Lorentz transforms via Clifford algebra
Robin K. S. Hankin
Source:vignettes/lorentz_clifford.Rmd
lorentz_clifford.Rmd
To cite the clifford package in publications please use
Hankin (2022b). 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
Snygg (2010). Consider the following
four-vector:
(fourvec <- c(1,5,3,2)) # a four-vector## [1] 1 5 3 2
u <- c(0.2,0.3,0.4) # a three-velocityWe wish to consider the effect of a Lorentz transformation of
s. This is done by the boost() function of the
lorentz package (Hankin
2022a):
(Bmat <- boost(u)) # Bmat = "B-matrix"## t x y z
## t 1.1867817 -0.23735633 -0.35603450 -0.47471266
## x -0.2373563 1.02576299 0.03864448 0.05152597
## y -0.3560345 0.03864448 1.05796672 0.07728896
## z -0.4747127 0.05152597 0.07728896 1.10305195The transformation itself is simply matrix multiplication:
Bmat %*% fourvec## [,1]
## t -2.017529
## x 5.110444
## y 3.165666
## z 2.220888We 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))## Element of a Clifford algebra, equal to
## + 1e_1 + 5e_2 + 3e_3 + 2e_4Algebraically 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
## [,1]
## [1,] -37We might use the slightly slicker and more efficient idiom
quad.form() from the quadform package:
quad.form(M,fourvec)## [,1]
## [1,] -37The Clifford equivalent would be scalprod() [remembering
to set the signature to 1]:
## [1] -37We 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:
## 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_12We 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_14But 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_34Above, 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:
## Element of a Clifford algebra, equal to
## + 1.0457 - 0.1135e_12 - 0.17025e_13 - 0.22699e_14The 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 \(\overline{s}=B^{-1}sB\):
## Element of a Clifford algebra, equal to
## - 2.0175e_1 + 5.1104e_2 + 3.1657e_3 + 2.2209e_4Comparing with the result from the lorentz package
Bmat %*% fourvec## [,1]
## t -2.0175
## x 5.1104
## y 3.1657
## z 2.2209we see agreement to within numerical precision. We can further verify that the squared interval is unchanged:
## [1] -37matching 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_1234In 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:
## [1] 1We 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_4Algebra 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 -1Thus we can see, for example, that e12*e13 = -e23 and
e13*e12 = +e23.