Lorentz Transformation as Rotation or Lorentz Boost

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:

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

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 \(\text{d}x\):

We can convert equations 5 and 6 into matrix notation:

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:

Thus we complete our Lorentz matrix in Eq. 7:

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:

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

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:

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

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

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

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

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

Lorentz boosts in the \(y\) and \(z\) directions can be derived analogously.