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Lorentz Transformations
Orthogonal Transformations
In Euclideal space R^2 (and Rn), it is valuable to find all the invertible linear changes of variable y = Rx that preserve the length of a vector ‖Rx‖ = ‖x‖
so that y^21 + y 22 = x^21 + x^22.
These are orthogonal transformations. Say
x 1 =ay 1 + by 2 x 2 =cy 1 + dy 2.
Then
x^21 + x^22 = (a^2 + c^2 )y 12 + 2(ab + cd)y 1 y 2 + (b^2 + d^2 )y 22.
We therefore want
a^2 + c^2 = 1, ab + cd = 0, and b^2 + d^2 = 1
There are four variables and only three conditions so we will have one free parameter. To satisfy the first condition it is natural to let a = cos θ and c = sin θ. For the second condition, let b = −c = − sin θ and d = a = cos θ. The third condition is also satisfied. This gives the matrix
R =
cos θ − sin θ sin θ cos θ
These are the rotations of ther plane R^2.
By a similar computation, these are also the only linear changes of variable that preserve the Laplace operator
∂^2 u ∂x^21
∂^2 u ∂x^22
∂^2 u ∂y 12
∂^2 u ∂y 22
Lorentz Transformations
It is also valuable to find all linear changes of variable
x′^ =αx + βt t′^ =γx + δt
that preserve the wave equation
∂^2 u ∂t^2
− c^2 ∂^2 u ∂x^2
∂^2 u ∂t′^2
− c^2 ∂^2 u ∂x′^2
where c is a constant (the speed of sound or light).
By the chain rule,
utt−c^2 uxx = (δ^2 −c^2 γ^2 )ut′t′+2(βδ−c^2 αγ)ux′t′+(β^2 −c^2 α^2 )ux′x′.
Thus we want
δ^2 − c^2 γ^2 = 1, βδ − c^2 αγ = 0, and β^2 − c^2 α^2 = −c^2
First pick γ and δ so that δ^2 −c^2 γ^2 = 1, and then let β = ±c^2 γ, α = ±δ. To preserve orientation we use the + signs. Since c^2 α^2 − β^2 = c^2 and cosh^2 σ − sinh^2 σ = 1, it is traditional to write α = cosh σ, β = c sinh σ. For any real σ the transforma- tion
x′^ = (cosh σ) x + (c sinh σ) t t′^ = (
c
sinh σ) x + (cosh σ) t
