# A (para)Vector of space and time

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 a scalar component $$a_0$$ to form a something called a paravector - a scalar value plus a vector value. Furthermore, we will see that this scalar component results in a amplitude-squared metric matching the metric of space-time in Special Relativity. Finally, we will see that we can rotate these space-time paravectors in space without affecting the temporal component, corresponding to changing our frame of reference within a single inertial frame.

Let's start by defining an arbitrary two-dimensional paravector as a vector plus a scalar component:

$\v{A} = a_0 + \v{a} = a_0 + \vcTwo{a}$

Note that a more general multivector in two dimensions would additionally have a term in $$\v{xy}$$. We are limiting the discussion here to a paravector only so we can gain some insight from the two dimensional case before we examine the three dimensional case more generically.

## The amplitude-squared of a 2D paravector

Multiplying this paravector by itself, we see:

\begin{aligned} \mathbf{A}^2 &= (a_0 + \v{a})(a_0 + \v{a}) \\ &= a_0^2 + \v{a}a_0 + a_0\v{a} + \v{a}^2 \\ &= a_0^{2} + \mathbf{a}^2 + 2a_0\v{a} \end{aligned}

which is not a scalar at all. So, unlike the case of a vector where the amplitude squared is just the square of the vector $$|\v{a}|^2 = \v{a}^2$$, for a paravector $$|\v{A}|^2 \ne \v{A}^2$$.

We need to find a conjugate of $$\v{A}$$, written $$\c{\v{A}}$$, such that $$|\v{A}|^2 = \v{A}\c{\v{A}}$$ results in a scalar value for a paravector while also simplifying, when the scalar component of $$\v{A}$$ is zero, to the familiar magnitude of a vector.

##### Amplitude squared of a paravector
See if you can show that the only possibility for such a conjugate of $$\v{A}$$ is: $$\c{\v{A}} = a_0 - \v{a}$$ so that the ampitude squared is the scalar value $$\v{A}\c{\v{A}} = a_0^2 - \v{a}^2$$ As a hint to get started, if we instead assume that the conjugate is a general linear function of the parts: $$\c{\v{A}} = f(\v{A}) = s_0 a_0 + s_a\v{a}$$ we can then solve to find the values of the real numbers that result in a scalar value.

Evaluating,

\begin{aligned} \v{A}f(\v{A}) &= (\v{a} + a_0)(s_a\v{a} + s_0 a_0) \\ &= s_a\v{a}^2 + s_0 a_0 \v{a} + a_0 s_a\v{a} + s_0 a_0^2 \\ &= s_a\v{a}^2 + a_0 \v{a}(s_0 + s_a) + s_0 a_0^2 \end{aligned}

This result will be a scalar value if and only if $$s_0 + s_a = 0$$. Our further requirement that for a plain vector $$|\v{a}|^2 = |\v{a}^2|$$ means that for the special case of $$\v{A} = \v{a} + 0$$ we have:

$\v{A}f(\v{A}) = (\v{a} + 0)(s_a\v{a} + s_0 0) = s_a\v{a}^2$

So we are further restricted to $$s_a = \pm 1$$ and therefore $$s_0 = \mp 1$$. We will stick to convention for now (and see later on why the convention is useful) and choose: $$s_a = -1$$ and $$s_0 = 1$$ so that

$f(\v{A}) = \c{\v{A}} = a_0 - \v{a}$

.

The definition of the conjugate, $$\c{\v{A}} = a_0 - \v{a}$$ gives us a metric for the amplitude squared of a scalar plus vector so that for any paravector $$\v{A} = a_0 + \v{a}$$,

$|\v{A}|^2 = \v{A}\c{\v{A}} = a_0^2 - \v{a}^2$

## The Clifford Conjugation

The conjugation we have found here is called the Clifford conjugation. For a more general and formal definition of the Clifford Conjugation as well as a derivation of the amplitude squared for a multivector, see Exploring the origin of Minkowski spacetime.

For our purposes here, it is enough to understand that the Clifford conjugation of any value of the geometric algebra is found by reversing all geometric products and negating all vectors. Let's see some examples.

