Quaternions and SR

  • Thread starter arcnets
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  • #1
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Hi all,
just discovered the LaTeX feature, so why not play around with it a bit.

Quaternions are (sort of) generalized complex numbers where you have one real unit and 3 imaginary units i, j, and k. The basic equation is
[tex]
i^2 = j^2 = k^2 = ijk = -1.
[/tex]
Now, if we have a 4-vector
[tex]
\vec{r} = (t, x, y, z),
[/tex]
we could write it as a quaternion
[tex]
R = t + ix + jy + kz.
[/tex]
Defining
[tex]
R_3 = ix + jy + kz,
[/tex]
and
[tex]
R_0 = t
[/tex]
we get
[tex]
R^2 = (t^2 - x^2 - y^2 - z^2) + 2R_{0}R_{3}.
[/tex]
Thus
[tex]
R^2{}_0 = S^2
[/tex]
Where S2 is a relativistic invariant.

Maybe it's just useless, I'm just playing around. Any suggestions?
 
Last edited:

Answers and Replies

  • #2
Hurkyl
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Even better, you can define [tex]\inline{\bar{R} := t-ix-jy-kz}[/tex], and by analogy with complex dot products we would compute the length squared of [tex]\inline{R}[/tex] as

[tex]
\begin{align}
R\bar{R} &= (t + ix + jy + kz) (t - ix - jy - kz) \\
&= t^2 - (ix + jy + kz)^2 \\
&= t^2 - x^2 - y^2 - z^2 - xy(ij + ji) - xz(ik + ki) - yz(jk + kj) \\
&= t^2 - x^2 - y^2 - z^2
\end{align}
[/tex]

Quaternions are fun things, I wish I knew more about them!
 
  • #3
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Originally posted by Hurkyl
Even better, you can define [tex]\inline{\bar{R} := t-ix-jy-kz}[/tex], and by analogy with complex dot products we would compute the length squared of [tex]\inline{R}[/tex] as

[tex]
\begin{align}
R\bar{R} &= (t + ix + jy + kz) (t - ix - jy - kz) \\
&= t^2 - (ix + jy + kz)^2 \\
&= t^2 - x^2 - y^2 - z^2 - xy(ij + ji) - xz(ik + ki) - yz(jk + kj) \\
&= t^2 - x^2 - y^2 - z^2
\end{align}
[/tex]

Quaternions are fun things, I wish I knew more about them!
Unit quaternions uniquely define orientation, as such they are great for rigid body dynamics!

Quaternions rock!
 
  • #4
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Wow!

Wow! Thanks for the encouragement.

There's also some nice websites on this...
 
  • #5
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Originally posted by Hurkyl
[tex]
R\bar{R} &= t^2 - x^2 - y^2 - z^2
[/tex]
You sure? I get
[tex]
R\bar{R} &= t^2 + x^2 + y^2 + z^2
[/tex]
 
  • #6
Hurkyl
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I like your answer better. :smile:
 
  • #7
Yay!

I get the same answer as arcnets. I see where you went wrong too. I too believe that this answer is much more likeable.

I love quaternions too! I remember way back when I first discovered complex numbers. Quaternions all the way!
There are also octonions. These are very useful in string theory, Yang-Mills gauge theories and explaining quarks and leptons. But most of all they're fun to play with!

Quaternion, Octinion, Sedenions...
 
  • #8
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Yes, but you give up useful properties as dimension increases. For instance, complex numbers are commutative and associative. Quaternions are not commutative (ij = -ji) but still associative. I guess octonions are neither.

I think it's fascinating in my above example that, when R is a quaternion, then R2 is still a quaternion, and it's time-like component is a relativistic invariant.

My original hope was that we could write some physical equations with quaternions instead of vectors. Especially, in electrodynamics...
 
  • #9
chroot
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Originally posted by arcnets
My original hope was that we could write some physical equations with quaternions instead of vectors. Especially, in electrodynamics...
This is exactly what Hamilton did some hundred and fifty years ago (IIRC). It's a nightmare. I'm sure you can find a book on Hamilton's model in your local university library. It was all replaced with vectors by Heaviside, and thank god for that.

- Warren
 
  • #10
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chroot,
of course that bursts my bubble a bit. Well let me tell why I thought quaternions might be a good idea.
When I learned SR, I wondered "Why 4 spacetime dimensions, why not 3 or 5?", and "Where does the strange metric come from?".
Then I learned Algebra, and when it came to division algebrae, it turned out that these can only have dimensions d=1,2,4,8,16... where d=4 (quaternions) is the largest one that is still associative. Plus, it has the right "metric" (loosely speaking).
So I thought, since quaternions offer an answer to both questions, they may be the 'natural' description for electrodynamics, SR, and so on.
Now you say it's a nightmare.
Well, I'll try to write down some basic equations (e.g. Maxwell's laws) with quaternions, and then let's see where it starts to become a nightmare. Or maybe someone else likes to try...

arcnets
(who sounds a bit like a troll here, but thinks he isn't since he can easily be persuaded...)
 
Last edited:
  • #11
chroot
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arcnets,

Nay, I applaud your efforts to learn the nightmare! It's always instructive to learn about those models that were discarded. If you have an interest in how quaternions were used, by all means, I encourage you to pursue the subject to its finale.

I believe Maxwell actually orginally formulated his theory of electromagnetism with Hamilton's quaternions -- 20 equations with 20 variables, IIRC -- and Heaviside rewrote them in vector notation, collapsing them to 4 equations in 2 variables.

(Differential forms allow the reduction of Maxwell's laws down to just two equations in two variables, btw.)

