Measurement, uncertainty and later bridges — Examples

Adopt as the declared physical content of a separate hypothetical constrained model the ideal planar point pendulum: one bob of inertial mass $m>0$, a fixed pivot at the origin of an inertial frame, and an ideal massless rigid rod of length $l>0$ constraining the bob to the vertical $e_1,e_3$ plane. Write its position as $x(t)=l(\sin \theta (t)e_1-\cos \theta (t)e_3)$ for frame time $t\in \mathbb{R}$, with a $C^2$ angle $\theta$ measured from the downward vertical; $\theta =0$ is the bottom position. Adopt the uniform acceleration field $-g{e}_3$, with $g>0$ a declared acceleration parameter of dimension $\mathrm{\ell }{\tau }^{-2}$, so the applied force is $-mg{e}_3$ using that same inertial mass. No separate gravitational mass or empirical mass-equality claim is introduced. The ideal rod reaction has zero tangential component. The adopted exact equation of this constrained model is $m{l}^2{\theta }^{\mathrm{\prime }\mathrm{\prime }}+mgl\sin \theta =0$, and small-angle idealization $T=2\pi \sqrt{l\mathrm{/}g}$ for the period $T$ of a small oscillation, whose linearization at $\theta =0$ is the quadratic mode of angular frequency $\omega =\sqrt{g\mathrm{/}l}$. The model and its small-angle idealization are declared data of the example, adopted from the cited treatments; the exact finite-amplitude period discrepancy is not derived here. This constrained system is not a force-branch operational observable model: for $N=1$, the rod and plane constraints give the configuration set $Q_{\mathrm{r}\mathrm{o}\mathrm{d}}=\{(x_1,x_2,x_3)\in {\mathbb{R}}^3:x_2=0,x_1^2+x_3^2=l^2\},$ a circle with empty interior, whereas Operational observable model of the force branch: parameters, preparations, observation times, readout, units and additive error requires an open configuration domain $Q\subseteq {\mathbb{R}}^{3N}$. This example supplies only the constrained angle equation and a hypothetical data-reduction map. No measurement is performed and no measured value is used.

The proposed data-reduction map. Use numerical input coordinates $l,T$ for the pendulum length in metres and period in seconds, respectively. Algebraically solving the adopted small-angle relation $T=2\pi \sqrt{l\mathrm{/}g}$ for the parameter $g$ gives the reconstruction function $h:U⟶\mathbb{R},U:=\{(l,T)\in {\mathbb{R}}^2:l>0,T>0\},h(l,T):=\frac{4{\pi }^{2}l}{T^2},$ which is a data-reduction map, not the prediction map $\mathrm{\Phi }$ of Operational observable model of the force branch: parameters, preparations, observation times, readout, units and additive error. For a specified oscillation amplitude $0<\alpha <\pi$ released from rest, let $T_{\mathrm{e}\mathrm{x}\mathrm{a}\mathrm{c}\mathrm{t}}(l,g,\alpha )$ denote the period of that declared ideal motion when a finite period exists. The finite-amplitude discrepancy is $\delta :=\mathrm{\mid }{T}_{\mathrm{e}\mathrm{x}\mathrm{a}\mathrm{c}\mathrm{t}}(l,g,\alpha )-2\pi \sqrt{l\mathrm{/}g}\mathrm{\mid }\ge 0$, in units of $\tau$; this example neither estimates nor bounds $\delta$. Any actual apparatus calibration would also require a calibration datum as in Calibration of a mechanical apparatus against the primitive standards of the framework: reference inputs, deviations, rules and calibrated prediction maps, including its reference deviations and calibration rule; this item supplies no reference observations or realized setting. In the calculations below $l$ and $T$ are numerical coefficients in metres and seconds, $h$ is the numerical coefficient in metres per second squared, and Euclidean norms are applied to those fixed numerical coordinates, not to sums of physical quantities of different dimensions; the constants depend on this coordinate convention.

Declared data. Fix a nominal input $\mu =(l_0,T_0)\in U$ with $l_0,T_0>0$ and radii $0<{\rho }_l<l_0$, $0<{\rho }_T<T_0$, and put $U_B:=(l_0-{\rho }_l,l_0+{\rho }_l)\times (T_0-{\rho }_T,T_0+{\rho }_T)\subseteq U,$ an open convex box. Let $X=(l,T):\mathrm{\Omega }\to U_B$ be a square-integrable random vector with $E[X]=\mu ,\mathrm{Cov}(X)=\left(\begin{array}{cc}{\sigma }_l^2& {\sigma }_{lT}\\ {\sigma }_{lT}& {\sigma }_T^2\end{array}\right)\mathrm{.}$ All entries are declared model data: no reading of an apparatus is asserted. With $h$ as above:

