Em Minkowski Mathematical Prerequisites

Let $4:=\{0,1,2,3\}$ and let $V:={\mathbb{R}}^4$ be the set of all functions $4\to \mathbb{R}$, with the pointwise vector-space structure of The vector space $F^X$ of all functions $X\to F$ with pointwise operations, and $F^n$ as the case $X=n=\{0,1,\dots ,n-1\}$; so $V$ is the function space $F^X$ of The vector space $F^X$ of all functions $X\to F$ with pointwise operations, and $F^n$ as the case $X=n=\{0,1,\dots ,n-1\}$ at $F=\mathbb{R}$ and $X=4$, and its elements are written $v=(v^0,v^1,v^2,v^3)$ with $v^a:=v(a)$ for $a\in \{0,1,2,3\}$, the components being indexed from $0$ and written with an upper index; the index conventions used throughout this item are fixed below. Let $A$ be the same set ${\mathbb{R}}^4$, whose elements are now called points, together with the map

$V\times A\to A,(v,p)\mapsto v+p,$

given by pointwise addition of functions. Then the following three conditions hold.

1. $(v+v^{\mathrm{\prime }})+p=v+(v^{\mathrm{\prime }}+p)$ for all $v,v^{\mathrm{\prime }}\in V$ and all $p\in A$.
2. $0+p=p$ for every $p\in A$.
3. For all $p,q\in A$ there is exactly one $v\in V$ with $v+p=q$, namely $v=q-p$.

Conditions 1 and 2 are read at each of the four coordinates from the field axioms of $\mathbb{R}$, and condition 3 holds because $q-p$ satisfies $(q-p)+p=q$ coordinatewise while any $v$ with $v+p=q$ agrees with $q-p$ at every coordinate (Vector space over a field). A triple consisting of a set $A$, a vector space $V$ and a map $V\times A\to A$ satisfying 1–3 is called an affine space with translation space $V$; the three conditions are its definition, so by 1–3 the set $A$ above is an affine space with translation space $V$.

For $a\in \{0,1,2,3\}$ let $e_a\in V$ be given by $e_a(b):=1$ for $b=a$ and $e_a(b):=0$ for $b\mathrm{\ne }a$. Then $(e_0,e_1,e_2,e_3)$ is an ordered basis of $V$ (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis): every $v\in V$ equals $v^0{e}_0+\cdots +v^3{e}_3$, because the two sides have the same value at each $b\in \{0,1,2,3\}$; and if $c^0{e}_0+\cdots +c^3{e}_3=0_V$ then evaluating at $b$ gives $c^b=0$, so the list is linearly independent. We write $v=v^0{e}_0+\cdots +v^3{e}_3$ and use the coordinate columns of Coordinate columns $[v{]}_{\mathcal{B}}$ and matrices $[T{]}_{\mathcal{B}}^{\mathcal{C}}$ of linear maps relative to ordered bases with respect to this ordered basis.

The Minkowski form. Define

$\eta (v,w):=-v^0{w}^0+v^1{w}^1+v^2{w}^2+v^3{w}^3,v,w\in V\mathrm{.}$

This is the standard bilinear form $⟨x,y⟩={\sum }_{i<n}{x}_i{y}_i$ of The standard bilinear form $⟨x,y⟩={\sum }_{i<n}{x}_i{y}_i$ on $F^n$ with the sign of the first coordinate reversed, together with the Euclidean inner product of The Euclidean inner product $⟨x,y⟩={\sum }_{k<n}{x}_k{y}_k$ on ${\mathbb{R}}^n$ on the last three coordinates; directly from the formula, for fixed $w$ the assignment $v\mapsto \eta (v,w)$ is a linear combination of the component functions $v\mapsto v^a$, hence linear, and similarly in $w$, so $\eta$ is a bilinear form, and $\eta (v,w)=\eta (w,v)$ for all $v,w$, so it is symmetric (Bilinear forms, and symmetric, skew-symmetric, and alternating bilinear forms). Its matrix in the ordered basis $(e_0,e_1,e_2,e_3)$ is

$(\eta (e_a,e_b){)}_{a,b}=\mathrm{diag}(-1,1,1,1),$

and this matrix is its own inverse, the product of $\mathrm{diag}(-1,1,1,1)$ with itself being the identity matrix (Rectangular matrix multiplication and the identity matrix $I_n$, including zero-sized shapes); in particular it is invertible, so $\eta$ is nondegenerate (The matrix, left and right radicals, rank, and nondegeneracy of a bilinear form on a finite-dimensional space).

