B) - N(A) dot(A, B) dot(B, C)
(%i8) ratcoef(NVABC,cross(A,B)),expand;
(%o8)                     - dot(A, B) triple(A, B, C)
(%i9)