1. $h$ is $C^{\mathrm{\infty }}$ on $U$, with $\frac{\mathrm{\partial }h}{\mathrm{\partial }l}(l,T)=\frac{4{\pi }^2}{T^2}=\frac{h(l,T)}{l},\frac{\mathrm{\partial }h}{\mathrm{\partial }T}(l,T)=-\frac{8{\pi }^{2}l}{T^3}=-\frac{2h(l,T)}{T},$ so $Dh(\mu )$ has the two components $h(\mu )\mathrm{/}{l}_0$ and $-2h(\mu )\mathrm{/}{T}_0$;
2. the Hessian on $U$ is $H_h(l,T)=\left(\begin{array}{cc}0& -{\displaystyle \frac{8{\pi }^2}{T^3}}\\ -{\displaystyle \frac{8{\pi }^2}{T^3}}& {\displaystyle \frac{24{\pi }^{2}l}{T^4}}\end{array}\right),$ and on $U_B$ its operator norm is at most the Frobenius bound $B:=\left(2\right(\frac{8{\pi }^2}{(T_0-{\rho }_T{)}^3}{)}^2+(\frac{24{\pi }^2(l_0+{\rho }_l)}{(T_0-{\rho }_T{)}^4}{)}^2{)}^{1\mathrm{/}2};$
3. the covariance of the linearization $T_{\mu }(x)=h(\mu )+Dh(\mu )\cdot (x-\mu )$ is the $1\times 1$ matrix $\mathrm{Cov}(T_{\mu }(X))=\frac{16{\pi }^4}{T_0^4}{\sigma }_l^2-\frac{64{\pi }^4{l}_0}{T_0^5}{\sigma }_{lT}+\frac{64{\pi }^4{l}_0^2}{T_0^6}{\sigma }_T^2,$ which for uncorrelated inputs (${\sigma }_{lT}=0$) reads $\mathrm{Cov}(T_{\mu }(X))=\frac{16{\pi }^4}{T_0^4}({\sigma }_l^2+\frac{4l_0^2}{T_0^2}{\sigma }_T^2)=h(\mu {)}^2(\frac{{\sigma }_l^2}{l_0^2}+\frac{4{\sigma }_T^2}{T_0^2});$
4. for the nonlinearity remainder $R:=h(X)-h(\mu )-Dh(\mu )\cdot (X-\mu )$ one has the pointwise bound $\mid R\mid \le \frac{B}{2}\mid X-\mu {\mid }^2$, hence $\mid E[R]\mid \le E\mid R\mid \le \frac{B}{2}E\mid X-\mu {\mid }^2$, and if $E\mid X-\mu {\mid }^4<+\mathrm{\infty }$ then $E[R^2]\le \frac{B^2}{4}E\mid X-\mu {\mid }^4$.

Since $E[X]=\mu$ gives $E[Dh(\mu )(X-\mu )]=0$, the mean of the reading $\widehat{g}:=h(X)$ satisfies $\mid E[\widehat{g}]-h(\mu )\mid \le \frac{B}{2}E\mid X-\mu {\mid }^2$, while the exact variance of the linearization is the matrix of clause 3. The actual variance has the explicit correction of Mean, mean-square and covariance bounds for second-order linearization error: with $s^2=\mathrm{Var}(T_{\mu }(X))$ and $M_4=E\mathrm{\mid }X-\mu {\mathrm{\mid }}^4$, $\mathrm{\mid }\mathrm{Var}(\widehat{g})-s^2\mathrm{\mid }\le B\sqrt{s^2{M}_4}+\frac{B^2}{4}{M}_4\mathrm{.}$ The box ensures $M_4\le ({\rho }_l^2+{\rho }_T^2{)}^2<\mathrm{\infty }$; the model discrepancy $\delta$ and the reference deviations of the calibration contribute further unquantified terms. The protocol is a proposal: no value of $l$, $T$, $g$, ${\sigma }_l$, ${\sigma }_T$, ${\sigma }_{lT}$, $\delta$ or $B$ is a measurement here.

