Lorentz Transformation as Rotation or Lorentz Boost

by Alexander FufaeV

Table of contents

Lorentz transformation as rotation This transformation rotates a coordinate system in spacetime.
Lorentz transformation as Lorentz boost This (special) Lorentz transformation transforms the time and location coordinate of an inertial system moving in a certain direction.

Lorentz transformation as rotation

A rotation in spacetime means that only the spatial vector $\boldsymbol{r} = (x,y,z)$ is rotated in three-dimensional space without changing the time coordinate $t$ during the transformation.

The Lorentz transformation (matrix $\Lambda $) is of the form:

Rotation in 4d spacetime

$$ \begin{align} \Lambda ~=~ \begin{bmatrix}1 & 0 \\ 0 & \mathcal{R}_{\varphi} \end{bmatrix} \end{align} $$

Here $\mathcal{R}_{\varphi}$ is the rotation matrix which rotates the spatial vector $\boldsymbol{r}$ by the angle $\varphi$.

Example: Rotation in three-dimensional space-time

This is how a two-dimensional rotation matrix $\mathcal{R}_{\varphi}$ looks like:

$$ \begin{align} \mathcal{R}_{\varphi} ~=~ \begin{bmatrix}\cos(\varphi) & -\sin(\varphi) \\ \sin(\varphi) & \cos(\varphi) \end{bmatrix} \end{align} $$

The Lorentz matrix is thus a 3x3 matrix and looks like this:

$$ \begin{align} \mathcal{R}_{\varphi} ~=~ \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos(\varphi) & -\sin(\varphi) \\ 0 & \sin(\varphi) & \cos(\varphi) \end{bmatrix} \end{align} $$

Example: Rotation in four-dimensional space-time

This is how a three-dimensional rotation matrix $\mathcal{R}_{\varphi}$ looks like, rotating the spatial vector $\boldsymbol{r}$ around the $x$ axis:

$$ \begin{align} \mathcal{R}_{\varphi} ~=~ \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos(\varphi) & -\sin(\varphi) \\ 0 & \sin(\varphi) & \cos(\varphi) \end{bmatrix} \end{align} $$

The Lorentz matrix is thus a 4x4 matrix and describes a rotation in a four-dimensional space-time:

$$ \begin{align} \mathcal{R}_{\varphi} ~=~ \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & \cos(\varphi) & -\sin(\varphi) \\ 0 & 0 & \sin(\varphi) & \cos(\varphi) \end{bmatrix} \end{align} $$

Lorentz transformation as Lorentz boost

Lorentz boost is a Lorentz transformation that transforms the spacetime coordinates of a system without performing a rotation.

Let's look at a Lorentz boost in $x$ direction. For this we take two inertial systems $S$ and $S'$ whose axes are parallel to each other (that means: the coordinate systems are not rotated relative to each other). $x$ and $x'$ axes are parallel, $y$ and $y'$ axes are parallel, as are $z$ and $z'$ axes.

The inertial system $S'$ moves with a constant relative velocity $\boldsymbol{v} = v_{\text x} \, \hat{\boldsymbol{e}}_{\text x} $ in $x$ direction past the inertial system $S$ in such a way that at time $t = t' = 0$ the two coordinate systems are on top of each other.

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From the point of view of the $S$-system, $S'$-system moves with a constant velocity in the positive $x$-direction.

Since $S'$ moves only in the $x$ direction, all the distances $\text{d}y = \text{d}y'$ and $\text{d}z = \text{d}z'$ perpendicular to the direction of motion remain unchanged. That means: In both inertial systems the $y$- and $z$-distance of two points is equal. So $\text{d}y'$ and $\text{d}z'$ are independent of the time interval $\text{d}t$ and spatial interval $\text{d}x$:

$$ \begin{align} \text{d}y' ~=~ 0\,(c\,\text{d}t) ~+~ 0\,\text{d}x ~+~ 1\, \text{d}y ~+~ 0 \, \text{d}z \end{align} $$

$$ \begin{align} \text{d}z' ~=~ 0\,(c\,\text{d}t) ~+~ 0\,\text{d}x ~+~ 0\, \text{d}y ~+~ 1\, \text{d}z \end{align} $$

We can convert equations 5 and 6 into matrix notation:

$$ \begin{align} \begin{bmatrix}c\,\text{d}t' \\ \text{d}x' \\ \text{d}y' \\ \text{d}z' \end{bmatrix} ~=~ \begin{bmatrix} \star & \star & \star & \star \\ \star & \star & \star & \star \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix}c\,\text{d}t \\ \text{d}x \\ \text{d}y \\ \text{d}z \end{bmatrix} \end{align} $$

