4
$\begingroup$

I'm using quaternions to describe 3D rotations which necessitates that the quaternion is normalized. I'm getting more interested in quaternions and I tried to check what the quaternion magnitude represents, though I couldn't find any information. Searching on the internet for possible literature only returns results for using quaternions in orientation, frame transformation or translation, for which only unit quaternions are used.

So the question is, can anyone tell what the magnitude of the quaternion represents or how and when it is used? If someone knows about relevant literature, that would be great.

Thanks

$\endgroup$
4
  • $\begingroup$ Hi Darren, are you familiar with fractals at all? The entire process hinges on whether or not the quaternion magnitude is less than some threshold value. paulbourke.net/fractals/quatjulia $\endgroup$ Jan 20 at 19:43
  • $\begingroup$ Knew about fractals but never saw them in 3D. Thanks for the link. $\endgroup$ Jan 20 at 23:26
  • $\begingroup$ Do you want me to write up an answer related to fractals? $\endgroup$ Jan 21 at 1:31
  • $\begingroup$ Sorry for late reply. Not really needed as I asked the questions out of curiosity, But if you want so I'll surely read it. Thanks $\endgroup$ Jan 26 at 10:04

2 Answers 2

4
$\begingroup$

Quaternions are an extension of complex numbers and, as for complex numbers, the magnitude or modulus is the "length" of the quaternion measured from the "origin".

You can also think to quaternions as vectors in a $4-$D space and the modulus of the quaternion is the modulus of the corresponding vector.

Also, the modulus of a quaternion is used to express the quaternion in polar form (in a similar way as complex numbers), as you can see here.

An introduction to quaternion with geometric application and more references is here.

$\endgroup$
1
  • $\begingroup$ Never though about quaternions in polar form Thanks for the links $\endgroup$ Jan 20 at 23:32
2
$\begingroup$

In terms of 3-d rotations, the quaternion magnitude doesn't mean anything, since it cancels out in the process of conjugation.

What I mean is that $qxq^{-1}=\frac{q}{|q|}|q|xq^{-1}=\frac{q}{|q|}x(q^{-1}||q|)=\frac{q}{|q|}x\left(\frac{q}{|q|}\right)^{-1}$ so it isn't material what the magnitude is.

We usually take it to be unit length to make the representation nicer.

$\endgroup$
1
  • 1
    $\begingroup$ ... and for novices it might be worth emphasizing that taking quaternions of unit norm to represent rotations, etc., works well because the product of two unit-norm quaternions is again a unit-norm quaternion... $\endgroup$ Jan 20 at 19:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.