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 \(\mathbf{B} = \mathbf{b}e^{\mathbf{bxy}\phi}\), 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:

\[ \begin{aligned} \mathbf{BC} &= \mathbf{b}e^{\mathbf{bxy}\phi_B}\mathbf{c}e^{\mathbf{cxy}\phi_C} \\ \end{aligned} \]

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:

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

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 \(\mathbf{C}\):

\[ \mathbf{BC} = e^{-\mathbf{y}\frac{\phi}{2}} \]

with the result that:

\[ \begin{aligned} \mathbf{CBABC} &= e^{\mathbf{y}\frac{\phi}{2}}(\mathbf{a} + t_a\mathbf{xy})e^{-\mathbf{y}\frac{\phi}{2}} \\ &= e^{\mathbf{y}\frac{\phi}{2}}\mathbf{a}e^{-\mathbf{y}\frac{\phi}{2}} + e^{\mathbf{y}\frac{\phi}{2}}t_a\mathbf{xy}e^{-\mathbf{y}\frac{\phi}{2}} \\ &= e^{\mathbf{y}\frac{\phi}{2}}\mathbf{a}e^{-\mathbf{y}\frac{\phi}{2}} + t_a\mathbf{xy}e^{-\mathbf{y}\phi} \end{aligned} \]

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. \(\mathbf{a} = x_a\mathbf{x}\), we can further simplify the rotation:

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

Plotting sinh, cosh and tanh

Wikimedia Commons

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 (with the speed of light ratio normalized to 1):

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

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. \(\mathbf{a} = y_a\mathbf{y}\). In this case:

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

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