# The Lorentz Transformation

In the previous section we saw that we can rotate a space-time multivector $$\v{A}$$ in space - using a rotation from one normal vector $$\v{b}$$ to another normal vector $$\v{c}$$ to form the rotation $$\v{bc} = e^{\v{xy}\frac{\theta}{2}}$$. We also saw that we can define a normal space-time multivector as $$\v{B} = \v{b}e^{\v{bt}\phi}$$, where $$\v{b}^2 = 1$$. In this section we will consider a space-time multivector $$\v{A}$$ being rotated in space-time - using a rotation from one normal multivector $$\v{B}$$ to another normal multivector $$\v{C}$$, 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} \v{BC} &= \v{b}e^{\v{bt}\phi_B}\v{c}e^{\v{ct}\phi_C} \\ \end{aligned}

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

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

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

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

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

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

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

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

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

with the result that:

\begin{aligned} \v{CBABC} &= e^{\v{xt}\frac{\phi}{2}}(\v{a} + t_a\v{t})e^{-\v{xt}\frac{\phi}{2}} \\ &= e^{\v{xt}\frac{\phi}{2}}\v{a}e^{-\v{xt}\frac{\phi}{2}} + e^{\v{xt}\frac{\phi}{2}}t_a\v{t}e^{-\v{xt}\frac{\phi}{2}} \\ &= e^{\v{xt}\frac{\phi}{2}}\v{a}e^{-\v{xt}\frac{\phi}{2}} + t_a\v{t}e^{-\v{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. $$\v{A} = x_a\v{x} + t_a\v{t}$$, we can further simplify the rotation:

\begin{aligned} \v{e^{\v{xt}\frac{\phi}{2}}Ae^{-\v{xt}\frac{\phi}{2}}} &= e^{\v{xt}\frac{\phi}{2}}x_a\v{x}e^{-\v{xt}\frac{\phi}{2}} + t_a\v{t}e^{-\v{xt}\phi} \\ &= x_a\v{x}e^{-\v{xt}\phi} + t_a\v{t}e^{-\v{xt}\phi} \\ &= x_a\v{x}(\cosh\phi - \v{y}\sinh\phi) + t_a\v{t}(\cosh\phi - \v{y}\sinh\phi) \\ &= \v{x}(x_a\cosh\phi - t_a\sinh\phi) + \v{t}(t_a\cosh\phi - x_a\sinh\phi) \\ &= \cosh\phi(x_a - t_a\tanh\phi)\v{x} + \cosh\phi(t_a - x_a\tanh\phi)\v{t} \end{aligned} Plotting sinh, cosh and tanh Wikimedia Commons

Note that $$\tanh\phi$$ has the property of ranging from -1 to 1 (see figure) 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 $$\v{A} = x_a\v{x} + t_a\v{t}$$ (with the speed of light ratio normalized to 1):

$\v{e^{\v{xt}\frac{\phi}{2}}Ae^{-\v{xt}\frac{\phi}{2}}} = \frac{1}{\sqrt{1-v^2}}(x_a - t_av)\v{x} + \frac{1}{\sqrt{1-v^2}}(t_a - x_av)\v{t}$

### 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. $$\v{A} = y_a\v{y} + t_a\v{t}$$. In this case:

\begin{aligned} \v{e^{\v{xt}\frac{\phi}{2}}Ae^{-\v{xt}\frac{\phi}{2}}} &= y_a\v{y}e^{\v{xt}\frac{\phi}{2}}e^{-\v{xt}\frac{\phi}{2}} + t_a\v{t}e^{-\v{xt}\phi} \\ &= y_a\v{y} + t_a\v{t}(\cosh\phi - \v{xt}\sinh\phi) \\ &= y_a\v{y} + t_a\v{t}\cosh\phi - t_a\v{x}\sinh\phi \\ &= y_a\v{y} + \frac{t_a}{\sqrt{1-v^2}}\v{t} - \frac{v t_a}{\sqrt{1-v^2}}\v{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 of $$e^{\v{xt}\phi}$$ of a general 2-dimensional multivector:

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

is then

$\v{e^{\v{xt}\frac{\phi}{2}}Ae^{-\v{xt}\frac{\phi}{2}}} = \frac{1}{\sqrt{1-v^2}}(x_a - t_av)\v{x} + y_a\v{y} + \frac{1}{\sqrt{1-v^2}}(t_a - x_av)\v{t}$

Now that we have managed to derive the Lorentz transformation as a geometric rotation in Spacetime, it is time to further our understanding of Geometric Algebra so that we can remove some of the cognitive load. Up until this point, we have been working with a preferred coordinate frame with a set of basis vectors $$\v{x}$$ and $$\v{y}$$, but before we generalise this to the 3 dimensions of space and one of time within which we appear to exist, it is worth seeing how we can represent the Lorentz transformation without referring to a preferred reference frame.