The Spin Chronicles (Part 23): Rotational dig

 Continuing from Part 22 we will now use the formulas to get to some real work.

"We dig dig dig dig dig dig dig
In our mine the whole day through
To dig dig dig dig dig dig dig
Is what we like to do

It ain't no trick to get rich quick
If you dig dig dig with a shovel or a pick
In a mine! In a mine! In a mine! In a mine!
Where a million diamonds shine!

We dig dig dig dig dig dig dig
From early morn till night
We dig dig dig dig dig dig dig up
Everything in sight"

We have the group G. By the way we, already noticed in  Part 13 that it is isomorphic to the group SL(2,C) of complex 2x2 matrices of determinant 1, the double cover of the connected component of the identity of the Lorentz group of special relativity. We will study now the action of G on the null cone of Minkowski space, on the sphere of its generating lines, and on the plane x,y obtained by stereographic projection. 

We dig dig dig dig dig dig dig up
Everything in sight

More precisely: we will study of the action of its one-parameter subgroups of the form g(t) = exp(tX), where t is a real parameter. The condition gν(g) = 1, defining G, implies X + ν(X) = 0 for generators X. Writing X as (X₀,X), and knowing that ν(X) = (X₀,-X), we deduce that the scalar part X₀ must be zero, while the complex vector part X is arbitrary. The real dimension of the space of complex vectors X is 6. Thus we will study 6 one-parameter subgroups, for X = e¹, e², e³, ie¹, ie², ie³.   The action we are studying is ς' = g ς τ(g). Here we recall the form of τ in the complex representation u=(u₀,u): τ(u) =  (u₀,u)*, where * stands for complex conjugation. Thus τ(e^k)=e^k, and τ(ie^k)=-ie^k, for k=1,2,3.

Let us start with the last one, namely with X = ie³. Then τ(exp(ite³)) = exp(-ite³). We will calculate the action of exp(ite_³) on basic vectors e⁰=1, e¹, e², e³. Evidently

e'⁰ = exp(ite³)e₀exp(-ite³) = e⁰,

and 

e'3 = exp(ite³)e³exp(-ite³) = e³.

Now, e¹ anticommutes with e³. Therefore

e¹exp(-ite³) = exp(ite³) e¹, and so

exp(ite³)e¹exp(-ite³) = exp(2ite³)e¹.

For the same reason

e'² = exp(ite³)e²exp(-ite³) = exp(2ite³)e².

But (e³)² =1, so the exponential can be easily calculated

exp(2ite³) = cos(2t)+i sin(2t) e³,

and so

e'² = (cos(2t)+i sin(2t) e³)e² = cos(2t)e² +i sin(2t) e³e².

Now it is good to remember that i = e¹e²e³,  therefore ie3e2 = e¹. The final result is:

e'² = cos(2t)e² +sin(2t) e¹.

The same way we obtain

e'¹ = cos(2t)e¹ - sin(2t) e².

This is how the basis vectors transform. Components transform by the transposed matrix. So if  ζ is a vector on the null cone, the transformed vector has coordinates:

ζ'₀ = ζ₀,
ζ'₁ = cos(2t)ζ₁ + sin(2t)ζ₂,
ζ'₂ = cos(2t)ζ₂ - sin(2t)ζ₁,
ζ'₃ = ζ₃.

We return now to the formulas from part 18:

ζ₀ = (1+x·x)/2,

ζ₁ = x₁,

ζ₂ = x₂,

ζ₃ = (1-x·x)/2.

Here x stands for the plane coordinates x₁,x₂.  Evidently ζ'₀+ζ'₃ = ζ₀+ζ₃ = 1, therefore we get

x'₁ = cos(2t)x₁ + sin(2t)x₂,
x'₂ = cos(2t)x₂ - sin(2t)x₁.

We have a simple rotation of the points of the plane by the angle 2t. The stars on the sky are simply rotating around the Earth axis. Quite normal. Or, perhaps, it is the Earth that is rotating? Who knows ...? The math for the rotation is the same in both cases.

Exercise 1. Use the same reasoning to deduce what happens for g(t) = exp(t e³). Thus no "i" in the exponential. First things will be easy, but there will be a little bit more needed at the very end.

  P.S. 10-12-24 Reply to Bjab's comment:

My input for Wolfram Alpha:

FullSimplify[{{cos(θ)*sin(φ), -r*sin(θ)*sin(φ), r*cos(θ)*cos(φ)},{sin(θ)*sin(φ), r*sin(φ)*cos(θ),r*sin(θ)*cos(φ)},{cos(φ),0,-r*sin(φ)}}^(-1)]

The answer:

It looks correct, but evidently Wolfram Alpha did not simplify sin^2+sin^2, which we have to do by hand. I used Mathematica. The answer is