# A Vector in Spacetime

So far we have considered operations on vectors only - linear combinations of the two basis vectors, $$\mathbf{x}$$ and $$\mathbf{y}$$. Here we will consider a more general combination of a vector together with an $$\mathbf{xy}$$ component to form a "multivector". Furthermore, we will see that this extra component behaves geometrically as we expect time to behave in the space-time of Special Relativity and, by extension, in our every day experience. We will then see that we can rotate these space-time multivectors in space without affecting the temporal component, corresponding to changing your frame of reference within a single inertial frame.

For clarity, let's define $$\mathbf{t}$$ as

$\mathbf{t} = \mathbf{xy}$

noting that

$\mathbf{t}^2 = \mathbf{xyxy} = -1$

and

$\mathbf{xt} = \mathbf{xxy} = -\mathbf{xyx} = -\mathbf{tx} \\ \mathbf{yt} = -\mathbf{ty}$

We can then define an arbitrary two-dimensional space-time multivector as a vector plus a component of $$\mathbf{t}$$:

$\mathbf{A} = \mathbf{a} + t_a\mathbf{t}$

## The magnitude-squared of a space-time multivector

Multiplying this multivector by itself, we see:

\begin{aligned} \mathbf{A}^2 &= (\mathbf{a} + t_a\mathbf{t})(\mathbf{a} + t_a\mathbf{t}) \\ &= \mathbf{a}^2 + t_a\mathbf{at} + t_a\mathbf{ta} + t_a^2\mathbf{tt} \\ &= \mathbf{a}^2 + t_a\mathbf{at} - t_a\mathbf{at} - t_a^2 \\ &= \mathbf{a}^2 - t_a^2 \\ &= r_a^2 - t_a^2 \end{aligned}

If you are familiar with Special Relativity, you will recognise this as the Minkowski space-time metric. Note, we have used above some of the earlier properties of rotations to equate $$\mathbf{ta} = -\mathbf{at}$$, but in full for clarity:

\begin{aligned} \mathbf{ta} &= \mathbf{t}r_a \mathbf{x} e^{\mathbf{xy}\theta_a} \\ &= - r_a \mathbf{xt} e^{\mathbf{xy}\theta_a} \\ &= -r_a \mathbf{x}e^{\mathbf{xy}\theta_a}\mathbf{t} \\ &= -\mathbf{at} \end{aligned}

So the square of a space-time multivector results in a scalar value, a value which matches the Minkowski space-time metric.

$|\mathbf{A}|^2 = \mathbf{A}^2 = r_a^2 - t_a^2$

## Normalised Space-Time Multivectors

An arbitrary space-time multivector $$\mathbf{A}$$ is normal if

$\mathbf{A}^2 = 1$

Using the trigonometric identity that

$\cosh^2\phi - \sinh^2\phi = 1$

then $$\mathbf{A}$$ will be a normalised multivector if $$\mathbf{a}^2 = 1$$ and:

$\mathbf{A} = \mathbf{a}\cosh \phi + \mathbf{t}\sinh \phi$

We can verify this by expanding:

\begin{aligned} \mathbf{A}^2 &= (\mathbf{a}\cosh \phi + \mathbf{t}\sinh \phi)(\mathbf{a}\cosh \phi + \mathbf{t}\sinh \phi) \\ &= \mathbf{a}^2\cosh^2\phi + \mathbf{at}\cosh\phi\sinh\phi + \mathbf{ta}\cosh\phi\sinh\phi + \mathbf{tt}\sinh^2\phi \\ &= \cosh^2\phi + \mathbf{at}\cosh\phi\sinh\phi - \mathbf{at}\cosh\phi\sinh\phi - \sinh^2\phi \\ &= 1 \end{aligned}

We can simplify our definition of a normal space-time multivector by looking back to Euler's formula and geometric algebra where we saw that if $$\mathbf{a}^2 = 1$$, then

