### Quaternion raised to the power of a Quaternion

greenspun.com : LUSENET : quaternions : One Thread

Hi ; to all makers and contributors of this Forum; (I present my thanks to the ones, who read and try to answer my question.)

Question;

if ; x=Quaternion[a,b,c,d] , We know that we can express Exp[x] or (e^x) by an advanced math program, such as Mathematica (3.0 or over).

All I want to know is that ; can we express (or is it defined mathematically) ----> x^y ? (where x , y are quaternions)? That is;(Quaternion^Quaternion) OR (Quaternion raised to the power of a Quaternion). I couln't be able to find the answer in any place (including Mathematica 3.0)

On the other hand we know that (complex^complex) kind of expressions are possible and can be evaluated easily by any of the advanced math programs (including mathematica 2.1).

I have made the following simple program in mathematica 3.0

-------------------------------------------------------------------<< "Algebra`Quaternions`" x := Quaternion[4, 6, -12, -4] y := Conjugate[x] q1 := x^y q2 := y^x q3 := x^x q4 := y^y Simplify[Expand[(q1 + q2)/(q3 + q4)]] -------------------------------------------------------------------

Do you know what it gives at the end ? answer : 1 (one) !!!!! (Unbelieveable)

no matter what x=Quaternion[a,b,c,d] takes for (a,b,c,d) (a,b,c,d any). It gives a result of UNITY

You may try it yourselves by just copying and pasting on any (mathematica 3.0 or 4.0). I don't believe the result of this small program and I think this result is a blunder made by me. After some people try it and find the same unbelieveable result, please tell me where I have made the mistake.

With best wishes Kent

-- Kent Roler (kentroler@hotmail.com), January 20, 2003

Hello Kent:

Here is what I do to calculate q^n, n being an integer. I calculate Log(q), multiply that by n, then take the exponent:

Exp[Log[q] . 3] would give the cube

If two quaternions, q and p, happened to be pointing in exactly the same direction so that their crossproduct was zero, the same technique could be applied:

Exp[Log[q] . p]

What if they were not in the same direction? Calculate how much of Log[q] is in the direction of p by taking the even product (Log[q[.p + p.Log[q])/2. You also need to calculate how much is orthogonal: (Log[q].p - p.Log[q])/2. Take Exp of each and add them together. If Log[q] is almost parallel to p, then the result is almost like the previous one.

That is what I would do if I ran the zoo,

doug

-- Doug Sweetser (sweetser@alum.mit.edu), February 18, 2003.

Part of the difficulty in defining a quaternion power of a quaternion is that addition may not be commutative when it occurs in a quaternion power. Consider a simple counter-example:

exp(s + ix + jy + kz) =exp(s) x exp(ix) x exp(jy) x exp(kz) =exp(s) x [cos(x) + i*sin(x)] x [cos(y) + j*sin(y)] x [cos(z) + k*sin(z)]

But on the other hand:

exp(s + jy + ix + kz) =exp(s) x [cos(y) + j*sin(y)] x [cos(x) + i*sin(x)] x [cos(z) + k*sin(z)]

These expressions are of like form. Assume we can equate like parts. Then we would get the following multiplicative commutation:

[cos(x) + i*sin(x)] x [cos(y) + j*sin(y)] = [cos(y) + j*sin(y)] x [cos(x) + i*sin(x)] , which expands to:

cos(x)cos(y) + i*sin(x)cos(y) + cos(x)*j*sin(y) + ij*sin(x)sin(y) = cos(y)cos(x) + j*sin(y)cos(x) + cos(y)*i*sin(x) + ji*sin(y)sin(x)

Commuting products with real factors and inserting the Hamilton postulates ij=k and ji= -k gives:

cos(x)cos(y) + i*sin(x)cos(y) + j*cos(x)cos(y) + k*sin(x)sin(y) = cos(x)cos(y) + i*sin(x)cos(y) + j*cos(x)sin(y) - k*sin(x)sin(y) .

Cancelling like terms on both sides gives:

k*sin(x)sin(y) = -k*sin(x)sin(y) .

Thus k = -k . This contradicts Hamilton's postulates.

So we can conclude that additive terms in a quaternionic power do not exhibit commutativity unconditionally. (Cyclic permutations however may hold.)

Best regards,

Matthew

-- Matthew McCann PE (mmccann@franciscan.edu), March 13, 2004.

Hello Matthew:

I will have to disagree semantically that there is any "difficulty in defining a quaternion power of a quaternion." We have one definition, it works all the time. The issue you raise is a nice feature that applies to multiplying two exponentials that happen to commute. That convenience of adding exponents does not apply to quaternions directly.

If you can figure out those parts of two quaternions that are codirectional, then that codirectional part will have the nice addition property. There will also be an anti-commuting contribution that points in a different direction (always with zero scalar because it is the cross product term). One needs to weight the two contributions: codirectional plus cross product.

doug

-- Douglas Sweetser (sweetser@alum.mit.edu), March 13, 2004.

Hi, Doug!

