Worked instance of the operational vocabulary: an illustrative bounded record for the longitudinal Doppler ratio and its parameter window

Fix $\omega >0$ and let $k$ be the covector whose frame-$x$ components are $(-\omega \mathrm{/}c,\omega \mathrm{/}c,0,0)$, so $\varpi =\omega \mathrm{/}c>0$ and $s=e_1$ with $\mathrm{\mid }s\mathrm{\mid }=1$; by [F4] the covector $k$ is a future null wave covector with frame-$x$ data $(\omega ,e_1)$; equip it with the affine phase $\theta (p):=k(p-o)$, whose differential is $k$. For each real $\beta \in (-1,1)$ put $x^{\mathrm{\prime }}:=B(\beta e_1)\circ x$ and $f_a:=B(\beta e_1{)}^{-1}{e}_a$; by [F5] the map $B(\beta e_1)$ is proper orthochronous with inverse $B(-\beta e_1)$, and [F5] applied to $-\beta e_1$ gives that $B(-\beta e_1)$ also preserves $\eta$, the future cone and the orientation, so $(f_0,f_1,f_2,f_3)=(B(-\beta e_1)e_0,\dots ,B(-\beta e_1)e_3)$ is again $\eta$-orthonormal, future and positively oriented and $(o;f_0,f_1,f_2,f_3)$ is an inertial frame by [F6], with chart $x^{\mathrm{\prime }}$ because $E_f=E\circ B(\beta e_1{)}^{-1}$ gives $E_f^{-1}(p-o)=x^{\mathrm{\prime }}(p)$. The transition is $x^{\mathrm{\prime }}=B(\beta e_1)x$ with parameter $\beta e_1$ of modulus $\mathrm{\mid }\beta \mathrm{\mid }<1$, so [F2] clause (i) with $s=e_1$ gives the frame-$x^{\mathrm{\prime }}$ data $({\omega }^{\mathrm{\prime }},s^{\mathrm{\prime }})$ of the same covector $k$ with ${\omega }^{\mathrm{\prime }}={\gamma }_{\beta }(1-\beta )\omega$ and ${\gamma }_{\beta }=(1-{\beta }^2{)}^{-1\mathrm{/}2}$, this being the reading of the rest observer of $x^{\mathrm{\prime }}$ by clause (ii); and ${\gamma }_{\beta }(1-\beta )=(1-\beta {)}^{1\mathrm{/}2}(1+\beta {)}^{-1\mathrm{/}2}=((1-\beta )\mathrm{/}(1+\beta ){)}^{1\mathrm{/}2}$ by [F8], the positive square root. Hence the dimensionless ratio $F(\beta ):={\omega }^{\mathrm{\prime }}\mathrm{/}\omega$ equals $((1-\beta )\mathrm{/}(1+\beta ){)}^{1\mathrm{/}2}$ for every $\beta \in (-1,1)$.

The map $F$ is strictly decreasing on $(-1,1)$: for $-1<{\beta }_a<{\beta }_b<1$ the difference of the two radicands is $\frac{1-{\beta }_a}{1+{\beta }_a}-\frac{1-{\beta }_b}{1+{\beta }_b}=\frac{2({\beta }_b-{\beta }_a)}{(1+{\beta }_a)(1+{\beta }_b)}>0$, since the numerator and both denominator factors are positive by [F9], and applying the strictly increasing map $u\mapsto u^{1\mathrm{/}2}$ of [F9] to the two positive radicands gives $F({\beta }_a)>F({\beta }_b)$.

For every real $r>0$ the equation $F(\beta )=r$ is equivalent to $\frac{1-\beta }{1+\beta }=r^2$, because $u\mapsto u^{1\mathrm{/}2}$ is injective on positives and $(u^{1\mathrm{/}2}{)}^2=u$ by [F8]; solving, $\beta =\frac{1-r^2}{1+r^2}=:\beta (r)$, which lies in $(-1,1)$ because $1+\beta (r)=\frac{2}{1+r^2}>0$ and $1-\beta (r)=\frac{2r^2}{1+r^2}>0$; and $F(\beta (r))=r$ again by [F8]. Thus $F$ is a strictly decreasing bijection of $(-1,1)$ onto $(0,\mathrm{\infty })$ with inverse $\beta (\cdot )$, and $\beta (\cdot )$ is itself strictly decreasing on $(0,\mathrm{\infty })$.

