Spacetime Primitives — Examples

The displacement $v=(1,1,0,0)$ satisfies $\eta (v,v)=-1+1=0$ and $v\mathrm{\ne }0$, so $v$ is null by [L2]; since $v^0=1=\mid v_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid$, the vector lies in $\mathrm{cl}(C^{+})$ and is future null by [L3].

The displacement $w=(2,1,0,0)$ satisfies $\eta (w,w)=-4+1=-3<0$, so $w$ is timelike by [L2]; since $w^0=2>1=\mid w_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid$, it lies in $C^{+}$ and is future timelike by [L3].

The displacement $z=(1,2,0,0)$ satisfies $\eta (z,z)=-1+4=3>0$, so $z$ is spacelike by [L2].

The displacement $u=(-1,1,0,0)$ satisfies $\eta (u,u)=-1+1=0$ and $u\mathrm{\ne }0$, so $u$ is null by [L2]; since $u^0=-1=-\mid u_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid$, it lies in $\mathrm{cl}(C^{-})$ and is past null by [L3].

The frame vector $e_0$ has coordinates $(1,0,0,0)$ by [L4], so $\eta (e_0,e_0)=-1<0$ and $e_0^0=1>0=\mid (e_0{)}_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid$: $e_0$ is future timelike; negating, $-e_0$ has $\eta (-e_0,-e_0)=-1<0$ and $(-e_0{)}^0=-1<0$, so $-e_0$ is past timelike.

For any displacement $y$, membership in $\mathrm{cl}(C^{+})$ is exactly the condition $y^0\ge \mid y_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid$ by [L3], which forces $\eta (y,y)=-(y^0{)}^2+\mid y_{\mathrm{v}\mathrm{e}\mathrm{c}}{\mid }^2\le 0$ by [L1]; hence no future causal displacement is spacelike, and the coordinate inequalities of [L3] reproduce the classifications computed in steps 1.1 through 1.5 for the five displayed displacements.

Evaluating the form on $u$ against $v$, $\eta (u,v)=-2v^0+v^1$, so the orthogonality condition $\eta (u,v)=0$ is $v^1=2v^0$ and $u^{\perp }$ is that hyperplane by [L2].

On $u^{\perp }$, substituting $v^1=2v^0$ into the quadratic form gives $\eta (v,v)=-(v^0{)}^2+4(v^0{)}^2+(v^2{)}^2+(v^3{)}^2=3(v^0{)}^2+(v^2{)}^2+(v^3{)}^2$, a sum of nonnegative terms; it vanishes exactly when $v^0=v^2=v^3=0$, and then also $v^1=2v^0=0$, so $v=0$. Hence the restriction of $\eta$ to $u^{\perp }$ is positive definite.

Every nonzero $v\in u^{\perp }$ is spacelike by step 2.1 and [L3]; moreover $u$ itself is future timelike by [L4], since $\eta (u,u)=-3<0$ and $u^0=2>1=\mid u_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid$. Thus the orthogonal complement of this timelike displacement is genuinely positive definite, in contrast with the null case.

For the general clause of the Example, let $\tilde{u}=({\tilde{u}}^0,{\tilde{u}}_{\mathrm{v}\mathrm{e}\mathrm{c}})$ be any timelike displacement and let $w=(w^0,w_{\mathrm{v}\mathrm{e}\mathrm{c}})\mathrm{\ne }0$ satisfy $\eta (\tilde{u},w)=0$. Timelikeness gives $\mid {\tilde{u}}^0\mid >\mid {\tilde{u}}_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid$ by [L1] and [L3], and orthogonality gives ${\tilde{u}}_{\mathrm{v}\mathrm{e}\mathrm{c}}\cdot w_{\mathrm{v}\mathrm{e}\mathrm{c}}={\tilde{u}}^0{w}^0$. Suppose, for contradiction, that $\eta (w,w)\le 0$; then $\mid w_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid \le \mid w^0\mid$, and $w_{\mathrm{v}\mathrm{e}\mathrm{c}}\mathrm{\ne }0$, because $w_{\mathrm{v}\mathrm{e}\mathrm{c}}=0$ would give ${\tilde{u}}^0{w}^0=0$ and hence $w=0$, contradicting $w\mathrm{\ne }0$. Cauchy–Schwarz [L5] now yields the contradictory chain $\mid {\tilde{u}}^0\mid \mid w^0\mid =\mid {\tilde{u}}_{\mathrm{v}\mathrm{e}\mathrm{c}}\cdot w_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid \le \mid {\tilde{u}}_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid \mid w_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid <\mid {\tilde{u}}^0\mid \mid w_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid \le \mid {\tilde{u}}^0\mid \mid w^0\mid$, where the strict inequality uses $\mid {\tilde{u}}_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid <\mid {\tilde{u}}^0\mid$ and $w_{\mathrm{v}\mathrm{e}\mathrm{c}}\mathrm{\ne }0$ and the last inequality uses $\mid w_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid \le \mid w^0\mid$. Hence $\eta (w,w)>0$, that is, every nonzero displacement orthogonal to a timelike displacement is spacelike by [L3].