### Conjugating vectors

As we've already seen, if $$\v{A} = a_0 + \v{a}$$, then

$\c{\v{A}} = a_0 - \v{a}$

In this case, there are no products of vectors, so it is only the negation of vectors that is relevant. Note that we get the same result when represeting $$\v{a}$$ with respect to a specific orthonormal basis, so if $$\v{A} = a_0 + \vcTwo{a}$$, then

$\c{\v{A}} = a_0 - a_x \v{x} - a_y\v{y} = a_0 - \v{a}$

### Conjugating products of vectors

If a rotor is defined by the product of two normal vectors, $$\v{b}$$ and $$\v{c}$$, $$\v{R} = \v{bc}$$, then

$\c{\v{R}} = \c{\v{bc}} = \c{\v{c}}\c{\v{b}} = (\v{-c})(\v{-b}) = \v{cb}$

So in this case it is reversing the geometric product which is significant, while the negation of vectors cancels itself (in this case where there are an even number of vectors). Note that we get the same result if we instead represent the rotor as an exponent,

$\v{R} = \v{bc} = \cos(\theta) + \v{xy}\sin(\theta) = e^{\v{xy}\theta}$

where $$\theta$$ is the angle between the two normal vectors, then

$\c{\v{R}} = e^{\v{yx}\theta} = e^{-\v{xy}\theta}$

### Conjugating arbitrary terms

If we define another paravector (or arbitrary multi-vector), $$\v{B}$$ and use it to define a product of three terms, $$\v{BAR}$$, then the Clifford conjugate of that product is the reverse product of the individual conjugations:

$\c{\v{BAR}} = \c{\v{R}}\c{\v{A}}\c{\v{B}}$

This is technically known as an anti-automorphism and it ensures that the Clifford conjugate can always be applied a second time to get back the original value:

$\c{(\c{\v{R}}\c{\v{A}}\c{\v{B}})} = \v{BAR}$

which is also known more technically as an involution.

## Normalised paravectors

An arbitrary paravector $$\v{A}$$ is normal if

$|\v{A}|^2 = \v{A}\c{\v{A}} = a_0^2 - \v{a}^2 = 1$

Noting that this is similar to the trigonometric identity

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

allows us to choose a normalized vector $$\v{a}$$ and a value of $$\phi$$ such that

$\v{A} = \cosh \phi + \v{a}\sinh \phi$

We can now verify that $$\v{A}$$ is normal as long as $$\v{a}$$ is normal by expanding:

\begin{aligned} \v{A\c{A}} &= (\cosh \phi + \v{a}\sinh \phi)(\cosh \phi - \v{a}\sinh \phi) \\ &= \cosh^2\phi - \v{a}\cosh\phi\sinh\phi + \v{a}\cosh\phi\sinh\phi - \v{a}^2\sinh^2\phi \\ &= \cosh^2\phi - \sinh^2\phi \\ &= 1 \end{aligned}

We can simplify our definition of a normal paravector even further by looking back to Euler's formula and geometric algebra where we saw that for a normal vector $$\v{a}$$,

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

With this simpler definition of a normal paravector, $$\v{A} = e^{\v{a}\phi}$$, it becomes trivial to show that is is indeed normal:

\begin{aligned} |\v{A}|^2 &= \v{A}\c{\v{A}} \\ &= e^{\v{a}\phi}e^{-\v{a}\phi} \\ &= 1 \end{aligned}

## Rotating a paravector in space only

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

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

we find that we cannot simply multiply a paravector by the rotor to obtain a new paravector, as:

\begin{aligned} \v{AR} &= \v{A}e^{\v{xy}\theta} \\ &= (a_0 + \v{a})e^{\v{xy}\theta} \\ &= a_0 e^{\v{xy}\theta} + \v{a}e^{\v{xy}\theta} \\ \end{aligned}

includes a bivector ($$\v{xy}$$) term (if you expand the first term in the above result). The similar problem arises if we try to simply multiply an arbitrary three-dimensional vector by our rotor. It turns out that a much more general way to apply rotations (and other operations) is use the Clifford conjugate! If we halve the angle between the two vectors defining our rotor, so that:

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

and apply it twice - once from the left as the conjugate and once from the right, we find:

\begin{aligned} \v{\c{R}AR} &= e^{\v{xy}\frac{\theta}{2}}(a_0 + \v{a})e^{-\v{xy}\frac{\theta}{2}} \\ &= e^{\v{xy}\frac{\theta}{2}}a_0e^{-\v{xy}\frac{\theta}{2}} + e^{\v{xy}\frac{\theta}{2}}\v{a}e^{-\v{xy}\frac{\theta}{2}} \\ &= a_0 + \v{a}e^{-\v{xy}\theta} \\ \end{aligned}

That is, when applying the spatial rotation in this way, the vector component of $$\v{A}$$ is rotated by $$\theta$$, while the scalar part is not affected at all.

The fact that $$\v{\c{R}AR}$$ is a rotation can be verified simply by confirming that the amplitude-squared is unchanged:

\begin{aligned} |\v{\c{R}AR}|^2 &= \v{\c{R}AR}\c{\v{\c{R}AR}} \\ &= \v{\c{R}AR}\v{\c{R}\c{A}R} \\ &= \v{A}\c{\v{A}} \end{aligned}

But what if, rather than defining a rotation as a normalised combination of a scalar and a bivector, we instead define a rotation as a normalized combination of a scalar and a vector? As we will see, it turns out to be equivalent to the Lorentz transformation of Special Relativity!