Spacetime Primitives

Let $V$ be a real vector space, with addition $+$, zero $0$, and additive inverses (Vector space over a field). An affine space over $V$ is a nonempty set $A$ together with a map

$+:A\times V⟶A,(p,v)⟼p+v,$

such that

1. $(p+v)+w=p+(v+w)$ and $p+0=p$ for all $p\in A$ and $v,w\in V$; and
2. for all $p,q\in A$ there is exactly one $v\in V$ with $p+v=q$.

The map in clause 1 is the action of $V$ on $A$. The unique vector supplied by clause 2 is written $q-p$ and is called the displacement from $p$ to $q$; thus $q-p$ is the unique $v\in V$ with $p+v=q$. An affine space over the real vector space $V$ is also called a real affine space modelled on $V$, and then $V$ is its space of displacements.

The following identities are consequences of the two clauses, with the displayed justification:

• $p+(q-p)=q$ for all $p,q\in A$: this restates the defining property of $q-p$ in clause 2.
• $p-p=0$ for all $p\in A$: taking $q=p$ in the previous display gives $p+(p-p)=p=p+0$, and clause 2 applied to the pair $(p,p)$ gives $p-p=0$.
• $p+0=p$ for all $p\in A$: this is clause 1.
• $(q-p)+(r-q)=r-p$ for all $p,q,r\in A$: $p+[(q-p)+(r-q)]=[p+(q-p)]+(r-q)=q+(r-q)=r=p+(r-p)$, so clause 2 applied to the pair $(p,r)$ identifies the two vectors.
• For each $p\in A$ the map $V\to A$, $v\mapsto p+v$, is a bijection: for every $q\in A$, clause 2 supplies $v=q-p$ with $p+v=q$, proving surjectivity; it is injective because $p+v=p+w$ gives $p+(v-w)=[p+v]+(-w)=p+w-w=p=p+0$, so clause 2 applied to $(p,p)$ gives $v-w=0$, that is, $v=w$.

Affine coordinate systems. Suppose $V$ is $4$-dimensional and $E:{\mathbb{R}}^4\to V$ is a linear isomorphism (Invertible linear maps, linear isomorphisms, and inverse linear maps). For a chosen point $o\in A$, define

$x(p):=E^{-1}(p-o),p\in A\mathrm{.}$

This is a bijection $x:A\to {\mathbb{R}}^4$: it is injective and surjective because $E^{-1}$ is a bijection and $p\mapsto p-o$ is a bijection of $A$ with $V$ by the bijection just proved, applied with that $o$. Such maps are the affine coordinate systems on $A$; clause 2 is exactly what makes them well defined, because it is clause 2 that attaches to each ordered pair $(o,p)$ a unique displacement $p-o$.

Let $V$ be a $4$-dimensional real vector space (Vector space over a field, Finite-dimensional vector space, and its dimension ${\dim }_{F}V$; infinite-dimensional means having no finite basis). A Lorentz form on $V$ is a symmetric bilinear form $\eta :V\times V\to \mathbb{R}$ (Bilinear forms, and symmetric, skew-symmetric, and alternating bilinear forms) such that

1. $\eta$ is nondegenerate: both radicals ${\mathrm{rad}}_L(\eta )=\{u:\eta (u,v)=0\text{ for every }v\}$ and ${\mathrm{rad}}_R(\eta )=\{v:\eta (u,v)=0\text{ for every }u\}$ are $\{0\}$ (The matrix, left and right radicals, rank, and nondegeneracy of a bilinear form on a finite-dimensional space); and
2. $\eta$ has inertia $(3,1,0)$: in some (equivalently, in every) basis of $V$ in which the matrix of $\eta$ is diagonal, exactly three diagonal entries are positive, exactly one is negative, and none is zero (Positive and negative definiteness, the inertia $(p,q,r)$, rank $p+q$, and signature $p-q$ of a real symmetric bilinear or quadratic form).

The inertia of a symmetric bilinear form is the triple $(p,q,r)$ of numbers of positive, negative and zero diagonal entries of any diagonalizing basis, and it is independent of the diagonalizing basis by Sylvester's law of inertia (Positive and negative definiteness, the inertia $(p,q,r)$, rank $p+q$, and signature $p-q$ of a real symmetric bilinear or quadratic form); hence clause 2 is a property of $\eta$ itself and not of a chosen basis. The associated quadratic form of $\eta$ is

$q(v):=\eta (v,v),v\in V\mathrm{.}$

A Lorentz vector space of signature $(-,+,+,+)$ is a pair $(V,\eta )$ consisting of a $4$-dimensional real vector space $V$ and a Lorentz form $\eta$ on $V$; the name records that the diagonal normal form of $\eta$ has one negative entry followed by three positive entries. No positive-definite structure on $V$ is assumed or implied by this definition, and $q$ takes both negative and positive values on nonzero vectors: a basis in which $\eta$ is diagonal with diagonal entries $-1,1,1,1$ exists by clause 2, and on its first vector $q$ is $-1$, while on each of the other three it is $+1$.