For a general pair of inertial frames, substituting $E^{\mathrm{\prime }}(u)=E(Mu)$, equivalently $E^{\mathrm{\prime }}=E\circ M$, into $x^{\mathrm{\prime }}=E^{\mathrm{\prime }-1}(p-o^{\mathrm{\prime }})$ and using $p-o^{\mathrm{\prime }}=E(x(p)-x(o^{\mathrm{\prime }}))$ gives $x^{\mathrm{\prime }}(p)=M^{-1}(x(p)-x(o^{\mathrm{\prime }}))=M^{-1}x(p)-M^{-1}x(o^{\mathrm{\prime }})$, an affine map with invertible linear part $M^{-1}$ by [L1] and [L6].

For the restricted basis, $\eta (e_0^{\mathrm{\prime }},e_0^{\mathrm{\prime }})=-{\cosh }^{2}r+{\sinh }^{2}r=-1$ and $\eta (e_1^{\mathrm{\prime }},e_1^{\mathrm{\prime }})=-{\sinh }^{2}r+{\cosh }^{2}r=1$ by [L3] and [L2], while $\eta (e_0^{\mathrm{\prime }},e_1^{\mathrm{\prime }})=-\cosh r\sinh r+\sinh r\cosh r=0$; also $\eta (e_i^{\mathrm{\prime }},e_j^{\mathrm{\prime }})={\delta }_{ij}$ for $i,j\in \{2,3\}$ and the mixed entries vanish. Hence $(e_0^{\mathrm{\prime }},e_1^{\mathrm{\prime }},e_2^{\mathrm{\prime }},e_3^{\mathrm{\prime }})$ is $\eta$-orthonormal by [L1].

The matrix $M$ of the restricted change of basis is the displayed block matrix, and $\det M={\cosh }^{2}r-{\sinh }^{2}r=1$ by [L2]; its inverse is the displayed block matrix, since multiplying the two blocks gives the identity using ${\cosh }^{2}r-{\sinh }^{2}r=1$.

By ${\cosh }^{2}r-{\sinh }^{2}r=1$ and $\cosh r>0$ one has $\cosh r=\sqrt{1+{\sinh }^{2}r}>\mid \sinh r\mid$ for every real $r$; with $\eta (e_0^{\mathrm{\prime }},e_0^{\mathrm{\prime }})=-1<0$ from step 1.2, the vector $e_0^{\mathrm{\prime }}$ is future timelike by [L4]. Since $\det M=1>0$ by step 1.3, the primed basis lies in the same positive class as the positively oriented original basis by [L5]; hence $(o^{\mathrm{\prime }};e_0^{\mathrm{\prime }},e_1^{\mathrm{\prime }},e_2^{\mathrm{\prime }},e_3^{\mathrm{\prime }})$ is an inertial frame by [L1].

With $o^{\mathrm{\prime }}=o$ and $x(o)=0$ by [L1], the transition law of step 1.1 gives $x^{\mathrm{\prime }}(o)=M^{-1}(x(o)-x(o^{\mathrm{\prime }}))=0$, and for the event $p=x^{-1}(1,0,0,0)$, whose coordinates in the first chart are $x(p)=(1,0,0,0)$, it gives $x^{\mathrm{\prime }}(p)=M^{-1}(1,0,0,0)=(\cosh r,-\sinh r,0,0)$ by the inverse computed in step 1.3. This completes the worked computation of the transition and its two displayed values.

By [L1] and [L3], $\eta (f_0,f_0)=\eta (-f_0,-f_0)=-1$, so $f_0,-f_0\in T$. They lie in opposite components: if both lay in the same component, its convexity [L2] would put $0=\frac{1}{2}{f}_0+\frac{1}{2}(-f_0)$ in $T$, contradicting $\eta (0,0)=0$. Thus $g_0:=f_0$ when $f_0\in C^{+}$ and $g_0:=-f_0$ otherwise is always future. Set $g_i:=f_i$ for $i=1,2,3$. Retaining $f_0$ retains the basis, while negating it preserves its norm and all vanishing off-diagonal pairings, since $\eta (-f_0,f_i)=-\eta (f_0,f_i)=0$; hence $(g_0,g_1,g_2,g_3)$ is $\eta$-orthonormal. Its change-of-basis matrix is the identity when $f_0$ is future and $\mathrm{diag}(-1,1,1,1)$ when $f_0$ is past. Their determinants are $1$ and $-1$, respectively, so [L4] shows that the orientation is retained in the first case and reversed in the second. In particular, past-directed negative input becomes future-directed positive input.

In the two-dimensional illustration, $f_0=(-1,0)$ has $\eta (f_0,f_0)=-1<0$ and $f_0^0=-1<0$, so $f_0$ is past timelike by the explicit two-dimensional convention in the Example; $-f_0=(1,0)$ has $\eta (-f_0,-f_0)=-1$ by [L3], $(-f_0{)}^0=1>0=\mid (-f_0{)}_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid$, so $-f_0$ is future timelike by that same convention; and $\eta (-f_0,f_1)=0$, $\eta (f_1,f_1)=1$ are direct evaluations. Hence $(-f_0,f_1)$ is an $\eta$-orthonormal pair with future first vector and with change-of-basis determinant $-1$, so it is positively oriented exactly when the original pair was negatively oriented.

