Future causal reachability, absence of causal loops, time monotonicity and the straight-worldline proper-time maximum

Let $a<b$ and let $x:[a,b]\to A$ be a piecewise $C^1$ future causal path with $D=x(b)-x(a)$, and fix an inertial chart, writing $\mathrm{x}=({\mathrm{x}}^0,z)$ for the coordinate path of $x$; by [L6] the coordinate path is piecewise $C^1$ with the same partition and with one-sided tangents ${\mathrm{x}}^{\mathrm{\prime }}=E^{-1}{x}^{\mathrm{\prime }}$, which are nonzero future causal because $x^{\mathrm{\prime }}$ is and $E^{-1}$ is linear with the coordinate criterion of [L3] read in this basis. Applying [L1] to the coordinate path on each piece: every coordinate derivative is absolutely improperly integrable with signed improper integral equal to the corresponding endpoint increment, so the coordinates of $D$ are the sums of these increments and $D\in \overline{C^{+}}$ by the description $\overline{C^{+}}=\{v:v^0\ge \mathrm{\mid }{v}_s\mathrm{\mid }\}$ of [L3]; indeed [L1] gives $D$ nonzero future causal with $D^0>0$ and $\mathrm{\mid }{D}_s\mathrm{\mid }\le D^0$. If some one-sided tangent of $x$ is future timelike, [L1] gives $D$ future timelike, that is $D\in C^{+}$ by [L3]. In particular $D\mathrm{\ne }0$ for every such path with $a<b$, so the third clause of (a) holds; this is (a).

For (b) first: $p\le p$ because $p-p=0\in \overline{C^{+}}$; if $p\le q$ and $q\le r$, then $r-p=(q-p)+(r-q)\in \overline{C^{+}}$ by [L3], so $p\le r$. Thus $\le$ is reflexive and transitive. Also $p\in R^{+}(p)$ by [L4] and $p\in J^{+}(p)$ because $0\in \overline{C^{+}}$ by [L3]. If $q\in R^{+}(p)$ with $q\mathrm{\ne }p$, then a piecewise $C^1$ future causal path joins $p$ to $q$, so its displacement $D=q-p$ lies in $\overline{C^{+}}$ by step 1.1, that is $q\in J^{+}(p)$ by [L4]; hence $R^{+}(p)\subseteq J^{+}(p)$. Conversely let $q\in J^{+}(p)$ with $q\mathrm{\ne }p$; then $D=q-p$ is a nonzero element of $\overline{C^{+}}$, hence future causal by [L3], and the straight path $\sigma (s)=p+sD$, $s\in [0,1]$, has the constant tangent ${\sigma }^{\mathrm{\prime }}\equiv D$ by [L6], a nonzero future causal vector, so $\sigma$ is a piecewise $C^1$ future causal path from $p$ to $q$; therefore $q\in R^{+}(p)$ and the straight path from $p$ to $q$ is a future causal path. Conversely, if the straight path from $p$ to $q$ with $q\mathrm{\ne }p$ is a future causal path, its constant tangent $D$ is future causal and nonzero, so $D\in \overline{C^{+}}$ and $q\in J^{+}(p)$ by [L4]; thus $J^{+}(p)=R^{+}(p)$ and the straight-segment criterion holds for $q\mathrm{\ne }p$.

For (b) second: if $q\in I^{+}(p)$ then $D=q-p\in C^{+}$ by [L4], so $D$ is future timelike and nonzero, and the straight path from $p$ to $q$ has the constant tangent $D$, so it is a future timelike path; conversely, if the straight path from $p$ to $q$ is future timelike then its tangent $D$ is future timelike, that is $D\in C^{+}$, so $q\in I^{+}(p)$ by [L4]: this proves the first displayed equality. For the second, every $q\in I^{+}(p)$ is reachable from $p$ by the straight path, which has a future timelike tangent at every point, so $I^{+}(p)$ is contained in the third set; conversely, if $q$ is reachable from $p$ by a piecewise $C^1$ future causal path having a future timelike one-sided tangent at some point, then $q\mathrm{\ne }p$ or $q=p$; in the first case step 1.1 gives $D\in C^{+}$, so $q\in I^{+}(p)$, while in the second case there is a path from $p$ to itself with a future timelike tangent, which step 1.1 turns into $0\in C^{+}$, a contradiction; hence the third set is contained in $I^{+}(p)$. Finally $p\mathrm{\notin }{I}^{+}(p)$ because $0\mathrm{\notin }{C}^{+}$ by [L4], and $p$ is not in the first or third set because the straight path from $p$ to $p$ has the constant tangent $0$, which is not future timelike, and a path from $p$ to itself with a future timelike tangent cannot exist as just shown.