- Warren
 
  • #12
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Well, first idea: Product of two space-like quaternions. Let
[tex]
A = ia_1 + ja_2 + ka_3
[/tex]
[tex]
B = ib_1 + jb_2 + kb_3
[/tex]
Thus,
[tex]
AB = -a_1b_1 + ija_1b_2 + ika_1b_3 + jia_2b_1 - a_2b_2
[/tex]
[tex]
+ jka_2b_3 + kia_3b_1 + kja_3b_2 - a_3b_3
[/tex]
[tex]
= -(a_1b_1 + a_2b_2 +a_3b_3) + i(a_2b_3-a_3b_2) + j(a_3b_1-a_1b_3) + k(a_1b_2-a_2b_1).
[/tex]
Thus, we can define dot and cross products like this:
[tex]
A \cdot B := -(AB)_0
[/tex]
[tex]
A \times B := (AB)_3
[/tex]
Now, doesn't this look good? - I guess the nightmare starts when the timelike component of A or B is not zero...

(Edit: TeX is hard. How do I make a new line inside TeX?)
 
Last edited:
  • #13
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I forgot to write the conclusion
[tex]
AB = A \times B - A \cdot B.
[/tex]
Which is valid only if A, B are space-like.
Anyway...
Originally posted by chroot
20 equations with 20 variables
Yes I understand that's not very practical and thus not very desirable. BTW, were these variables all physical ones?
 
Last edited:
  • #14
chroot
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Originally posted by arcnets
(Edit: TeX is hard. How do I make a new line inside TeX?)
Use the "align*" environment, like this:

[tex]
\begin{align*}
AB = &-a_1b_1 + ija_1b_2 + ika_1b_3 + jia_2b_1 - a_2b_2\\
&+ jka_2b_3 + kia_3b_1 + kja_3b_2 - a_3b_3\\
= &-(a_1b_1 + a_2b_2 +a_3b_3) + i(a_2b_3-a_3b_2)\\
&+ j(a_3b_1-a_1b_3) + k(a_1b_2-a_2b_1).
\end{align*}
[/tex]

(Click the image to see its code.) The &'s denote the position in each line to align vertically. \\ represents a newline.

And yeah, it's got a little learning curve to it, but so does any computer software with such flexibility and power.

- Warren
 
Last edited:
  • #15
Without Quaternions however, we would not have known the simple nature of vectors. The modern vector calculations stemed from Quaternions.
 
  • #16
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Originally posted by chroot
Use the "align*" environment
Thank you!

Dorian Gray,
from what you and others said here, quaternions seem to have been used prior to vectors in history. Interesting...
 
  • #17
selfAdjoint
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Maxwell toyed with quaternions, but didn't use them in developing his theory. Maxwell's friend and colleague Tait was the big quaternion fan and wrote a huge tome expressing three dimensional physics in quaternions.

Three-dimnensional vector analysis was developed from quaternions in the early twentieth century by several workers (the fact that different people worked on it accounts for some of the confusion of notation and terminology in the older books) In the US it was Josiah Willard Gibbs of statistical mechanics fame who was the key player.

BTW, have you noticed that the algebra of quaternions is isomorphic to the algebra of Pauli spin matrices?
 
  • #18
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selfAdjoint,
thanks for your information on more historical background. I guess research has already covered this and I should do some reading...
BTW, have you noticed that the algebra of quaternions is isomorphic to the algebra of Pauli spin matrices?
Yes, i've seen that in a book.

Here's another formula:
[tex]
AB = a_0b_0 + a_0B_3 + A_3b_0 - A_3 \cdot B_3 + A_3 \times B_3.
[/tex]
(Where index 0 means timelike and 3 means spacelike.)
All terms are symmetric except the last which is antisymmetric, so we can write
[tex]
AB - BA = 2(A_3 \times B_3)
[/tex]
and thus we get a simple expression for the vector product:
[tex]
A_3 \times B_3 = \frac{1}{2}(AB - BA).
[/tex]

(Edit: sign error removed)
 
Last edited:
  • #19
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Likewise,
[tex]
\frac{1}{2}(AB + BA) = a_0b_0 + a_0B_3 + A_3b_0 - A_3 \cdot B_3.
[/tex]
Or, maybe better:
[tex]
\frac{1}{2}(AB + BA) = a_0B + Ab_0 - a_0b_0 - A_3 \cdot B_3.
[/tex]
OK, this formula for scalar product is not so nice, but still far from 'nightmare'.

(Edit: 2nd formula inserted)
 
Last edited:
  • #20
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OK, now for some physics.
If we have a 4-vector potential, written as a quaternion:
[tex]
A = \phi + ia_x + ja_y + ka_z
[/tex]
and write the 4-differentiation as a quaternion
[tex]
\nabla = \frac{\partial}{\partial t} + i\frac{\partial}{\partial x} + j\frac{\partial}{\partial y} + k\frac{\partial}{\partial z}
[/tex]
then we know
[tex]
\nabla_0a_0 - \nabla_3 \cdot A_3 = 0.
[/tex]
Thus, using the 1st equation in my last post,
[tex]
\frac{1}{2}(\nabla A + A \nabla}) = \nabla_0A_3 + \nabla_3a_0.
[/tex]
Well, Maxwell says that the electric field E is in fact
[tex]
E = -\nabla_0A_3 - \nabla_3a_0
[/tex]
Thus,
[tex]
E = -\frac{1}{2}(\nabla A + A \nabla}).
[/tex]

Similarly,
[tex]
B = \frac{1}{2}(\nabla A - A \nabla}).
[/tex]
 
Last edited:
  • #21
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...which leads to the conclusion
[tex]
E + B = -A\nabla.
[/tex]
Note: There's not a field tensor in all this. Electric and magnetic field quaternions are just being added. Giving a simple quaternion product. Can this be correct? And if not, where did I go wrong?
 

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