The time component $c\,\text{d}t'$ and the spatial component $\text{d}x'$ also do not depend on the distances $\text{d}y$ and $\text{d}z$ perpendicular to the direction of motion. But $c\,\text{d}t'$ and $\text{d}x'$ can depend on each other:

$$ \begin{align} c\, \text{d}t' ~=~ A\,(c\,\text{d}t) ~+~ B\,\text{d}x ~+~ 0\, \text{d}y ~+~ 0 \, \text{d}z \end{align} $$

$$ \begin{align} \text{d}x' ~=~ D\,(c\,\text{d}t) ~+~ E\,\text{d}x ~+~ 0\, \text{d}y ~+~ 0\, \text{d}z \end{align} $$

Thus we complete our Lorentz matrix in Eq. 7:

$$ \begin{align} \begin{bmatrix}c\,\text{d}t' \\ \text{d}x' \\ \text{d}y' \\ \text{d}z' \end{bmatrix} ~=~ \begin{bmatrix} A & B & 0 & 0 \\ D & E & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix}c\,\text{d}t \\ \text{d}x \\ \text{d}y \\ \text{d}z \end{bmatrix} \end{align} $$

The coefficients $A$, $B$, $D$ and $E$, which tell us how the time and spatial coordinates transform, still have to be found out. To do this, we exploit the invariance of the infinitesimal line element 2. The line element $\text{d}s'^2 ~=~ -c^2\,\text{d}t'^2 + \text{d}x'^2 +\text{d}y'^2 + \text{d}z'^2$ in the inertial system $S'$, expressed with distances of the system $S$, looks like this:

$$ \begin{align} \text{d}s'^2 ~=~ -(A\,c\,\text{d}t + B\text{d}x)^2 + (D\,c\,\text{d}t + E\, \text{d}x)^2 + \text{d}y^2 + \text{d}z^2 \end{align} $$

Let us convert Eq. 11 to a more appropriate form:

$$ \begin{align} \text{d}s'^2 ~=~ -(A^2 - D^2) \, c^2 \, \text{d}t^2 ~+~ (E^2 - B^2) \, \text{d}x^2 ~+~ (2DE - 2AB)\,c\,\text{d}t\,\text{d}x ~+~ \text{d}y^2 ~+~ \text{d}z^2 \end{align} $$

Equation 12 must be according to the invariance of the line element, equal to the line element $\text{d}s^2 ~=~ -c^2\,\text{d}t^2 + \text{d}x^2 +\text{d}y^2 + \text{d}z^2$. Equating and comparing their coefficients yields the following system of equations with four unknowns:

$$ \begin{align} -(A^2 ~-~ D^2) ~&=~ -1 \\\\
E^2 ~-~ B^2 ~&=~ 1 \\\\
2DE ~-~ 2AB ~&=~ 0 \end{align} $$

By solving this system of equations, you find that $D = B $ and $E = A$. From the first equation follows:

$$ \begin{align} A ~=~ \sqrt{1 + B^2} \end{align} $$

Let's set $A := \gamma $ (usual notation) and express $B$ with $\gamma $: $B := -\beta \, \gamma $:

$$ \begin{align} \gamma ~=~ \sqrt{1 + \beta^2\,\gamma^2} \end{align} $$

Here, $\beta$ is a constant yet to be determined. Rearrangement for $\gamma$ results in:

$$ \begin{align} \gamma ~=~ \frac{1}{ \sqrt{1 - \beta^2} } \end{align} $$

The constant $\beta$ cannot be arbitrary, of course, because physically it must be dimensionless for the unit to be correct. Furthermore, for the non-relativistic case (i.e. $v \ll c$), the constant $\beta $ must approach zero. Both conditions are satisfied by $ \beta = \frac{v}{c}$:

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Lorentz factor explodes when the relative velocity tends to the speed of light.

Lorentz factor

$$ \begin{align} \gamma ~=~ \frac{1}{ \sqrt{1 - \frac{v^2}{c^2}} } \end{align} $$

The matrix in Eq. 10 for the Lorentz boost in the $x$ direction is thus given by:

Lorentz boost in x-direction

$$ \begin{align} \Lambda_{\text x} ~=~ \begin{bmatrix} \gamma & -\beta\,\gamma & 0 & 0 \\ -\beta\,\gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \end{align} $$

Lorentz boosts in the $y$ and $z$ directions can be derived analogously.

The Lorentz rotations and Lorentz boosts can of course be combined by applying the corresponding matrices to the spacetime coordinates (represented by a four vector) of a system one after another.