# Redshift Academy   Wolfram Alpha: Search by keyword: Astronomy

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- Units, Constants and Useful Formulas

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- ```Quaternions
-----------

Quaternions are a number system that extends the
complex numbers.

They are represented in the form:

q = a + bi + cj + dk

Where i2 = j2 = k2 = ijk = -1

They are cyclic, i.e.

i        ij = k  ji = -k
^ \       jk = i  ik = -j
/   v      ki = j  etc.
k <- j

Proof:

ijk = -1

iijk = -i

-jk = -i

jk = 1

Quaternions can be decomposed into scalar and vector
parts:

q = (a,b,c,d) = (a,v)
^ -----
|   ^
scalar  \
vector

A quaternion (0,v) is called a PURE quaternion.

A UNIT quaternion is a quaternion of modulus one.
(q = q/√(a2 + b2 + c2 + d2)

The Quaternion Group, Q8
------------------------

Quaternions form a group, Q8 = {1,-1,i,-i,j,-j,k,-k}

Operations:  + or .

Closure: q1 + q2/ = q3

q1.q2 = q3

Inverse:  q1 + (-q1) = 0

q1.q1* = 1

Identity: 0 + q1 = q1

1.q1 = q1

Associativity:   q1 + (q2 + q3) = (q1 + q2) + q3)

(q1q2)q3 = q1(q2)q3q2)

Commutativity:  q1 + q2 = q2 + q1

q1.q2 ≠ q2.q1

[i,j] = ij - ji = k - (-k) = 2k

Subgroups:  {1,-1,i,-i} {1,-1,j,-j} {1,-1,k,-k}

{1,-1} = center (kernel)

{1}

Quaternion Multiplication
-------------------------

q1 = (a,b,c,d)

q2 = (e,f,g,h)

q1q2 = (ae - bf - cg - dh,
af + be + ch - dg,
ag - bh + ce + df,
ah + bg - cf + de)

This can also be written in terms of real matrices
as follows:

-          -  - -
| a -b -c -d || e |
| b  a -d  c || f |
| c  d  a -b || g |
| d -c  b  a || h |
-          -  - -

Example:

(1,2,3,6)(0,1,0,0)

-          -  - -     -  -
| 1 -2 -3 -6 || 0 |   | -2 |
| 2  1 -6  3 || 1 | = |  1 |
| 3  6  1 -2 || 0 |   |  6 |
| 6 -3  2  1 || 0 |   | -3 |
-          -  - -     -  -

Individual elements are given by:

-       -
| 1 0 0 0 |
(1,0,0,0) = | 0 1 0 0 |
| 0 0 1 0 |
| 0 0 0 1 |
-       -

-         -
| 0 -1 0  0 |
(0,1,0,0) = | 1  0 0  0 |
| 0  0 0 -1 |
| d  0 1  0 |
-         -

etc.

Extracting the Dot and Cross Products
-------------------------------------

Let q1 = (a,x) and q2 = (b,y)

q1q2 = (ae - (bf + cg + dh),
af + be + ch - dg,
ag - bh + ce + df,
ah + bg - cf + de)

(ae - x.y,ay + ex + x ^ y)

Vector Rotation (Perpendicular Axis)
------------------------------------

n ^ v
^
|
|    v'      ☉ means out of screen
|   /
|  /
| /
|/θ
n ☉-----------> v

v' = vcosθ + (n ^ v)sinθ

Quaternion Rotation (Perpendicular Axis)
----------------------------------------

q = (0,v)

q is the PURE quaternion corresponds to the
vector being rotated.

q' = (0,v')

qn = (0,n)

n is the unit vector coming out of the screen
corresponding to the axis of rotation, and qn
is the correponding pure quaternion.  We need
to figure out what n ^ v equals.  Try,

qnq = (0,n)(0,v) = (0 - n.v, 0v + 0n + n ^ v)

= (0,n ^ v)

= n ^ v

Therefore, analagous to the vector forn, we can
write:

q' = qcosθ + (qnq)sinθ

= (cosθ + qnsinθ)q

Now,

qn2 = (0,n)(o,n)

= (0 - n.n,n ^ n)

= (-|n|2,0)

= (-1,0)

= -1

So qn is analagous to i so we can write the
quaternion form of Euler's equation as:

exp(qnθ) = (cosθ + qnsinθ) ... 1.

