---
title: "Lorentz transforms using Clifford algebra"
output: html_document
---

```{r setup, include=FALSE}
knitr::opts_chunk$set(echo = TRUE)
library("clifford")
library("emulator")
```

(throughout, we use units in which $c=1$.  Notation follows Snygg)

<p style="text-align: right;">
![](`r system.file("help/figures/clifford.png", package = "clifford")`){width=10%}
</p>





Consider the following four-vector:

```{r,define4vec}
(fourvec <- c(1,5,3,2))  # a four-vector
u <- c(0.2,0.3,0.4)  # a three-velocity
```

We wish to consider the effect of a Lorentz transformation of `s`. This is done by the `boost()` function of the `lorentz` package:

```{r,b1b2}
library("lorentz")
(Bmat <- boost(u))  # Bmat = "B-matrix"
```

The transformation itself is simply matrix multiplication:

```{r,matmultb}
Bmat %*% fourvec
```

We will effect this operation using Clifford algebra.  Conceptually I
am following Snygg but using a somewhat modified notation for
consistency with the `clifford` and `lorentz` packages.

## Lorentz transforms in terms of rapidity

The general form for a Lorentz transform of speed $u$ in the $x$-direction is

$$
\begin{pmatrix}
\overline{t}\\
\overline{x}
\end{pmatrix}
=
\begin{pmatrix}
\gamma&-\gamma v\\
-\gamma v&\gamma
\end{pmatrix}
\begin{pmatrix}
t\\x\end{pmatrix}
$$

where $\gamma=(1-u^2)^{-1/2}$.  Writing $\cosh\phi=\gamma$ and noting
that $\phi$ is real (sometimes $\phi$ is known as the _rapidity_) we
get

$$
\begin{pmatrix}
\cosh\phi&-\sinh\phi\\
-\sinh\phi &\cosh\phi
\end{pmatrix}
$$
for the transformation, and we can see that the matrix has unit determinant.

## Lorentz transforms in Clifford algebra

Above we considered the four-vector $s=(1,5,3,2)$.  In Clifford
formalism this appears as

```{r,cliff1}
(scliff <- as.1vector(fourvec))
```

Algebraically this would be
$1\mathbf{e}_1+5\mathbf{e}_2+3\mathbf{e}_3+2\mathbf{e}_4$ (Snygg would
write $1\mathbf{e}_0+5\mathbf{e}_1+3\mathbf{e}_2+2\mathbf{e}_3$; we
cannot use that notation here because basis vectors are numbered from
1 in the package, not zero).  The metric would be
 
$$
\begin{pmatrix}
 1&0&0&0\\
 0&-1&0&0\\
 0&0&-1&0\\
 0&0&0&-1
\end{pmatrix}
$$
   
[NB in relativity, the word "signature" refers to the eigenvalues of
the metric, so the signature of the above matrix would be $(1,3)$ [or
sometimes ${+}{-}{-}{-}$], because it has one positive and three
negative eigenvalues.  In package idiom, "signature" means the number
of basis vectors that square to $+1$ and $-1$, so we would implement
this metric using a signature of $(1,3)$].

The squared interval for our four-vector would be given by
 
```{r,squaredinterval}
M <- diag(c(1,-1,-1,-1))
t(fourvec) %*% M %*% fourvec
```
 
We might use the slightly slicker and more efficient idiom
`quad.form()` from the `emulator` package:

```{r,useemulator}
quad.form(M,fourvec)
```

The Clifford equivalent would be `scalprod()` [remembering to set the
signature to 1]:

```{r,usescalprod}
signature(1,3)
scalprod(scliff,scliff)
```

