Lorentz Transformations

Let $(V,\eta )$ be a $4$-dimensional Lorentz vector space (Lorentz forms of inertia (3,1,0) on a 4-dimensional real vector space) equipped with a time orientation $C^{+}$ (Time orientation, future and past timelike/null/causal vectors, and the closed cones), and let $(A,V,\eta ,C^{+})$ be the corresponding oriented time-oriented affine Minkowski space (Oriented time-oriented affine Minkowski space (the spacetime primitives)).

(a) Linear Lorentz maps. A linear Lorentz map is a linear map $L:V\to V$ (Linear map between vector spaces over the same field) such that

$\eta (Lv,Lw)=\eta (v,w)\text{for all }v,w\in V\mathrm{.}$

The set of all linear Lorentz maps is denoted $O(V,\eta )$.

(b) Matrix criterion. Let $(e_0,e_1,e_2,e_3)$ be an ordered $\eta$-orthonormal basis of $V$ (Every Lorentz form admits an ordered basis with matrix diag(-1,1,1,1)), so that ${\eta }_0:=\mathrm{diag}(-1,1,1,1)$ is the matrix of $\eta$ in that basis, and let $M$ be the matrix of $L$ in that basis (Coordinate columns $[v{]}_{\mathcal{B}}$ and matrices $[T{]}_{\mathcal{B}}^{\mathcal{C}}$ of linear maps relative to ordered bases). Then $L$ is a linear Lorentz map if and only if

$M^T{\eta }_{0}M={\eta }_0,$

where the product is matrix multiplication (Rectangular matrix multiplication and the identity matrix $I_n$, including zero-sized shapes). Every linear Lorentz map is a linear isomorphism of $V$ (Invertible linear maps, linear isomorphisms, and inverse linear maps), and the criterion itself does not depend on the $\eta$-orthonormal basis used. The three assertions of this clause are proved in Lorentz maps are isomorphisms of determinant ±1 that preserve causal character and permute the cones; the Lorentz sets are groups ↗.

(c) Proper and orthochronous maps; the Lorentz group. A linear Lorentz map $L$ is proper if $\det L=+1$ (The determinant of an endomorphism of a finite-dimensional vector space: its matrix determinant in an ordered basis in positive dimension, and $1$ on the zero space) and orthochronous if $L(C^{+})=C^{+}$, a condition on the image of the set $C^{+}$. The set of proper orthochronous linear Lorentz maps is denoted $S{O}^{+}(V,\eta )$. The set $O(V,\eta )$ is called the (full) Lorentz group of $(V,\eta )$ and $S{O}^{+}(V,\eta )$ the proper orthochronous Lorentz group; that these sets are groups under composition is proved in Lorentz maps are isomorphisms of determinant ±1 that preserve causal character and permute the cones; the Lorentz sets are groups ↗, which also shows that $L$ is orthochronous if and only if $L(C^{+})$ is contained in $C^{+}$.

(d) Affine Lorentz maps. An affine Lorentz map is a map $F:A\to A$ for which there exist $p_0\in A$ and a linear Lorentz map $L$ with

$F(p_0+v)=F(p_0)+Lv\text{for all }v\in V\mathrm{.}$

Equivalently $F(p)=F(p_0)+L(p-p_0)$ for all $p\in A$, and $L$ is then unique: it is recovered as $Lv=F(p+v)-F(p)$ for every $p\in A$, so it is independent of the base event. $F$ is called proper (respectively orthochronous) when its linear part $L$ is. If $x$ is the chart of an inertial frame (Inertial frames and inertial charts of the affine Minkowski model), then

$x(F(p))=Mx(p)+b\text{for all }p\in A,$

where $M$ is the matrix of $L$ in the frame's ordered basis and $b:=x(F(o))-Mx(o)\in {\mathbb{R}}^4$; the two sides are column vectors in ${\mathbb{R}}^4$.

