Lorentz transforms via Clifford algebra

In clifford: Arbitrary Dimensional Clifford Algebras

knitr::opts_chunk$set(echo = TRUE)
library("clifford")
library("quadform")
library("lorentz")
library("jordan")

knitr::include_graphics(system.file("help/figures/clifford.png", package = "clifford"))
knitr::include_graphics(system.file("help/figures/jordan.png", package = "jordan"))
knitr::include_graphics(system.file("help/figures/quadform.png", package = "quadform"))

To cite the clifford package in publications please use @hankin2022_clifford. In this short document I show how Clifford algebra may be used to effect Lorentz transforms, and showcase the clifford R package. Throughout, we use units in which $c=1$. Notation follows @snygg2010. Consider the following four-vector:

(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

(Bmat <- boost(u))  # Bmat = "B-matrix"

The transformation itself is simply matrix multiplication:

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

(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). Also note that the vectors appear in implementation-specific order, as per disordR discipline [@hankin2022_disordR]. 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

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 quadform package:

quad.form(M,fourvec)

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

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:

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:

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:

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:

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():

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:

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:

B*rev(B)

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

zap(rev(B)*scliff*B)

Comparing with the result from the lorentz package

Bmat %*% fourvec

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

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.

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:

zap(Buvw*rev(Buvw))

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

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:

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.

signature(Inf) # restore default, to avoid interference with other vignettes

References

clifford documentation built on June 8, 2025, 10:56 a.m.