Let $(V,\eta )$ be a Lorentz vector space of signature $(-,+,+,+)$ (Lorentz forms of inertia (3,1,0) on a 4-dimensional real vector space), with associated quadratic form $q(v)=\eta (v,v)$. A vector $v\in V$ is

• timelike when $q(v)<0$;
• null, or lightlike, when $v\mathrm{\ne }0$ and $q(v)=0$;
• spacelike when $q(v)>0$;
• causal when it is timelike or null.

The zero vector $0\in V$ satisfies $q(0)=0$ and is consequently none of timelike, null or spacelike: the null case explicitly excludes it, and the timelike and spacelike cases exclude it by $q(0)=0$ together with the strict inequalities. The causal character of a nonzero vector $v$ is the sign of the real number $q(v)$, and the three names are attached to the three signs as above. When no confusion arises, the pairing $\eta$ itself is written $⟨v,w⟩:=\eta (v,w)$.

Each nonzero vector has exactly one causal character. For a nonzero $v\in V$ the real number $q(v)$ satisfies exactly one of $q(v)<0$, $q(v)=0$, $q(v)>0$ by trichotomy of the order on $\mathbb{R}$, so $v$ is timelike, null or spacelike, and no two of the three alternatives can hold at once. The causal vectors are therefore exactly the nonzero vectors that are timelike or null, and the zero vector is causal in no sense introduced here.

Let $\eta$ be a Lorentz form on a $4$-dimensional real vector space $V$. There is an ordered basis $(e_0,e_1,e_2,e_3)$ of $V$ with $\eta (e_a,e_b)=\mathrm{diag}(-1,1,1,1{)}_{ab}$ for all $a,b$. Such a basis is called $\eta$-orthonormal; in every $\eta$-orthonormal ordered basis exactly one basis vector is timelike and the other three are spacelike, and the count is the inertia $(3,1,0)$ of $\eta$.

Let $(V,\eta )$ be a Lorentz vector space of signature $(-,+,+,+)$ (Lorentz forms of inertia (3,1,0) on a 4-dimensional real vector space). Two vectors $u,v\in V$ are $\eta$-orthogonal, written $u\perp v$, when $\eta (u,v)=0$. For a subset $S\subseteq V$ the orthogonal complement of $S$ is

$S^{\perp }:=\{v\in V:\eta (v,s)=0\text{ for every }s\in S\},$

and for a vector $u$ one writes $u^{\perp }:=\{u{\}}^{\perp }$.

Elementary properties. For every subset $S\subseteq V$:

• $S^{\perp }$ is a linear subspace of $V$: if $v,w\in S^{\perp }$ and $\lambda \in \mathbb{R}$, then for every $s\in S$ bilinearity gives $\eta (v+\lambda w,s)=\eta (v,s)+\lambda \eta (w,s)=0$, so $v+\lambda w\in S^{\perp }$. In particular the zero vector lies in $S^{\perp }$ and $S^{\perp }$ is never empty.
• $S\subseteq (S^{\perp }{)}^{\perp }$: if $s\in S$, then $\eta (v,s)=0$ for every $v\in S^{\perp }$ by the definition of $S^{\perp }$, and symmetry of $\eta$ gives $\eta (s,v)=0$ for every $v\in S^{\perp }$, that is, $s\in (S^{\perp }{)}^{\perp }$.
• $S^{\perp }=(\mathrm{span}S{)}^{\perp }$: the inclusion $S\subseteq \mathrm{span}S$ gives $(\mathrm{span}S{)}^{\perp }\subseteq S^{\perp }$; conversely, if $v\in S^{\perp }$ and $u={\sum }_i{\lambda }_i{s}_i$ is a finite linear combination with $s_i\in S$, then linearity in the second argument gives $\eta (v,u)={\sum }_i{\lambda }_i\eta (v,s_i)=0$, so $v\in (\mathrm{span}S{)}^{\perp }$.

In particular, for every $u\in V$ one has $u\in u^{\perp }$ exactly when $\eta (u,u)=0$, that is, exactly when $u$ is null or $u=0$ (Timelike, null, spacelike and causal vectors of a Lorentz vector space).

