Worked Bell-type translated congruence: constant launch-frame separation without Born rigidity

With $q_1(t)=\frac{1}{2}\alpha t^2$ one has $q_1^{\mathrm{\prime }}(t)=\alpha t$ and $q_1^{\mathrm{\prime }\mathrm{\prime }}(t)=\alpha$, so the field of the Example is $n=\gamma (1,\alpha t\mathrm{/}c,0,0)$ and $s=\gamma (\alpha t\mathrm{/}c,1,0,0)$, where $\gamma =(1-{\alpha }^2{t}^2\mathrm{/}{c}^2{)}^{-1\mathrm{/}2}$ is defined and positive on the subluminal domain $\mathrm{\mid }t\mathrm{\mid }<c\mathrm{/}\alpha$. Evaluating the diagonal form gives $\eta (n,n)={\gamma }^2(-1+{\alpha }^2{t}^2\mathrm{/}{c}^2)=-1$, $\eta (s,s)={\gamma }^2(-{\alpha }^2{t}^2\mathrm{/}{c}^2+1)=1$ and $\eta (n,s)={\gamma }^2(-\alpha t\mathrm{/}c+\alpha t\mathrm{/}c)=0$; hence $n$ is a future unit timelike field and $s$ a unit rest vector.

Direct computation of the contraction: with $n_b={\eta }_{bc}{n}^c=(-\gamma ,\gamma \beta ,0,0)$ and $s=\gamma (\beta ,1,0,0)$, and with $n$ depending only on $t=x^0\mathrm{/}c$, one has ${\mathrm{\partial }}_1{n}_b={\mathrm{\partial }}_2{n}_b={\mathrm{\partial }}_3{n}_b=0$ and ${\mathrm{\partial }}_0=\frac{1}{c}\frac{d}{dt}$; by [L4], ${\mathrm{\partial }}_0{n}_0=-{\gamma }^3\beta {\beta }^{\mathrm{\prime }}\mathrm{/}c=-{\gamma }^3{\alpha }^{2}t\mathrm{/}{c}^3$ and ${\mathrm{\partial }}_0{n}_1=\frac{1}{c}\frac{d}{dt}(\gamma \beta )={\gamma }^3{\beta }^{\mathrm{\prime }}\mathrm{/}c={\gamma }^3\alpha \mathrm{/}{c}^2$. Hence $s^a{s}^b{\mathrm{\partial }}_a{n}_b=(\gamma \beta {)}^2{\mathrm{\partial }}_0{n}_0+(\gamma \beta )(\gamma ){\mathrm{\partial }}_0{n}_1={\gamma }^5{\beta }^2(-\beta {\beta }^{\mathrm{\prime }})\mathrm{/}c+{\gamma }^5\beta {\beta }^{\mathrm{\prime }}\mathrm{/}c={\gamma }^5\beta {\beta }^{\mathrm{\prime }}(1-{\beta }^2)\mathrm{/}c={\gamma }^3\beta {\beta }^{\mathrm{\prime }}\mathrm{/}c={\gamma }^3{\alpha }^{2}t\mathrm{/}{c}^3$, where $1-{\beta }^2={\gamma }^{-2}$ and only the cross term $(a,b)=(0,1)$ occurs because ${\mathrm{\partial }}_1{n}_0=0$; this reproduces the supplier formula of [L1]. Since $\alpha >0$ and $\gamma \ge 1$ on the domain, $\theta (s,s)=0$ exactly at $t=0$ and is nonzero for every $t\mathrm{\ne }0$, with opposite signs on the two sides of $0$. By [L2] the congruence is therefore not Born rigid on any neighbourhood of any $t\mathrm{\ne }0$, and at $t=0$ the rigidity condition fails immediately on both sides.

The translated members at a fixed chart time $t$ have positions $q(t)+{\xi }_1$ and $q(t)+{\xi }_2$ in the spatial coordinates, so their Euclidean separation is $\mathrm{\mid }{\xi }_1-{\xi }_2\mathrm{\mid }$, independent of $t$; equivalently, in the three-dimensional spatial coordinates the congruence is a translation family at each instant even though $\theta \mathrm{\ne }0$ off $t=0$. Thus constant instantaneous separation of the labels and Born rigidity are different requirements, which is the content of [L1].

Conclusion and scope: the computation exhibits the exact expansion ${\gamma }^3{\alpha }^{2}t\mathrm{/}{c}^3$, nonzero for $t\mathrm{\ne }0$ and zero at $t=0$, so the translated congruence is not Born rigid while its labels keep constant separation $\mathrm{\mid }{\xi }_1-{\xi }_2\mathrm{\mid }$. No statement is made about a material rod, a thread or a fracture: a breaking or stress conclusion would require an elasticity model, a constitutive law and a failure criterion, none of which is part of the kinematic model, and no such conclusion is drawn.