# Lorentz Transformation as Rotation or Lorentz Boost

## Table of contents

## 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**

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

## 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.

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 \(\text{d}x\):

**Equation for infinitesimal distance dy**

**Equation for infinitesimal distance dz**

We can convert equations 5

and 6

into matrix notation:

**Spacetime coordinates of two systems linked together 1**

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:

**Equation for an infinitesimal time component**

**Equation for an infinitesimal dx-element**

Thus we complete our Lorentz matrix in Eq. 7

:

**Four-vector of one system is Lorentz matrix applied to four-vector of another system**

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:

**Infinitesimal line element with unknown coefficients**

Let us convert Eq. 11

to a more appropriate form:

**Line element with unknown coefficients transformed**

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:

**System of equations with four unknown coefficients of the Lorentz matrix**

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:

**A equals root of 1 plus B squared**

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

**Gamma factor**

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

**Formula for gamma factor with beta**

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 \( \beta = \frac{v}{c}\):

**Lorentz factor**

The matrix in Eq. 10

for the Lorentz boost in the \(x\) direction is thus given by:

**Lorentz boost in x-direction**

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.