Worked four-dimensional causal diamond and the boost-invariant area of its planar section

Let $r=x^{-1}(u)$ be an event with coordinates $u$. By [L3] the displacement $r-p$ has coordinates $u$, so [L1] gives $r\in J^{+}(p)$ iff $u^0\ge \mathrm{\mid }{u}_s\mathrm{\mid }$; likewise the displacement $q-r$ has coordinates $(2-u^0,-u^1,-u^2,-u^3)$ of spatial norm $\mathrm{\mid }{u}_s\mathrm{\mid }$, so $r\in J^{-}(q)$ iff $q\in J^{+}(r)$ iff $2-u^0\ge \mathrm{\mid }{u}_s\mathrm{\mid }$. Hence the diamond is $\{x^{-1}(u):\mathrm{\mid }{u}_s\mathrm{\mid }\le u^0\le 2-\mathrm{\mid }{u}_s\mathrm{\mid }\}$, and the two strict versions give $I^{+}(p)\cap I^{-}(q)=\{x^{-1}(u):\mathrm{\mid }{u}_s\mathrm{\mid }<u^0<2-\mathrm{\mid }{u}_s\mathrm{\mid }\}$ by the strict clauses of [L1]. For the interior: the triangle inequality gives $\mid \mathrm{\mid }{u}_s\mathrm{\mid }-\mathrm{\mid }{v}_s\mathrm{\mid }\mid \le \mathrm{\mid }{u}_s-v_s\mathrm{\mid }\le \mathrm{\parallel }u-v{\mathrm{\parallel }}_2$, so each of the functions $u\mapsto u^0-\mathrm{\mid }{u}_s\mathrm{\mid }$ and $u\mapsto 2-u^0-\mathrm{\mid }{u}_s\mathrm{\mid }$ changes by at most $2\mathrm{\parallel }u-v{\mathrm{\parallel }}_2$. At a point where both are positive, the coordinate ball of radius half their smaller value keeps both positive, i.e. lies in the interior of the diamond; if $u^0=\mathrm{\mid }{u}_s\mathrm{\mid }$ or $u^0=2-\mathrm{\mid }{u}_s\mathrm{\mid }$, every neighbourhood contains points with $u^0$ decreased or increased slightly, which violate $u^0\ge \mathrm{\mid }{u}_s\mathrm{\mid }$ or $u^0\le 2-\mathrm{\mid }{u}_s\mathrm{\mid }$, so such points are not interior; hence the interior is exactly the displayed open set, which is $I^{+}(p)\cap I^{-}(q)$; this is (a).

In the section $u^2=u^3=0$ the description of step 1.1 reads $\mathrm{\mid }{u}^1\mathrm{\mid }\le u^0\le 2-\mathrm{\mid }{u}^1\mathrm{\mid }$, whose boundary consists of the four segments joining $(0,0)$ to $(1,1)$, $(1,1)$ to $(2,0)$, $(2,0)$ to $(1,-1)$ and $(1,-1)$ to $(0,0)$: the square with those four vertices (successive edge columns have squared Euclidean length $2$ and scalar product $0$). Its diagonals are the displacement $(2,0,0,0)$ from $p$ to $q$, of interval magnitude $\sqrt{-\eta }=\sqrt{4}=2$ m, and the displacement $(0,2,0,0)$ from $x^{-1}(1,-1,0,0)$ to $x^{-1}(1,1,0,0)$, of interval magnitude $\sqrt{\eta }=\sqrt{4}=2$ m, both by [L2]; relative to its centre $(1,0)$, the four triangles have edge columns chosen from $(\pm 1,0)$ and $(0,\pm 1)$, each of area $1\mathrm{/}2$ by the stated coordinate-area definition, so their total area is $2$ m${}^2$, also $\frac{1}{2}\cdot 2\cdot 2$; this is (b).

By [L4] the chart $x^{\mathrm{\prime }}$ of the boosted frame satisfies $x^{\mathrm{\prime }}=B(b)\circ x$, so the $x^{\mathrm{\prime }}$-coordinates of the events of the $1+1$ section are the images under the linear map $B(b)$ of their $x$-coordinates, and the coordinate image of the square of step 2.1 is the parallelogram spanned by the images of the two diagonals: $B(b)(2,0)=2\gamma (1,-b)$ and $B(b)(0,2)=2\gamma (-b,1)$. Its centre is $B(b)(1,0)=\gamma (1,-b)$, its half-diagonal vectors are $a:=\gamma (1,-b)$ and $c:=\gamma (-b,1)$, and it is the union of the four triangles $(0,\pm a,\pm c)$ in centred coordinates, each of area $\frac{1}{2}\mathrm{\mid }\det (a,c)\mathrm{\mid }$; hence its area is $2\mathrm{\mid }\det (a,c)\mathrm{\mid }=2{\gamma }^2(1\cdot 1-(-b)(-b))=2{\gamma }^2(1-b^2)=2$ m${}^2$, since ${\gamma }^2(1-b^2)=1$; this holds for every $\mathrm{\mid }b\mathrm{\mid }<1$ and is (c).

Clause (a) is step 1.1, clause (b) is step 2.1, clause (c) is step 3.1, and clause (d) follows from [L5] applied to the straight segment from $p$ to $q$, whose displacement has $\eta =-4$ m${}^2$ and interval magnitude $2$ m, giving proper time $2\mathrm{/}c$ s; nothing in the computation identifies the coordinate area $2$ m${}^2$ with a physical time or clock reading, and no measurement is reported.