For (d): let $[a^{\mathrm{\prime }},b^{\mathrm{\prime }}]\subseteq [a,b]$ and let $y$ be a piecewise $C^1$ future causal path. If $a^{\mathrm{\prime }}=b^{\mathrm{\prime }}$, the coordinate increment is zero. Otherwise the restriction $y{\mathrm{\mid }}_{[a^{\mathrm{\prime }},b^{\mathrm{\prime }}]}$ is again a piecewise $C^1$ future causal path, so step 1.1 gives $D^{\mathrm{\prime }}:=y(b^{\mathrm{\prime }})-y(a^{\mathrm{\prime }})\in \overline{C^{+}}$; writing $\mathrm{y}$ for the coordinate path of $y$ in the chart $x$, one has $({\mathrm{x}}^0\circ y)(b^{\mathrm{\prime }})-({\mathrm{x}}^0\circ y)(a^{\mathrm{\prime }})=D^{\mathrm{\prime }0}$ by [L6], and $D^{\mathrm{\prime }0}\ge \mathrm{\mid }{D}_s^{\mathrm{\prime }}\mathrm{\mid }\ge 0$ by [L3]; hence $x^0\circ y$ is nondecreasing. If $y$ is nonconstant on $[a^{\mathrm{\prime }},b^{\mathrm{\prime }}]$ then $y{\mathrm{\mid }}_{[a^{\mathrm{\prime }},b^{\mathrm{\prime }}]}$ is nonconstant, so step 1.1 gives $D^{\mathrm{\prime }}\mathrm{\ne }0$, and a nonzero element of $\overline{C^{+}}$ has positive time coordinate by [L3], that is $({\mathrm{x}}^0\circ y)(b^{\mathrm{\prime }})>({\mathrm{x}}^0\circ y)(a^{\mathrm{\prime }})$; this is (d).

For (c): if $p\le q$ and $q\le p$ then $D=q-p\in \overline{C^{+}}$ and $p-q=-D\in \overline{C^{+}}$, so $D\in \overline{C^{-}}=-\overline{C^{+}}$ and $D\in \overline{C^{+}}\cap \overline{C^{-}}=\{0\}$ by [L3], that is $p=q$. Conversely, if antisymmetry fails, that is $p\mathrm{\ne }q$ with $p\le q$ and $q\le p$, then step 2.1 makes the straight paths from $p$ to $q$ and from $q$ to $p$ future causal, and their concatenation $z:[0,2]\to A$, $z(s)=p+s(q-p)$ for $s\in [0,1]$ and $z(s)=q+(s-1)(p-q)$ for $s\in [1,2]$, is a closed piecewise $C^1$ future causal path with $z(0)=z(2)=p$ and $z(1)=q\mathrm{\ne }p$; conversely, if a nonconstant closed piecewise $C^1$ future causal path $x:[a,b]\to A$ with $x(b)=x(a)=p$ exists, choose $s\in (a,b)$ with $x(s)\mathrm{\ne }p$, which exists because $x$ is nonconstant on $[a,b]$ and $x(a)=x(b)$; applying step 1.1 to the restrictions to $[a,s]$ and $[s,b]$ gives $x(s)-p\in \overline{C^{+}}$ and $p-x(s)\in \overline{C^{+}}$, that is $p\le x(s)$ and $x(s)\le p$ with $x(s)\mathrm{\ne }p$, so antisymmetry fails. Hence the antisymmetry of $\le$ is equivalent to the absence of a nonconstant closed piecewise $C^1$ future causal path; this is (c).