(e) Convention. In this framework, unless stated otherwise, the term Lorentz transformation means a proper orthochronous affine Lorentz map. Improper or non-orthochronous Lorentz maps are always named explicitly.

The composition of two affine Lorentz maps is affine Lorentz with linear part the composition of the linear parts, and the identity and the inverse of an affine Lorentz map are again such maps; these facts are part of Lorentz maps are isomorphisms of determinant ±1 that preserve causal character and permute the cones; the Lorentz sets are groups ↗.

Let $(o;e_0,e_1,e_2,e_3)$ be an inertial frame of the model (Inertial frames and inertial charts of the affine Minkowski model), and let $x=(x^0,x^1,x^2,x^3)$ be its coordinates, so that a vector is written $x=x^0{e}_0+x^1{e}_1+x^2{e}_2+x^3{e}_3$ and the spatial part is the coordinate column $x_{\mathrm{v}\mathrm{e}\mathrm{c}}=(x^1,x^2,x^3)$ of ${\mathbb{R}}^3$ with its standard Euclidean structure (The Euclidean inner product $⟨x,y⟩={\sum }_{k<n}{x}_k{y}_k$ on ${\mathbb{R}}^n$).

For $\mathit{\beta }\in {\mathbb{R}}^3$ with $\mathrm{\mid }\mathit{\beta }\mathrm{\mid }<1$, where $\mathrm{\mid }\mathit{\beta }\mathrm{\mid }$ is the Euclidean norm, set

$\gamma (\mathit{\beta }):=(1-\mathrm{\mid }\mathit{\beta }{\mathrm{\mid }}^2{)}^{-1\mathrm{/}2}\ge 1,$

with equality exactly for $\mathit{\beta }=\mathbf{0}$ and strict inequality for $\mathit{\beta }\mathrm{\ne }\mathbf{0}$; here the power is the rational power (Rational powers $a^r$ of a positive base, Laws of rational exponents). The passive Lorentz boost with parameter $\mathit{\beta }$ is the linear map $B(\mathit{\beta }):V\to V$ (Linear map between vector spaces over the same field) defined by its action on coordinates: for $\mathit{\beta }\mathrm{\ne }\mathbf{0}$,

$x^{\mathrm{\prime }0}=\gamma (x^0-\mathit{\beta }\cdot x_{\mathrm{v}\mathrm{e}\mathrm{c}}),x_{\mathrm{v}\mathrm{e}\mathrm{c}}^{\mathrm{\prime }}=x_{\mathrm{v}\mathrm{e}\mathrm{c}}+\frac{\gamma -1}{\mathrm{\mid }\mathit{\beta }{\mathrm{\mid }}^2}(\mathit{\beta }\cdot x_{\mathrm{v}\mathrm{e}\mathrm{c}})\mathit{\beta }-\gamma \mathit{\beta }{x}^0,$

and for $\mathit{\beta }=\mathbf{0}$, $B(\mathbf{0}):={\mathrm{id}}_V$. Here $x^{\mathrm{\prime }}$ denotes the coordinates of $B(\mathit{\beta })$ applied to the vector with coordinates $x$, and $\mathit{\beta }\cdot x_{\mathrm{v}\mathrm{e}\mathrm{c}}$ is the Euclidean scalar product. Equivalently, in the ordered basis $(e_0,e_1,e_2,e_3)$ the matrix of $B(\mathit{\beta })$ is

$\left(\begin{array}{cc}\gamma & -\gamma {\mathit{\beta }}^T\\ -\gamma \mathit{\beta }& I_3+{\displaystyle \frac{\gamma -1}{\mathrm{\mid }\mathit{\beta }{\mathrm{\mid }}^2}}\mathit{\beta }{\mathit{\beta }}^T\end{array}\right),$

that is, $B^0{}_0=\gamma$, $B^0{}_i=B^i{}_0=-\gamma {\beta }_i$ for $i=1,2,3$, and $B^i{}_j={\delta }^i{}_j+\frac{\gamma -1}{\mathrm{\mid }\mathit{\beta }{\mathrm{\mid }}^2}{\beta }_i{\beta }_j$ for $i,j=1,2,3$, with the spatial block replaced by $I_3$ when $\mathit{\beta }=\mathbf{0}$. In the $1+1$ case $\mathit{\beta }=\beta e_1$ this reads $x^{\mathrm{\prime }0}=\gamma (x^0-\beta x^1)$ and $x^{\mathrm{\prime }1}=\gamma (x^1-\beta x^0)$. The vector $\mathit{\beta }$ is called the boost parameter; it is dimensionless and mathematically independent of any unit convention.

