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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^{T}ηx
- - - -
= | t x y z || -t |
- - | x |
| y |
| z |
- -
= x^{μ}η_{μν}x^{ν}
= x^{μ}x_{ν}
= t^{2} - x^{2} - y^{2} - z^{2}
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}_{ν}x^{ν}
Lorentz transformations leave the inner product invariant.
Thus,
x'.x' = x.x
The set of all matrices obeying Λ^{T}ηΛ = 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})^{T}x
Therefore,
~ ^{ }~
x^{T}x -> ((Λ^{-1})^{T}x))^{T}Λx
Using the property that (AB)^{T} = B^{T}A^{T} we get:
~
-> x^{T}Λ^{-1}Λx
~ ~
Check x^{T}x -> x^{T}Λ^{-1}Λx:
- - - - - -
| 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/(γ^{2} - β^{2}γ^{2}))| βγ γ 0 0 |
^{ } | 0 0 1 0 |
^{ } | 0 0 0 1 |
^{ } - -
Now 1/(γ^{2} - β^{2}γ^{2}) = 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^{T}x = (ηx)^{T}x
Again, using the property that (AB)^{T} = B^{T}A^{T} we get:
~
x^{T}x = x^{T}η^{T}x
^{ } = x^{T}ηx since η is always symmetric.
Now consider the transformation:
x^{T}ηx -> (Λx)^{T}ηΛx
Again, using the property that (AB)^{T} = B^{T}A^{T} we get:
x^{T}ηx -> x^{T}Λ^{T}ηΛx
^{ } -> x^{T}Λx i.f.f. Λ^{T}ηΛ = 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 (≡ δ^{i}_{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^{T}ηΛ = 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 = γ^{2} - β^{2}γ^{2} = 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 |
- - - -
- -
| -(γ^{2} - β^{2}γ^{2}) 0 0 0 |
| 0 (γ^{2} - β^{2}γ^{2}) 0 0 |
| 0 0^{ } 1 0 |
| 0 0^{ } 0 1 |
- -
Now γ^{2} - β^{2}γ^{2} = 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}_{ν}x^{ν}
Where Λ^{μ}_{ν} is the matrix
- -
| Λ^{0}_{0} Λ^{0}_{1} Λ^{0}_{2} Λ^{0}_{3} |
| Λ^{1}_{0} Λ^{1}_{1} Λ^{1}_{2} Λ^{1}_{3} |
| Λ^{2}_{0} Λ^{2}_{1} Λ^{2}_{2} Λ^{2}_{3} |
| Λ^{3}_{0} Λ^{3}_{1} Λ^{3}_{2} Λ^{3}_{3} |
- -
and x^{μ} and x^{ν} are 4-vectors.
For a tensor with 2 indeces this becomes:
T^{μν} -> (T')^{μν} = Λ^{μ}_{σ}Λ^{ν}_{ρ}T^{σρ}
(We can think of this as (A')^{μ}(B')^{ν} = Λ^{μ}_{σ}A^{σ}Λ^{ν}_{ρ}B^{ρ})
Thus, for example:
η_{σρ} = η_{μν}Λ^{μ}_{σ}Λ^{ν}_{ρ}
Therefore, each index gets its own transformation.
Consider:
T^{μν} = Λ^{μ}_{σ}Λ^{ν}_{ρ}T^{σρ}
Λ^{μ}_{σ} and Λ^{ν}_{ρ} are transformation matrices. The tensor
components then transform as:
T^{μν} = ΣΣΛ^{μ}_{σ}Λ^{ν}_{ρ}T^{σρ}
^{σ}^{ρ}
Where σ amd ρ run from 0 to 3 for 4D spacetime.
