# The Lorentz Transformation

In the previous section we saw that we can rotate a space-time multivector $$\mathbf{A}$$ in space - using a rotation from one normal vector $$\mathbf{b}$$ to another normal vector $$\mathbf{c}$$ to form the rotation $$\mathbf{bc} = e^{\mathbf{xy}\frac{\theta}{2}}$$. We also saw that we can define a normal space-time multivector as $$\mathbf{B} = \mathbf{b}e^{\mathbf{bt}\phi}$$, where $$\mathbf{b}^2 = 1$$. In this section we will consider a space-time multivector $$\mathbf{A}$$ being rotated in space-time - using a rotation from one normal multivector $$\mathbf{B}$$ to another normal multivector $$\mathbf{C}$$, and see that the result is the standard Lorentz transformation.

## A simplified Space-Time rotation

If we were to do a generic space-time multivector rotation, the rotation would be:

\begin{aligned} \mathbf{BC} &= \mathbf{b}e^{\mathbf{bt}\phi_B}\mathbf{c}e^{\mathbf{ct}\phi_C} \\ \end{aligned}

but we are going to simplify this in two ways. First by assuming that the vector component of both $$\mathbf{B}$$ and $$\mathbf{C}$$ are equal, and second that they are equal to the basis vector $$\mathbf{x}$$. In which case we get a much simplified rotation:

\begin{aligned} \mathbf{BC} &= \mathbf{x}e^{\mathbf{xt}\phi_B}\mathbf{x}e^{\mathbf{xt}\phi_C} \\ &= e^{-\mathbf{xt}\phi_B}\mathbf{xx}e^{\mathbf{xt}\phi_C} \\ &= e^{\mathbf{xt}(\phi_C-\phi_B)} \\ &= e^{-\mathbf{xt}\phi} \end{aligned}

where $$\phi = \phi_B - \phi_C$$ is the angle between $$\mathbf{B}$$ and $$\mathbf{C}$$.

Similar to spatial rotations of space-time multivectors, $$\mathbf{BC}$$ will not work as a rotation by simple multiplication:

$(\mathbf{ABC})^2 = \mathbf{ABCABC} \neq \mathbf{A}^2$

but if we left and right multiply then the magnitude-squared remains constant, as it must under a rotation:

\begin{aligned} (\mathbf{CBABC})^2 &= \mathbf{CBABCCBABC} \\ &= \mathbf{CBAABC} \\ &= \mathbf{CBBCA}^2 \\ &= \mathbf{A}^2 \\ \end{aligned}

Given that we are now applying tho rotation twice, we define $$\phi$$ as double the angle between $$\mathbf{B}$$ and $$\mathbf{C}$$:

$\mathbf{BC} = e^{-\mathbf{xt}\frac{\phi}{2}}$

with the result that:

\begin{aligned} \mathbf{CBABC} &= e^{\mathbf{xt}\frac{\phi}{2}}(\mathbf{a} + t_a\mathbf{t})e^{-\mathbf{xt}\frac{\phi}{2}} \\ &= e^{\mathbf{xt}\frac{\phi}{2}}\mathbf{a}e^{-\mathbf{xt}\frac{\phi}{2}} + e^{\mathbf{xt}\frac{\phi}{2}}t_a\mathbf{t}e^{-\mathbf{xt}\frac{\phi}{2}} \\ &= e^{\mathbf{xt}\frac{\phi}{2}}\mathbf{a}e^{-\mathbf{xt}\frac{\phi}{2}} + t_a\mathbf{t}e^{-\mathbf{xt}\phi} \end{aligned}

## The Lorentz Transformation

### A vector in the direction of rotation

If we first look at the case where the vector being rotated is in the same direction as the rotation itself, ie. $$\mathbf{A} = x_a\mathbf{x} + t_a\mathbf{t}$$, we can further simplify the rotation:

\begin{aligned} \mathbf{CBABC} &= e^{\mathbf{xt}\frac{\phi}{2}}x_a\mathbf{x}e^{-\mathbf{xt}\frac{\phi}{2}} + t_a\mathbf{t}e^{-\mathbf{xt}\phi} \\ &= x_a\mathbf{x}e^{-\mathbf{xt}\phi} + t_a\mathbf{t}e^{-\mathbf{xt}\phi} \\ &= x_a\mathbf{x}(\cosh\phi - \mathbf{y}\sinh\phi) + t_a\mathbf{t}(\cosh\phi - \mathbf{y}\sinh\phi) \\ &= \mathbf{x}(x_a\cosh\phi - t_a\sinh\phi) + \mathbf{t}(t_a\cosh\phi - x_a\sinh\phi) \\ &= \cosh\phi(x_a - t_a\tanh\phi)\mathbf{x} + \cosh\phi(t_a - x_a\tanh\phi)\mathbf{t} \end{aligned}

Noting that $$\tanh\phi$$ has the property of ranging from -1 to 1 (graph) and that $$\cosh\phi$$ and $$\tanh\phi$$ can be related by:

$\cosh\phi = \frac{1}{\sqrt{1-\tan^2\phi}}$

So with the substitution of $$v = \tanh\phi$$, we're left with the Lorentz transformation of $$\mathbf{A} = x_a\mathbf{x} + t_a\mathbf{t}$$ (with the speed of light ratio normalized to 1):

$\mathbf{CBABC} = \frac{1}{\sqrt{1-v^2}}(x_a - t_av)\mathbf{x} + \frac{1}{\sqrt{1-v^2}}(t_a - x_av)\mathbf{t}$

### A vector perpendicular to the direction of rotation

Next let us look at the case where the vector being rotated is perpendicular to the rotation, ie. $$\mathbf{A} = y_a\mathbf{y} + t_a\mathbf{t}$$. In this case:

\begin{aligned} \mathbf{CBABC} &= y_a\mathbf{y}e^{\mathbf{xt}\frac{\phi}{2}}e^{-\mathbf{xt}\frac{\phi}{2}} + t_a\mathbf{t}e^{-\mathbf{xt}\phi} \\ &= y_a\mathbf{y} + t_a\mathbf{t}(\cosh\phi - \mathbf{xt}\sinh\phi) \\ &= y_a\mathbf{y} + t_a\mathbf{t}\cosh\phi - t_a\mathbf{x}\sinh\phi \\ &= y_a\mathbf{y} + \frac{t_a}{\sqrt{1-v^2}}\mathbf{t} - \frac{v t_a}{\sqrt{1-v^2}}\mathbf{x} \\ \end{aligned}

So the component perpendicular to the direction of the rotation is not affected by the transformation, just as with the Lorentz transformation.

### A general 2-dimensional space-time vector

The space-time rotation between \mathbf{xt} of a general 2-dimensional multivector:

$\mathbf{A} = x_a\mathbf{x} + y_a\mathbf{y} + t_a\mathbf{t}$

is then

$\mathbf{CBABC} = \frac{1}{\sqrt{1-v^2}}(x_a - t_av)\mathbf{x} + y_a\mathbf{y} + \frac{1}{\sqrt{1-v^2}}(t_a - x_av)\mathbf{t}$