Lorentz transforms via Clifford algebra

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))

## Element of a Clifford algebra, equal to
## + 1e_1 + 5e_2 + 3e_3 + 2e_4

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 (Hankin 2022c). 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

##      [,1]
## [1,]  -37

We might use the slightly slicker and more efficient idiom quad.form() from the quadform package:

quad.form(M,fourvec)

##      [,1]
## [1,]  -37

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

signature(1,3)
scalprod(scliff,scliff)

## [1] -37

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

## Element of a Clifford algebra, equal to
## scalar ( 1 )

We may verify that rapidities add:

B <- function(phi){cosh(phi/2) + sinh(phi/2)*e(1:2)}
B(0.26) * B(1.33)

## Element of a Clifford algebra, equal to
## + 1.333011 + 0.8814299e_12

B(0.26 + 1.33) # should match

## Element of a Clifford algebra, equal to
## + 1.333011 + 0.8814299e_12

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

## [1] 1

zap(B3(0.3,k1)*B3(1.9,k1))  # zap() kills terms with small coefficients

## Element of a Clifford algebra, equal to
## + 1.668519 + 0.3071989e_12 - 0.507546e_13 + 1.196654e_14

zap(B3(0.3+1.9,k1)) # should match previous line (up to numerical accuracy)

## Element of a Clifford algebra, equal to
## + 1.668519 + 0.3071989e_12 - 0.507546e_13 + 1.196654e_14

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))

## Element of a Clifford algebra, equal to
## + 3.716216 - 0.479412e_12 - 0.653413e_13 + 0.277158e_23 + 3.722481e_14 -
## 1.071824e_24 + 0.691205e_34

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))

## Element of a Clifford algebra, equal to
## + 1.0457 - 0.1135e_12 - 0.17025e_13 - 0.22699e_14

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

B*rev(B)

## Element of a Clifford algebra, equal to
## scalar ( 1 )

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

zap(rev(B)*scliff*B)

## Element of a Clifford algebra, equal to
## - 2.0175e_1 + 5.1104e_2 + 3.1657e_3 + 2.2209e_4

Comparing with the result from the lorentz package

Bmat %*% fourvec

##      [,1]
## t -2.0175
## x  5.1104
## y  3.1657
## z  2.2209

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)

## [1] -37

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)

## Element of a Clifford algebra, equal to
## + 1.2914 + 0.45284e_12 + 0.69409e_13 - 0.14103e_23 + 0.07821e_14 + 0.07767e_24
## + 0.02902e_34 - 0.04011e_1234

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))

## [1] 1

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

n <- as.1vector(c(1,0,0,0))
zap(rev(Buvw) * n * Buvw)

## Element of a Clifford algebra, equal to
## + 2.3891e_1 + 0.98367e_2 + 1.9312e_3 + 0.10269e_4

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

##       C       e12     e13     e14     e23     e24     e34     e1234  
## C     1       +1e12   +1e13   +1e14   +1e23   +1e24   +1e34   +1e1234
## e12   +1e12   1       -1e23   -1e24   -1e13   -1e14   +1e1234 +1e34  
## e13   +1e13   +1e23   1       -1e34   +1e12   -1e1234 -1e14   -1e24  
## e14   +1e14   +1e24   +1e34   1       +1e1234 +1e12   +1e13   +1e23  
## e23   +1e23   +1e13   -1e12   +1e1234 -1      +1e34   -1e24   -1e14  
## e24   +1e24   +1e14   -1e1234 -1e12   -1e34   -1      +1e23   +1e13  
## e34   +1e34   +1e1234 +1e14   -1e13   +1e24   -1e23   -1      -1e12  
## e1234 +1e1234 +1e34   -1e24   +1e23   -1e14   +1e13   -1e12   -1

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