Redshift Academy


The Lorentz Group
-----------------

The Lorentz group is the SO(3,1) symmetry group of Special 
Relativity.  It is the extension of the rotation group SO(3) 
that mixes spatial directions to the group that mixes both 
space and time.  The Lorentz group is a Lie group.

x.x = x^Tηx

       -       -  -  -
    = | t x y z || -t |
       -       - |  x |
                 |  y |
                 |  z |
                  -  -
    = x^μη_μνx^ν

    = x^μx_ν

    = t² - x² - y² - z²

The Lorentz transform can be thought as a 4 x 4 matrix
boosting or rotating the coordinates in inertial frame, S, 
to inertial frame, S'.  Thus,

x^μ' = Λ^μ_νx^ν

Lorentz transformations leave the inner product invariant.
Thus,

x'.x' = x.x

The set of all matrices obeying Λ^TηΛ = η form the LORENTZ 
GROUP SO(3,1).  

Proof:

Consider the transformation of a vector, x:

x -> Λx  

Define the DUAL representation which transforms as follows:       

~       ~
x -> (Λ^-1)^Tx

Therefore,

~          ~
x^Tx -> ((Λ^-1)^Tx))^TΛx

Using the property that (AB)^T = B^TA^T we get:

      ~
   -> x^TΛ^-1Λx

      ~      ~
Check x^Tx -> x^TΛ^-1Λx:

     - -           - -           -           -
    | a |     ~   | e |         |  γ  -βγ 0 0 |        
x = | b | and x = | f | and Λ = | -βγ  γ  0 0 |
    | c |         | g |         |  0   0  1 0 |
    | d |         | h |         |  0   0  0 1 |
     - -           - -           -           -

Now,

                 -         -
                | γ  βγ 0 0 |
Λ^-1 = (1/(γ² - β²γ²))| βγ γ  0 0 | 
                | 0  0  1 0 |
                | 0  0  0 1 |
                 -         -

Now 1/(γ² - β²γ²) = 1

Therefore, expanding both sides gives:
                   
                     - -  -         - -            -  - -
                    | e |^T| γ  βγ 0 0 ||  γ  -βγ 0 0 || a |
ae + bf + cg + dh = | f || βγ γ  0 0 || -βγ  γ  0 0 || b |
                    | g || 0  0  1 0 ||  0   0  1 0 || c |
                    | h || 0  0  0 1 ||  0   0  0 1 || d |
                     - -  -         - -            -  - -

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

                  = LHS  Q.E.D.

          ~                  ~     ~
Note:  If x is complex then (x)^T)^* ≡ x^†

Define,

~
x = ηx where η is the metric.

Therefore,

~
x^Tx = (ηx)^Tx

Again, using the property that (AB)^T = B^TA^T we get:

~
x^Tx = x^Tη^Tx

   = x^Tηx  since η is always symmetric.  

Now consider the transformation:

x^Tηx -> (Λx)^TηΛx

Again, using the property that (AB)^T = B^TA^T we get:

x^Tηx -> x^TΛ^TηΛx

    -> x^TΛx i.f.f. Λ^TηΛ = η

Because elements with determinants ≠ +1 do not form a subgroup
of SO(3,1), the determinant of Λ is required to equal +1. 

In 3D Euclidean space η is the 3 x 3 identity matrix (≡ δⁱ_j).

In 4D flat spacetime η is the 4 x 4 Minkowski metric. 

In 4D curved spacetime the situation is a little more 
complicated.  Lorentz transformations only refer to 
transformations in the context of flat spacetime. In 
curved spacetime the assumption is that the spacetime 
can be considered to be locally flat.  In other words, 
the metric around any given point is arbitrarily close 
to the Minkowski metric within a small area around that 
point.  Essentially, there is a 4-vector in the tangent 
space at each spacetime point through which the particle's 
world line passes.  