The declared model. In the force branch Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance take $N=1$ label of mass $m>0$, time window $I_0=\mathbb{R}$, configuration domain $Q={\mathbb{R}}^3$ and the zero force map $F(t,x,v):=0$. It is continuous and state-Lipschitz with constant $0$, so it satisfies the standing hypotheses of the branch. With $e_1$ the first standard basis vector, let $\mathrm{\Theta }=Z=\mathbb{R}$, with both labels read in velocity units $\mathrm{\ell }{\tau }^{-1}$ (Dimensions and SI unit conventions of the non-relativistic classical mechanics framework). Use the constant family $F_{\theta }=0$ for every $\theta \in \mathrm{\Theta }$ and the preparation map $\zeta :Z\to \mathbb{R}\times {\mathbb{R}}^3\times {\mathbb{R}}^3$, $\zeta (z):=(0,0,z{e}_1)$ at time $t_0=0$. Then $x_{\theta ,z}(t)=(zt,0,0)$ is the global motion for every pair. Fix an observation time $t_1>0$ and the position readout $R(\theta ,z,(x,w)):=x\cdot e_1$, so the full prediction map is $\mathrm{\Phi }(\theta ,z)=z{t}_1$. For $v\in \mathbb{R}$ choose the diagonal pair $(\theta ,z)=(v,v)$ and abbreviate $x_v:=x_{v,v}$ and $\mathrm{\Phi }(v):=\mathrm{\Phi }(v,v)=x_v(t_1)\cdot e_1$. This explicit choice permits a one-label formula without identifying the parameter argument with the preparation argument of Operational observable model of the force branch: parameters, preparations, observation times, readout, units and additive error. The readout has dimension $\mathrm{\ell }$ and is read in metres; the force family does not depend on its parameter label.

The motion. The curve $x_v(t):=(vt,0,0)$, $t\in \mathbb{R}$, is a motion of the declared model with datum $\zeta (v)$: its coordinate velocity is $w_v(t)=(v,0,0)$ (Coordinate velocity and acceleration of a worldline in an inertial frame, and the spatial velocity and acceleration vectors), its coordinate acceleration vanishes, and $m{\ddot{x}}_v=0=F$ identically (The associated first-order system, motions, maximal motions and the Newtonian flow of a force model). Hence the prediction map is $\mathrm{\Phi }(v)=v{t}_1$, of dimension $\mathrm{\ell }$ with $[t_1]=\tau$.

The repeated protocol. Fix $n\ge 1$ and adopt the IID sampling hypothesis and the second-moment hypothesis of Repeated-observation protocols of the force branch: adopted IID and second-moment sampling hypotheses and the sample mean for the errors ${\epsilon }_1,\dots ,{\epsilon }_n$ of $n$ repetitions of the protocol at the fixed pair $(\theta ,z)=(v,v)$, on a common probability space. Suppose additionally that $E[{\epsilon }_1]=0$ and $\mathrm{Var}({\epsilon }_1)={\sigma }^2\in [0,+\mathrm{\infty })$; these two statements are extra model hypotheses, not part of either definition. Then the observations are $Y_k=v{t}_1+{\epsilon }_k$ and $\mu :=E[Y_1]=v{t}_1,{\sigma }^2=\mathrm{Var}(Y_1),\bar{Y}=\frac{1}{n}{\sum }_{k=1}^n{Y}_k\mathrm{.}$ Consequently, by Finite IID sample-mean variance and Chebyshev bound, $E\mid \bar{Y}-v{t}_1{\mid }^2=\frac{{\sigma }^2}{n},\mathbb{P}(\mid \bar{Y}-v{t}_1\mid \ge a)\le \frac{{\sigma }^2}{n{a}^2}(a>0)\mathrm{.}$

A declared illustration. With the declared model data $v:=2\mathrm{m}{\mathrm{s}}^{-1}$, $t_1:=3\mathrm{s}$, ${\sigma }^2:=4{\mathrm{m}}^2$ and $n:=16$, the nominal prediction is $v{t}_1=6\mathrm{m}$ and the bound reads $\mathbb{P}(\mid \bar{Y}-6\mathrm{m}\mid \ge 1\mathrm{m})\le \frac{4}{16\cdot 1}=\frac{1}{4}$. These numbers are hypothetical illustration of the formula; they are not measured values, and no apparatus is asserted to realize the protocol.

Fix one particle label ($N=1$), the force branch and a model as in Operational observable model of the force branch: parameters, preparations, observation times, readout, units and additive error, and fix two inertial frames $F$, $F^{\mathrm{\prime }}$ of the model with adapted coordinates $(t,x)$ and $(t^{\mathrm{\prime }},x^{\mathrm{\prime }})$. By Galilean transformation of worldlines: velocity and acceleration laws, frame-independence of zero coordinate acceleration, and the group of frame changes there are $b\in \mathbb{R}$, $a_0,v\in {\mathbb{R}}^3$ and $R\in SO(3)$ with $t^{\mathrm{\prime }}(p)=t(p)+b,x^{\mathrm{\prime }}(p)=Rx(p)+vt(p)+a_0\text{for every event }p\mathrm{.}$

The protocol in $F$. Take a protocol of Operational observable model of the force branch: parameters, preparations, observation times, readout, units and additive error with a single observation time $t_1\in I_0$ and the position readout $\mathrm{\Phi }(\theta ,z):=x_{\theta ,z}(t_1)\in {\mathbb{R}}^3$ of dimension $\mathrm{\ell }$, read in metres (Dimensions and SI unit conventions of the non-relativistic classical mechanics framework); the observation is $Y=\mathrm{\Phi }(\theta ,z)+\epsilon$ for an error random element $\epsilon :\mathrm{\Omega }\to {\mathbb{R}}^3$ on a probability space $(\mathrm{\Omega },\mathfrak{F},\mathbb{P})$.