Consistency of the definition. The displayed formulas assign to every coordinate pair $(x^0,x_{\mathrm{v}\mathrm{e}\mathrm{c}})$ a pair $(x^{\mathrm{\prime }0},x_{\mathrm{v}\mathrm{e}\mathrm{c}}^{\mathrm{\prime }})$; writing the result as the matrix product above and using that ${\mathbb{R}}^4$ is identified with coordinates through A finite list $v:n\to V$ is an ordered basis if and only if every $x\in V$ equals ${\sum }_{i<n}{\lambda }_i{v}_i$ for exactly one $\lambda :n\to F$; those scalars are the coordinates of $x$ in that ordered basis (see Coordinate columns $[v{]}_{\mathcal{B}}$ and matrices $[T{]}_{\mathcal{B}}^{\mathcal{C}}$ of linear maps relative to ordered bases), the map $v\mapsto E(M(E^{-1}(v)))$ is the unique linear map with that matrix. The case $\mathit{\beta }=\mathbf{0}$ is declared separately because the expression $(\gamma -1)\mathrm{/}\mathrm{\mid }\mathit{\beta }{\mathrm{\mid }}^2$ is undefined there. That the two cases fit together is the limit computation recorded next.

The limit as $\mathit{\beta }\to \mathbf{0}$. For $\mathit{\beta }\mathrm{\ne }\mathbf{0}$ the identity ${\gamma }^2-1=\mathrm{\mid }\mathit{\beta }{\mathrm{\mid }}^2{\gamma }^2$ gives

$\frac{\gamma -1}{\mathrm{\mid }\mathit{\beta }{\mathrm{\mid }}^2}=\frac{{\gamma }^2}{\gamma +1}\mathrm{.}$

Since $\mathrm{\mid }\mathit{\beta }{\mathrm{\mid }}^2\le 3\mathrm{/}4$ forces $1\le \gamma \le 2$, the bounds $\mathrm{\mid }{\beta }_i\mathrm{\mid }\le \mathrm{\mid }\mathit{\beta }\mathrm{\mid }$ and $\mathrm{\mid }{\beta }_i{\beta }_j\mathrm{\mid }\le \mathrm{\mid }\mathit{\beta }{\mathrm{\mid }}^2$ (from the Euclidean norm convention, Absolute value in an ordered field, Basic properties of the absolute value) show that $\gamma \to 1$ as $\mathit{\beta }\to \mathbf{0}$; here continuity of the positive-base power at fixed exponent $-1\mathrm{/}2$ is Continuity and derivatives of positive-base real powers, with real and rational powers identified by The exponential definition of real powers agrees with the existing rational powers. Hence $(\gamma -1)\mathrm{/}\mathrm{\mid }\mathit{\beta }{\mathrm{\mid }}^2\to 1\mathrm{/}2$, so $B^i{}_j={\delta }^i{}_j+\frac{1}{2}{\beta }_i{\beta }_j+o(\mathrm{\mid }\mathit{\beta }{\mathrm{\mid }}^2)$ and $B(\mathit{\beta })\to I$ entrywise, without a hidden division by zero. Here $o(\mathrm{\mid }\mathit{\beta }{\mathrm{\mid }}^2)$ denotes a remainder whose ratio to $\mathrm{\mid }\mathit{\beta }{\mathrm{\mid }}^2$ tends to $0$ through nonzero parameters: the ratio of each spatial remainder is bounded by $\mathrm{\mid }{\gamma }^2\mathrm{/}(\gamma +1)-1\mathrm{/}2\mathrm{\mid }\to 0$. The time-space entries obey $\mathrm{\mid }\gamma {\beta }_i\mathrm{\mid }\le 2\mathrm{\mid }\mathit{\beta }\mathrm{\mid }\to 0$, and the time-time entry tends to $1$ by $\gamma \to 1$. The definition asserts no Lorentz property: that $B(\mathit{\beta })$ is a proper orthochronous Lorentz map with inverse $B(-\mathit{\beta })$ is proved in The passive boost is a proper orthochronous Lorentz map with inverse B(-beta), and its collinear composition law in Collinear boost composition and the relativistic velocity-addition law. The physical reading of $c\mathit{\beta }$ as a relative velocity is not part of this definition (Linear and affine Lorentz maps, and the Lorentz group).

