biVector forum

Finding the rotor

Vectorspace GA
#1

I have 6 vectors in 3D space, namely p,q,a,b,c,d. For example p can be p=(0,0,1). I would like to solve the following equations to find R given these 6 vectors:

R(-p)R’ = p (where R’ is R_dagger)
R(-q)R’ = q
…
R(-d)R’ = d

which can be multiplied to give 1 equation, namely:

RpqabcdR’ = pqabcd

Further:

Rpqabcd = pqabcdR

[R,pqabcd] = 0 (where the commutator is defined [A,B] = AB - BA)

I am a bit confused since this equation is solved by R = 1 (1 being the identity by setting θ = 0), but the individual equations cannot be solved by R = 1 since R(-p)R’ = p gives -p = p ie all vectors would have to be 0.

Can I use R = pqabcd?

If so how can I compute this efficiently and fast computationally?

The point of this is that if an R can be found then the set of p,q,a,b,c,d represent a non-chiral set and if R does not exist then the set is chiral. Basically, we start from the set and apply the parity operator which gives a minus sign to all the vectors in the set and try to find a single rotation to bring them all back to the original set. It is like looking at your right hand in the mirror and checking to see if it can be super-imposed on your real right hand - it cant.

Thanks,

Takis

#2

In 3D, we can write R = exp(Pi/2 j a), where j is the unit pseudoscalar and a is the rotation axis vector. All vectors orthogonal to a will be affected, but the vectors parallel to a will be invariant. Therefore, there is no such a rotor for 6 arbitrary vectors.

#3

I forgot to mention that we are allowed any arbitrary permutation between a,b,c,d and separately p with q. In general these are 4 vectors and the symmetry group is the Lorentz group that includes boosts, rotations and S2xS4 (the permutation groups).

#4

I am not following you. What do you mean by 4 vectors in 3D? Paravectors? And how your question is related to permutations and the Lorentz group?

#5

Sorry that was a typo it meant to say 4D (Minkowski). Essentially 2 anti parallel incoming momenta p and q collide to form 4 particles in the final state represented by a,b,c and d. Let the symmetry group of our system be defined as the Lorentz group with S_2 on p and q and S_4 on the final state momenta. Then an event is defined non-chiral iff after parity inverting (p,q,a,b,c,d) i can map it back to its original with actions from the symmetry group. Note in the latter 6 tuple every input is a 4 vector, so if g is an element of the symmetry group the g(p,q,a,b,c,d)=(gp,gq,ga,gb,gc,gd) and we are careful to remember that if g is solely one of the elements of one of the permutation groups then it only acts on the initial or final states depending on whether it is S_2 or S_4 accordingly. An event that cannot be mapped back to itself after parity by actions of the symmetry group is called chiral. These chiral events have a distinct parity handedness. Because their configuration is not the same as their mirror (parity) configuration modulo the symmetry group.

An example could be p = (E_p, 0, 0, p) with energy E_p and magnitude of 3 momentum p processed by the incoming particle labelled with 4 momentum p. Similarly in the center of mass frame q = (E_q, 0, 0, -p).

So the task is to generate randomly events on the computer represented by 6 4-vectors and be able to label them chiral or non-chiral.

#6

You can solve such a problem using Cl3 (3D geometric algebra). The formalism is simpler and gives a clear 3D geometric interpretation.