The previous section derived the Lorentz Transformation as a geometric rotation in two-dimensional Spacetime. Rather than simply generalising this to three dimensions, we will now take the opportunity to further our understanding of Geometric Algebra so that we can remove some of the cognitive load of dealing with lots of coordinates.

One of the most important aspects of Geometric Algebra is that we can stop thinking in terms of coordinates of a specific reference frame, and rather represent everything as vectors.

We will begin by revisiting the two-dimensional rotation in space and move towards the Lorentz Transformation with three spatial and one temporal dimension.

- Rotations in 2D Space revisited
- Rotations in 3D Space

Up until now we have represented a two-dimensional rotation defined by two normal vectors \(\v{a}\) and \(\v{b}\) as \(e^{\v{xy}\theta}\), where \(\theta\) is the angle separating the two vectors (in the anti-clockwise direction), and \(\v{xy}\) is the unit element of the two dimensional plane being rotated. In this section we will see that we can also define this same rotation, without reference to a specific coordinate frame with base vectors \(\v{x}\) and \(\v{y}\), as:

\[ \v{ab} = e^{\v{P}\theta} \]

where \(\v{P}\) is equivalent to \(\v{xy}\), representing the unit element of area in the plane, but which will be defined without reference to \(\v{x}\) or \(\v{y}\).

Unlike the two dimensional case, we will see that in three dimensions, the product of two normal vectors, \(\v{ab}\) with a third arbitrary vector \(\v{d}\) does not generally result in a rotation, as the arbitrary vector \(\v{d}\) does not necessarily lie on the same plane as \(\v{a}\) and \(\v{b}\). Instead we will need to right and left multiply by \(\v{ab}\) and its reverse \(\v{ba}\). We introduce the concept of a rotor \( \v{R} = \v{ab} = e^{\v{P}\frac{\theta}{2}} \) and its reverse as \( \v{\tilde{R}} = \v{ba} = e^{-\v{P}\frac{\theta}{2}} \) to see that we can rotate the component of \(\v{d}\) which lies on the plane while leaving the component of \(\v{d}\) which is perpendicular to the plane unaffected:

\[ \v{\tilde{R}dR} = \v{d_\parallel}e^{\v{P}\theta} + \v{d_\perp} \]

where \(\v{d_\parallel}\) and \(\v{d_\perp}\) are the components of \(\v{d}\) parallel and perpendicular to the plane \(\v{P}\) respectively.