Quaternion Lorentz Transformations

Short description: Linear transformation of spacetime coordinates

In special relativity, a Lorentz transformation is a real linear transformation of the spacetime coordinates t, x, y, z that preserves the Minkowski invariant or spacetime interval $$c^2{t}^2-x^2-y^2-z^2$$ Using the 4x4 metric tensor $\eta$, this is $$X^T\eta X=\left[\begin{array}{cccc}ct& x& y& z\end{array}\right]\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& -1& 0& 0\\ 0& 0& -1& 0\\ 0& 0& 0& -1\\ \end{array}\right]\left[\begin{array}{c}ct\\ x\\ y\\ z\\ \end{array}\right]$$ Here $X^T$ is the matrix transpose of $X$.

One way to do Lorentz transformations is to let $X^{\prime }=AX$, where $A$ is a 4x4 real matrix that makes $$c^{2}t{\text{'}}^2-x{\text{'}}^2-y{\text{'}}^2-z{\text{'}}^2=c^2{t}^2-x^2-y^2-z^2$$ This is so if $A^T\eta A=\eta$.

Another way to do Lorentz transformations is to let the spacetime coordinates be represented by a 2x2 hermitian matrix

$$X=\left[\begin{array}{cc}ct+z& x-iy\\ x+iy& ct-z\end{array}\right]$$ The determinant of $X$ is its Minkowski invariant. Let A be a 2x2 matrix with determinant 1 and let $A^{†}$ be the hermitian conjugate of A (the complex conjugate of the transpose of A). Then $X^{\prime }=A^{†}XA$ has the same determinant as $X$ since the determinant of a product is the product of the determinants. Also, $X^{\prime }$ is hermitian since the hermitian conjugate of a product is the product of the hermitian conjugates in reverse order and since $X=X^{†}$. So this is a Lorentz transformation.

The method we will discuss in this article is Lorentz transformations using the complex quaternions. This method is equivalent to the method using 2x2 matrices. We will discuss this later. The complex quaternions have the advantages of being more transparent and simpler to work with.

Definition

The complex quaternions have the form $\text{<mi fromhbox="1">Q</mi>}=a+b\text{<mi fromhbox="1">I</mi>}+c\text{<mi fromhbox="1">J</mi>}+d\text{<mi fromhbox="1">K</mi>}$ for complex a, b, c, and d. The quaternion basis elements I, J, and K satisfy

$$\text{<mi fromhbox="1">I</mi>}\text{<mi fromhbox="1">I</mi>}=\text{<mi fromhbox="1">J</mi>}\text{<mi fromhbox="1">J</mi>}=\text{<mi fromhbox="1">K</mi>}\text{<mi fromhbox="1">K</mi>}=\text{<mi fromhbox="1">I</mi>}\text{<mi fromhbox="1">J</mi>}\text{<mi fromhbox="1">K</mi>}=-1$$

From these, using associativity, it follows that $$\text{<mi fromhbox="1">I</mi>}\text{<mi fromhbox="1">J</mi>}=-\text{<mi fromhbox="1">J</mi>}\text{<mi fromhbox="1">I</mi>}=\text{<mi fromhbox="1">K</mi>}\text{<mi fromhbox="1">J</mi>}\text{<mi fromhbox="1">K</mi>}=-\text{<mi fromhbox="1">K</mi>}\text{<mi fromhbox="1">J</mi>}=\text{<mi fromhbox="1">I</mi>}\text{<mi fromhbox="1">K</mi>}\text{<mi fromhbox="1">I</mi>}=-\text{<mi fromhbox="1">I</mi>}\text{<mi fromhbox="1">K</mi>}=\text{<mi fromhbox="1">J</mi>}$$

The real quaternions can be used to do spatial rotations, but not to do Lorentz transformations with a boost. But if a, b, c, and d are allowed to be complex, they can.