and,

q' = exp(qnθ)q

Example:

q = (0,0,1) = k

qn = (0,1,0) = j

θ = π/2

z
|
^ q
|
|
-------->------- y
/        qn
/
v q'
/
x

q' = exp(θqn)q

= exp(πj/2)k

= jk

= i

= (1,0,0)

Vector Rotation (Arbitrary Axis)
-------------------------------- v = v∥ + v⊥

v' = v∥' + v⊥'

= v∥ + v⊥'

= v∥ + cosθv⊥ + sinθ(n ^ v⊥)

= v∥ + cosθ(v - v∥) + sinθ(n ^ v⊥)

n = nxi + nyj + nzk is a unit vector corresponding
to the arbitrary axis of rotation.

(n ^ v⊥) = n ^ (v - v∥)

= n ^ v - n ^ v∥

= n ^ v

v' = (1 - cosθ)v∥ + cosθv + sinθ(n ^ v)

Now,

v∥ = (v.n)n projection of v onto n

v' = (1 - cosθ)(v.n)n + cosθv + sinθ(n ^ v)

This is the RODRIGUES ROTATION FORMULA.

Example:

n = (0,0,1) v = (1,0,0) θ = π/2

v' = (1 - cosθ)(v.n)n + cosθv + sinθ(n ^ v)

= 0 + 0 + (0,0,1) ^ (1,0,0)

= (0,1,0)

Quaternion Rotation (Arbitrary Axis)
------------------------------------

q = (0,v)

q∥ = (0,v∥)

q⊥ = (0,v⊥)

q⊥' = (0,v⊥')

qn = (0,n)

Where,

qn = (0,nxi,nyj,nzk) is now a pure quaternion
corresponding to the arbitrary axis of rotation,
n.  As before,

q' = q∥ + q⊥'

= q∥ + exp(θ.qn)q⊥

= q∥ + (cosθ + sinθnxi + sinθnyj + sinθnkk)q⊥

Now exp(θ.qn)q⊥ = q⊥exp(-θ.qn)

Proof:

(cosθ,sinθn)(0,v⊥) = (0,v⊥)(cosθ,-sinθn)

(0,cosθv⊥ + sinθ(n ^ v⊥)) = (0,cosθv⊥ - sinθ(v⊥ ^ n))

(0,cosθv⊥ + sinθ(n ^ v⊥)) = (0,cosθv⊥ + sinθ(n ^ v⊥)) Q.E.D.

Likewise,

exp(θ.qn)q∥ = q∥exp(θ.qn)

Proof:

[q1,q2] = 2(v1 ^ v2)

q' = exp(θqn/2)exp(-θqn/2)q∥ + exp(θqn/2)exp(θqn/2)q⊥

Using exp(-θqn/2)q∥ = q∥exp(-θqn/2) and exp(θ.qn)q⊥ =
q⊥exp(-θ.qn) from above gives:

q' = exp(θqn/2)q∥exp(-θqn/2) + exp(θqn/2)q⊥exp(-θqn/2)

= exp(θqn/2)(q∥ + q⊥)exp(-θqn/2)

= exp(θqn/2)qexp(-θqn/2)

In terms of sin and cos this is:

g = exp(θqn/2)

= cos(θ/2) + sin(θ/2)nxi + sin(θ/2)nyj + sin(θ/2)nkk

g* = exp(-θqn/2)

= cos(θ/2) - sin(θ/2)nxi - sin(θ/2)nyj - sin(θ/2)nkk

Therefore, we can write:

q' = gqg* = gqg-1

Example:

qn = (0,0,0,1) q = (0,1,0,0) θ = π/2

g = cos(π/4) + sin(π/4)nzk

= √2/2 + (√2/2)k

= (1 + k)/√2

q' = gqg*

= [(1,0,0,1)/√2](0,1,0,0)[(1,0,0,-1)/√2]

= (0,0,1,0)

Why the conjugate?  Consider:

q' = gq

= [(1,0,0,1)/√2](0,1,0,0)

= (0,1/√2,1/√2,0)

This is a rotation of 45°.  Therefore, the product
gq does not accomplish the full rotation. by itself.