\newcommand{\ei}{\mathbf e}

We seek a boost $B\in{\mathcal C}_{1,3}$ such that
$\overline{s}=B^{-1}sB$ (juxtaposition indicating geometric product).
We will start with a boost in the $x$-direction with rapidity $\phi$.
This would be $B=\cosh(\phi/2)+{\mathbf e}_{12}\sinh(\phi/2)$.  We
note that $B^{-1}=\cosh(\phi/2)-{\mathbf e}_{12}\sinh(\phi/2)$.
Numerically:

```{r}
phi <- 2.1234534   # just a made-up random value
B <- cosh(phi/2) + sinh(phi/2)*e(1:2) 
Binv <- rev(B) # cosh(phi/2)- sinh(phi/2)*e(1:2)
B*Binv
```

We may verify that rapidities add:
```{r}
B <- function(phi){cosh(phi/2) + sinh(phi/2)*e(1:2)}
B(0.26) * B(1.33)
B(0.26 + 1.33) # should match
```


We may formally write $B=\exp({\mathbf e}_{12}\phi/2)$ on the grounds that
 
$$
\begin{eqnarray}
\exp({\mathbf e}_{12}x) &=&1+\mathbf{e}_{12}x + \frac{(\mathbf{e}_{12}x)^2}{2!} + 
\frac{(\mathbf{e}_{12}x)^3}{3!}+\frac{(\mathbf{e}_{12}x)^4}{4!}+\cdots\\
&=& (1+x^2/2+x^4/4!+\cdots) + \mathbf{e}_{12}(x+\frac{x^3}{3!}+\cdots)\\
&=& \cosh x + \mathbf{e}_{12}\sinh x
\end{eqnarray}
$$
 
and note that this exponential obeys the usual rules for the regular
exponential function $e^x,x\in\mathbb{R}$.  More generally, if we have
a transform of rapidity $\phi$ and direction cosines $k_x,k_y,k_z$
then the transform would be

$$
B_{xyz}=
\cosh(\phi/2)
+k_x\mathbf{e}_{12}\sinh(\phi/2)
+k_y\mathbf{e}_{13}\sinh(\phi/2)
+k_z\mathbf{e}_{14}\sinh(\phi/2)
$$

and we can use standard Clifford algebra (together with the fact that
$k_x^2+k_y^2+k_z^2=1$) to demonstrate the transformations.
Numerically:

```{r}
B3 <- function(phi,k){cosh(phi/2) + (
     +k[1]*sinh(phi/2)*e(c(1,2))
     +k[2]*sinh(phi/2)*e(c(1,3))
     +k[3]*sinh(phi/2)*e(c(1,4))
   )}
k <- function(kx,ky){c(kx, ky, sqrt(1-kx^2-ky^2))}
kx <- +0.23
ky <- -0.38


k1 <- k(kx=0.23, ky=-0.38)
sum(k1^2) # verify; should be = 1
zap(B3(0.3,k1)*B3(1.9,k1))  # zap() kills terms with small coefficients
zap(B3(0.3+1.9,k1)) # should match previous line (up to numerical accuracy)
```

But if the two boosts have different direction cosines, the result is
more complicated:

```{r}
k2 <- k(-0.5,0.1)
zap(B3(2.4,k1) * B3(1.9,k2))
```

Above, we see new terms not present in the pure boosts which
correspond to rotation.

Now we consider a general four-vector
$s=s^1\mathbf{e}_1+s^2\mathbf{e}_2+s^3\mathbf{e}_3+s^4\mathbf{e}_4$
and calculate $B^{-1}sB$.  This is made easier if we use the facts
that $\mathbf{e}_{12}$ commutes with $\mathbf{e}_3$ and $\mathbf{e}_4$
as well as scalars, and anticommutes with $\mathbf{e}_1$ and
$\mathbf{e}_2$.  Noting that $\exp(\mathbf{e}_{12})$ is a linear
combination of a scalar and $\mathbf{e}_{12}$ we have

