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Re: Quaternion problem-> visualization
*To*: mathgroup at smc.vnet.net
*Subject*: [mg67471] Re: Quaternion problem-> visualization
*From*: Roger Bagula <rlbagula at sbcglobal.net>
*Date*: Tue, 27 Jun 2006 03:14:42 -0400 (EDT)
*References*: <e7ap5q$952$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
Staffan Langin wrote:
>Hello,
>
>I'm a total Mathematica newbee so please bear with me :-). I'm trying to let
>Mathematica analytically evaluate a second order derivative for me. The
>function has several parameters with some of them being of
>"Quaternion-type". I want to instruct Mathematica that the commutative rule
>doesn't apply when evaluation the function. Is that possible? Thanks in
>advance.
>
>Staffan Langin
>
>
>
Apparently ther is a reason a Clifford Torus
has the name Clifford attached!
If you take the two complex plane circles:
z1=Cos[t]+I*Sin[t]
z2=Cos[p]+I*Sin[p]
and form :
q=z1+J*z2
You get:
Quaternion[Cos[t],Sin[t],Cos[p],Sin[p]]
The visualization in 3d is:
Clear[x1, y1, z1]
w = Sqrt[2] - Sin[p]
x1 = 2*Cos[t]/w
y1 = 2*Sin[t]/w
z1 = 2*Cos[p]/w
g1 = ParametricPlot3D[{x1, y1, z1}, {p, 0.001, 2*Pi - 0.001}, {t,
0, 2*Pi}, PlotPoints -> {40, 40}, Axes -> False, Boxed -> False,
PlotRange -> All]
Show[g1, ViewPoint -> {-3.370, -0.295, -0.076}]
Show[g1, ViewPoint -> {0.002, 0.045, 3.383}]
Show[g1, ViewPoint -> {0.209, 3.361, 0.329}]
The really good thing is that the double derivative in two variables
comes out as the negative
of the original quaternion ( it's a special case):
<< Algebra`Quaternions`
i = {{0, 1}, {-1, 0}};
j = {{0, I}, {I, 0}};
k = {{I, 0}, {0, -I}};
e = IdentityMatrix[2];
q[t_, x_, y_, z_] := e*t + x*i + j*y + k*z;
(* Clifford torus quaternion*)
qc = q[Cos[t], Sin[t], Cos[p], Sin[p]]
qcs = q[Cos[t], -Sin[t], -Cos[p], -Sin[p]]
FullSimplify[ExpandAll[qc.qcs]]
dqt = D[qc, {p, 2}]
dqp = D[qc, {t, 2}]
q2_out = Quaternion[Re[dqt[[1, 1]]], Re[dqt[[1, 2]]], Re[dqt[[1,
2]]], Im[dqt[[1, 1]]]] + Quaternion[Re[dqp[[1, 1]]], Re[dqp[[
1, 2]]], Re[dqp[[1, 2]]], Im[dqp[[1, 1]]]]
(* if the angles are all real*)
-dqt - dqp == qc
True
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