Domains under the coordinate change. The transformed configuration set at time $t^{\mathrm{\prime }}$ is $Q_{t^{\mathrm{\prime }}}^{\mathrm{\prime }}:=RQ+v(t^{\mathrm{\prime }}-b)+a_0$, so the transformed time-configuration domain $\{(t^{\mathrm{\prime }},x^{\mathrm{\prime }}):t^{\mathrm{\prime }}\in I_0+b,x^{\mathrm{\prime }}\in Q_{t^{\mathrm{\prime }}}^{\mathrm{\prime }}\}$ is open (it is the image of $I_0\times Q$ under the invertible affine frame change). It need not be a product with one fixed configuration set, for example when a boost translates bounded $Q$. The following formulas always describe the coordinates of the same worldline and recorded events. They define a force-branch protocol in $F^{\mathrm{\prime }}$ only when the transformed model has an admissible fixed-domain presentation, in particular when $Q={\mathbb{R}}^3$; no fixed-domain model is asserted in the general moving-domain case.

The coordinate readout in $F^{\mathrm{\prime }}$. The same event at which the readout is taken has $F^{\mathrm{\prime }}$-time $t_1+b$; reading the same label position in $F^{\mathrm{\prime }}$ at that time defines the transformed prediction map on the same parameter/preparation labels, ${\mathrm{\Phi }}^{\mathrm{\prime }}(\theta ,z):=x_{\theta ,z}^{\mathrm{\prime }}(t_1+b)=R\mathrm{\Phi }(\theta ,z)+v{t}_1+a_0,$ and the $F^{\mathrm{\prime }}$-coordinates of the recorded point are $Y^{\mathrm{\prime }}:=RY+v{t}_1+a_0={\mathrm{\Phi }}^{\mathrm{\prime }}(\theta ,z)+R\epsilon ,$ so the additive-error form is preserved with the rotated error ${\epsilon }^{\mathrm{\prime }}:=R\epsilon$ and the transformed open error domain $V^{\mathrm{\prime }}:=RV$, which contains $0$ and contains every value of ${\epsilon }^{\mathrm{\prime }}$.

Transform of the error law. Suppose $\epsilon$ is square-integrable with mean $E[\epsilon ]=0$ and covariance matrix $\mathrm{\Sigma }$ in the sense of Mean vector and covariance matrix of a square-integrable random vector. Then ${\epsilon }^{\mathrm{\prime }}=R\epsilon$ has mean $0$ and covariance $\mathrm{Cov}({\epsilon }^{\mathrm{\prime }})=R\mathrm{\Sigma }{R}^{\mathsf{T}},$ and $\mid {\epsilon }^{\mathrm{\prime }}\mid =\mid \epsilon \mid$ in the Euclidean norm because $R$ is orthogonal. In particular an additive error declared in one frame need not have the same law in another: its law is the pushforward under $R$, which equals the original law precisely when that law is invariant under this rotation, and the prediction map acquires the known affine term $v{t}_1+a_0$. A protocol must declare the frame of its readout and of its error model.

Let a repeated-observation protocol of Repeated-observation protocols of the force branch: adopted IID and second-moment sampling hypotheses and the sample mean satisfy its IID sampling hypothesis and its second-moment hypothesis, and let $A$ be an event of a single observation with $0<\mathbb{P}(A)=p<1$. Then every finite sample $(Y_1,\dots ,Y_n)$ of the protocol contains at least one outcome in $A$ for every $n\ge 1$; equivalently, if a finite sample contains no outcome in $A$, the declared sampling model is refuted.

The construction is a finite probability computation; it contains no actual apparatus, no performed preparation, no data and no observed sequence; [L4] supplies only declared operational model data. Its physical reading is the elementary consistency statement of Repeated-observation protocols of the force branch: adopted IID and second-moment sampling hypotheses and the sample mean: a model assigning probability $p$ per trial to an outcome class is compatible with a finite sample that misses it, and the probability of the miss is $(1-p{)}^n$ when the trials are independent.

What the counterexample does not show is that the declared model is true, or that some other model is false: consistency of one finite sample with a model is not evidence for it, and comparing models requires a specified test — data space, model family, null set, rejection region and size — as in Model testing for the force branch: what a test specifies, what rejection and retention establish, and what is not claimed in its role as a scope item of the companion page.