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Next: Episode I Up: Lorentz Contraction Previous: Vector exponential

Visualizing products

Addition of quaternions is not so much different from vector addition, but it would be nice to have a lore of logarithms by which products could be represented by sums. Due to the noncommutativity of the matrices involved, it could hardly be expected that simple addition would suffice. Nevertheless the noncommutativity is fairly mild with a trigonometric law of composition. For rotations, the result is a calculus of curved vectors on a spherical surface, whose analogue is what interests us here.

Scalar exponents give purely commutative results, so principal interest lies in the behavior of vector exponents, which results in a restriction to unimodular matrices. The product of two unimodular matrices is unimodular, so the system is closed; but a unimodular matrix is not a vector. Is there some way to have both? The answer lies in the observation that all vectors anticommute, for the matrices of SL(2,R) as well as for Hamilton's quaternions. Therefore vor vectors ${\bf u}$ and ${\bf v}$,

$\displaystyle e^{\phi {\bf u}} {\bf v}$ = $\displaystyle {\bf v}e^{-\phi {\bf u}},$ (65)

and by a little cleverness,
$\displaystyle e^{\phi {\bf u}} {\bf v}$ = $\displaystyle e^{\frac{\phi}{2}{\bf u}} {\bf v}e^{-\frac{\phi}{2} {\bf u}}.$ (66)

Since similarity transformation preserves traces, we conclude that not only is $\vert{\bf v}\vert$ unaltered by a unimodular multiplier, but also ${\rm Trace}({\bf v})$. The difference between the trace and the determinant is the squared vector norm, leaving the useful result that vectors lying on a surface of given norm transform into vectors of the same norm as a consequence of SL(2,R) multiplication. Therefore, there is a replacement for the unit sphere whose geometry was used in the study of rotations.

We need such a surface, that planes of all orientations will intersect it, letting exponential-representing arcs lie in actual intersecting planes, and so be joinable tip to tail. The hyperboloid of one sheet, the locus of vectors of squared norm -1, satisfies that requirement.


next up previous contents
Next: Episode I Up: Lorentz Contraction Previous: Vector exponential
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2000-03-17