Thanks, I see what you mean. I looked up Hamilton's discussion in his Lectures On Quaternions, and he agrees with your remark about the nice law of exponents holding only for commuting terms. Incidentally, Hamilton defines exp(q) by the Taylor series,

1 + q + q^2/2! + q^3/3! + q^4/4! + ... = exp(q)

Best regards,

Matthew

-- Matthew McCann PE (mmccann@franciscan.edu), March 14, 2004.

Hello Matthew:

Using Hamilton's Lectures, now that is retro! I have used the power series to test simple trig functions. The world of math must be consistent, and it has always worked. It works because all quaternion commute with themselves.

This might have a deep consequence for group theory and the standard model (here's one coming in from left field!). The continuous group SU(2) can be represented by the unit quaternions, which are quaternions of the form: (0, a, b, c). The a, b, and c can take on any darn value at all, BIG or ultra small - except all zero - and the norm of that number is 1. The lie algebra su(2) clearly has the right number of degrees of freedom, namely three. A quaternion that would always commute with this would have the form: (d, a, b, c)/norm(d, a, b, c). This number has a norm of 1 also because I normalized it :-) This number could represent the continuous group U(1) because its lie algebra u(1) should only have one degree of freedom. People usually use complex numbers with a norm of 1 which commute with other complex numbers. A representation of a quaternion with its four degrees of freedom could be: (q/norm q) (q - q*), U(1)xSU(2), which is the symmetry of the electroweak force.

Ain't that wild?

doug

-- Douglas Sweetser (sweetser@alum.mit.edu), March 16, 2004.

Hello to everybody, we found your forum few days ago. If it is useful for the discussion, note that we use, and derived, the following expressions (left and right) for the quaternion power of a quaternion q^p (q and p unitary quaternions)

q^p Left = q^p0 x (cosh(phi)+ sinh(phi) uq x up) = (q^p0) x fqp q^p Right= (cosh(phi)+ sinh(phi) up x uq)x q^p0 = fqp* x (q^p0)

where: x = quaternion product * = conjugate p = p0 + (p1 i + p2 j + p3 k) q = q0 + (q1 i + q2 j + q3 k) uq = 0 + (q1 i + q2 j +q3 k) / sin(arccos(q0)) up = 0 + (p1 i + p2 j +p3 k) / sin(arccos(p0)) phi= (arccos(q0))sin(arccos(p0)) fqp = (cosh(phi)+ sinh(phi) uq x up)

q^p does not meet in general the commutative law and the law of the sum of exponents for powers of equal bases (i.e. q^p x q^s ÅÇ q^s x q^p ÅÇ q^(p + s)) due to the presence of quaternion fqp. But, if uq x up = +/-1, i.e. q and p are co-directional quaternions, fqp = fqp* and the expressions reduce to:

q^p = (exp(+/-phi))q^p0

best regards

Giorgio Balzarotti and Gianluca Marsiglia 28 January 2005

giorgio.balzarotti@galileoavionica.it gianluca.marsiglia@galileoavionica.it

-- Giorgio Balzarotti (giorgio.balzarotti@galileoavionica.it), January 28, 2005.

I'm sorry that editing of my mail appear bad. Giorgio

-- Giorgio Balzarotti (giorgio.balzarotti@galileoavionica.it), January 28, 2005.

Hi to everybody, our previous contibution to this forum is actually unworth because of the bad editing of the posted message: that's why we're going to submit again our response to the question of the quaternion power of quaternions, hoping to share your opinions on our results. Thanks. Giorgio Balzarotti, Gianluca Marsiglia.

Original message:

<< Hello to everybody, we found your forum few days ago. If it is useful for the discussion, note that we use, and derived, the following expressions (left and right) for the quaternion power of a quaternion q^p (q and p unitary quaternions)

q^p Left = q^p0 x (cosh(phi)+ sinh(phi) uq x up) = (q^p0) x fqp q^p Right= (cosh(phi)+ sinh(phi) up x uq)x q^p0 = fqp* x (q^p0)

where: x = quaternion product * = conjugate

p = p0 + (p1 i + p2 j + p3 k) q = q0 + (q1 i + q2 j + q3 k)

uq = 0 + (q1 i + q2 j +q3 k) / sin(arccos(q0)) up = 0 + (p1 i + p2 j +p3 k) / sin(arccos(p0))

phi= (arccos(q0))sin(arccos(p0)) fqp = (cosh(phi)+ sinh(phi) uq x up)

q^p does not meet in general the commutative law and the law of the sum of exponents for powers of equal bases (i.e. q^p x q^s =/ q^s x q^p =/ q^(p + s) ) due to the presence of quaternion fqp.

But, if uq x up = +/-1, i.e. q and p are co-directional quaternions, fqp = fqp* and the expressions reduce to:

q^p = (exp(+/-phi))q^p0

Best regards. Giorgio Balzarotti, giorgio.balzarotti@galileoavionica.it Gianluca Marsiglia, gianluca.marsiglia@galileoavionica.it >>

-- Gianluca Marsiglia (gianluca.marsiglia@galileoavionica.it), February 01, 2005.