Check Λ^TηΛ = η: 

 -           -  -        -  -           -
|  γ  -βγ 0 0 || -1 0 0 0 ||  γ  -βγ 0 0 |      
| -βγ  γ  0 0 ||  0 1 0 0 || -βγ  γ  0 0 |
|  0   0  1 0 ||  0 0 1 0 ||  0   0  1 0 |
|  0   0  0 1 ||  0 0 0 1 ||  0   0  0 1 |
 -           -  -        -  -           -

detΛ = γ² - β²γ² = 1

 -           -  -          -
|  γ  -βγ 0 0 || -γ  βγ 0 0 |      
| -βγ  γ  0 0 || -βγ γ  0 0 |
|  0   0  1 0 ||  0  0  1 0 |
|  0   0  0 1 ||  0  0  0 1 |
 -           -  -          -
 -                         -
| -(γ² - β²γ²)   0       0 0 |     
|      0     (γ² - β²γ²) 0 0 |
|      0        0     1 0 |
|      0        0     0 1 |
 -                         -

Now γ² - β²γ² = 1.  Therefore, this reduces to:

 -        -
| -1 0 0 0 |     
|  0 1 0 0 |  Q.E.D.
|  0 0 1 0 |
|  0 0 0 1 |
 -        -

Tensor Form
-----------

We have seen that:

x^μ -> (x')^μ = Λ^μ_νx^ν

Where Λ^μ_ν is the matrix 

 -             -
| Λ⁰₀ Λ⁰₁ Λ⁰₂ Λ⁰₃ |
| Λ¹₀ Λ¹₁ Λ¹₂ Λ¹₃ |
| Λ²₀ Λ²₁ Λ²₂ Λ²₃ |
| Λ³₀ Λ³₁ Λ³₂ Λ³₃ |
 -             -

and x^μ and x^ν are 4-vectors.

For a tensor with 2 indeces this becomes:

T^μν -> (T')^μν = Λ^μ_σΛ^ν_ρT^σρ

(We can think of this as (A')^μ(B')^ν = Λ^μ_σA^σΛ^ν_ρB^ρ)

Thus, for example:

η_σρ = η_μνΛ^μ_σΛ^ν_ρ

Therefore, each index gets its own transformation.  
Consider:

T^μν = Λ^μ_σΛ^ν_ρT^σρ

Λ^μ_σ and Λ^ν_ρ are transformation matrices.  The tensor 
components then transform as:

T^μν = ΣΣΛ^μ_σΛ^ν_ρT^σρ
      ^σ^ρ

Where σ amd ρ run from 0 to 3 for 4D spacetime.

  = Λ^μ₀Λ^ν₀T⁰⁰ + Λ^μ₀Λ^ν₁T⁰¹ + Λ^μ₁Λ^ν₀T¹⁰ + Λ^μ₁Λ^ν₁T¹¹ ... Λ^μ_σΛ^ν_ρT^σρ 

This is referred to as the INDEX FORM.

We can also write this in MATRIX FORM as:

T' = ΛTΛ^T

Proof:

To get the matrix form we can rearrange the terms
but we need to ensure that like indeces are adjacent 
to each other before contraction.  Therefore,  

T^μν = Λ^μ_σΛ^ν_ρT^σρ ≡ Λ^μ_σT^σρΛ^ν_ρ ≡ T^σρΛ^μ_σΛ^ν_ρ

T^μν = Λ^μ_σΛ^ν_ρT^σρ 

  = Λ^ν_ρΛ^μ_σT^σρ 

  = Λ^ν_ρ(ΛT)^μρ

  = Λ^ν_ρB^μρ  where B = ΛT

  = (ΛB^T)^νμ 

  = (BΛ^T)^μν

  = (ΛTΛ^T)^μν

Therefore,

T' = ΛTΛ^T

Inverse Notation
----------------

Λ^TηΛ = η

Multiply from left by η^-1 to get:

η^-1Λ^TηΛ = η^-1η

Therefore,

η^-1Λ^TηΛ = I

Multiply from the right by Λ^-1 to get:

η^-1Λ^TηΛΛ^-1 = IΛ^-1

Therefore,

η^-1Λ^Tη = Λ^-1

Lets look at the LHS and add back in the indeces.
Noting that (η^μν)^-1 = η_μν and that matrix 
multiplication requires getting repeated indeces 
adjacent to each other.

η_αν(Λ^μ_ν)^Tη^μβ -> η_ανΛ^ν_μη^μβ

            = Λ_α^β

This works for the Lorentz transformation because 
η has the property η = η^T = η^-1.