For (e), normalization: since $q\in I^{+}(p)$, step 2.2 says the straight path from $p$ to $q$ is future timelike; in particular $D=q-p$ is a future timelike vector, so $\eta (D,D)<0$ and $T=\sqrt{-\eta (D,D)}>0$, and $D\in C^{+}$ by [L4]. Choose the inertial frame with origin event $p$, so that $x(p)=0$ and the coordinates of $D$ in this frame are those of $x(q)$. The vector $v:=D\mathrm{/}T$ satisfies $\eta (v,v)=\eta (D,D)\mathrm{/}{T}^2=-1$ and $v\in C^{+}$ (a positive multiple of a future timelike vector), so $v$ is future unit timelike; write $v=(v^0,v_{\mathrm{v}\mathrm{e}\mathrm{c}})$ in coordinates, so that $v^0>0$, $\mathrm{\mid }{v}_{\mathrm{v}\mathrm{e}\mathrm{c}}\mathrm{\mid }<v^0$ and $(v^0{)}^2-\mathrm{\mid }{v}_{\mathrm{v}\mathrm{e}\mathrm{c}}{\mathrm{\mid }}^2=1$ by [L3] and [L5]. Put $\beta :=v_{\mathrm{v}\mathrm{e}\mathrm{c}}\mathrm{/}{v}^0$; then $\mathrm{\mid }\beta \mathrm{\mid }<1$ and $B:=B(\beta )$ is proper orthochronous by [L7], so $Bv$ is future unit timelike, and its time coordinate is $\gamma (v^0-\beta \cdot v_{\mathrm{v}\mathrm{e}\mathrm{c}})=\gamma ((v^0{)}^2-\mathrm{\mid }{v}_{\mathrm{v}\mathrm{e}\mathrm{c}}{\mathrm{\mid }}^2)\mathrm{/}{v}^0=\gamma \mathrm{/}{v}^0$ with $\gamma =(1-\mathrm{\mid }\beta {\mathrm{\mid }}^2{)}^{-1\mathrm{/}2}=v^0$; hence $(Bv{)}^0=1$, and since $\eta (Bv,Bv)=-1$ the spatial part of $Bv$ satisfies $\mathrm{\mid }(Bv{)}_{\mathrm{v}\mathrm{e}\mathrm{c}}{\mathrm{\mid }}^2=((Bv{)}^0{)}^2-1=0$ by [L5], so $Bv=e_0$ and $BD=T{e}_0$. The ordered basis $e_a^{\mathrm{\prime }}:=B^{-1}{e}_a$ is $\eta$-orthonormal, future and positively oriented because $B^{-1}$ is a proper orthochronous Lorentz map and such maps preserve $\eta$, the cone $C^{+}$, the orientation and the volume form by [L7] and [L8], so $(p;e_0^{\mathrm{\prime }},e_1^{\mathrm{\prime }},e_2^{\mathrm{\prime }},e_3^{\mathrm{\prime }})$ is an inertial frame; its chart $x_1$ satisfies $x_1=M_B\circ x$ (same origin $p$), where $M_B:=E^{-1}\circ B\circ E$, because with $E_1(u)={\sum }_a{u}^a{e}_a^{\mathrm{\prime }}=B^{-1}(E(u))$ one has $E_1^{-1}=E^{-1}B$ and $x_1(r)=E_1^{-1}(r-p)=E^{-1}(B(r-p))=E^{-1}(B(E(x(r))))=M_{B}x(r)$ for the matrix $M_B$ of $B$ by [L8] and [L6]; therefore $x_1(q)-x_1(p)=E^{-1}(BD)=(T,0,0,0)$.

