Wolfram Alpha:

```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 = xTηx

-       -  -  -
= | t x y z || -t |
-       - |  x |
|  y |
|  z |
-  -
= xμημνxν

= xμxν

= 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,

xμ' = Λμνxν

Lorentz transformation leave the inner product invariant.
Thus,

x'.x' = x.x

The set of all matrices obeying ΛTηΛ = 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,

~             ~
xTx -> ((Λ-1)Tx))TΛx

Using the property that (AB)T = BTAT we get:

~
-> xTΛ-1Λx

~      ~
Check xTx -> xTΛ-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,

~
xTx = (ηx)Tx

Again, using the property that (AB)T = BTAT we get:

~
xTx = xTηTx

= xTηx  since η is always symmetric.

Now consider the transformation:

xTηx -> (Λx)TηΛx

Again, using the property that (AB)T = BTAT we get:

xTηx -> xTΛTηΛx

-> xTΛx i.f.f. ΛTηΛ = 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 (≡ δij).

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 ΛTηΛ = 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:

xμ -> (x')μ = Λμνxν

Where Λμν is the matrix

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

and xμ and xν are 4-vectors.

For a tensor with 2 indeces this becomes:

Tμν -> (T')μν = ΛμσΛνρTσρ

(We can think of this as (A')μ(B')ν = ΛμσAσΛνρBρ)

Thus, for example:

ησρ = ημνΛμσΛνρ

Therefore, each index gets its own transformation.
Consider:

Tμν = ΛμσΛνρTσρ

Λμσ and Λνρ are transformation matrices.  The tensor
components then transform as:

Tμν = ΣΣΛμσΛνρTσρ
σρ

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

= Λμ0Λν0T00 + Λμ0Λν1T01 + Λμ1Λν0T10 + Λμ1Λν1T11 ... ΛμσΛνρTσρ

This is referred to as the INDEX FORM.

We can also write this in MATRIX FORM as:

T' = ΛTΛT

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,

Tμν = ΛμσΛνρTσρ ≡ ΛμσTσρΛνρ ≡ TσρΛμσΛνρ

Tμν = ΛμσΛνρTσρ

= ΛνρΛμσTσρ

= Λνρ(ΛT)μρ

= ΛνρBμρ  where B = ΛT

= (ΛBT)νμ

= (BΛT)μν

= (ΛTΛT)μν

Therefore,

T' = ΛTΛT

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

ΛTηΛ = η

Multiply from left by η-1 to get:

η-1ΛTηΛ = η-1η

Therefore,

η-1ΛTηΛ = I

Multiply from the right by Λ-1 to get:

η-1ΛTηΛΛ-1 = IΛ-1

Therefore,

η-1ΛTη = Λ-1

Lets look at the LHS and add back in the indeces.
Noting that (ημν)-1 = ημν and that matrix
multiplication requires getting repeated indeces

ηαν(Λμν)Tημβ -> ηανΛνμημβ

= Λαβ

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

Therefore, in index notation (Λ-1)μν ≡ Λνμ

Note that this is different to (ΛT)μν ≡ Λνμ

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

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

-        -
| 1 0 0 0 |
ΛμσΛμν = | 0 1 0 0 | = δσν
| 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:

Aμ = ΛμνAν

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:

Aν = ΛνμAμ

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:

ημνΛμσΛνρ = ησρ

The infinitesimal Lorentz transformation can be constructed
as:

Λμσ = δμσ + ωμσ and Λνρ = δνρ + ωνρ where ω is a small
quantity and δνσ is the Kronecker delta in tensor form.

Note:  In spacetime δab or δab don't make sense.  Since
ηabδcc = ηac, δcc acts as the identity matrix.

ημν(δμσ + ωμσ)(δνρ + ωνρ) = ημν(δμσδνρ + δμσωνρ + ωμσδνρ + O((ω)2)

= ημν(δμσδνρ + δμσωνρ + ωμσδνρ)

= ημν(δμσδνρ + δμσωνρ + δνρωμσ)

= ημνδμσδνρ + ημνδμσωνρ + ημνδνρωμσ

Now δμσ = ημρηρσ and δνρ = ηνσησρ therefore ημνημρηρσηνσησρ = ησρ

ημν(δμσ + ωμσ)(δνρ + ωνρ) = ησρ + ωσρ + ωρσ

For this to be true ωσρ + ωρσ = 0.  Therefore, ωσρ = -ωρσ.
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:

(Mρσ)μν = i(ηρμησν - ησμηρν)

Where ρ and σ indicate which generator and μ and ν are the
matrix row and column.  Thus, Mσρ = -Mρσ 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:

ηβν(Mρσ)μβ = i(ηρμησβηβν - ησμηρβηβν)

(Mρσ)μν = i(ηρμδσν - ησμδρν)

Therefore, for example:

-       -                            -          -
| 0 i 0 0 |                          | 1  0  0  0 |
(M01)μν = | 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)μν = i| 1 0 0 0 |
| 0 0 0 0 |
| 0 0 0 0 |
-       -

Mρσ are referred to as the GENERATORS of the Lorentz
transformation.

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

ωμν = (1/2)Ωρσ(Mρσ)μν

Where Ωρσ 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.  Ωρσ is also antisymmetric
in the indeces.