Therefore, in index notation (Λ^-1)^μ_ν ≡ Λ_ν^μ

Note that this is different to (Λ^T)^μ_ν ≡ Λ^ν_μ

       -           -  
    |  γ  -βγ 0 0 |     
Λ^μ_ν = | -βγ  γ  0 0 |
    |  0   0  1 0 |
    |  0   0  0 1 |
       -           -  

       -         -  
    | γ  βγ 0 0 |     
Λ_μ^σ = | βγ γ  0 0 |
    | 0  0  1 0 |
    | 0  0  0 1 |
       -         - 
      
         -        -
     | 1 0 0 0 |
Λ_μ^σΛ^μ_ν = | 0 1 0 0 | = δ^σ_ν
     | 0 0 1 0 |
     | 0 0 0 1 |
         -        -

If we define a contravariant 4-vector as (-ct,x,y,z) then
the vector transforms as:

A^μ = Λ^μ_νA^ν

A boost in the x direction is:

 -           -  -   -     -           -
|  γ  -βγ 0 0 || -ct |   | -γ(ct + βx) |
| -βγ  γ  0 0 ||   x | = |  γ(x + βct) |    
|  0   0  1 0 ||   y |   |      y      |
|  0   0  0 1 ||   z |   |      z      |
 -           -  -   -     -           -

If we define a covariant 4-vector as (ct,x,y,z) then the 
vector transforms as:

A_ν = Λ_ν^μA_μ

A boost in the x direction is:

 -          -  -  -     -          -
| γ  βγ  0 0 || ct |   | γ(ct + βx) |
| βγ  γ  0 0 ||  x | = | γ(x + βct) |    
| 0   0  1 0 ||  y |   |      y     |
| 0   0  0 1 ||  z |   |      z     |
 -          -  -  -     -          -

The product of 
 -                          -  -          -
| -γ(ct + βx) γ(x + βct) y z || γ(ct + βx) |
 -                          - | γ(x + βct) |
                              |     y      |
                              |     z      |
                               -          -

is:

-(γ(ct + βx))² + (γ(x + βct))² + y² + z²

γ²[-t² - β²x² - 2βxt + x² + β²t² + 2βxt] + y² + x²

γ²[-t²(1 - β²) + x²(1 - β²)] + y² + x²

γ²(1 - β²)[-t² + x²] + y² + x²

-t² + x² + y² + x² as expected (the invariant interval).

Infinitesimal Group Generators
------------------------------

We now look at the generators used to create the actual
transformations, Λ, that form the group.  Consider:

η_μνΛ^μ_σΛ^ν_ρ = η_σρ

The infinitesimal Lorentz transformation can be constructed
as:

Λ^μ_σ = δ^μ_σ + ω^μ_σ and Λ^ν_ρ = δ^ν_ρ + ω^ν_ρ where ω is a small 
quantity and δ^ν_σ is the Kronecker delta in tensor form.

Note:  In spacetime δ^ab or δ_ab don't make sense.  Since 
η_abδ^c_c = η_ac, δ^c_c acts as the identity matrix. 

η_μν(δ^μ_σ + ω^μ_σ)(δ^ν_ρ + ω^ν_ρ) = η_μν(δ^μ_σδ^ν_ρ + δ^μ_σω^ν_ρ + ω^μ_σδ^ν_ρ + O((ω)²)

                       = η_μν(δ^μ_σδ^ν_ρ + δ^μ_σω^ν_ρ + ω^μ_σδ^ν_ρ)

                       = η_μν(δ^μ_σδ^ν_ρ + δ^μ_σω^ν_ρ + δ^ν_ρω^μ_σ)

                       = η_μνδ^μ_σδ^ν_ρ + η_μνδ^μ_σω^ν_ρ + η_μνδ^ν_ρω^μ_σ

Now δ^μ_σ = η^μρη_ρσ and δ^ν_ρ = η^νση_σρ therefore η_μνη^μρη_ρση^νση_σρ = η_σρ

η_μν(δ^μ_σ + ω^μ_σ)(δ^ν_ρ + ω^ν_ρ) = η_σρ + ω_σρ + ω_ρσ

For this to be true ω_σρ + ω_ρσ = 0.  Therefore, ω_σρ = -ω_ρσ.  
Thus, the matrices of the infinitesimal generators of the 
Lorentz transformations are antisymmetric.  