For (f): let $F$ be a proper orthochronous affine Lorentz map, so $F(r)=F(p_0)+L(r-p_0)$ with $L$ a linear Lorentz map, and $F(r+v)=F(r)+Lv$ for all $r\in A$ and $v\in V$ by [L8]. For a piecewise $C^1$ curve $z:[a,b]\to A$ the composition $F\circ z$ is piecewise $C^1$ on the same partition with $(F\circ z{)}^{\mathrm{\prime }}=L\circ z^{\mathrm{\prime }}$: the difference quotients satisfy $(F(z(s+h))-F(z(s)))\mathrm{/}h=L((z(s+h)-z(s))\mathrm{/}h)$, and $L$ is continuous and linear, so the limit is $L(z^{\mathrm{\prime }}(s))$ by [L6] and [L8]. Since $L$ and $L^{-1}$ preserve the causal character of every nonzero vector and the cone $C^{+}$ by [L8], the composition $F\circ z$ is a future causal path, respectively a future timelike path, exactly when $z$ is; with the definition of $R^{+}$ in [L4] this gives $F(R^{+}(p))=R^{+}(F(p))$, and with $J^{+}=R^{+}$ from step 2.1 it gives $F(J^{+}(p))=J^{+}(F(p))$. For the chronological future, $q\in I^{+}(p)$ iff the straight path from $p$ to $q$ is future timelike by step 2.2; its image under $F$ is the straight path from $F(p)$ to $F(q)$ with constant tangent $L(q-p)$ by [L6], and $L(q-p)$ is future timelike exactly when $q-p$ is by [L8], so $F(I^{+}(p))=I^{+}(F(p))$ (the reverse inclusion follows by applying the same argument to $F^{-1}$, which is again a proper orthochronous affine Lorentz map). Finally $F(q)-F(p)=L(q-p)$, so $q-p\in \overline{C^{+}}$ iff $L(q-p)\in \overline{C^{+}}$ by the cone preservation, that is $p\le q$ iff $F(p)\le F(q)$; this is (f).

For the inequality of (e): let $y:[a,b]\to A$ be a piecewise $C^1$ future timelike worldline with $y(a)=p$ and $y(b)=q$, with weak pieces $a=a_1<b_1=\cdots <b_m=b$, and let $y_1:=x_1\circ y$ be its coordinate path in the chart $x_1$ of step 3.2. By [L6] and the affinity of $x_1$ from step 3.2, $y_1$ is piecewise $C^1$ on the same partition with coordinate tangents $y_1^{\mathrm{\prime }}=E^{-1}(B{y}^{\mathrm{\prime }})$; the vectors $B{y}^{\mathrm{\prime }}\in V$ are future timelike because $y^{\mathrm{\prime }}$ is and $B$ preserves $\eta$ and the cone by [L8]; writing $y_1=(y_1^0,z_1)$ with spatial part $z_1$, [L3] gives $y_1^{0^{\mathrm{\prime }}}\ge \mathrm{\mid }{z}_1^{\mathrm{\prime }}\mathrm{\mid }\ge 0$ and $y_1^{0^{\mathrm{\prime }}}>0$ on each open piece. For such numbers $0\le \mathrm{\mid }{z}_1^{\mathrm{\prime }}\mathrm{\mid }\le y_1^{0^{\mathrm{\prime }}}$ one has $\sqrt{(y_1^{0^{\mathrm{\prime }}}{)}^2-\mathrm{\mid }{z}_1^{\mathrm{\prime }}{\mathrm{\mid }}^2}\le y_1^{0^{\mathrm{\prime }}}$ with equality exactly when $z_1^{\mathrm{\prime }}=0$. Applying [L1] to the coordinate path $y_1$ on each piece $(a_i,b_i)$ gives finite nonnegative improper integrals with ${\int }_{(a_i,b_i)}\sqrt{(y_1^{0^{\mathrm{\prime }}}{)}^2-\mathrm{\mid }{z}_1^{\mathrm{\prime }}{\mathrm{\mid }}^2}\le {\int }_{(a_i,b_i)}{y}_1^{0^{\mathrm{\prime }}}=y_1^0(b_i)-y_1^0(a_i)$, and summing over the finitely many pieces [L2] and telescoping across the continuous junctions give $c\mathrm{\Delta }\tau (y)={\sum }_i{\int }_{(a_i,b_i)}\sqrt{(y_1^{0^{\mathrm{\prime }}}{)}^2-\mathrm{\mid }{z}_1^{\mathrm{\prime }}{\mathrm{\mid }}^2}\le y_1^0(b)-y_1^0(a)=x_1^0(q)-x_1^0(p)=T$, because $x_1(q)-x_1(p)=(T,0,0,0)$ by step 3.2; hence $\mathrm{\Delta }\tau (y)\le T\mathrm{/}c$.

