Fermi-Walker transport preserves the metric, and explicit rigid and non-rigid observer congruences

Preservation (a): write $s$ for the given parameter and ${\tau }^{\mathrm{\prime }}(s)>0$ for its proper-time derivative. By [F1], $X^{\mathrm{\prime }}={\tau }^{\mathrm{\prime }}{c}^{-2}(\eta (A,X)U-\eta (U,X)A)$ and likewise for $Y$. Bilinearity gives $\frac{d}{ds}\eta (X,Y)=\eta (X^{\mathrm{\prime }},Y)+\eta (X,Y^{\mathrm{\prime }})={\tau }^{\mathrm{\prime }}{c}^{-2}[\eta (A,X)\eta (U,Y)-\eta (U,X)\eta (A,Y)+\eta (A,Y)\eta (U,X)-\eta (U,Y)\eta (A,X)]=0$. The two pairs cancel, so the scalar pairing is constant on the interval by A function continuous on an interval $I$ whose derivative vanishes at every interior point of $I$ is constant on $I$; consequently two such functions with the same derivative differ by a constant; the fields are $C^1$, including continuity at any included endpoints. Norms and orthogonality are therefore preserved. For the unique transported fields with prescribed tetrad initial values, all pairings remain $\mathrm{diag}(-1,1,1,1{)}_{ab}$, and the time leg is $U\mathrm{/}c$ because [F1] makes $U\mathrm{/}c$ transported with the same initial value. Hence the family remains an observer-adapted tetrad.

Rindler coordinates (b), part 1: the map $\mathrm{\Phi }(\theta ,\rho ,y,z)=(\rho \sinh \theta ,\rho \cosh \theta ,y,z)$ is smooth on $\mathbb{R}\times (0,\mathrm{\infty })\times {\mathbb{R}}^2$ by [F4] and [F8], and its image is contained in $W$ because $\cosh \theta >\mathrm{\mid }\sinh \theta \mathrm{\mid }$ for every $\theta$; conversely, for $(x^0,x)$ with $x>\mathrm{\mid }{x}^0\mathrm{\mid }$ the numbers $\rho :=\sqrt{x^2-(x^0{)}^2}>0$ and $\theta :=\frac{1}{2}\log ((x+x^0)\mathrm{/}(x-x^0))\in \mathbb{R}$ are well defined, and the identities $\rho \sinh \theta =x^0$, $\rho \cosh \theta =x$ follow from $e^{\pm \theta }=\sqrt{(x\pm x^0)\mathrm{/}(x\mp x^0)}$ and ${\cosh }^2-{\sinh }^2=1$; hence $\mathrm{\Phi }$ is a bijection of its domain onto $W$. Its $(\theta ,\rho )$ derivative block has determinant $\rho ({\cosh }^2\theta -{\sinh }^2\theta )=\rho >0$ and the transverse block is the identity; the inverse-function theorem in [F10] therefore makes its inverse smooth locally at every point of $W$, hence smooth on $W$. Differentiating the four component functions gives $d{x}^0=\sinh \theta d\rho +\rho \cosh \theta d\theta$ and $dx=\cosh \theta d\rho +\rho \sinh \theta d\theta$, so the pullback metric is $-(\sinh \theta d\rho +\rho \cosh \theta d\theta {)}^2+(\cosh \theta d\rho +\rho \sinh \theta d\theta {)}^2+d{y}^2+d{z}^2=d{\rho }^2-{\rho }^{2}d{\theta }^2+d{y}^2+d{z}^2$, the cross terms cancelling and the ${\sinh }^2-{\cosh }^2=-1$ identity producing the term $-{\rho }^{2}d{\theta }^2$.

