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Old 04-18-2004, 07:45 AM   #1 (permalink)
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Location: Ireland
Maths: Quaternion Multiplication

I'm trying to learn the formula for multiplying quaternions, but I don't understand it. I can use it alright, I just don't see any pattern to it. For all I can see it is just an abitrary collection of terms. (unlike addition and subtraction which are obvious)


( w + x i + y j + z k ) *
(w' + x'i + y'j + z'k )

=
(ww' - xx' - yy' - zz')
+(wx' + xw' + yz' - zy')i
+(wy' + yw' + zx' - xz')j
+(wz' + zw' + xy' - yx')k


Does anyone know a pattern or a pnemonic of some kind that would halp me remember this? Because I'm not in the mood for for learning off a 16 term formula! (yes. I'm lazy).

Thanks.
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Old 04-18-2004, 08:18 AM   #2 (permalink)
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-1 = i^2 = j^2 = k^2
i = jk = - kj
j = ki = -ik
k = ij = -ji

Edit2: The above 4 lines are all you need to know. What follows are mnemonics and tricks and other ways to help you remember the above 4 lines.

start with:
i j k
and look at the identies:
i=j k
_ j=k i
(I rotated the i around from the left to the far right)
_ _ k=i j
(I rotated the i and j around from the left to the far right)

any "base" element (i j or k) is equal to the product, in order, of the "next two".

Any two different "base" elements are negated when you multiply them together the opposite way.

Any "base" element squared is negative one.

Edit:
Some more math that might make it make more sense.
if x is i j or k, then:
x^2 / x = -1 / x
x = - 1/x
1/x = -x
so
1/i = -i
1/j = -j
1/k = -k
(or, in english, the multiplicative inverse of i is negative i)

start with
i = j k (eqn 1)
divide by k on the right:
i / k = j k / k
i * -k = j
- i k = j
j = - i k (eqn 2)

divide eqn 1 by j on the left: (same as multiplying by -j)
1/j * i = 1/j * j * k
1/j * i = k
-j * i = k
k = - j i

Multiply eqn 2 by j:
-1 = j^2 = - i k j = - j i k
i k j = j i k = 1
if
i k j = 1
then
-i * i * k j = -i
i = -k j (eqn 3)
and you can use eqn 3 to prove the other two identities.

Not sure if this makes things make more sense. =)
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Last edited by JHVH : 10-29-4004 BC at 09:00 PM. Reason: Time for a rest.

Last edited by Yakk; 04-18-2004 at 07:27 PM..
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Old 04-18-2004, 07:06 PM   #3 (permalink)
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start with yakk's opening four lines, and then just multiply the two quats together as though they are polynomials (that is, do term by term multiplication). then just simplify the result using the four rules above. lastly, just factor out the i, j, and k and that's all there is to it. so really all you have to memorize are the first four rules.
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Old 04-18-2004, 07:26 PM   #4 (permalink)
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Location: Ontario, Canada
thanks Phukraut. =)

I was playing around after the first 4 lines, probably should have made that clear. I'll edit my post to reduce the confusion, including phukraut's advice.

The interesting part is:
-1 = i^2 = j^2 = k^2
i = jk
(-1)*x = x*(-1) for all x

that is sufficient: you can derive all the other rules from just that. (the last one is just saying that -1 commutes with every element in the Quaternion).

I've heard that Quaternion's represent 3 dimensional rotations in some natural way. I understand how this might work theoretically, but have never really poked at it. Anyone want to chime up? =)
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Last edited by JHVH : 10-29-4004 BC at 09:00 PM. Reason: Time for a rest.
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Old 04-19-2004, 06:15 AM   #5 (permalink)
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hey np. if anyone think quats are complicated, check out octonions! these things must have have value in high dimension physics rotations (in string theory? don't know too much about that).

http://mathworld.wolfram.com/Octonion.html
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Old 04-19-2004, 11:32 AM   #6 (permalink)
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Location: Ireland
Thanks for your help Yakk and phukraut. Tis appreciated!

Yakk, if you want to see how quaternion rotations work, I've scanned in two pages which may be of interest to you:

http://www.akacentral.com/quaternions.htm
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Last edited by CSflim; 04-19-2004 at 11:34 AM..
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