Minkowski quaternions

A Minkowski quaternion, adopting the convention of P. A. M. Dirac, has the form:

$$\text{<mi fromhbox="1">X</mi>}=t+ix\text{<mi fromhbox="1">I</mi>}+iy\text{<mi fromhbox="1">J</mi>}+iz\text{<mi fromhbox="1">K</mi>}$$

Here $i$ is the square root of -1 and c=1 henceforth.

The reason for this is that its norm is the Minkowski invariant $t^2-x^2-y^2-z^2$. The norm is defined as

$$\mathbf{𝐍}(a+b\mathbf{𝐈}+c\mathbf{𝐉}+d\mathbf{𝐊})=a^2+b^2+c^2+d^2$$

and has the important property that the norm of a product is the product of the norms, making the complex quaternions a composition algebra. A real non-zero quaternion always has real positive norm, but a non-zero complex quaternion can have a norm with any complex value, including zero.

As discussed in biquaternions, a biquaternion $\text{<mi fromhbox="1">Q</mi>}=a+b\text{<mi fromhbox="1">I</mi>}+c\text{<mi fromhbox="1">J</mi>}+d\text{<mi fromhbox="1">K</mi>}$ with complex $a,b,c,d$ has two kinds of conjugates:

• The biconjugate is

$$Q^{*}=a-b\mathbf{𝐈}-b\mathbf{𝐉}-d\mathbf{𝐊},$$

• The complex conjugate is

$$\overline{Q}=\overline{a}+\overline{b}\mathbf{𝐈}+\overline{c}\mathbf{𝐉}+\overline{d}\mathbf{𝐊}$$ The overbar $\stackrel{}{¯}$ denotes complex conjugation. The biconjugate of a product is the product of the biconjugates in reverse order. The operations denoted by the asterisk superscript and by the overbar are defined as in biquaternions.

For a Minkowski quaternion

$$\stackrel{*}{\stackrel{‾}{\mathbf{𝐗}}}=\mathbf{𝐗}$$

As can be seen from the definition, this is a necessary and sufficient condition for a complex quaternion $\mathbf{𝐗}$ to be a Minkowski quaternion.

Also needed is the identity $$\mathbf{𝐗}{\mathbf{𝐗}}^{*}=\mathbf{𝐗}\overline{\mathbf{𝐗}}=t^2-x^2-y^2-z^2$$

Lorentz transformations

General form

Let $\mathbf{𝐐}$ be a complex quaternion of norm one and let $\mathbf{𝐗}$ be a Minkowski quaternion. Then

$${\mathbf{𝐗}}^{\prime }=\stackrel{*}{\stackrel{‾}{\mathbf{𝐐}}}\text{<mi fromhbox="1">X</mi>}\text{<mi fromhbox="1">Q</mi>}=\stackrel{‾}{(\stackrel{*}{\stackrel{‾}{\mathbf{𝐐}}}\text{<mi fromhbox="1">X</mi>}\text{<mi fromhbox="1">Q</mi>}}{)}^{*}$$

Because of the second equality, ${\mathbf{𝐗}}^{\prime }$ is a Minkowski quaternion. And if $\mathbf{𝐐}$ has norm 1, then the norm of ${\mathbf{𝐗}}^{\prime }$ equals the norm of $\mathbf{𝐗}$. This is then a linear transformation of one Minkowski quaternion into another Minkowski quaternion having the same Minkowsky invariant. Therefore it is a Lorentz transformation.

Spatial rotations and Lorentz boosts

Let $\mathbf{𝐧}$ be the real direction quaternion

$$\mathbf{𝐧}=n_1\mathbf{𝐈}+n_2\mathbf{𝐉}+n_3\mathbf{𝐊}\text{ such that }{n}_1^2+n_3^2+n_3^2=1$$

Spatial rotations are represented by

$$\mathbf{𝐑}=\exp (-\frac{\theta }{2}\mathbf{𝐧})=\cos \left(\frac{\theta }{2}\right)\mathbf{-}\mathbf{𝐧}\sin \left(\frac{\theta }{2}\right)$$

$\mathbf{𝐑}$ has norm 1 and so represents a Lorentz transformation. It does not change the scalar part and so must be a rotation.