Rigid rotation (c), part 1: $K={\mathrm{\partial }}_0+(\mathrm{\Omega }\mathrm{/}c)(-x^2{\mathrm{\partial }}_1+x^1{\mathrm{\partial }}_2)$ is a smooth vector field on the whole chart, and $\eta (K,K)=-1+{\mathrm{\Omega }}^2{r}^2\mathrm{/}{c}^2<0$ exactly on the stated region $U_{\mathrm{\Omega }}$ (and everywhere when $\mathrm{\Omega }=0$) by [F2]; its components in the inertial chart are $K=(1,-\frac{\mathrm{\Omega }}{c}{x}^2,\frac{\mathrm{\Omega }}{c}{x}^1,0)$, so $K$ is future where it is timelike, and $\gamma (r)=(1-{\mathrm{\Omega }}^2{r}^2\mathrm{/}{c}^2{)}^{-1\mathrm{/}2}$ is smooth and positive on $U_{\mathrm{\Omega }}$ by [F8], so $n=\gamma K$ satisfies $\eta (n,n)={\gamma }^2(-1+{\mathrm{\Omega }}^2{r}^2\mathrm{/}{c}^2)=-1$ and is future unit, that is, a congruence on $U_{\mathrm{\Omega }}$ with the stated observer field. Computing the symmetric derivative of the components ${\eta }_{ab}{K}^b=(-1,-\frac{\mathrm{\Omega }}{c}{x}^2,\frac{\mathrm{\Omega }}{c}{x}^1,0)$ gives ${\mathrm{\partial }}_a(\eta K{)}_b+{\mathrm{\partial }}_b(\eta K{)}_a=0$: the only nonzero first partials are ${\mathrm{\partial }}_1(\eta K{)}_2=\mathrm{\Omega }\mathrm{/}c$ and ${\mathrm{\partial }}_2(\eta K{)}_1=-\mathrm{\Omega }\mathrm{/}c$, which cancel in the symmetrization, so $K$ is a Killing field.

Bell setup (d): for $q$ of class $C^2$ with $\mathrm{\mid }{q}_1^{\mathrm{\prime }}\mathrm{\mid }<c$, the functions $\gamma =(1-q_1^{\mathrm{\prime }2}\mathrm{/}{c}^2{)}^{-1\mathrm{/}2}$ and $\beta =q_1^{\mathrm{\prime }}\mathrm{/}c$ are $C^1$ in $t$ by [F8], so the field $n(x)=\gamma (1,\beta ,0,0)$ evaluated at $t=x^0\mathrm{/}c$ is a $C^1$ future unit field on any open subset $\mathrm{\Omega }$ of the chart with $\mathrm{\Omega }\subseteq \{\mathrm{\mid }{q}_1^{\mathrm{\prime }}(x^0\mathrm{/}c)\mathrm{\mid }<c\}$; it is tangent to the translated worldlines $x_{\xi }(t)=(ct,q(t)+\xi )$, since their coordinate derivative $(c,q_1^{\mathrm{\prime }},0,0)=c(1,\beta ,0,0)$ is a positive multiple of $n$. The longitudinal unit vector $s=\gamma (\beta ,1,0,0)$ satisfies $\eta (s,s)={\gamma }^2(-{\beta }^2+1)=1$ and $\eta (n,s)={\gamma }^2(-\beta +\beta )=0$, so $s$ is a unit rest vector of $n$.