$e^{\mathbf{a}\phi} = \cosh\phi + \mathbf{a}\sinh\phi$

This allows us to simplify $$\mathbf{A}$$ as follows:

\begin{aligned} \mathbf{A} &= \mathbf{a}\cosh\phi + \mathbf{t}\sinh\phi \\ &= \mathbf{a}(\cosh\phi + \mathbf{at}\sinh\phi) & \text{, since }\mathbf{a}^2 = 1\\ &= \mathbf{a}e^{\mathbf{at}\phi} \end{aligned}

With this simpler definition of a normal space-time multivector, seeing that it is indeed normal is trivial:

\begin{aligned} \mathbf{A}^2 &= \mathbf{a}e^{\mathbf{at}\phi}\mathbf{a}e^{\mathbf{at}\phi}\\ &= \mathbf{a}e^{\mathbf{at}\phi}e^{-\mathbf{at}\phi}\mathbf{a} \\ &= 1 \end{aligned}

## Time is unchanged under spatial rotations

If we define a spatial rotation using two normalised vectors, $$\mathbf{b}$$ and $$\mathbf{c}$$, where the difference in orientation between the two is $$\theta$$, so that:

$\mathbf{bc} = e^{\mathbf{xy}\theta}$

we find that it no longer defines a rotation when simply applied to an arbitrary multivector, because the magnitude-squared after the multiplication is not the same as before (Note, in the following section we will again use the earlier properties of rotations to equate $$\mathbf{cba} = \mathbf{abc}$$):

\begin{aligned} (\mathbf{Abc})^2 &= \mathbf{AbcAbc} \\ &= (\mathbf{a} + t_a\mathbf{t})\mathbf{bc}(\mathbf{a} + t_a\mathbf{t})\mathbf{bc} \\ &= (\mathbf{a} + t_a\mathbf{t})(\mathbf{bca} + t_a\mathbf{bct})\mathbf{bc} \\ &= (\mathbf{a} + t_a\mathbf{t})(\mathbf{acb} + t_a\mathbf{tbc})\mathbf{bc} \\ &= (\mathbf{a} + t_a\mathbf{t})(\mathbf{a}\mathbf{cbbc} + t_a\mathbf{t}\mathbf{bcbc}) \\ &= (\mathbf{a} + t_a\mathbf{t})(\mathbf{a} + t_a\mathbf{t}e^{\mathbf{xy}2\theta}) \\ &\neq \mathbf{A}^2 \end{aligned}

So instead we halve the angle of rotation, defining:

$\mathbf{bc} = e^{\mathbf{xy}\frac{\theta}{2}}$

and apply it twice - once from the left and once from the right: $$\mathbf{cbAbc}$$.

The fact that $$\mathbf{cbAbc}$$ is a rotation can be verified simply by confirming the magnitude-squared is now unchanged:

\begin{aligned} (\mathbf{cbAbc})^2 &= \mathbf{cbAbccbAbc} &\\ &= \mathbf{cb}\mathbf{A}^2\mathbf{bc} \\ &= \mathbf{A}^2\mathbf{cbbc} &\text{, as }\mathbf{A}^2\text{is scalar}\\ &= \mathbf{A}^2 \end{aligned}

Now, evaluating the actual rotation, we find:

\begin{aligned} \mathbf{cbAbc} &= \mathbf{cb}(\mathbf{a} + t_a\mathbf{t})\mathbf{bc} \\ &= \mathbf{cbabc} + t_a\mathbf{cbtbc} \\ &= \mathbf{abcbc} + t_a\mathbf{tcbbc} \\ &= \mathbf{a}e^{\mathbf{xy}\theta} + t_a\mathbf{t} \end{aligned}

That is, when applying the spatial rotation in this way, the vector component of $$\mathbf{A}$$ is rotated by $$\theta$$, while the time component of the multi-vector is not affected at all. We can change position and orientation within the one inertial frame without affecting the measured time.

But what if, rather than defining a rotation between two spatial vectors, we instead define a rotation between a spatial vector and the pseudo-scalar temporal component $$\mathbf{t}$$ - a two-dimensional rotation in space-time?