Let $(o;e_0,e_1,e_2,e_3)$ be an inertial frame of the model (Inertial frames and inertial charts of the affine Minkowski model) and let $S:=\mathrm{span}(e_1,e_2,e_3)$ be its spatial three-space with $\delta$ the restriction of $\eta$ to $S$ (The Euclidean spatial structure and cross product of an inertial frame). Then $\delta$ is a positive-definite symmetric bilinear form, $(e_1,e_2,e_3)$ is $\delta$-orthonormal, and $S=\{w\in V:\eta (e_0,w)=0\}$.

A spatial rotation of the frame is a linear map $R:V\to V$ (Linear map between vector spaces over the same field) such that

1. $R(e_0)=e_0$;
2. $R(S)\subseteq S$, so that the restriction $R{\mathrm{\mid }}_S$ is an endomorphism of $S$;
3. $\delta (Rw,Rz)=\delta (w,z)$ for all $w,z\in S$; and
4. the restriction $R{\mathrm{\mid }}_S$ has determinant $+1$ computed in the ordered basis $(e_1,e_2,e_3)$ (The determinant of an endomorphism of a finite-dimensional vector space: its matrix determinant in an ordered basis in positive dimension, and $1$ on the zero space).

Condition 3 makes $R{\mathrm{\mid }}_S$ injective, since $Rw=0$ with $w\in S$ forces $\delta (w,w)=\delta (Rw,Rw)=0$ and hence $w=0$ by positive definiteness; $S$ being finite-dimensional, $R{\mathrm{\mid }}_S$ is then an automorphism of $S$, so in fact $R(S)=S$. The extension of such an $R{\mathrm{\mid }}_S$ by $R(e_0)=e_0$ is also written $R$ and called a spatial rotation of the frame.

Matrix form and the group $SO(3)$. Via the $\delta$-orthonormal basis $(e_1,e_2,e_3)$, a linear map of $V$ fixing $e_0$ and preserving $S$ corresponds to the block matrix $\mathrm{diag}(1,A)$ with $A\in M_3(\mathbb{R})$; under this correspondence the maps of conditions 1-4 are exactly those with