Rindler congruence (b), part 2: in the coordinates $(\theta ,\rho ,y,z)$ the field is $n={\rho }^{-1}{\mathrm{\partial }}_{\theta }$; its acceleration in these coordinates is computed from the inertial components of $n$, namely $n=\cosh \theta e_0+\sinh \theta e_1$ (because ${\rho }^{-1}{\mathrm{\partial }}_{\theta }=\cosh \theta {\mathrm{\partial }}_0+\sinh \theta {\mathrm{\partial }}_1$ in the inertial chart, as ${\mathrm{\partial }}_{\theta }=\rho (\cosh \theta {\mathrm{\partial }}_0+\sinh \theta {\mathrm{\partial }}_1)$): the coordinate partial derivatives of the components are ${\mathrm{\partial }}_{x^0}\theta =x\mathrm{/}{\rho }^2$ and ${\mathrm{\partial }}_x\theta =-x^0\mathrm{/}{\rho }^2$, whence ${\mathrm{\nabla }}_{{\mathrm{\partial }}_0}n=(x\mathrm{/}{\rho }^2){\mathrm{\partial }}_{\rho }$ and ${\mathrm{\nabla }}_{{\mathrm{\partial }}_1}n=-(x^0\mathrm{/}{\rho }^2){\mathrm{\partial }}_{\rho }$; therefore $a={\mathrm{\nabla }}_{n}n=\cosh \theta \cdot (x\mathrm{/}{\rho }^2){\mathrm{\partial }}_{\rho }-\sinh \theta \cdot (x^0\mathrm{/}{\rho }^2){\mathrm{\partial }}_{\rho }=(x\cosh \theta -x^0\sinh \theta ){\rho }^{-2}{\mathrm{\partial }}_{\rho }={\rho }^{-1}{\mathrm{\partial }}_{\rho }$. Its norm is $\eta (a,a)={\rho }^{-2}\eta ({\mathrm{\partial }}_{\rho },{\mathrm{\partial }}_{\rho })={\rho }^{-2}$ by the pullback metric of step 1.2. The metric has no $d\theta d\rho$, $d\theta dy$ or $d\theta dz$ cross terms, so $n$ is orthogonal to ${\mathrm{\partial }}_{\rho },{\mathrm{\partial }}_y,{\mathrm{\partial }}_z$; the slices $\theta =const$ have tangent space $\ker (n^{\mathrm{♭}})$ (they are the level sets of the coordinate $\theta$), hence the rest distribution is integrable, and by [F3] its vorticity vanishes; since the induced metric $d{\rho }^2+d{y}^2+d{z}^2$ on the slices is independent of $\theta$ and $h(n,\cdot )=0$, while $[n,{\mathrm{\partial }}_{\rho }]=(1\mathrm{/}\rho )n$ and $[n,{\mathrm{\partial }}_y]=[n,{\mathrm{\partial }}_z]=0$, the Lie-derivative formula ${\mathcal{L}}_{n}h(X,Y)=n(h(X,Y))-h([n,X],Y)-h(X,[n,Y])$ gives zero on this rest basis, $({\mathcal{L}}_{n}h)(\chi ,{\chi }^{\mathrm{\prime }})=0$ for $\chi ,{\chi }^{\mathrm{\prime }}\in \{{\mathrm{\partial }}_{\rho },{\mathrm{\partial }}_y,{\mathrm{\partial }}_z\}$ and the congruence is Born rigid by [F3]. Finally $[n,{\mathrm{\partial }}_{\rho }]=[{\rho }^{-1}{\mathrm{\partial }}_{\theta },{\mathrm{\partial }}_{\rho }]=-{\mathrm{\partial }}_{\rho }({\rho }^{-1}){\mathrm{\partial }}_{\theta }={\rho }^{-2}{\mathrm{\partial }}_{\theta }=+(1\mathrm{/}\rho )n\mathrm{\ne }0$ by [F6], so the frame $(n,{\mathrm{\partial }}_{\rho },{\mathrm{\partial }}_y,{\mathrm{\partial }}_z)$, whose members are orthonormal by the metric displayed in step 1.2, is not the coordinate frame of any chart (for coordinate frames all brackets vanish).

