Experimental Interface — Examples

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.

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 the ordered basis $(f_0,f_1,f_2,f_3)=(B(-\beta e_1)e_0,\dots ,B(-\beta e_1)e_3)$ is again $\eta$-orthonormal, future and positively oriented and $(o;f_0,f_1,f_2,f_3)$ is an inertial frame by [F6]; its chart is $x^{\mathrm{\prime }}$, because the frame isomorphism $E_f=E\circ B(\beta e_1{)}^{-1}$ gives $E_f^{-1}(p-o)=B(\beta e_1)(E^{-1}(p-o))=x^{\mathrm{\prime }}(p)$, and $f_0$ is future unit timelike.

The transition between the inertial charts $x$ and $x^{\mathrm{\prime }}$ is $x^{\mathrm{\prime }}=B(\beta e_1)x$ with parameter $\beta e_1$ of modulus $\mathrm{\mid }\beta \mathrm{\mid }<1$; [F2] clause (i) with $s=e_1$ therefore gives the frame-$x^{\mathrm{\prime }}$ data $({\omega }^{\mathrm{\prime }},s^{\mathrm{\prime }})$ of the same covector $k$ with ${\omega }^{\mathrm{\prime }}={\gamma }_{\beta }(1-\beta )\omega$, where ${\gamma }_{\beta }=(1-{\beta }^2{)}^{-1\mathrm{/}2}$; 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], so the dimensionless ratio $F(\beta ):={\omega }^{\mathrm{\prime }}\mathrm{/}\omega$ is the positive real number displayed.

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 that positive difference 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 $\beta (\cdot )$.

Let $S:=\mathrm{\Theta }:=(-1,1)$, $\theta :=\mathrm{i}\mathrm{d}$, and let $D(\beta )$ be the finite run description comprising the frames $x$ and $x^{\mathrm{\prime }}$ with their charts, the real boost parameter $\beta$, the wave covector $k$ with its affine phase, and the observer vector $f_0$ with reception event $o$; let $F$ be the map of step 1.2. Then $M=(S,\mathrm{\Theta },\theta ,D,F)$ is an operational model in the sense of [F1] with $d=1$: each component of $F(\beta )$ is the rest-observer frequency ratio $F(\beta )=-ck(f_0)\mathrm{/}\omega$ in the sense of [F3], assigned to $D(\beta )$ by clauses (i) and (ii) of [F2] under the adopted idealization [A1], with the frame, sign and unit conventions declared, so condition $(\ast )$ holds.

Declare the comparison rule that a parameter $\beta$ is consistent with the one-point record $R$ exactly when its prediction $F(\beta )$ lies in the record's declared closed interval $[z_1-{\epsilon }_1,z_1+{\epsilon }_1]$; by [F1] clause (f) such a rule must be declared separately, and this is the declared rule of the present algebraic reading. By strict decrease and the inverse of steps 1.3 and 1.4, the consistency set is $C(R,M)=\{\beta \in (-1,1):F(\beta )\in [z_1-{\epsilon }_1,z_1+{\epsilon }_1]\}=[\beta (z_1+{\epsilon }_1),\beta (z_1-{\epsilon }_1)]$, a nonempty closed interval.

With $z_1=9\mathrm{/}10$ and ${\epsilon }_1=1\mathrm{/}100$ the declared interval is $[89\mathrm{/}100,91\mathrm{/}100]$, and the two distinct parameters ${\beta }_1:=\beta (91\mathrm{/}100)=1719\mathrm{/}18281$ and ${\beta }_2:=\beta (89\mathrm{/}100)=2079\mathrm{/}17921$ satisfy $F({\beta }_1)=91\mathrm{/}100$ and $F({\beta }_2)=89\mathrm{/}100$ by step 1.4, hence both lie in $C(R,M)$; they are distinct because ${\beta }_2-{\beta }_1=\frac{2079\cdot 18281-1719\cdot 17921}{17921\cdot 18281}=\frac{7200000}{327613801}>0$, the denominator being positive and the numerator a positive integer difference. Since $R$ is a one-point data record of [F1] clause (c) with positive declared uncertainty, the uniqueness assertion of the Statement refuted is false: a declared bounded one-point record need not determine the parameter of an operational model. This is an algebraic consistency computation on declared illustrative values; it asserts no measurement, calibration, probability model, statistical inference or empirical conclusion about any model or postulate.