We can write a basis of these six 4 × 4 antisymmetric 
matrices corresponding to the 3 rotations and 3 boosts 
as:

(M^ρσ)^μν = i(η^ρμη^σν - η^σμη^ρν)

Where ρ and σ indicate which generator and μ and ν are the 
matrix row and column.  Thus, M^σρ = -M^ρσ and the 6 matrices 
are:

M⁰¹ = -M¹⁰, M⁰² = -M²⁰, M⁰³ = -M³⁰, M¹¹ = -M¹¹, M¹² = -M²¹, M¹³ = -M³¹

In accordance with:

00 10 20 30
01 11 21 31
02 12 22 32
03 13 23 33

For example,

(M⁰¹)⁰¹ = i(η⁰⁰η¹¹ - η¹⁰η⁰¹) = -i

(M⁰¹)¹⁰ = i(η⁰¹η¹⁰ - η¹¹η⁰⁰) = i

If we continue we get the following:

       -        -
      | 0 -i 0 0 | 
(M⁰¹)^μν = | i  0 0 0 | 
      | 0  0 0 0 |
      | 0  0 0 0 |
       -        -

If we use these matrices for anything practical (for
example, if we want to multiply them together, or act 
on some field) we will typically need to lower one 
index.  We can do this using the Minkowskic metric as
follows:

η_βν(M^ρσ)^μβ = i(η^ρμη^σβη_βν - η^σμη^ρβη_βν)

(M^ρσ)^μ_ν = i(η^ρμδ^σ_ν - η^σμδ^ρ_ν)

Therefore, for example:

       -       -                            -          -
      | 0 i 0 0 |                          | 1  0  0  0 |
(M⁰¹)^μ_ν = | i 0 0 0 | after multiplying by η = | 0 -1  0  0 |
      | 0 0 0 0 |                          | 0  0 -1  0 |
      | 0 0 0 0 |                          | 0  0  0 -1 |
       -       -                            -          -

Or,

        -       -                                   
       | 0 1 0 0 |                                 
(M⁰¹)^μ_ν = i| 1 0 0 0 | 
       | 0 0 0 0 |                                 
       | 0 0 0 0 |                                 
        -       -          

M^ρσ are referred to as the GENERATORS of the Lorentz 
transformation.  

We are now in a position to write ω_σρ from before as:

ω^μ_ν = (1/2)Ω_ρσ(M^ρσ)^μ_ν

Where Ω_ρσ consists of 6 numbers corresponding to the 
6 generators.  They tell us what kind of Lorentz 
transformation we are performing (i.e., rotate by 
θ = π/7 about the z-direction and boost at speed 
v = 0.2c in the x direction.  Ω_ρσ is also antisymmetric 
in the indeces.

It is customary to define the rotation and boost 
generators using index notation as follows:

J_i = (1/2)ε_ijkM_jk

K_i = M_0i

The factor of (1/2) is needed to avoid double counting
as we sum over both j and k.  This is not an issue for
K_i because the sum is only over one index.  Thus, for 
J₁ we get:

J₁ = (1/2)[ε₁₂₃M₂₃ + ε₁₃₂M₃₂]

Since ε₁₂₃ = -ε₁₃₂ and M₂₃ = -M₃₂, the second term is 
additive.  

We can write:

Rotation about x:

      -        -
     | 0 0 0  0 | 
J₁ = i| 0 0 0  0 | ≡ (M²³)^μ_ν
     | 0 0 0 -1 |
     | 0 0 1  0 |
      -        -

Rotation about y:

      -       -
     | 0  0 0 0 | 
J₂ = i| 0  0 0 1 | ≡ (M¹³)^μ_ν
     | 0  0 0 0 |
     | 0 -1 0 0 |
      -       -

Rotation about z:

      -       -
     | 0 0  0 0 | 
J₃ = i| 0 0 -1 0 | ≡ (M¹²)^μ_ν
     | 0 1  0 0 |
     | 0 0  0 0 |
      -       -

Boost along x:

      -       -
     | 0 1 0 0 | 
K₁ = i| 1 0 0 0 | ≡ (M⁰¹)^μ_ν
     | 0 0 0 0 |
     | 0 0 0 0 |
      -       -

Boost along y:

      -       -
     | 0 0 1 0 | 
K₂ = i| 0 0 0 0 | ≡ (M⁰²)^μ_ν
     | 1 0 0 0 |
     | 0 0 0 0 |
      -       -

Boost along z:

      -       -
     | 0 0 0 1 | 
K₃ = i| 0 0 0 0 | ≡ (M⁰³)^μ_ν
     | 0 0 0 0 |
     | 1 0 0 0 |
      -       -

Lorentz generators can be added together, or multiplied 
by real numbers, to get more Lorentz generators. For 
example,

             -             -
            | 0  B_x  B_y  B_z |
B.K + θ.J = | B_x  0 -θ_z  θ_y |  
            | B_y  θ_z  0 -θ_x |
            | B_z -θ_y  θ_x  0 |
             -             -

Differential Operators
----------------------

Consider a boost along x.

     -          -  
    |  γ -βγ 0 0 |
Λ = | -βγ γ  0 0 | where β = v/c
    |  0  0  1 0 |
    |  0  0  0 1 | 
     -          -  

Λ = 1 + B_xf'(Λ_x) + ...

  = I + B_x∂Λ_x/∂B_x + ...

The derivative of the matrix is the matrix of the entries 
differentiated with respect to the same variable.  Thus,

          -         -
        |  0 -γ 0 0 |
∂Λ/∂B_x = | -γ  0 0 0 | 
        |  0  0 0 0 |
        |  0  0 0 0 |
          -         -

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

      = -γK_x

Therefore, we can write:

B_x(γ) = I - iγβK_x

Therefore, by comparison:

∂B/∂β = -γK_x

When β is small (= δ), γ ~ 1 and we get:

B_x(γ) = I - δK_x

Differential operators can also be used to represent 
infinitesimal generators.

x' = xcosθ - ysinθ  and y' = xsinθ + ycosθ 

For small θ's this becomes:

x' = x - ydθ and y' = xdθ + y

An arbitrary differentiable function F(x,y) then 
transforms as:

F(x',y') = F(x - ydθ,xdθ + y)

Using dF = (∂F/∂x)dθ + (∂F/∂y)dθ we get:

F(x',y') = F(x,y) + x(∂F/∂y)dθ - y(∂F/∂x)dθ

         = F(x,y) + {x(∂F/∂y) - y(∂F/∂x)}dθ

Therefore, we can associate infinitesimal rotations 
with the operator:

R = x(∂F/∂y) - y(∂F/∂x)

In general we can write:

M^ρσ = i(x^ρ∂^σ - x^σ∂^ρ)

Or, in terms of J and K:

J_x ≡ i(y∂_z - z∂_y)

J_y ≡ i(z∂_x - x∂_z) 

J_z ≡ i(x∂_y - y∂_x)

-K_x ≡ i(x∂_t + t∂_x)

-K_y ≡ i(y∂_t + t∂_y)

-K_z ≡ i(z∂_t + t∂_z)

Consider a rotation about the z-axix.  As noted before, 
the differential of a matrix is the differential of the 
indivividual elements.  Thus, for a rotation about the 
z-axis:

 -              -      -               -
| 1   0    0   0 |    | 0   0     0   0 |
| 0 cosθ -sinθ 0 | -> | 0 -sinθ -cosθ 0 |
| 0 sinθ  cosθ 0 |    | 0  cosθ -sinθ 0 |
| 0   0    0   1 |    | 0   0     0   0 |
 -              -      -               -

Therefore,

 -               -  - -     -              -
| 0   0     0   0 || t |   |        0       |
| 0 -sinθ -cosθ 0 || x | = | -xsinθ - ycosθ |
| 0  cosθ -sinθ 0 || y |   |  xcosθ - ysinθ |
| 0   0     0   0 || z |   |        0       |
 -               -  - -     -              -

The RHS is equivalent to:

i∂_x(xcosθ - ysinθ) = icosθ and i∂_y(xcosθ - ysinθ) = -isinθ

Therefore,

i(x∂_y - y∂_x) = -ixsinθ - iycosθ

i∂_x(xsinθ + ycosθ) = isinθ and i∂_y(xsinθ + ycosθ) = icosθ

Therefore,

i(x∂_y - y∂_x) = ixcosθ - iysinθ

Setting θ = 0 yields:

  -        -  
 | 0 0  0 0 |  
i| 0 0 -1 0 | = J_z
 | 0 1  0 0 |   
 | 0 0  0 0 | 
  -        -  

Similarly, for a boost in the x direction:

 -                 -      -                 -
|  coshζ -sinhζ 0 0 |    |  sinhζ -coshζ 0 0 |
| -sinhζ  coshζ 0 0 | -> | -coshζ  sinhζ 0 0 |
|    0      0   1 0 |    |    0      0   0 0 |       
|    0      0   0 1 |    |    0      0   0 0 |  
 -                 -      -                 -

Therefore,

 -                 -  - -     -                -
|  sinhζ -coshζ 0 0 || t |   |  tsinhζ - xcoshζ | 
| -coshζ  sinhζ 0 0 || x | = | -tcoshζ + xsinhζ | 
|    0      0   0 0 || y |   |         0        |
|    0      0   0 0 || z |   |         0        |
 -                 -  - -     -                -

i∂_t(tcoshζ - xsinhζ) = icoshζ and i∂_x(tcoshζ - xsinhζ) = -isinhζ

Therefore,

i(t∂_x + x∂_t) = -itsinhζ + ixcoshζ

i∂_t(-tsinhζ + xcoshζ) = -isinhζ and i∂_x(-tsinhζ + xcoshζ) = icoshζ

Therefore,

i(t∂_x + x∂_t) = itcoshζ - ixsinhζ

Setting ζ = 0 yields:

  -       -  
 | 0 1 0 0 |  
i| 1 0 0 0 | = K_x
 | 0 0 0 0 |   
 | 0 0 0 0 | 
  -       -  

These are the classical generators of angular momentum 
generalized to include time.  They illustrate how to 
pass between matrix and vector representations of elements 
of the Lie algebra.

We showed before that (M^ρσ)^μν = i(η^ρμη^σν - η^σμη^ρν).  If we 
multiply both sides by η_μν we get:

M^ρσ = i(η^ρσ - η^σρ)

Now, also from before,

M^ρσ = i(x^ρ∂^σ - x^σ∂^ρ)

Therefore, we can conclude that:

η^ρσ ≡ x^ρ∂^σ and η^σρ ≡ x^σ∂^ρ

The Lorentz Algebra
-------------------

It can be shown that the generators obey the LORENTZ ALGEBRA
relations:

[M^ρσ,M^αβ] = i(η^σαM^ρβ - η^ραM^σβ + η^ρβM^σα - η^σβM^ρα) ... A.

Proof:

[M^ρσ,M^αβ] = i²{(x^ρ∂^σ - x^σ∂^ρ)(x^α∂^β - x^β∂^α) - (x^α∂^β - x^β∂^α)(x^ρ∂^σ - x^σ∂^ρ)}

      = -{(x^ρ∂^σx^α∂^β - x^ρ∂^σx^β∂^α - x^σ∂^ρx^α∂^β + x^σ∂^ρx^β∂^α) 
             - (x^α∂^βx^ρ∂^σ - x^α∂^βx^σ∂^ρ - x^β∂^αx^ρ∂^σ + x^β∂^αx^σ∂^ρ)}

      = -{(x^ρη^σα∂^β - x^ρη^σβ∂^α - x^ση^ρα∂^β + x^ση^ρβ∂^α) 
             - (x^αη^βρ∂^σ - x^αη^βσ∂^ρ - x^βη^αρ∂^σ + x^βη^ασ∂^ρ)}

      = -{x^ρη^σα∂^β - x^ρη^σβ∂^α - x^ση^ρα∂^β + x^ση^ρβ∂^α 
             - x^αη^βρ∂^σ + x^αη^βσ∂^ρ + x^βη^αρ∂^σ - x^βη^ασ∂^ρ}

      = -{η^σα(x^ρ∂^β - x^β∂^ρ) - η^ρα(x^σ∂^β - x^β∂^σ) + η^ρβ(x^σ∂^α - x^α∂^σ) 
             - η^σβ(x^ρ∂^α - x^α∂^ρ)} 

Using the fact that () = M/i = -iM this becomes:

[M^ρσ,M^αβ] = i{η^σαM^ρβ - η^ραM^σβ + η^ρβM^σα - η^σβM^ρα} 

We could also have derived this using i(η^ρμδ^σ_ν - η^σμδ^ρ_ν) as:

[M^ρσ,M^αβ]^μ_ν = (M^ρσ)^μ_λ(M^αβ)^λ_ν - (M^αβ)^μ_λ(M^ρσ)^λ_ν

      = i²{(η^ρμδ^σ_λ - η^σμδ^ρ_λ)(η^αλδ^β_ν - η^βλδ^α_ν) 
                - (η^αμδ^β_λ - η^βμδ^α_λ)(η^ρλδ^σ_ν - η^σλδ^ρ_ν)}

      = -{(η^ρμδ^σ_λη^αλδ^β_ν - η^ρμδ^σ_λη^βλδ^α_ν - η^σμδ^ρ_λη^αλδ^β_ν + η^σμδ^ρ_λη^βλδ^α_ν)
                - (η^αμδ^β_λη^ρλδ^σ_ν - η^αμδ^β_λη^σλδ^ρ_ν - η^βμδ^α_λη^ρλδ^σ_ν + η^βμδ^α_λη^σλδ^ρ_ν)}

Contracting the middle δ and η gives:

[M^ρσ,M^αβ]^μ_ν = -{(η^ρμη^σαδ^β_ν - η^ρμη^σβδ^α_ν - η^σμη^ραδ^β_ν + η^σμη^ρβδ^α_ν)
                - (η^αμη^βρδ^σ_ν - η^αμη^βσδ^ρ_ν - η^βμη^αρδ^σ_ν + η^βμη^ασδ^ρ_ν)}

Rearranging and factoring gives:

[M^ρσ,M^αβ]^μ_ν = -{η^σα(η^ρμδ^β_ν - η^βμδ^ρ_ν) - η^σβ(η^ρμδ^α_ν - η^αμδ^ρ_ν) 
                - η^ρα(η^σμδ^β_ν - η^βμδ^σ_ν) + η^ρβ(η^σμδ^α_ν - η^αμδ^σ_ν)} 

Again, using the fact that () = M/i = -iM this becomes:

[M^ρσ,M^αβ]^μ_ν = i{η^σα(M^ρβ)^μ_ν - η^σβ(M^ρα)^μ_ν - η^ρα(M^σβ)^μ_ν + η^ρβ(M^σα)^μ_ν} 

The LIE ALGEBRA for the Lorentz group SO(3,1) can be obtained 
using A. as:

(M^ρσ)^μν = i(η^ρμη^σν - η^σμη^ρν)

[J_i,J_j] = iε^ijkJ_k

[J_i,K_j] = iε^ijkK_k

[K_i,K_j] = -iε^ijkJ_k

Note:  Boosts alone cannot be a subgroup of SO(1,3) since 
combinations involving boosts result in rotations!

Generation of a Lorentz Transform
---------------------------------

Now that we have described the generators of the Lorentz
group we need to explore how these generators actually 
produce a specific finite transformation, Λ

We will go through the exercise for a boost in the x 
direction and a rotation around the x axis.  

The boost generator for the x direction, K₁, is:

       -       -
     | 0 1 0 0 |
K₁ = i| 1 0 0 0 |
     | 0 0 0 0 |
     | 0 0 0 0 |
       -       -

A boost along x is constructed as n consecutive boosts 
of δ.  Thus,

ζ = nδ

Λ = lim_n->∞(I + (iζ/n)K₁)ⁿ 

  = exp(iζK₁) from the lemma lim_n->∞(1 + A/n)ⁿ = exp(A)

But exp(iζK₁) also can be written as the Taylor series:

Λ = exp(iζK₁) = 1 + iζK₁ + (iζK₁)²/2! - (iζK₁)³/3! + (iζK₁)⁴/4! 

                - (iζB₁)⁵/5! + (iζB₁)⁶/6!  

Now, let,

       -         -
     |  0 -1 0 0 |
iK₁ = | -1  0 0 0 | = B₁
     |  0  0 0 0 |
     |  0  0 0 0 |
       -         -
So,
       -       -
    | 1 0 0 0 |
B₁² = | 0 1 0 0 | 
    | 0 0 0 0 |
    | 0 0 0 0 |
       -       -

It is easy to show:

B₁ = B₁³ = B₁⁵ ...

B₁² = B₁⁴ = B₁⁶ ...