Index conventions. The components $v^a$ of a vector are the numbers in its expansion $v=v^0{e}_0+\cdots +v^3{e}_3$ in the ordered basis $(e_0,\dots ,e_3)$, and for a linear map $T:V\to V$ one writes $(Tv{)}^a$ for the components of $Tv$ and $T^a{}_b$ for the matrix entries, so that $(Tv{)}^a={\sum }_b{T}^a{}_b{v}^b$. Write ${\eta }_{ab}:=\eta (e_a,e_b)$, so that $({\eta }_{ab})=\mathrm{diag}(-1,1,1,1)$, and let $({\eta }^{ab})$ denote the inverse matrix of $({\eta }_{ab})$, that is, ${\sum }_b{\eta }^{ab}{\eta }_{bc}=1$ when $a=c$ and ${\sum }_b{\eta }^{ab}{\eta }_{bc}=0$ when $a\mathrm{\ne }c$. The lowered components of $v$ are $v_a:={\sum }_b{\eta }_{ab}{v}^b$, and the operation is inverted by $v^a={\sum }_b{\eta }^{ab}{v}_b$, because $\mathrm{diag}(-1,1,1,1)$ is its own inverse; explicitly $v_0=-v^0$ and $v_i=v^i$ for $i=1,2,3$. In these conventions $\eta (v,w)={\sum }_{a,b}{\eta }_{ab}{v}^a{w}^b={\sum }_a{v}_a{w}^a$ for all $v,w\in V$.

Causal classification. Let $v\in V$ with $v\mathrm{\ne }0$. Then $v$ is timelike when $\eta (v,v)<0$, null when $\eta (v,v)=0$, and spacelike when $\eta (v,v)>0$; it is causal when $\eta (v,v)\le 0$. If the nonzero $v$ is causal, then $v^0\mathrm{\ne }0$: the inequality $-(v^0{)}^2+(v^1{)}^2+(v^2{)}^2+(v^3{)}^2\le 0$ gives $(v^0{)}^2\ge (v^1{)}^2+(v^2{)}^2+(v^3{)}^2$, and $v^0=0$ would force $(v^1{)}^2+(v^2{)}^2+(v^3{)}^2=0$, hence $v^1=v^2=v^3=0$ and $v=0$, contrary to the hypothesis. A nonzero causal $v$ is therefore future-directed when $v^0>0$ and past-directed when $v^0<0$, and exactly one of these two alternatives holds for a nonzero real $v^0$ (Order on the reals). The future causal cone is

$C^{+}:=\{v\in V:\eta (v,v)\le 0,v^0>0\},$

whose elements are exactly the future-directed causal vectors; the zero vector is not in $C^{+}$, because it fails $v^0>0$.

Orientation. The standard orientation of $V$ is the orientation class of the ordered basis $(e_0,e_1,e_2,e_3)$ (Orientation of a finite-dimensional real vector space).

Lorentz maps. A Lorentz map is a linear map $L:V\to V$ (Linear map between vector spaces over the same field) that is invertible (Invertible linear maps, linear isomorphisms, and inverse linear maps) and satisfies

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

Writing $M:=[L{]}_{(e_0,\dots ,e_3)}\in M_4(\mathbb{R})$ for the matrix of $L$ in the ordered basis $(e_0,\dots ,e_3)$ (Coordinate columns $[v{]}_{\mathcal{B}}$ and matrices $[T{]}_{\mathcal{B}}^{\mathcal{C}}$ of linear maps relative to ordered bases), $M^{\mathsf{T}}$ for its transpose (The transpose $A^{\mathsf{T}}$ of a matrix), and $\eta$ also for the matrix $\mathrm{diag}(-1,1,1,1)$, the condition is the matrix identity

