Inertial frames and inertial charts of the affine Minkowski model

Let $(A,V,\eta ,C^{+})$ be an oriented time-oriented affine Minkowski space (Oriented time-oriented affine Minkowski space (the spacetime primitives)), with orientation fixed as a positive ray of the determinant line ${\mathrm{\Lambda }}^{4}V$ (Determinant-line orientations of finite-dimensional real vector spaces) and with the corresponding class of positively oriented ordered bases of $V$ (Orientation of a finite-dimensional real vector space); by Orientations and positive basis classes agree in positive dimension the positive ray selects exactly one such class for $\dim V=4>0$.

An inertial frame of the model is a tuple $(o;e_0,e_1,e_2,e_3)$ consisting of an event $o\in A$ and an ordered basis $(e_0,e_1,e_2,e_3)$ of $V$ that is

1. $\eta$-orthonormal: $\eta (e_a,e_b)=\mathrm{diag}(-1,1,1,1{)}_{ab}$ for all $a,b$ (Every Lorentz form admits an ordered basis with matrix diag(-1,1,1,1));
2. future: $e_0\in C^{+}$, i.e. $e_0$ is future timelike; and
3. positively oriented: the ordered basis lies in the positive-basis class selected by the fixed orientation.

The event $o$ is the origin of the frame and $e_0,e_1,e_2,e_3$ are its basis vectors.

Inertial charts. For an inertial frame $(o;e_0,e_1,e_2,e_3)$, define

$E:{\mathbb{R}}^4⟶V,E(u):={\sum }_{a=0}^3{u}^a{e}_a\mathrm{.}$

Then $E$ is a linear isomorphism (Invertible linear maps, linear isomorphisms, and inverse linear maps): it is linear by construction, and it is bijective because $(e_0,e_1,e_2,e_3)$ is an ordered basis, every $v\in V$ having exactly one coordinate list $u$ with $v={\sum }_a{u}^a{e}_a$ (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). The inertial chart of the frame is the map

$x:A⟶{\mathbb{R}}^4,x(p):=E^{-1}(p-o),$

and its values $x^a(p)$ are the inertial coordinates of the event $p$ in the frame. Since $E$ and $v\mapsto p+v$ are bijections, $x$ is a bijection $A\to {\mathbb{R}}^4$; and $x(p)=0$ exactly when $p-o=0_V$, that is, exactly when $p=o$.

Transition law. Let $(o^{\mathrm{\prime }};e_0^{\mathrm{\prime }},e_1^{\mathrm{\prime }},e_2^{\mathrm{\prime }},e_3^{\mathrm{\prime }})$ be another inertial frame with chart $x^{\mathrm{\prime }}$, and let $M=(M_a^b)$ be the change-of-basis matrix defined by $e_a^{\mathrm{\prime }}={\sum }_b{M}_a^b{e}_b$, a $4\times 4$ real matrix that is invertible because both families are ordered bases. Then $E^{\mathrm{\prime }}(u)={\sum }_a{u}^a{e}_a^{\mathrm{\prime }}=E(Mu)$ for all $u\in {\mathbb{R}}^4$, for the matrix-vector product $Mu$ in the coordinates of the first frame, because ${\sum }_a{u}^a{\sum }_b{M}_a^b{e}_b={\sum }_b({\sum }_a{M}_a^b{u}^a)e_b$; hence $E^{\mathrm{\prime }}=E\circ M$ and $E^{\mathrm{\prime }-1}=M^{-1}\circ E^{-1}$, so for every event $p$,

$x^{\mathrm{\prime }}(p)=M^{-1}(x(p)-x(o^{\mathrm{\prime }}))=M^{-1}x(p)-M^{-1}x(o^{\mathrm{\prime }})\mathrm{.}$

In particular the transition between two inertial charts is an affine map of ${\mathbb{R}}^4$ with invertible linear part $M^{-1}$ and constant part $-M^{-1}x(o^{\mathrm{\prime }})$, and it is a bijection of ${\mathbb{R}}^4$ with inverse $u\mapsto Mu+x(o^{\mathrm{\prime }})$, which is the transition in the reverse direction.

The inertial atlas. The inertial atlas of the model is the family of all inertial frames together with their inertial charts. Its charts are available at every event by Existence of future-oriented inertial frames at every event ↗, which supplies for every $o\in A$ an inertial frame with origin $o$; and any two of its charts are related by the affine transition displayed above.