General form of a Lorentz transformation

[rbwang1225]

I got the general form of a Lorentz transformation in a GR book,

${\textbf{x'}}$ = ${\textbf{x}}$ + (γ-1)$\frac{{\textbf{x}•\textbf{v}}}{v^2}\textbf{v}$-γ$\textbf{v}$t

t' = γ(t-$\frac{{\textbf{x}•\textbf{v}}}{c^2}$)

from frame S to frame S' moving with relative velocity $\textbf{v}$.

How to derive this?

 

[Bill_K]

The Lorentz transformation along the x-axis is

x' = γ(x - vt)
y' = y
z' = z
t' = γ(t - vx/c²)

This is

x_∥' = γ(x_∥ - vt)
x_⊥' = x_⊥
t' = γ(t - 1/c² v·x_∥)

Put x_∥ = v v·x/v² and x = x_⊥ + x_∥ and you've got it.

 

[rbwang1225]

I don't understand what you mean. What is x = x + x?

 

[PhilDSP]

x_∥ represents the variable in the coordinate system for the spatial dimension that is oriented parallel to the vector pointing from the emitter to the detector (or detector to emitter)

x_⊥ represents either of the variables in the other 2 spatial coordinates that are oriented perpendicular to the vector pointing from the emitter to the detector (or detector to emitter)

In other words, if x is parallel to the vector then x_∥ -> x while x_⊥ -> y and z
Otherwise the relationship holds more generally

 

[Chestermiller]

There is another way of deriving this result. Suppose that the relative velocity of the S' frame of reference relative to the S frame of reference has components in all three coordinate directions. You can first rotate the cartesian spatial coordinates about the z axis to bring the velocity vector into a new x-z plane. Then you can apply a second rotation about the new y-axis to orient the relative velocity vector along a second new x axis. Then you can apply the boost in the second new coordinate system. After applying the boost, you can apply the inverse (= transpose) of the previous (combined) rotation matrices, and arrive at the transformed coordinates.

I like Bill K's approach better, because it is more elegant. He resolves an arbitrary position vector x into components perpendicular and parallel to the relative velocity vector. Only the component parallel to the velocity vector receives the boost. The component perpendicular to the velocity vector comes through unscathed.

Incidentally, this is not the most general form of the Lorentz Transformation. You can rigidly reorient the spatial coordinate axes of the S' frame of reference in any way you wish (by additional rotations) and still satisfy the condition that the relative velocity of the S frame of reference relative to the S' frame of reference has a magnitude equal to that of the S' frame of reference relative to the S frame of reference.