Boosts are represented by

$$\mathbf{𝐁}=\exp (-i\frac{\alpha }{2}\mathbf{𝐧})=\text{cosh}(\frac{\alpha }{2})-i\mathbf{𝐧}\text{sinh}(\frac{\alpha }{2})$$

$\mathbf{𝐁}$ also has norm 1 and so also represents a Lorentz transformation. It does not change the vector part normal to $\mathbf{𝐧}$ and so must be a Lorentz boost.

Expressing the exponentials as circular or hyperbolic trigonometric functions is basically De Moivre's formula.

It is immediately seen that $\mathbf{𝐁}$ and $\mathbf{𝐑}$ have the conjugate and norm properties

$$R{R}^{*}=1R=\overline{R}N(\mathbf{𝐑})=1$$

$$B={\overline{B}}^{*}{B}^{*}=\overline{B}B{B}^{*}=1B\overline{B}=1N(\mathbf{𝐁})=1$$

Here $N(\mathbf{𝐑})$ and $N(\mathbf{𝐁})$ are the respective norms of $R$ and $B$. If a complex quaternion has one of these sets of conjugate and norm properties, it must have the corresponding form given. Also note that $\mathbf{𝐑}\mathbf{𝐑}$ has the same form as $\mathbf{𝐑}$ except that $\theta /2$ is replaced by $\theta$ and that $\mathbf{𝐁}\mathbf{𝐁}$ has the same form as $\mathbf{𝐁}$ except that $\alpha /2$ is replaced by $\alpha$. Useful identities for representing a Lorentz transformation as a boost followed by a rotation or vice versa are

$$(\overline{RB}{)}^{*}(RB)=B^2(BR)(\overline{BR}{)}^{*}=B^2$$

The general spatial rotations and Lorentz boosts can be worked out by letting $\mathbf{𝐗}=t+i\mathbf{𝐫}$ where $\mathbf{𝐫}=x\mathbf{𝐈}+y\mathbf{𝐉}+z\mathbf{𝐊}$ and then repeatedly using the identity for the product of vectors

$$\mathbf{𝐧}\mathbf{𝐫}=-(\mathbf{𝐧},\mathbf{𝐫})+\mathbf{𝐧}\mathbf{𝐱}\mathbf{𝐫}$$

$$\mathbf{𝐫}\mathbf{𝐧}=-(\mathbf{𝐧},\mathbf{𝐫})-\mathbf{𝐧}\mathbf{𝐱}\mathbf{𝐫}$$

$$\mathbf{𝐧}\mathbf{𝐫}\mathbf{𝐧}=-\mathbf{𝐧}(\mathbf{𝐧},\mathbf{𝐫})-\mathbf{𝐧}\mathbf{𝐱}(\mathbf{𝐧}\mathbf{𝐱}\mathbf{𝐫})$$

Here $(\mathbf{𝐧},\mathbf{𝐫})$ is the scalar product of $\mathbf{𝐧}$ and $\mathbf{𝐫}$ and $\mathbf{𝐧}\mathbf{𝐱}\mathbf{𝐫}$ is their cross product.

Examples

Let $\mathbf{𝐧}=\mathbf{𝐈}$. Then the boost $\mathbf{𝐁}$ in the x direction gives the familiar coordinate transformations: $$t^{\prime }=\text{cosh}(\alpha )t-\text{sinh}(\alpha )x$$ $$x^{\prime }=\text{cosh}(\alpha )x-\text{sinh}(\alpha )t{y}^{\prime }=y{z}^{\prime }=z$$

Now let $\mathbf{𝐧}=\mathbf{𝐊}$. The spatial rotation $\mathbf{𝐑}$ is then a rotation about the z axis and gives the again familiar coordinate transformations:

$$x^{\prime }=x\cos (\theta )-y\sin (\theta )$$

$$y^{\prime }=x\sin (\theta )+y\cos (\theta )$$

$$t^{\prime }=t{z}^{\prime }=z$$

2x2 matrices

By a simple identification, we show that Lorentz transformations using complex quaternions are equivalent to Lorentz transformations using 2x2 matrices. The complex quaternions have the advantages of being more transparent and simpler to work with.