Quaternion Rotation Matrix
--------------------------

A quaternion rotation:

q' = gqg* = gqg-1

With

q = w + xi + yj + zk and qq* = 1 = w2 + x2 + y2 + z2

Can be algebraically manipulated (not proven here)
into a matrix rotation q' = Rq, where R is the
rotation matrix given by:

-                                          -
| 1 - 2y2 - 2z2    2xy - 2zw     2xz + 2yw   |
R = |   2xy + 2zw    1 - 2x2 - 2z2   2yz - 2xw   |
|   2xz - 2yw     2yz + 2xw    1 - 2x2 - 2y2 |
-                                          -

This is a rotation around the vector n = (x,y,z)
(or equivalently qn = nxi + nyj + nzk) by an angle
2θ.  We will demonstrate this using the previous
example where qn = (0,0,0,1) q = (0,1,0,0) θ = π/2
∴ 2θ = π.  Plugging the numbers in gives:

-       -  - -     -  -
| -1  0 0 || 1 |   | -1 |
|  0 -1 0 || 0 | = |  0 |
|  0  0 1 || 0 |   |  0 |
-       -  - -     -  -

This is indeed a rotation of π!

The 2:1 nature is apparent since both qn and -qn
map to the same R, i.e. the z2 is always positive.

Relation Between SU(2) and SO(3)
--------------------------------

Instead of the 4 x 4 real matrices shown before,
we can also write quaternions as 2 x 2 complex
matrices as follows:

q = a + bi +cj + dk

= (a + bi) + (c + di)j  ij = k

= z + wj

We can then write this as a matrix in the same
way that we can write a complex number as a
matrix.

-      -     -                 -
| z  -w  | = | a + bi  -(c + di) |
| w*  z* |   | c + di    a - bi  |
-      -     -                 -

-----------------------------------------------------

Digression:

Complex Numbers as Matrices
---------------------------

Consider the 2 complex numbers.

(a + bi)(c + di) = ac - bd + (ad + bc)i

This can be written as the matrix:

-    -  - -
| a -b || c |
| b  a || d |
-    -  - -

The matrix on the LHS enables us to write.

-   -           -    -
1 = | 1 0 | and i = | 0 -1 |
| 0 1 |         | 1  0 |
-   -           -    -

Euler's formula can then be written as:

exp(iθ) = cosθ + isinθ

-    -      -          -
exp(| 0 -θ |) = | cosθ -sinθ |
| θ  0 |    | sinθ  cosθ |
-    -      -          -

Differentiating both sides w.r.t. θ gives:

iexp(iθ) = -sinθ + icosθ

-    -  -          -     -           -
| 0 -1 || cosθ -sinθ | = | -sinθ -cosθ |
| 1  0 || sinθ  cosθ |   |  cosθ -sinθ |
-    -  -          -     -           -

-----------------------------------------------------

Example:

(1,2,3,6)(0,1,0,0)

-                  -  -                  -
| (1 + 2i) -(3 + 6i) || (0 + i)  -(0 + 0i) |
| (3 - 6i)  (1 - 2i) || (0 - 0i)  (0 - i)  |
-                  -  -                  -

-                  -
= | (-2 + i) -(6 - 3i) |
| (6 + 3i) (-2 - i)  |
-                  -

-  -
| -2 |
= |  1 |
|  6 |
| -3 |
-  -

Which is exactly the same result from before.

The following matrices satisfy these conditions:

-   -       -    -       -   -       -    -
I = | 1 0 | i = | 0 -1 | j = | 0 i | k = | -i 0 |
| 0 1 |     | 1  0 |     | i 0 |     |  0 i |
-   -       -    -       -   -       -    -

= -iσ2        = -iσ1       = -iσ3

It is easy to check that the matrices satisfy the
quaternion identites.  In addition, all have
determinnt = 1 and i† = -i, j† = -j, k† = -k.
These are the Pauli matrices of SU(2) that are
isomorphic to the quaternions of unit norm.

Given the 2:1 correspndence between rotations of
R and qqg-1, it implies that there is a 2:1
homomorphism from SU(2) to SO(3).  This has very
important consequences in the physics of spin.

The unit quaternions can also be thought of as
the group corresponding to the 3-sphere, S3, that
gives the group Spin(3), which is isomorphic to
SU(2) and also the double cover of SO(3).``` 