$M^{\mathsf{T}}\eta M=\eta \mathrm{.}$

Expanding $Lv$ and $Lw$ in the basis gives $\eta (Lv,Lw)={\sum }_{c,d}(M^{\mathsf{T}}\eta M{)}_{cd}{v}^c{w}^d$, whereas $\eta (v,w)={\sum }_{c,d}{\eta }_{cd}{v}^c{w}^d$, and the two sides agree for all $v,w$ exactly when $(M^{\mathsf{T}}\eta M{)}_{cd}={\eta }_{cd}$ for every $c,d$, since inserting the basis vectors $v=e_c$, $w=e_d$ gives the converse (Rectangular matrix multiplication and the identity matrix $I_n$, including zero-sized shapes). Equivalently, the matrix of $\eta$ in the basis $(L{e}_0,\dots ,L{e}_3)$ is again $\mathrm{diag}(-1,1,1,1)$; this is the change-of-basis congruence of A basis change by $P$ changes the matrix of a bilinear form from $A$ to $P^{\mathsf{T}}AP$ for the basis change from $(e_0,\dots ,e_3)$ to its image under the invertible map $L$. The invertibility requirement in the definition of a Lorentz map is in fact automatic: if $L$ satisfies the displayed identity and $Lv=0$, then $\eta (v,w)=\eta (Lv,Lw)=0$ for every $w$, so $v$ lies in the left radical of $\eta$, which is $\{0\}$ because $\eta$ is nondegenerate; thus $L$ is injective, and an injective linear endomorphism of the four-dimensional space $V$ is surjective by Rank-nullity: ${\dim }_{F}V=\mathrm{nullity}T+\mathrm{rank}T$ together with If ${\dim }_{F}V=n$ and $U$ is a linear subspace of $V$, then $U$ is finite-dimensional, ${\dim }_{F}U\le n$, and ${\dim }_{F}U=n$ if and only if $U=V$, so $L$ is a bijection, and its inverse $S:=L^{-1}$ is linear: for all $x,y\in V$ and all reals $a,b$, linearity of $L$ gives $L(aS(x)+bS(y))=ax+by$, and applying $S$ gives $S(ax+by)=aS(x)+bS(y)$; thus $L$ is invertible (Invertible linear maps, linear isomorphisms, and inverse linear maps). The map $L$ is proper when $\det L=1$ (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 when $L(C^{+})=C^{+}$.

Frames. A frame of $A$ is a point $o\in A$ together with an ordered basis $(u_0,u_1,u_2,u_3)$ of $V$ such that $\eta (u_a,u_b)=\mathrm{diag}(-1,1,1,1{)}_{ab}$ for all $a,b$ and $u_0\in C^{+}$. The coordinates of $p\in A$ in that frame are the unique $(x^0,x^1,x^2,x^3)\in {\mathbb{R}}^4$ with $p=o+x^0{u}_0+x^1{u}_1+x^2{u}_2+x^3{u}_3$; uniqueness holds because $(u_0,u_1,u_2,u_3)$ is an ordered basis, and $p-o$ is the unique element of $V$ carrying $o$ to $p$.

Tensors and components. For nonnegative integers $p,q$, a tensor of type $(p,q)$ on $V$ is a map $(V^{\ast }{)}^p\times V^q\to \mathbb{R}$ that is linear in each argument separately, where $V^{\ast }$ is the algebraic dual of Linear functionals and the algebraic dual $V^{\ast }=\mathcal{L}(V,F)$. An empty product is a singleton, so a $(0,0)$ tensor is a scalar. Covectors are the $(0,1)$ case, a vector $v$ defines the $(1,0)$ tensor $\theta \mapsto \theta (v)$, and a two-form is a skew-symmetric $(0,2)$ tensor. These objects are maps, with arrays recording their components in a chosen basis.

Fix a frame $(o,(u_a))$ and define its linear coordinate map $B:{\mathbb{R}}^4\to V$ by $B\xi :={\sum }_a{\xi }^a{u}_a$. It is invertible because $(u_a)$ is a basis, and $\eta (B\xi ,B\zeta )=\eta (\xi ,\zeta )$ by its defining pairings. For a Lorentz map $L:V\to V$, put $N:=[L{]}_{(u_a)}$, identifying its matrix with the linear map $B^{-1}LB$ on ${\mathbb{R}}^4$. With origin fixed and new basis $u_a^{\mathrm{\prime }}:=L^{-1}{u}_a$, the coordinates of the same point obey $x^{\mathrm{\prime }}=Nx$: indeed $p-o=Bx=L^{-1}B{x}^{\mathrm{\prime }}$ is equivalent to this identity. The basis is orthonormal since $L^{-1}$ preserves $\eta$, and it is a frame when $L$ is orthochronous, since then $L^{-1}{u}_0\in C^{+}$.