The quaternion basis elements $\mathbf{𝐈},\mathbf{𝐉},\mathbf{𝐊}$ can be represented as the 2x2 matrices $-i{\sigma }_x,-i{\sigma }_y,-i{\sigma }_z$, respectively. Here the ${\sigma }_i$ are the 2x2 Pauli spin matrices. These have the same multiplication table. This representation is not unique. For instance, without changing the multiplication table, the sign of any two can be reversed, or the ${\sigma }_i$ can be cyclically permuted, or a similarity transformation can be done so that the ${\sigma }_i$ are replaced by $S^{-1}{\sigma }_{i}S$.

Everything that follows is by simple replacement of $\mathbf{𝐈},\mathbf{𝐉},\mathbf{𝐊}$ by $-i{\sigma }_x,-i{\sigma }_y,-i{\sigma }_z$. Except for X, lower case letters q, r, b, and ${\sigma }_i$ are used for 2x2 matrices.

The Minkowski 2x2 matrix then has the form $$X=t+x{\sigma }_x+y{\sigma }_y+z{\sigma }_z=\left(\begin{array}{cc}t+z& x-iy\\ x+iy& t-z\end{array}\right)$$

Let an arbitrary 2x2 matrix have the form $q=a+b{\sigma }_x+c{\sigma }_y+d{\sigma }_z$, where a, b, c, and d are complex.

• The analog of the biconjugate is $q^{*}=a-b{\mathit{\sigma }}_x-c{\mathit{\sigma }}_y-d{\mathit{\sigma }}_z,$
• The analog of the complex conjugate is $\overline{q}=\overline{a}-\overline{b}{\mathit{\sigma }}_x-\overline{c}{\mathit{\sigma }}_y-\overline{d}{\mathit{\sigma }}_z$
• The analog of the biconjugate of the complex conjugate is the hermitean conjugate (conjugate transpose) since the ${\sigma }_i$ are hermitean 2x2 matrices:

$${\overline{q}}^{*}=q^{†}=\overline{a}+\overline{b}{\mathit{\sigma }}_x+\overline{c}{\mathit{\sigma }}_y+\overline{d}{\mathit{\sigma }}_z$$

• The analog of the norm is $N(q)=a^2-b^2-c^2-d^2$. This is also its determinant $\left|\begin{array}{cc}a+d& b-ic\\ b+ic& a-d\end{array}\right|$
• The Lorentz transformation is $X^{\prime }={\overline{q}}^{*}Xq=q^{†}Xq$ for a 2x2 matrix q that has norm 1 (determinant 1).

A direction can be represented as $\mathbf{𝐧}\cdot \mathit{\sigma }=n_1{\sigma }_x+n_2{\sigma }_y+n_3{\sigma }_z$ where $n_1^2+n_2^2+n_3^2=1$

The spatial rotation is $r=\exp (i\frac{\theta }{2}\mathbf{𝐧}\cdot \mathit{\sigma })$ so ${\overline{r}}^{*}\equiv r^{†}=\exp (-i\frac{\theta }{2}\mathbf{𝐧}\cdot \mathit{\sigma })$

The Lorentz boost is $b=\exp (-\frac{\alpha }{2}\mathbf{𝐧}\cdot \mathit{\sigma })$ so ${\overline{b}}^{*}\equiv b^{†}=\exp (-\frac{\alpha }{2}\mathbf{𝐧}\cdot \mathit{\sigma })$

See also

• Biquaternion
• Quaternion
• Lorentz transformation
• Spacetime algebra