# Polar Representations of Quaternions

Posted by doug
QMN1009.2352
Doug Sweetser
latex pdf:
Mathematica Notebook:
abstract:

Complex numbers have a polar representation. Three complex numbers that share the same real number are subgroups of a quaternion. Therefore a polar representation of a quaternion must exist. The amplitude is the absolute value of the whole quaternion. The imaginary number i expands to i, j, k for quaternions. The remaining question is how to handle the angle. Two ways work. The first is to take the inverse cosine of the scalar over the absolute value of the quaternion. The second method takes the inverse tangent of the absolute values of the 3-vector over the scalar. A right triangle is animated so the connection between velocity and the polar representation is more apparent.

Calculate the various parts that go in to the polar representation:

$A = (1, 2, 3, 4) \quad eq. 1$

$|A|=\sqrt{30} \quad eq. 2$

$\theta =\arccos \left(\frac{A+A^*}{2|A|}\right)=\arccos \left(\frac{scalar(A)}{|A|}\right)=\arccos \left(\frac{1}{\sqrt{30}}\right) \quad eq. 3$

$\vec{I}=\frac{vector(A)}{|vector(A)|}=\left(0,\frac{2}{\sqrt{29}},\frac{3}{\sqrt{29}},\frac{4}{\sqrt{29}}\right) \quad eq. 4$

The polar representation of a quaternion is the magnitude of the quaternion times the exponential of angle theta in the I direction:

$|A| \exp \left(\theta \vec{I}\right)=(1,2,3,4) \quad eq. 5$

Bingo, bingo, exactly right. Write it out:

$q=\sqrt{q q^*}\exp \left(\arccos \left(\frac{q+q^*}{2\sqrt{q q^*}}\right)\frac{q-q^*}{\sqrt{-\left(q-q^*\right)^2}}\right) \quad eq. 6$

A short paper by Drahoslava Janovska \unicode{0301}, and Gerhard Opfer gave used an Arctan instead of an Arccos, and a different angle for the polar representation (available at http://www.waset.org/journals/waset/v47/v47-29.pdf). Try their suggestion at the bottom of the polar decomposition.

$\alpha =\arctan \left(\frac{\left|A-A^*\right|}{\left|A+A^*\right|}\right)=\arctan \left(\sqrt{29}\right) \quad eq. 7$

$|A| \exp \left(\alpha \vec{I}\right)=(1,2,3,4) \quad eq. 8$

Bingo, bingo.

$q=\sqrt{q q^*}\exp \left(\arctan \left(\frac{\sqrt{-\left(q-q^*\right)^2}}{\left(q+q^*\right)}\right)\frac{q-q^*}{\sqrt{-\left(q-q^*\right)^2}}\right) \quad eq. 9$

What is the difference between the arctan and the arccosine? A cosine is the near side over the hypotenuse. A tangent is the far over the near side. The choice for the definiton of \theta reflects the difference between the two trig functions.

$N\left[ArcCos\left[\frac{1}{\sqrt{30}}\right]\right]$

$N\left[ArcTan\left[\sqrt{29}\right]\right]$

$1.38719$

$1.38719$

Why does the polar representation matter? The euclidean representation are 4 independent numbers: think of an unending grid. The polar representation connects the numbers together, but what does it mean? Consider the polar representation of events in spacetime, with time as the scalar and displacement as the 3-vector. The |A| part is positive definite, the distance covered in spacetime. The exponential is an angle times an imaginary number. For a quaternion, the imaginary number is a 3-vector, so can point in an arbitrary direction in 3D space. The ratio of change in space to change in time is the velocity.

All of the trig relationships are about right triangles inside a unit circle. Both the unit circle and a right triangle can be animated:

Circles and right triangles look like straight lines in spacetime, although they can look right in the three complex planes. The interval between the origin and the last blue event is timelike. The slope of the blue line is fairly close to 45 degrees so this represents a fast moving particle. A classical particle would make a very narrow triangle.\quad For this future spacelike right triangle, the observer is in red, never moving, while the particle in motion gets farther away. When the particle in motion reaches the unit circle, there is what I call a bling when many events appear on the screen for a moment which connects the observer to the particle in motion. Two of the three sides of the right triangle could be transversed by real particles.

A triangle formed from spacelike events appears similar, only the interpretation changes. Only one of the world lines could be done by a real particle. A triangle in 3D space appears for only an instant, and has nothing to do with the polar representation because there is no change in time.

Document Description
# of pages:
2
# of figures:
1
Change Log:
2010 Nov. 3: Added animated right triangle.
Tags
Physics Tag:
quantum field theory
Math Tag:
complex numbers
quaternions
polar representation
Programming Tag:
command line quaternions