A nonzero vector can be orthogonal to itself. Take a basis $e_0,e_1,e_2,e_3$ of $V$ in which the matrix of $\eta$ is diagonal with diagonal entries $-1,1,1,1$; such a basis exists by clause 2 of the definition of a Lorentz form. Then $v:=e_0+e_1$ is nonzero while $\eta (v,v)=\eta (e_0,e_0)+2\eta (e_0,e_1)+\eta (e_1,e_1)=-1+0+1=0$, so $v\in v^{\perp }$ with $v\mathrm{\ne }0$. This is exactly what fails for a positive-definite inner product and is why orthogonal complements in a Lorentz vector space need not be positive definite.

Let $(V,\eta )$ be a Lorentz vector space of signature $(-,+,+,+)$ (Lorentz forms of inertia (3,1,0) on a 4-dimensional real vector space), with causal terminology as in Timelike, null, spacelike and causal vectors of a Lorentz vector space, and let

$T:=\{v\in V:\eta (v,v)<0\}$

be its timelike set. Components and closures refer to the topology obtained by using any ordered basis to identify $V$ with Euclidean ${\mathbb{R}}^4$; The timelike set has exactly two convex components, described in orthonormal coordinates ↗, step 5.1, proves that the resulting open sets are independent of the basis. By that lemma the set $T$ has exactly two connected components, and each of them is nonempty, open, convex and path-connected. A time orientation of $(V,\eta )$ is a choice of one of these two components. Fixing such a choice, write $C^{+}$ for the chosen component and

$C^{-}:=T\setminus C^{+}$

for the other component; thus $T=C^{+}\bigsqcup C^{-}$ is a disjoint union and $C^{+}$, $C^{-}$ are exactly the two components supplied by the lemma. Relative to this choice a vector $v\in V$ is

• future timelike when $v\in C^{+}$;
• past timelike when $v\in C^{-}$;
• future null when $v\mathrm{\ne }0$, $\eta (v,v)=0$ and $v\in \mathrm{cl}(C^{+})$;
• past null when $v\mathrm{\ne }0$, $\eta (v,v)=0$ and $v\in \mathrm{cl}(C^{-})$;
• future causal when it is future timelike or future null;
• past causal when it is past timelike or past null.

The sets $\mathrm{cl}(C^{+})$ and $\mathrm{cl}(C^{-})$ are the future closed cone and the past closed cone of the time-oriented space $(V,\eta ,C^{+})$. By The timelike set has exactly two convex components, described in orthonormal coordinates ↗ each of them is a convex cone, their intersection is $\{0\}$, and a vector is causal exactly when it is future causal or past causal, in which case exactly one of these two alternatives holds: no nonzero vector is both future causal and past causal. The timelike vectors are exactly the points of $T=C^{+}\cup C^{-}$, and $0\mathrm{\notin }T$ because $\eta (0,0)=0$; since a null vector is nonzero with $\eta (v,v)=0$, no vector is both timelike and null, so the classes above are mutually exclusive where the definitions say they are: no vector is both future and past causal, and no timelike vector is null.

Let $(V,\eta )$ be a Lorentz vector space of signature $(-,+,+,+)$ and let $(e_0,e_1,e_2,e_3)$ be an $\eta$-orthonormal ordered basis. Write $v=v^0{e}_0+v^1{e}_1+v^2{e}_2+v^3{e}_3$ for the coordinates of $v$ (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) and

$\mid v_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid :=\sqrt{(v^1{)}^2+(v^2{)}^2+(v^3{)}^2}\mathrm{.}$

Then $\eta (v,v)=-(v^0{)}^2+\mid v_{\mathrm{v}\mathrm{e}\mathrm{c}}{\mid }^2$, and the timelike set $T:=\{v\in V:\eta (v,v)<0\}$ satisfies

$T=T_{+}\cup T_{-},T_{+}:=\{v:v^0>\mid v_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid \},T_{-}:=\{v:v^0<-\mid v_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid \}\mathrm{.}$

The sets $T_{+}$ and $T_{-}$ are nonempty, disjoint, open in $V$ (for the metric topology of the distance $d(u,w):=\parallel c(u)-c(w){\parallel }_2$ induced by the coordinates $c(v):=(v^0,v^1,v^2,v^3)$), convex and path-connected, and they are exactly the two connected components of $T$. In every $\eta$-orthonormal ordered basis the two displayed coordinate sets coincide with the two components of $T$ in some order, and they may be interchanged relative to a previously fixed naming of the components. Moreover

$\mathrm{cl}(T_{+})=\{v:v^0\ge \mid v_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid \},\mathrm{cl}(T_{-})=\{v:v^0\le -\mid v_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid \},\mathrm{cl}(T_{+})\cap \mathrm{cl}(T_{-})=\{0\}\mathrm{.}$