$A^{T}A=I_3\text{and}\det A=+1.$

Indeed, the matrix of the adjoint of an endomorphism of the inner product space $(S,\delta )$ in a $\delta$-orthonormal basis is the transpose of its matrix (In orthonormal bases, the matrix of the adjoint is the conjugate transpose of the matrix), and an endomorphism of a finite-dimensional inner product space preserves the inner product exactly when $T^{\ast }T=I$ (For an endomorphism in finite dimension, preserving lengths, preserving inner products, carrying orthonormal bases to orthonormal bases, and $T^{\ast }T=I$ are equivalent, Linear isometries, and orthogonal or unitary operators on finite-dimensional inner product spaces). Writing $SO(3):=\{A\in M_3(\mathbb{R}):A^{T}A=I_3,\det A=+1\}$, the orthogonal operators on $S$ form a group and each of them has determinant $\pm 1$ (Orthogonal and unitary operators form groups, and their determinants have modulus one), and $\det (S_1{S}_2)=\det (S_1)\det (S_2)$ for composable endomorphisms (For endomorphisms $S$ and $T$ of one finite-dimensional vector space, $\det (ST)=\det (S)\det (T)$); hence the determinant-one part is a subgroup, the spatial rotations of the frame form a subgroup of the general linear group of $V$, that is, of the group of invertible linear maps $V\to V$ under composition (Invertible linear maps, linear isomorphisms, and inverse linear maps), and the map $R\mapsto A(R)$ sending a rotation to the matrix of $R{\mathrm{\mid }}_S$ in the basis $(e_1,e_2,e_3)$ is a group isomorphism onto $SO(3)$. The determinant in condition 4 is independent of the ordered basis of $S$ (The determinant of an endomorphism of a finite-dimensional vector space: its matrix determinant in an ordered basis in positive dimension, and $1$ on the zero space). Thus, for fixed $\eta$, the rotation subgroup depends only on the time axis $e_0$: it fixes $e_0$ and acts on $S=e_0^{\perp }$ by determinant-one $\delta$-isometries. Changing the origin or the spatial orthonormal basis while retaining $e_0$ leaves this subgroup unchanged; changing $e_0$ can change which maps are spatial rotations. No claim is made here that spatial rotations are Lorentz maps; that is proved in Unique boost-rotation factorization of proper orthochronous Lorentz maps.

Let $(V,\eta )$ be a $4$-dimensional Lorentz vector space with a chosen time orientation $C^{+}$, and let $L:V\to V$ be a linear map with $\eta (Lv,Lw)=\eta (v,w)$ for all $v,w\in V$ (Linear and affine Lorentz maps, and the Lorentz group). Then:

(i) $L$ is a linear isomorphism of $V$; for every ordered $\eta$-orthonormal basis the matrix $M$ of $L$ satisfies $M^T{\eta }_{0}M={\eta }_0$ with ${\eta }_0=\mathrm{diag}(-1,1,1,1)$, and conversely every invertible matrix $M$ with $M^T{\eta }_{0}M={\eta }_0$ is the matrix of a linear Lorentz map in that basis.

(ii) $\det L=\pm 1$. The composition of two Lorentz maps is a Lorentz map, the identity is a Lorentz map, and the inverse of a Lorentz map is a Lorentz map; the sets $SO(V,\eta ):=\{L\in O(V,\eta ):\det L=1\}$, $O^{\uparrow }(V,\eta ):=\{L\in O(V,\eta ):L(C^{+})=C^{+}\}$ and $S{O}^{+}(V,\eta )=SO(V,\eta )\cap O^{\uparrow }(V,\eta )$ are subgroups of $O(V,\eta )$, which itself is a subgroup of the general linear group $GL(V)$.

(iii) $\eta (Lv,Lv)=\eta (v,v)$ for all $v\in V$; hence $L$ preserves the causal character of every nonzero vector (timelike, null or spacelike) and maps the set of null vectors onto itself. If $L$ is proper, then the oriented volume form is preserved: $ϵ(L{v}_0,\dots ,L{v}_3)=ϵ(v_0,\dots ,v_3)$ for all $v_0,\dots ,v_3\in V$ (The oriented volume form epsilon and the permutation symbols of an inertial frame); if $\det L=-1$ the two sides differ by the sign $-1$.

(iv) $L(C^{+})$ is one of the two components $C^{+},C^{-}$ of the timelike set, so exactly one of $L(C^{+})=C^{+}$ and $L(C^{+})=C^{-}$ holds; $L$ is orthochronous in the first case and non-orthochronous in the second, and $L(C^{+})=C^{+}$ holds if and only if $L(C^{+})$ is contained in $C^{+}$.

(v) The affine Lorentz maps of Linear and affine Lorentz maps, and the Lorentz group form a group under composition: the composition and the inverse of affine Lorentz maps are affine Lorentz maps, and every affine Lorentz map is a bijection of $A$.