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

Ji = (1/2)εijkMjk

Ki = M0i

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
Ki 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

We can write:

-        -
| 0 0 0  0 |
J1 = i| 0 0 0  0 | ≡ (M23)μν
| 0 0 0 -1 |
| 0 0 1  0 |
-        -

-       -
| 0  0 0 0 |
J2 = i| 0  0 0 1 | ≡ (M13)μν
| 0  0 0 0 |
| 0 -1 0 0 |
-       -

-       -
| 0 0  0 0 |
J3 = i| 0 0 -1 0 | ≡ (M12)μν
| 0 1  0 0 |
| 0 0  0 0 |
-       -

Boost along x:

-       -
| 0 1 0 0 |
K1 = i| 1 0 0 0 | ≡ (M01)μν
| 0 0 0 0 |
| 0 0 0 0 |
-       -

Boost along y:

-       -
| 0 0 1 0 |
K2 = i| 0 0 0 0 | ≡ (M02)μν
| 1 0 0 0 |
| 0 0 0 0 |
-       -

Boost along z:

-       -
| 0 0 0 1 |
K3 = i| 0 0 0 0 | ≡ (M03)μν
| 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   Bx  By  Bz |
B.K + θ.J = | Bx  0  -θz  θy |
| By  θz  0  -θx |
| Bz -θy  θx  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 + Bxf'(Λx) + ...

= I + Bx∂Λx/∂Bx + ...

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

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

= -γKx

Therefore, we can write:

Bx(γ) = I - iγβKx

Therefore, by comparison:

∂B/∂β = -γKx

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

Bx(γ) = I - δKx

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:

Mρσ = i(xρ∂σ - xσ∂ρ)

Or, in terms of J and K:

Jx ≡ i(y∂z - z∂y)

Jy ≡ i(z∂x - x∂z)

Jz ≡ i(x∂y - y∂x)

-Kx ≡ i(x∂t + t∂x)

-Ky ≡ i(y∂t + t∂y)

-Kz ≡ 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 | = Jz
| 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 | = Kx
| 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 (Mρσ)μν = i(ηρμησν - ησμηρν).  If we
multiply both sides by ημν we get:

Mρσ = i(ηρσ - ησρ)

Now, also from before,

Mρσ = i(xρ∂σ - xσ∂ρ)

Therefore, we can conclude that:

ηρσ ≡ xρ∂σ and ησρ ≡ xσ∂ρ

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

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

[Mρσ,Mαβ] = i(ησαMρβ - ηραMσβ + ηρβMσα - ησβMρα) ... A.

Proof:

[Mρσ,Mαβ] = i2{(xρ∂σ - xσ∂ρ)(xα∂β - xβ∂α) - (xα∂β - xβ∂α)(xρ∂σ - xσ∂ρ)}

= -{(xρ∂σxα∂β - xρ∂σxβ∂α - xσ∂ρxα∂β + xσ∂ρxβ∂α)
- (xα∂βxρ∂σ - xα∂βxσ∂ρ - xβ∂αxρ∂σ + xβ∂αxσ∂ρ)}

= -{(xρησα∂β - xρησβ∂α - xσηρα∂β + xσηρβ∂α)
- (xαηβρ∂σ - xαηβσ∂ρ - xβηαρ∂σ + xβηασ∂ρ)}

= -{xρησα∂β - xρησβ∂α - xσηρα∂β + xσηρβ∂α
- xαηβρ∂σ + xαηβσ∂ρ + xβηαρ∂σ - xβηασ∂ρ}

= -{ησα(xρ∂β - xβ∂ρ) - ηρα(xσ∂β - xβ∂σ) + ηρβ(xσ∂α - xα∂σ)
- ησβ(xρ∂α - xα∂ρ)}

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

[Mρσ,Mαβ] = i{ησαMρβ - ηραMσβ + ηρβMσα - ησβMρα}

We could also have derived this using i(ηρμδσν - ησμδρν) as:

[Mρσ,Mαβ]μν = (Mρσ)μλ(Mαβ)λν - (Mαβ)μλ(Mρσ)λν

= i2{(ηρμδσλ - ησμδρλ)(ηαλδβν - ηβλδαν)
- (ηαμδβλ - ηβμδαλ)(ηρλδσν - ησλδρν)}

= -{(ηρμδσληαλδβν - ηρμδσληβλδαν - ησμδρληαλδβν + ησμδρληβλδαν)
- (ηαμδβληρλδσν - ηαμδβλησλδρν - ηβμδαληρλδσν + ηβμδαλησλδρν)}

Contracting the middle δ and η gives:

[Mρσ,Mαβ]μν = -{(ηρμησαδβν - ηρμησβδαν - ησμηραδβν + ησμηρβδαν)
- (ηαμηβρδσν - ηαμηβσδρν - ηβμηαρδσν + ηβμηασδρν)}

Rearranging and factoring gives:

[Mρσ,Mαβ]μν = -{ησα(ηρμδβν - ηβμδρν) - ησβ(ηρμδαν - ηαμδρν)
- ηρα(ησμδβν - ηβμδσν) + ηρβ(ησμδαν - ηαμδσν)}

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

[Mρσ,Mαβ]μν = i{ησα(Mρβ)μν - ησβ(Mρα)μν - ηρα(Mσβ)μν + ηρβ(Mσα)μν}

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

(Mρσ)μν = i(ηρμησν - ησμηρν)

[Ji,Jj] = iεijkJk

[Ji,Kj] = iεijkKk

[Ki,Kj] = -iεijkJk

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 = limn->∞(I + (iζ/n)K1)n

= exp(iζK1) from the lemma limn->∞(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 = limn->∞(I + (iθ/n)J1)n

= exp(iθJ1) from the lemma limn->∞(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)
```