_{ } = Λ^{μ}_{0}Λ^{ν}_{0}T^{00} + Λ^{μ}_{0}Λ^{ν}_{1}T^{01} + Λ^{μ}_{1}Λ^{ν}_{0}T^{10} + Λ^{μ}_{1}Λ^{ν}_{1}T^{11} ... Λ^{μ}_{σ}Λ^{ν}_{ρ}T^{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}_{σ}Λ^{ν}_{ρ}T^{σρ} ≡ Λ^{μ}_{σ}T^{σρ}Λ^{ν}_{ρ} ≡ T^{σρ}Λ^{μ}_{σ}Λ^{ν}_{ρ}
T^{μν} = Λ^{μ}_{σ}Λ^{ν}_{ρ}T^{σρ}
_{ } = Λ^{ν}_{ρ}Λ^{μ}_{σ}T^{σρ}
_{ } = Λ^{ν}_{ρ}(ΛT)^{μρ}
_{ } = Λ^{ν}_{ρ}B^{μρ} where B = ΛT
_{ } = (ΛB^{T})^{νμ}
_{ } = (BΛ^{T})^{μν}
_{ } = (ΛTΛ^{T})^{μν}
Therefore,
T' = ΛTΛ^{T}
Inverse Notation
----------------
Λ^{T}ηΛ = η
Multiply from left by η^{-1} to get:
η^{-1}Λ^{T}ηΛ = η^{-1}η
Therefore,
η^{-1}Λ^{T}ηΛ = I
Multiply from the right by Λ^{-1} to get:
η^{-1}Λ^{T}ηΛΛ^{-1} = IΛ^{-1}
Therefore,
η^{-1}Λ^{T}η = 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})^{T}η^{μβ} -> η_{αν}Λ^{ν}_{μ}η^{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}^{σ}Λ^{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}_{ν}A^{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}^{μ}A_{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} + y^{2} + z^{2}
γ^{2}[-t^{2} - β^{2}x^{2} - 2βxt + x^{2} + β^{2}t^{2} + 2βxt] + y^{2} + x^{2}
γ^{2}[-t^{2}(1 - β^{2}) + x^{2}(1 - β^{2})] + y^{2} + x^{2}
γ^{2}(1 - β^{2})[-t^{2} + x^{2}] + y^{2} + x^{2}
-t^{2} + x^{2} + y^{2} + x^{2} 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_{μν}Λ^{μ}_{σ}Λ^{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_{ab}δ^{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}_{σ}δ^{ν}_{ρ} + δ^{μ}_{σ}ω^{ν}_{ρ} + ω^{μ}_{σ}δ^{ν}_{ρ} + O((ω)^{2})
_{ } = η_{μν}(δ^{μ}_{σ}δ^{ν}_{ρ} + δ^{μ}_{σ}ω^{ν}_{ρ} + ω^{μ}_{σ}δ^{ν}_{ρ})
_{ } = η_{μν}(δ^{μ}_{σ}δ^{ν}_{ρ} + δ^{μ}_{σ}ω^{ν}_{ρ} + δ^{ν}_{ρ}ω^{μ}_{σ})
_{ } = η_{μν}δ^{μ}_{σ}δ^{ν}_{ρ} + η_{μν}δ^{μ}_{σ}ω^{ν}_{ρ} + η_{μν}δ^{ν}_{ρ}ω^{μ}_{σ}
Now δ^{μ}_{σ} = η^{μρ}η_{ρσ} and δ^{ν}_{ρ} = η^{νσ}η_{σρ} therefore η_{μν}η^{μρ}η_{ρσ}η^{νσ}η_{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^{ρμ}η^{σν} - η^{σμ}η^{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:
M^{01} = -M^{10}, M^{02} = -M^{20}, M^{03} = -M^{30}, M^{11} = -M^{11}, M^{12} = -M^{21}, M^{13} = -M^{31}
In accordance with:
00 10 20 30
01 11 21 31
02 12 22 32
03 13 23 33
For example,
(M^{01})^{01} = i(η^{00}η^{11} - η^{10}η^{01}) = -i
(M^{01})^{10} = i(η^{01}η^{10} - η^{11}η^{00}) = i
If we continue we get the following:
_{ } - -
_{ } | 0 -i 0 0 |
(M^{01})^{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^{ρμ}η^{σβ}η_{βν} - η^{σμ}η^{ρβ}η_{βν})
(M^{ρσ})^{μ}_{ν} = i(η^{ρμ}δ^{σ}_{ν} - η^{σμ}δ^{ρ}_{ν})
Therefore, for example:
_{ } - - - -
_{ } | 0 i 0 0 | | 1 0 0 0 |
(M^{01})^{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 |
(M^{01})^{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_{ijk}M_{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
J_{1} we get:
J_{1} = (1/2)[ε_{123}M_{23} + ε_{132}M_{32}]
Since ε_{123} = -ε_{132} and M_{23} = -M_{32}, the second term is
additive.
We can write:
Rotation about x:
_{ } - -
_{ } | 0 0 0 0 |
J_{1} = i| 0 0 0 0 | ≡ (M^{23})^{μ}_{ν}
_{ } | 0 0 0 -1 |
_{ } | 0 0 1 0 |
_{ } - -
Rotation about y:
_{ } - -
_{ } | 0 0 0 0 |
J_{2} = i| 0 0 0 1 | ≡ (M^{13})^{μ}_{ν}
_{ } | 0 0 0 0 |
_{ } | 0 -1 0 0 |
_{ } - -
Rotation about z:
_{ } - -
_{ } | 0 0 0 0 |
J_{3} = i| 0 0 -1 0 | ≡ (M^{12})^{μ}_{ν}
_{ } | 0 1 0 0 |
_{ } | 0 0 0 0 |
_{ } - -
Boost along x:
_{ } - -
_{ } | 0 1 0 0 |
K_{1} = i| 1 0 0 0 | ≡ (M^{01})^{μ}_{ν}
_{ } | 0 0 0 0 |
_{ } | 0 0 0 0 |
_{ } - -
Boost along y:
_{ } - -
_{ } | 0 0 1 0 |
K_{2} = i| 0 0 0 0 | ≡ (M^{02})^{μ}_{ν}
_{ } | 1 0 0 0 |
_{ } | 0 0 0 0 |
_{ } - -
Boost along z:
_{ } - -
_{ } | 0 0 0 1 |
K_{3} = i| 0 0 0 0 | ≡ (M^{03})^{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_{x}f'(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^{ρμ}η^{σν} - η^{σμ}η^{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^{σα}M^{ρβ} - η^{ρα}M^{σβ} + η^{ρβ}M^{σα} - η^{σβ}M^{ρα}) ... A.