Therefore,

Λ = exp(ζB₁) = 1 + (ζ + ζ³/3! + ζ⁵/5! ...)B₁

                + (ζ²/2! + ζ⁴/4! + ζ⁶/6! ...)B₁²

            = 1 - B₁² + (ζ + ζ³/3! + ζ⁵/5! ...)B₁

                + (1 + ζ²/2! + ζ⁴/4! + ζ⁶/6! ...)B₁²

            = 1 - B₁² + B₁sinh(ζ) + B₁²cosh(ζ)

           -      -      -      -      -      -
        | 1 0 0 0 |   | 1 0 0 0 |   | 0 0 0 0 |
I - B₁² = | 0 1 0 0 | - | 0 1 0 0 | = | 0 0 0 0 |
        | 0 0 1 0 |   | 0 0 0 0 |   | 0 0 1 0 |
        | 0 0 0 1 |   | 0 0 0 0 |   | 0 0 0 1 |
           -      -      -      -      -      -

                -                     -
              |  cosh(ζ) -sinh(ζ) 0 0 |
Λ = exp(ζB₁) = | -sinh(ζ)  cosh(ζ) 0 0 |
              |     0        0    1 0 |
              |     0        0    0 1 |
                -                     -

               -           -
             |  γ  -βγ 0 0 |
           = | -βγ  γ  0 0 |
             |  0   0  1 0 |
             |  0   0  0 1 |
               -           -

The rotation generator around the x axis, J₁, is:

      -        -
     | 0 0 0  0 | 
J₁ = i| 0 0 0  0 | 
     | 0 0 0 -1 |
     | 0 0 1  0 |
      -        -

A rotation by a finite angle, θ, is constructed as n
consecutive rotations of θ/n.  Thus,

Λ = lim_n->∞(I + (iθ/n)J₁)ⁿ 

  = exp(iθJ₁) from the lemma lim_n->∞(1 + A/n)ⁿ = exp(A)

Again this can be expanded as a Taylor series:

Λ = exp(iθJ₁) = 1 + iθJ₁ + (iθJ₁)²/2! + (iθJ₁)³/3! + (iθJ₁)⁴/4! 

               + (iθJ₁)⁵/5! + (iθJ₁)⁶/6!  

Now, let,

       -        -
     | 0 0  0 0 |
iJ₁ = | 0 0  0 0 | = R₁
     | 0 0  0 1 |
     | 0 0 -1 0 |
       -        -
So,
       -         -
    | 0 0  0  0 |
R₁² = | 0 0  0  0 | 
    | 0 0 -1  0 |
    | 0 0  0 -1 |
       -         -

It is easy to show:

R₁ = -R₁³ = R₁⁵ ...

R₁² = -R₁⁴ = R₁⁶ ...

Therefore,

Λ = exp(θR₁) = 1 + θR₁ + θ²R₁²/2! - θ³R₁/3! - θ⁴R₁²/4! 

               + θ⁵R₁/5! + θ⁶R₁²/6!  

           = 1 + (θ - θ³/3! + θ⁵/5! ...)R₁ 

               - (-θ²/2! + θ⁴/4! - θ⁶/6! ...)R₁²  

           = 1 + R₁² + (θ - θ³/3! + θ⁵/5! ...)R₁ 

               - (1 - θ²/2! + θ⁴/4! - θ⁶/6! ...)R₁²  

           = 1 - R₁² + R₁sin(θ) - R₁²cos(θ)

           -      -      -        -      -      -
        | 1 0 0 0 |   | 0 0  0  0 |   | 1 0 0 0 |
I + R₁² = | 0 1 0 0 | + | 0 0  0  0 | = | 0 1 0 0 |
        | 0 0 1 0 |   | 0 0 -1  0 |   | 0 0 0 0 |
        | 0 0 0 1 |   | 0 0  0 -1 |   | 0 0 0 0 |
           -      -      -        -      -      -

                -                  -
              | 1 0   0      0     |
Λ = exp(θR₁) = | 0 1   0      0     |
              | 0 0  cos(θ) sin(θ) |
              | 0 0 -sin(θ) cos(θ) |
                -                  -
Summary
-------

In summary, a finite Lorentz transformation can be written 
as:

Λ = exp(iθ.J + iζ.K)