Re: determining a spinor from two vectors

[John Blackburne]

Interesting, I've not seen it done like that before.

First note that a^2 = b^2, as they are vectors with the same length, and a^2 and b^2 are scalars so commute with everything

You Know RR* = R*R = 1, so
bR = (RaR*)R = Ra(R*R) = Ra

To verify that b(a+b) = (a+b)a:

b(a + b) = ba + b^2 = ba + a^2 = a^2 + ba = (a + b)a

To verify that these satisfy the expression for R, bR = Ra:

bR =
b(b(a+b)) =
b^2 (a+b) =
a^2 (a+b) =
(a+b) a^2 =
((a+b)a)a =
Ra

This is not the way I'm used to doing it though: I usually use normalised rotors, though they require a bit more calculation.

On 29 Jun 2012, at 14:18, alex wrote:

>
> i am interested in finding a Spinor relating two arbitrary vectors in a euclidean space, in terms of those vectors.
> in the book 'geometric computing with clifford algebras', the problem of determining the Rotor that relates two vectors casually solved,
>
> given:
> b = RaR*
> then R = b(a+b) = (a+b)a
>
> 'as is readily verified' . i am having trouble verifying it, could someone show how this is done?

[alex]

awesome.  i have a couple questions.

This is a verification, which is what i asked, but even with this, i cant figure out how they created that expression. do you have any insights?
is there a similar expression, but for a spinor, RR* !=1 ?

thanks!
alex

[John Blackburne]

It comes from observing that to generate a rotor to rotate through angle t you need an operator R which rotates through t/2 which is applied twice, on either side. E.g. one expression for R is

R = e^(it/2)

where i is the bivector associated with the plane of the (simple) rotation and t is the angle.

So in a sense R is a rotation through half the angle, so from a to the bisector of a and b. And (a + b) is parallel to the bisector. Therefore b(a + b) and (a + b)a are both rotations through the correct angle. They are not normalised but this is easily done if needed. If not normalised R* must be the inverse, so RR* = R*R = 1.

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[alex]

i see. thank you,
alex

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[moppi]

...I just realized that in my previous message, when we get:

(a+b)/2 = (a.c)c^-1

we can simply do the following algebraic step:

(a+b)/2 = (a.(kc))(kc)^-1

for some arbitrary real scalar k. This is essentially the property of invariance to scalar multiplication of the projection operator. And this proves that the right term is equal to c multiplied by some other arbitrary real scalar, say K:

(a+b)/2 = Kc

at this point we can arbitrarily set K=1/2 and the expected result follows.