$$
\begin{eqnarray}
B^{-1}sB &=& \exp(-\mathbf{e}_{12}\phi/2)(s^1\mathbf{e}_1+s^2\mathbf{e}_2+s^3\mathbf{e}_3+s^4\mathbf{e}_4)\exp(\mathbf{e}_{12}\phi/2)\\
&=& (\mathbf{e}_1s^1+\mathbf{e}_2s^2)\exp(\mathbf{e}_{12}\phi/2)\exp(\mathbf{e}_{12}\phi/2)+\mathbf{e}_3s^3 + \mathbf{e}_4s^4\\
&=& \mathbf{e}_1(s^1\cosh\phi-s^2\sinh\phi)+\mathbf{e}_2(s^2\cosh\phi-s^1\sinh\phi)
+\mathbf{e}_3s^3+\mathbf{e}_4s^4
\end{eqnarray}$$

as required (it matches the matrix version).  If we have two boosts
$B_1$ and $B_2$ then the combined boost is either $B_1B_2$ (for $B_1$
followed by $B_2$) or $B_2B_1$ (for $B_2$ followed by $B_1$).
Numerical methods are straightforward as I will demonstrate below.

## Numerical methods: Lorentz transforms using the Clifford package

Above we considered boost `Bmat`, and here I will show the effect of
this boost in terms of Clifford objects, using a specialist function
`f()`:

```{r,speedtocliff}
f <- function(u){
	phi <- acosh(gam(u))               # rapidity
	k <- cosines(u)                    # direction cosines
 	return(
	       cosh(phi/2)                 # t
	+ k[1]*sinh(phi/2)*basis(c(1,2))   # x
	+ k[2]*sinh(phi/2)*basis(c(1,3))   # y
	+ k[3]*sinh(phi/2)*basis(c(1,4))   # z
	)
}
```

Thus we can express the Lorentz transform as a Clifford object:

```{r,dasadx}
u <- as.3vel(-c(0.2,0.3,0.4))  # negative (passive transform)
options(digits=5)
(B <- f(u))
```

The first thing to do is to verify that the inverse of `B` behaves as
expected:

```{r,checkrev}
B*rev(B)
```

Then we can apply the transformation $\overline{s}=B^{-1}sB$:

```{r,dsdasdfds}
zap(rev(B)*scliff*B)
```

Comparing with the result from the `lorentz` package

```{r,comparelorentz}
Bmat %*% fourvec
```

we see agreement to within numerical precision.  We can further verify
that the squared interval is unchanged:

```{r,checkinterval}
jj <- rev(B)*scliff*B
scalprod(jj,jj)
```
matching the untransformed square interval.


## Multiple boosts

Successive Lorentz boosts can induce a rotation as well as a
translation.

```{r,defv}
u <- as.3vel(c(0.2, 0.3,  0.4))
v <- as.3vel(c(0.5, 0.0, -0.4))
w <- as.3vel(c(0.0, 0.7,  0.1))
Buvw <- f(u)*f(v)*f(w)
zap(Buvw)
```

In the above, note that Clifford object `Buvw` has a nonzero scalar
component, and also a nonzero `e_1234` component.  However, it
represents a consistent Lorentz transformation:

```{r,checkbuvw}
zap(Buvw*rev(Buvw))
```

We can now apply this transform to a four-velocity:

```{r,testiton1000}
n <- as.1vector(c(1,0,0,0))
zap(rev(Buvw) * n * Buvw)
```


## Algebra of Clifford representations


We can shed some light on this representation of Lorentz transforms as follows:

```{r,algecliff}
signature(1,3)
L <- list(
    C     = basis(numeric()),
    e12   = basis(c(1,2)), e13 = basis(c(1,3)),
    e14   = basis(c(1,4)), e23 = basis(c(2,3)),
    e24   = basis(c(2,4)), e34 = basis(c(3,4)),
    e1234 = basis(1:4)
) 

out <- noquote(matrix("",8,8))
rownames(out) <- names(L)
colnames(out) <- names(L)
for(i in 1:8){
  for(j in 1:8){
    out[i,j] <- gsub('[_ ]','',as.character(L[[i]]*L[[j]]))
  }
}
options("width" = 110)
out
```

Thus we can see, for example, that `e12*e13 = -e23` and `e13*e12 = +e23`.