For a fixed vector $v$, covector $\theta$ and two-form $F$, let $\widehat{v}:=B^{-1}v$, ${\widehat{\theta }}_a:=\theta (u_a)$ and ${\widehat{F}}_{ab}:=F(u_a,u_b)$ be their old component columns and matrix; primes denote the corresponding components in $(u_a^{\mathrm{\prime }})$. Then

${\widehat{v}}^{\mathrm{\prime }}=N\widehat{v},{\widehat{\theta }}^{\mathrm{\prime }}=(N^{-1}{)}^{\mathsf{T}}\widehat{\theta },{\widehat{F}}^{\mathrm{\prime }}=(N^{-1}{)}^{\mathsf{T}}\widehat{F}{N}^{-1}\mathrm{.}$

The vector law follows from $B^{\mathrm{\prime }}=L^{-1}B=B{N}^{-1}$. Substituting $u_a^{\mathrm{\prime }}={\sum }_b{u}_b(N^{-1}{)}^b{}_a$ into $\theta (u_a^{\mathrm{\prime }})$ and $F(u_a^{\mathrm{\prime }},u_b^{\mathrm{\prime }})$ gives the other two laws by linearity and bilinearity. Equivalently, the coordinate functional $\theta \circ B^{\mathrm{\prime }}$ is $(\theta \circ B)\circ N^{-1}$, the transpose operation of The transpose or algebraic adjoint $T^{\ast }:W^{\ast }\to V^{\ast }$, $T^{\ast }(g)=g\circ T$, of a linear map $T:V\to W$, and the coordinate bilinear form obeys $F(B^{\mathrm{\prime }}\xi ,B^{\mathrm{\prime }}\zeta )=F(B{N}^{-1}\xi ,B{N}^{-1}\zeta )$. The geometric objects themselves are unchanged. In the standard frame $B=\mathrm{id}$, $N$ is the standard coordinate matrix of $L$; in a general frame it is $B^{-1}LB$.

Finally, a fixed positive real $c$ is separate extra structure: the coordinate $x^0$ has the dimension of length, and the time coordinate is $t:=x^0\mathrm{/}c$. No physical claim is made here; in particular the identification of $A$ with physical spacetime, of $C^{+}$ with the future light cone, and of $c$ with the speed of light in vacuum are later physical premises.

• Sign convention. The form has signature $(-1,1,1,1)$: with the index conventions of the Definition, $v^0$ is the time component and $v^1,v^2,v^3$ are the spatial components of $v$; the lowered time component $v_0=-v^0$ carries the opposite sign, while $v_i=v^i$ for $i=1,2,3$.
• What is not asserted. The affine space, the form, the cone and the transformation laws are mathematical data and conventions; the physical readings of these objects, and any statement that some physical trajectory or force equation has this form, are separate premises that this item does not supply.

Let $0\le \beta <1$ and put $\gamma :=(1-{\beta }^2{)}^{-1\mathrm{/}2}>0$ (Rational powers $a^r$ of a positive base, Laws of rational exponents). Define ${\mathrm{\Lambda }}_{\beta }:V\to V$ by $({\mathrm{\Lambda }}_{\beta }v{)}^0=\gamma (v^0-\beta v^1)$, $({\mathrm{\Lambda }}_{\beta }v{)}^1=\gamma (v^1-\beta v^0)$, $({\mathrm{\Lambda }}_{\beta }v{)}^2=v^2$, $({\mathrm{\Lambda }}_{\beta }v{)}^3=v^3$. Then ${\mathrm{\Lambda }}_{\beta }$ is linear (Linear map between vector spaces over the same field), ${\mathrm{\Lambda }}_{\beta }$ is a proper orthochronous Lorentz map, and more precisely: (a) $\eta ({\mathrm{\Lambda }}_{\beta }v,{\mathrm{\Lambda }}_{\beta }w)=\eta (v,w)$ for all $v,w\in V$; (b) $\det {\mathrm{\Lambda }}_{\beta }=1$; (c) ${\mathrm{\Lambda }}_{\beta }(C^{+})=C^{+}$; (d) ${\mathrm{\Lambda }}_{\beta }^{-1}={\mathrm{\Lambda }}_{-\beta }$. In particular a future causal vector has future causal image and $\eta ({\mathrm{\Lambda }}_{\beta }v,{\mathrm{\Lambda }}_{\beta }v)=\eta (v,v)$ for every $v$ (Order on the reals, Squaring is monotone on the nonnegatives). The hypothesis $0\le \beta <1$ is not removable: at $\beta =1$ the factor $\gamma$ is undefined.

