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?