Proof:
[M^{ρσ},M^{αβ}] = i^{2}{(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}∂^{σ}x^{α}∂^{β} - x^{ρ}∂^{σ}x^{β}∂^{α} - x^{σ}∂^{ρ}x^{α}∂^{β} + x^{σ}∂^{ρ}x^{β}∂^{α})
- (x^{α}∂^{β}x^{ρ}∂^{σ} - x^{α}∂^{β}x^{σ}∂^{ρ} - x^{β}∂^{α}x^{ρ}∂^{σ} + x^{β}∂^{α}x^{σ}∂^{ρ})}
^{ } = -{(x^{ρ}η^{σα}∂^{β} - x^{ρ}η^{σβ}∂^{α} - x^{σ}η^{ρα}∂^{β} + x^{σ}η^{ρβ}∂^{α})
- (x^{α}η^{βρ}∂^{σ} - x^{α}η^{βσ}∂^{ρ} - x^{β}η^{αρ}∂^{σ} + x^{β}η^{ασ}∂^{ρ})}
^{ } = -{x^{ρ}η^{σα}∂^{β} - x^{ρ}η^{σβ}∂^{α} - x^{σ}η^{ρα}∂^{β} + x^{σ}η^{ρβ}∂^{α}
- x^{α}η^{βρ}∂^{σ} + x^{α}η^{βσ}∂^{ρ} + x^{β}η^{αρ}∂^{σ} - x^{β}η^{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^{σα}M^{ρβ} - η^{ρα}M^{σβ} + η^{ρβ}M^{σα} - η^{σβ}M^{ρα}}
We could also have derived this using i(η^{ρμ}δ^{σ}_{ν} - η^{σμ}δ^{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}
^{ } = i^{2}{(η^{ρμ}δ^{σ}_{λ} - η^{σμ}δ^{ρ}_{λ})(η^{αλ}δ^{β}_{ν} - η^{βλ}δ^{α}_{ν})
- (η^{αμ}δ^{β}_{λ} - η^{βμ}δ^{α}_{λ})(η^{ρλ}δ^{σ}_{ν} - η^{σλ}δ^{ρ}_{ν})}
^{ } = -{(η^{ρμ}δ^{σ}_{λ}η^{αλ}δ^{β}_{ν} - η^{ρμ}δ^{σ}_{λ}η^{βλ}δ^{α}_{ν} - η^{σμ}δ^{ρ}_{λ}η^{αλ}δ^{β}_{ν} + η^{σμ}δ^{ρ}_{λ}η^{βλ}δ^{α}_{ν})
- (η^{αμ}δ^{β}_{λ}η^{ρλ}δ^{σ}_{ν} - η^{αμ}δ^{β}_{λ}η^{σλ}δ^{ρ}_{ν} - η^{βμ}δ^{α}_{λ}η^{ρλ}δ^{σ}_{ν} + η^{βμ}δ^{α}_{λ}η^{σλ}δ^{ρ}_{ν})}
Contracting the middle δ and η gives:
[M^{ρσ},M^{αβ}]^{μ}_{ν} = -{(η^{ρμ}η^{σα}δ^{β}_{ν} - η^{ρμ}η^{σβ}δ^{α}_{ν} - η^{σμ}η^{ρα}δ^{β}_{ν} + η^{σμ}η^{ρβ}δ^{α}_{ν})
- (η^{αμ}η^{βρ}δ^{σ}_{ν} - η^{αμ}η^{βσ}δ^{ρ}_{ν} - η^{βμ}η^{αρ}δ^{σ}_{ν} + η^{βμ}η^{ασ}δ^{ρ}_{ν})}
Rearranging and factoring gives:
[M^{ρσ},M^{αβ}]^{μ}_{ν} = -{η^{σα}(η^{ρμ}δ^{β}_{ν} - η^{βμ}δ^{ρ}_{ν}) - η^{σβ}(η^{ρμ}δ^{α}_{ν} - η^{αμ}δ^{ρ}_{ν})
- η^{ρα}(η^{σμ}δ^{β}_{ν} - η^{βμ}δ^{σ}_{ν}) + η^{ρβ}(η^{σμ}δ^{α}_{ν} - η^{αμ}δ^{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^{ρμ}η^{σν} - η^{σμ}η^{ρν})
[J_{i},J_{j}] = iε^{ijk}J_{k}
[J_{i},K_{j}] = iε^{ijk}K_{k}
[K_{i},K_{j}] = -iε^{ijk}J_{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, K_{1}, is:
- -
_{ } | 0 1 0 0 |
K_{1} = 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)K_{1})^{n}
= exp(iζK_{1}) from the lemma lim_{n->∞}(1 + A/n)^{n} = exp(A)
But exp(iζK_{1}) also can be written as the Taylor series:
Λ = exp(iζK_{1}) = 1 + iζK_{1} + (iζK_{1})^{2}/2! - (iζK_{1})^{3}/3! + (iζK_{1})^{4}/4!
