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Context

In electromagnetism, it is common to meet complex angles. Snell's law can produce them (with metals and/or total internal reflection), and when we plug them into Fresnel's equations, the results make physical sense: evanescent or inhomogeneous waves. The real and the imaginary parts of the complex wave vector are not collinear anymore: the wave propagates in one direction but decays in another. In other words, the planes of iso-phase and iso-amplitudes are not parallel anymore.

I am not sure yet that I can make use of complex axes of rotation, but it seems natural to consider them.

Question

Rotating by a complex angle around a real axis seems trivial from an algebraic point of view: we can represent it with a matrix of complex numbers. We should also be able to generalize it to rotations around complex axes: it's also just a matrix.

I can easily construct a 3D rotation matrix for a real axis and a real angle by creating a versor. But what happens when the versor is not a quaternion but a biquaternion? If $q$ is my complex versor and $v$ the complex vector that I wish to rotate, does $v'=qvq^{-1}$ still work? Or should I throw in some complex conjugations?

Edit 1: experiment

I wanted to compare a rotation matrix created without quaternion to a rotation matrix created with a quaternion. If these two matrices differ, then it means that the way I construct my quaternions is broken.

Without using quaternions, it is easy to create the matrix of a rotation around the axes $x$ $y$ and $z$, anything else is trickier. These are, of course, real axes of rotation, not complex ones. However, I can use complex angles there.

In this situation, then the two methods (with and without quaternion) give the same result. This is not a proof, but $v'=qvq^*$ seems to work when $q=\cos \theta + u \sin \theta$ with $\theta$ complex and $u$ real. Here, * denotes the biconjugate (flip the sign of the vector), not the complex conjugate. With real unit quaternions, $q^{-1}=q^*$. This may not work with biquaternions in general, even though it worked in this experiment.

Edit 2: getting closer

$q = (w, x, y, z)$ is a quaternion, its biconjugate is $q^*=(w,-x,-y,-z)$.

$qq^* = w^2 + x^2 + y^2 + z^2$ is a scalar, a priori complex.

If $qq^* \ne 0$ then $q$ has an inverse $q^{-1} = q^* / (qq^*)$.

A rotation of angle $2\theta$ around a real unit axis $u=(x, y, z)$ can be represented by the following quaternion. $$q = \cos \theta + \sin \theta \: (x, y, z)$$ $$qq^* = \cos^2 \theta + \sin^2 \theta (x^2+y^2+z^2)$$ The axis is real and unit, therefore $x^2+y^2+z^2$=1. This leaves $qq^* = \cos^2 \theta + \sin^2 \theta$ which equals 1 even when $\theta$ is complex.

Therefore, whenever the axis of rotation is real, even if the angle is complex, then $qq^*=1$ and $q^{-1} = q^*$. This means that the inverse of a rotation matrix is its transpose. The inverse is a rotation of angle $-\theta$ around the same axis. $$q^{-1} = q^* = \cos\theta-\sin\theta \: u = \cos(-\theta)+\sin(-\theta)u$$

However, when the axis is complex, then this does not work: $qq^*\ne 1$. Let us assume $qq^*\ne 0$. $$q^{-1} = \frac{1}{qq^*}(\cos(-\theta) + \sin(-\theta) \:u)$$ Is there an angle $\phi$ such that $q^{-1}=\cos \phi + \sin \phi \:u$? No, there are none: we cannot have $\cos^2+\sin^2=1$ both for $\theta$ and $\phi$ when $qq^* \ne 1$. This means that the inverse of a rotation by a complex angle around a complex axis is NOT a rotation (Ehm, see edit 3). Some scaling/shearing/something happened, which must be corrected. The matrix has an inverse, the quaternion has an inverse, but the inverse is not a rotation.

So, when the inverse of a rotation is not a rotation, are we even allowed to speak about rotations in the first place?

Only the complex axes $(x,y,z)$ for which $x^2+y^2+z^2=1$ seem appropriate if we want the inverse of a rotation to be a rotation as well. I have no idea what the shape of $x^2+y^2+z^2=1$ looks like.

Conclusion.

Rotating by a complex angle around a real axis works well.

Only a few complex axes (those for which $x^2+y^2+z^2=1$, even if the norm of said axis isn't 1 at all) are acceptable for rotations. By acceptable, I mean that the inverse of a rotation is a rotation, which is generally not the case with complex axes.

Edit 3, much later (2015-03-29)

I wrote in Edit 2 "the inverse of a rotation around a complex axis is not a rotation". Well, I haven't proven that actually: the only result I got was about quaternions. Maybe rotations around complex angles make sense but biquaternions don't represent them properly. (Btw, I solved my actual problem by avoiding trigonometry completely and avoiding the notion of rotation altogether, I'm only coming back to it out of curiosity).

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    $\begingroup$ Tl;dr. As a philosophic remark: various displacement vectors, axes, and angles are convenient intuitively-appealing presentations, but “correct” underlying objects are always symmetry groups. First, think of a group, only then – of a presentation of its elements. $\endgroup$ – Incnis Mrsi Nov 9 '14 at 15:19

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