Let $S:=\mathrm{\Theta }:=(-1,1)$, $\theta :=\mathrm{i}\mathrm{d}$, and let $D(\beta )$ comprise the frames $x$ and $x^{\mathrm{\prime }}$ with their charts, the real boost parameter $\beta$, the wave covector $k$ with its affine phase, the worldline $h:\mathbb{R}\to A$, $h(\rho )=o+\rho e_0$, with proper time ${\tau }_h(\rho )=\rho \mathrm{/}c$, and the observer vector $f_0$ with reception event $o$; the worldline is $C^{\mathrm{\infty }}$, its derivative is the future unit vector $e_0$, and the proper-time integral is $c^{-1}{\int }_0^{\rho }1du=\rho \mathrm{/}c$; then $M=(S,\mathrm{\Theta },\theta ,D,F)$ with the map $F$ of step 1.1 is an operational model in the sense of [F1] with $d=1$, because each component of $F(\beta )$ is assigned to $D(\beta )$ by [F2] clauses (i) and (ii) under the adopted idealization [A1], with the frame, sign and unit conventions declared, so condition $(\ast )$ holds. This is the model claimed in part (a) of the Example.

Since $F$ is strictly decreasing with inverse $\beta (\cdot )$ by steps 1.2 and 1.3, $F(\beta )\in [z-\epsilon ,z+\epsilon ]$ holds if and only if $\beta \in [\beta (z+\epsilon ),\beta (z-\epsilon )]$; the set is nonempty because $z-\epsilon >0$ puts $z\pm \epsilon$ in the range $(0,\mathrm{\infty })$ of $F$, and every element of it satisfies the declared bound of the record $R$ because its prediction lies in the declared closed interval, so $C(R,M)=[\beta (z+\epsilon ),\beta (z-\epsilon )]$ as claimed in part (b). For $\epsilon =0$ the two endpoints run together and $C(R,M)=\{\beta (z)\}$ by step 1.3, the singleton claimed; for $\epsilon >0$ we have $0<z-\epsilon <z+\epsilon$ and strict decrease of $\beta (\cdot )$ in step 1.3 gives the strict endpoint order $\beta (z-\epsilon )>\beta (z+\epsilon )$, with length $\beta (z-\epsilon )-\beta (z+\epsilon )=\frac{1-(z-\epsilon {)}^2}{1+(z-\epsilon {)}^2}-\frac{1-(z+\epsilon {)}^2}{1+(z+\epsilon {)}^2}=\frac{2(z+\epsilon {)}^2-2(z-\epsilon {)}^2}{[1+(z-\epsilon {)}^2][1+(z+\epsilon {)}^2]}=\frac{8z\epsilon }{[1+(z-\epsilon {)}^2][1+(z+\epsilon {)}^2]}>0$, where the numerator identity is algebra and the final sign uses $z>0$, $\epsilon >0$ and the positivity of the two denominator factors by [F9]. For $r>0$, the positive denominator in $\beta (r)=(1-r^2)\mathrm{/}(1+r^2)$ shows that $\beta (r)$ is positive, zero or negative according as $r<1$, $r=1$ or $r>1$; because $\beta (z+\epsilon )$ is the minimum of $C(R,M)$, the whole consistency interval is nonnegative exactly when $z+\epsilon \le 1$ and strictly positive exactly when $z+\epsilon <1$. These are precisely the sign qualifications in part (b); no sign restriction is imposed by its signed-velocity annotation.

For the illustrative values $z=9\mathrm{/}10$ and $\epsilon =1\mathrm{/}100$ we have $z+\epsilon =91\mathrm{/}100$ and $z-\epsilon =89\mathrm{/}100$, so by step 1.3 $\beta (91\mathrm{/}100)=\frac{1-8281\mathrm{/}10000}{1+8281\mathrm{/}10000}=\frac{1719}{18281}$ and $\beta (89\mathrm{/}100)=\frac{1-7921\mathrm{/}10000}{1+7921\mathrm{/}10000}=\frac{2079}{17921}$, both in $(-1,1)$, and $F(\beta (91\mathrm{/}100))=91\mathrm{/}100$, $F(\beta (89\mathrm{/}100))=89\mathrm{/}100$; the consistency set is $[1719\mathrm{/}18281,2079\mathrm{/}17921]$ by step 2.2, and its length is $\frac{8\cdot (9\mathrm{/}10)\cdot (1\mathrm{/}100)}{[1+7921\mathrm{/}10000][1+8281\mathrm{/}10000]}=\frac{7200000}{327613801}$ by step 2.2, using $17921\cdot 18281=327613801$. This is exact rational arithmetic on declared illustrative values; no measurement, probability model, sampling assumption or statistical inference is used.