_{ } - (iζB_{1})^{5}/5! + (iζB_{1})^{6}/6!
Now, let,
- -
_{ } | 0 -1 0 0 |
iK_{1} = | -1 0 0 0 | = B_{1}
_{ } | 0 0 0 0 |
_{ } | 0 0 0 0 |
- -
So,
- -
_{ } | 1 0 0 0 |
B_{1}^{2} = | 0 1 0 0 |
_{ } | 0 0 0 0 |
_{ } | 0 0 0 0 |
- -
It is easy to show:
B_{1} = B_{1}^{3} = B_{1}^{5} ...
B_{1}^{2} = B_{1}^{4} = B_{1}^{6} ...
Therefore,
Λ = exp(ζB_{1}) = 1 + (ζ + ζ^{3}/3! + ζ^{5}/5! ...)B_{1}
_{ } + (ζ^{2}/2! + ζ^{4}/4! + ζ^{6}/6! ...)B_{1}^{2}
_{ } = 1 - B_{1}^{2} + (ζ + ζ^{3}/3! + ζ^{5}/5! ...)B_{1}
_{ } + (1 + ζ^{2}/2! + ζ^{4}/4! + ζ^{6}/6! ...)B_{1}^{2}
_{ } = 1 - B_{1}^{2} + B_{1}sinh(ζ) + B_{1}^{2}cosh(ζ)
- - - - - -
_{ } | 1 0 0 0 | | 1 0 0 0 | | 0 0 0 0 |
I - B_{1}^{2} = | 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(ζB_{1}) = | -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, J_{1}, is:
_{ } - -
_{ } | 0 0 0 0 |
J_{1} = 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)J_{1})^{n}
= exp(iθJ_{1}) from the lemma lim_{n->∞}(1 + A/n)^{n} = exp(A)
Again this can be expanded as a Taylor series:
Λ = exp(iθJ_{1}) = 1 + iθJ_{1} + (iθJ_{1})^{2}/2! + (iθJ_{1})^{3}/3! + (iθJ_{1})^{4}/4!
_{ } + (iθJ_{1})^{5}/5! + (iθJ_{1})^{6}/6!
Now, let,
- -
_{ } | 0 0 0 0 |
iJ_{1} = | 0 0 0 0 | = R_{1}
_{ } | 0 0 0 1 |
_{ } | 0 0 -1 0 |
- -
So,
- -
_{ } | 0 0 0 0 |
R_{1}^{2} = | 0 0 0 0 |
_{ } | 0 0 -1 0 |
_{ } | 0 0 0 -1 |
- -
It is easy to show:
R_{1} = -R_{1}^{3} = R_{1}^{5} ...
R_{1}^{2} = -R_{1}^{4} = R_{1}^{6} ...
Therefore,
Λ = exp(θR_{1}) = 1 + θR_{1} + θ^{2}R_{1}^{2}/2! - θ^{3}R_{1}/3! - θ^{4}R_{1}^{2}/4!
_{ } + θ^{5}R_{1}/5! + θ^{6}R_{1}^{2}/6!
_{ } = 1 + (θ - θ^{3}/3! + θ^{5}/5! ...)R_{1}
_{ } - (-θ^{2}/2! + θ^{4}/4! - θ^{6}/6! ...)R_{1}^{2}
_{ } = 1 + R_{1}^{2} + (θ - θ^{3}/3! + θ^{5}/5! ...)R_{1}
_{ } - (1 - θ^{2}/2! + θ^{4}/4! - θ^{6}/6! ...)R_{1}^{2}
_{ } = 1 - R_{1}^{2} + R_{1}sin(θ) - R_{1}^{2}cos(θ)
- - - - - -
_{ } | 1 0 0 0 | | 0 0 0 0 | | 1 0 0 0 |
I + R_{1}^{2} = | 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(θR_{1}) = | 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)