If $(g_0,g_1,g_2,g_3)$ is positively oriented, retain it. Otherwise replacing $g_1$ by $-g_1$ preserves $\eta$-orthonormality by the sign computation of step 1.1, leaves the future first vector $g_0$ unchanged, and has change-of-basis matrix $\mathrm{diag}(1,-1,1,1)$ of determinant $-1$. There are exactly two orientation classes by [L4], so reversing the negative class gives the positive class. Interchanging $g_1$ and $g_2$ also has determinant $-1$ and preserves orthonormality and $g_0$, giving the same alternative correction.

In every initial time-orientation and orientation case, steps 1.1 and 2.1 produce an $\eta$-orthonormal, future, positively oriented ordered basis, which together with the chosen origin $o$ is an inertial frame by [L5]. Step 1.2 verifies the displayed two-dimensional illustration. Thus the recipe realizes the construction of Existence of future-oriented inertial frames at every event from arbitrary orthonormal input, including the past-directed negative example.

The coordinate matrix of the frame tuple $(e_0,e_1,e_2,e_3)$ is the identity matrix, whose determinant is $+1$; hence ${ϵ}_{0123}=+1$ by [L1]; this value is frame independent by [L4].

The tuple $(e_0,e_2,e_1,e_3)$ has coordinate matrix equal to the permutation matrix of the transposition $(12)$; its determinant is the single nonzero term of [L3], namely $\mathrm{sgn}(12)=-1$, so ${ϵ}_{0213}=-1$.

The tuple $(e_2,e_3,e_0,e_1)$ has coordinate matrix the permutation matrix of $\sigma =(02)(13)$, a product of two transpositions, so $\mathrm{sgn}\sigma =(-1{)}^2=+1$ and ${ϵ}_{2301}=+1$ by [L1] and [L3].

The tuple $(e_3,e_2,e_1,e_0)$ has coordinate matrix the permutation matrix of $\sigma =(03)(12)$, again a product of two transpositions, so ${ϵ}_{3210}=+1$ by [L1] and [L3].

If two of the indices $a,b,c,d$ coincide, then two columns of the coordinate matrix $[e_a,e_b,e_c,e_d]$ coincide, so its determinant is $0$ by [L2]; hence ${ϵ}_{abcd}=0$ for every tuple with a repeated index.

For the spatial symbol: the coordinate matrix of $(e_1,e_2,e_3)$ in the spatial ordered basis $(e_1,e_2,e_3)$ is the identity, giving ${ϵ}_{123}=+1$; the tuple $(e_2,e_1,e_3)$ is a transposition, giving ${ϵ}_{213}=-1$; and $(e_3,e_1,e_2)$ corresponds to the $3$-cycle $(132)$, an even permutation, giving ${ϵ}_{312}=+1$; repeated spatial indices give $0$ by the same column-coincidence argument as step 1.5.

Raising the indices with the diagonal inverse metric, exactly one term of the sum in [L1] survives for $(0,1,2,3)$, and ${ϵ}^{0123}={\eta }^{00}{\eta }^{11}{\eta }^{22}{\eta }^{33}{ϵ}_{0123}=(-1)(+1)(+1)(+1)(+1)=-1$; this is the value in every inertial frame, since both $ϵ$ and the diagonal inverse metric are frame independent by [L1] and [L4].

By [L1] fix an $\eta$-orthonormal ordered basis $(e_0,e_1,e_2,e_3)$ of $V$. Since it is a basis, $e_0\mathrm{\ne }0$, and by the displayed matrix $\eta (e_0,e_0)=-1<0$.

An inner product must satisfy $⟨v,v⟩\ge 0$ for every $v$, with equality only at $v=0$, by [L2]. The symmetric bilinear form $\eta$ fails this at the nonzero vector $e_0$, where $\eta (e_0,e_0)=-1<0$; hence $\eta$ is not positive definite and therefore is not an inner product, refuting the displayed statement. The contrast with [L3] shows that the failure comes from $\eta$ itself rather than from the definition of an inner product, which the Euclidean inner product satisfies.

Consequently $\eta$ does not induce a norm on $V$: the Euclidean norm of [L3] is defined as $\parallel x{\parallel }_2=\sqrt{⟨x,x⟩}$, and a norm arising from an inner product in this way requires the nonnegative quantity $⟨v,v⟩$ of [L2] for every $v$, whereas $\eta (e_0,e_0)=-1$ admits no real square root. In particular the positive-definite inner-product theory (Cauchy-Schwarz, norms and orthogonal projection) cannot be applied to $\eta$ without additional argument.

The displacement $u=(1,0,0,0)$ satisfies $\eta (u,u)=-1<0$, so $u$ is timelike by [L2]; since $u^0=1>0=\mid u_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid$, it is future timelike by [L3].