For the equality case of (e): $\mathrm{\Delta }\tau (y)=T\mathrm{/}c$ holds exactly when equality holds in every piece of step 4.1, that is when ${\int }_{(a_i,b_i)}{h}_i=0$ for the nonnegative continuous function $h_i:=y_1^{0^{\mathrm{\prime }}}-\sqrt{(y_1^{0^{\mathrm{\prime }}}{)}^2-\mathrm{\mid }{z}_1^{\mathrm{\prime }}{\mathrm{\mid }}^2}$ on each open piece. If $h_i(s_0)>0$ at some point of an open piece, continuity gives a compact subinterval $[u,v]$ of that piece on which $h_i\ge h_i(s_0)\mathrm{/}2>0$, so the improper integral dominates the proper integral over $[u,v]$ and is at least $h_i(s_0)(v-u)\mathrm{/}2>0$ by [L9], a contradiction; hence $h_i\equiv 0$ and $z_1^{\mathrm{\prime }}=0$ on each open piece. By [L10] each spatial coordinate of $y_1$ is constant on each open piece, and by continuity of $y_1$ the spatial coordinates are constant on all of $[a,b]$; since $x_1(p)=0$ and $x_1(q)=(T,0,0,0)$, the constant is $0$, so $y_1(s)=(\sigma (s),0,0,0)$ for the continuous function $\sigma :=y_1^0$ on $[a,b]$, which is $C^1$ on each open piece with ${\sigma }^{\mathrm{\prime }}=y_1^{0^{\mathrm{\prime }}}>0$; for $s<t$ in $[a,b]$ the increment $\sigma (t)-\sigma (s)$ is the sum of the increments $\sigma (\min (t,s_k))-\sigma (\max (s,s_{k-1}))$ over the finitely many pieces meeting $[s,t]$, each summand is nonnegative by [L11] applied on the closed piece (both arguments lie in that closed piece, where $\sigma$ is continuous, and ${\sigma }^{\mathrm{\prime }}>0$ on its interior), and at least one summand is strictly positive because its two arguments differ as $s<t$; hence $\sigma$ is strictly increasing, with $\sigma (a)=0$ and $\sigma (b)=T$; by [L12], for every $r\in [0,T]$ some $s\in [a,b]$ satisfies $\sigma (s)=r$, so $\sigma$ maps onto $[0,T]$. Therefore $y(s)=x_1^{-1}(\sigma (s),0,0,0)=p+\frac{\sigma (s)}{T}D$ by [L6], that is $y$ is the strictly increasing reparametrization $s\mapsto p+\phi (s)D$ of the straight segment from $p$ to $q$, where $\phi :=\sigma \mathrm{/}T$ is a continuous strictly increasing surjection $[a,b]\to [0,1]$, $C^1$ on the open pieces with ${\phi }^{\mathrm{\prime }}>0$. Conversely, if $y(s)=p+\phi (s)D$ with such a $\phi$, then $y^{\mathrm{\prime }}={\phi }^{\mathrm{\prime }}D$ on each open piece, so $\eta (y^{\mathrm{\prime }},y^{\mathrm{\prime }})={\phi }^{\mathrm{\prime }2}\eta (D,D)=-({\phi }^{\mathrm{\prime }}T{)}^2$ and the density is ${\phi }^{\mathrm{\prime }}T$; hence $c\mathrm{\Delta }\tau (y)=T{\sum }_i{\int }_{(a_i,b_i)}{\phi }^{\mathrm{\prime }}=T{\sum }_i(\phi (b_i)-\phi (a_i))=T(\phi (b)-\phi (a))=T$ by [L2] and [L9], giving $\mathrm{\Delta }\tau (y)=T\mathrm{/}c$ exactly for the strictly increasing reparametrizations of the straight segment: this is the equality case of (e).