# On quaternions

1. May 9, 2005

### Raparicio

Dear Friends,

I've read about quaternions, and they can be expresed in 4 terms or in angles like this:

$$a + ib + jc + kd = cos \theta + \vec{v} sin \theta$$

Quaternions are a generalization of the complex numbers, but in 3D. My question is about angles. For example, with a complex number, we can write:

$$a + ib = sin \theta + i \theta$$

By generalization, is this ok?

$$a + ib + jc + kd = cos \theta + i sin \theta + j sin \theta + k sin \theta$$

By other hand, we can take the angles in complex numbers like:

$$a + ib = cos \theta + i sin \theta$$ --> Only one angle

how much angles has a quaternion? has angle with the real part?

my best reggards.

Last edited: May 9, 2005
2. May 9, 2005

### neurocomp2003

quaternion represents 1 angle...it represents a matrix rotation

you can multiple quaternions to get another quaternion just like you
can multiple M*M

3. May 9, 2005

### James R

Raparicio:

I'm not sure you're correct.

I do not think that would work, since the possible values of $\cos \theta$ are limited to values between 1 and -1, while $a$, for example, has no such limit.

That is incorrect for general a and b, because you can see by inspection that it requires

$$a=\sin \theta, b=\theta$$

which would mean $a=\sin b$, which need not be true for a general complex number.

Again, that's wrong, because it implies that b=c=d, which need not be true for a general quaternion.

Actually, this isn't general enough. You need

$$a + ib = r [cos \theta + i sin \theta]$$

(i.e. an extra parameter, $r$, is requied).

4. May 9, 2005

### BobG

Almost correct.

If $$\vec{v}=ib + jc + kd$$
Then $$cos \theta + \vec{v} sin \theta=\cos \theta + ib \sin \theta + jc \sin \theta + kd \sin \theta$$

5. May 10, 2005

### Raparicio

exponencial

Aha!

And in exponential:

Then $$a·cos \theta · e^{ib \theta + jc \theta + kd \theta$$ ???

6. May 10, 2005

### BobG

I'm not positive if you're talking about quaternions in general.

However, what you're doing is normally used with unit quaternions. They'll always have a magnitude of 1. As nuerocomp2003 mentioned, they're used for vector transformation. The change the angles associated with the vector they're used on, but dont' change the magnitude of the vector they're used on.

With that in mind, I'm pretty sure you'll wind up simply with:

$$e^{(u \theta)}$$ where u is the unit vector whose i,j,k components are the direction cosines of the unit vector.

In other words, the $$a \cos \theta$$ part is not correct for a unit quaternion. The exponent part is correct.

7. May 11, 2005

### Raparicio

exactly

Exactly!!! (im talking about unit quaternions).