Let $J\subseteq \mathbb{R}$ be a nonempty open interval (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length) and let $z:J\to {\mathbb{R}}^4$ be of class $C^1$ ($C^k$ Euclidean maps and diffeomorphisms) with $\eta (z^{\mathrm{\prime }}(s),z^{\mathrm{\prime }}(s))<0$ and $z^0{}^{\mathrm{\prime }}(s)>0$ for every $s\in J$; $z^{\mathrm{\prime }}$ is the derivative of the vector-valued curve (The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral) and such a $z$ is called a $C^1$ future timelike curve. Put $g(s):=-\eta (z^{\mathrm{\prime }}(s),z^{\mathrm{\prime }}(s))>0$, fix $s_0\in J$, and define $\tau (s):=c^{-1}{\int }_{s_0}^s\sqrt{g(u)}du$ for $s\in J$. Then: (1) $g$ is continuous on $J$ and so is $s\mapsto \sqrt{g(s)}$, where $\sqrt{\cdot }$ is the nonnegative square root (Square roots exist: a unique $\sqrt{a}\ge 0$ with $(\sqrt{a}{)}^2=a$; the positives are $\{x^2:x\mathrm{\ne }0\}$, Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, A composite of continuous functions is continuous, with no side hypothesis of the kind the composition of limits needs); (2) $\tau$ is differentiable on $J$ with ${\tau }^{\mathrm{\prime }}(s)=c^{-1}\sqrt{g(s)}>0$, hence $\tau$ is strictly increasing on $J$, injective, and continuous (Every continuous function on an interval has a primitive; two primitives differ by a constant; and ${\int }_a^{b}f=G(b)-G(a)$ for any primitive $G$, On an interval $I$, for $f$ continuous on $I$ and differentiable at every interior point: $f^{\mathrm{\prime }}\ge 0$ throughout gives $f$ nondecreasing, $f^{\mathrm{\prime }}>0$ gives $f$ increasing, $f^{\mathrm{\prime }}\le 0$ and $f^{\mathrm{\prime }}<0$ give the two decreasing forms; conversely a nondecreasing $f$ has $f^{\mathrm{\prime }}\ge 0$ and a nonincreasing $f$ has $f^{\mathrm{\prime }}\le 0$ wherever it is differentiable, and no strict converse is claimed); (3) the inverse ${\tau }^{-1}:\tau [J]\to J$ is differentiable at every point $\tau (s)$ of the interval $\tau [J]$, with $({\tau }^{-1}{)}^{\mathrm{\prime }}(\tau (s))=c\mathrm{/}\sqrt{g(s)}$ (Continuous inverse theorem: a continuous injective $f$ on an interval $I$ is a bijection onto the order-convex set $f[I]$, and the inverse $g:f[I]\to I$ is continuous and strictly monotone in the same sense as $f$, Derivative of an inverse: if $f$ is continuous and injective on a nondegenerate interval $I$ and differentiable at $c\in I$ with $f^{\mathrm{\prime }}(c)\mathrm{\ne }0$, then the inverse $g$ is differentiable at $f(c)$ with $g^{\mathrm{\prime }}(f(c))=1\mathrm{/}{f}^{\mathrm{\prime }}(c)$; and if $f^{\mathrm{\prime }}(c)=0$ then $g$ is not differentiable at $f(c)$); (4) the reparametrized curve $z\circ {\tau }^{-1}:\tau [J]\to {\mathbb{R}}^4$ is differentiable at every $\tau (s)$, with tangent $U(\tau (s)):=c{z}^{\mathrm{\prime }}(s)\mathrm{/}\sqrt{g(s)}$ satisfying $\eta (U,U)=-c^2$ (The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g(c)$, then $f\circ g$ is differentiable at $c$ with $(f\circ g{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(g(c))g^{\mathrm{\prime }}(c)$, A vector-valued function has a limit, or is continuous, if and only if each of its components does; with the algebra of continuous vector-valued functions); (5) if instead $\eta (z^{\mathrm{\prime }}(s),z^{\mathrm{\prime }}(s))=0$ for all $s\in J$, then $g=0$, $\tau \equiv 0$ on $J$ and $\tau$ is not injective, so a null curve carries no proper-time parametrization (A function continuous on an interval $I$ whose derivative vanishes at every interior point of $I$ is constant on $I$; consequently two such functions with the same derivative differ by a constant); (6) if $z(s)=p+sv$ with $\eta (v,v)<0$ and $v^0>0$, then $\tau (s)=c^{-1}\sqrt{-\eta (v,v)}(s-s_0)$ and $U=cv\mathrm{/}\sqrt{-\eta (v,v)}$ is constant (A function differentiable at $c$ is continuous at $c$, The derivative $f^{\mathrm{\prime }}(c)={\lim }_{x\to c}\frac{f(x)-f(c)}{x-c}$ of $f:A\to \mathbb{R}$ at a point $c\in A$ that is a limit point of $A$, and differentiability on a set). The identification of such straight lines with inertial motion is a physical premise and is not asserted here.