Ambient rigid rotation (c), part 2: with $n=\gamma K$ and $n_b=\gamma K_b$, the symmetric derivative is ${\mathrm{\nabla }}_{(a}{n}_{b)}=K_{(a}{\mathrm{\partial }}_{b)}\gamma +\gamma {\mathrm{\nabla }}_{(a}{K}_{b)}$ by [F2], and the second term vanishes because $K$ is Killing (step 1.3); projecting with $h$ kills the first term, because $h^c{}_a{K}_c=K_a+n_a⟨n,K⟩=K_a-n_a\mathrm{/}\gamma =0$ (since $\eta (n,K)=\gamma \eta (K,K)=-1\mathrm{/}\gamma$ and $n_a\mathrm{/}\gamma =K_a$); hence $\theta =\gamma h^c{h}^d{\mathrm{\nabla }}_{(c}{K}_{d)}+0=0$ and the congruence is Born rigid by [F3]. For the vorticity put $\kappa :=\eta (K,\cdot )=(-1,-\frac{\mathrm{\Omega }}{c}{x}^2,\frac{\mathrm{\Omega }}{c}{x}^1,0)$ in components; then $d\kappa =2(\mathrm{\Omega }\mathrm{/}c)d{x}^1\wedge d{x}^2$ and $\kappa \wedge d\kappa =-2(\mathrm{\Omega }\mathrm{/}c)d{x}^0\wedge d{x}^1\wedge d{x}^2$ by [F5], the terms containing $(x^{1}d{x}^2-x^{2}d{x}^1)\wedge d{x}^1\wedge d{x}^2$ vanishing by alternation. Since $n^{\mathrm{♭}}=\gamma \kappa$ and $d\gamma \wedge \kappa \wedge \kappa =0$ by alternation, $n^{\mathrm{♭}}\wedge d(n^{\mathrm{♭}})={\gamma }^2\kappa \wedge d\kappa =-2{\gamma }^2(\mathrm{\Omega }\mathrm{/}c)d{x}^0\wedge d{x}^1\wedge d{x}^2$, which is nonzero at every point of $U_{\mathrm{\Omega }}$ when $\mathrm{\Omega }\mathrm{\ne }0$; by [F3] the vorticity $\omega$ is nowhere zero there and the ambient rest distribution is nowhere integrable, so no rest-simultaneity hypersurface exists locally. For $\mathrm{\Omega }=0$ the field is $n={\mathrm{\partial }}_0$, which is inertial and irrotational with rest hypersurfaces $x^0=const$.

Bell deformation (d): with $\beta =q_1^{\mathrm{\prime }}\mathrm{/}c$ and ${\beta }^{\mathrm{\prime }}=q_1^{\mathrm{\prime }\mathrm{\prime }}\mathrm{/}c$, the components of $n$ are $n=(n^0,n^1,n^2,n^3)=(\gamma ,\gamma \beta ,0,0)$, hence $n_b={\eta }_{bc}{n}^c=(-\gamma ,\gamma \beta ,0,0)$, and $n$ depends on the coordinates only through $t=x^0\mathrm{/}c$, so ${\mathrm{\partial }}_{1}n={\mathrm{\partial }}_{2}n={\mathrm{\partial }}_{3}n=0$ and ${\mathrm{\partial }}_0=\frac{1}{c}\frac{d}{dt}$ by [F2]. Using $\frac{d\gamma }{dt}={\gamma }^3\beta {\beta }^{\mathrm{\prime }}$ (from $\gamma =(1-{\beta }^2{)}^{-1\mathrm{/}2}$ and the chain rule) and $1+{\gamma }^2{\beta }^2={\gamma }^2$ one gets ${\mathrm{\partial }}_0{n}_0=-{\gamma }^3\beta {\beta }^{\mathrm{\prime }}\mathrm{/}c$ and ${\mathrm{\partial }}_0{n}_1=\frac{1}{c}\frac{d}{dt}(\gamma \beta )=\frac{{\gamma }^3{\beta }^{\mathrm{\prime }}\mathrm{/}c}{1}$; contracting with $s=\gamma (\beta ,1,0,0)$ gives $s^a{s}^b{\mathrm{\partial }}_a{n}_b=(\gamma \beta {)}^2(-{\gamma }^3\beta {\beta }^{\mathrm{\prime }}\mathrm{/}c)+(\gamma \beta )(\gamma )({\gamma }^3{\beta }^{\mathrm{\prime }}\mathrm{/}c)=\frac{{\gamma }^5\beta {\beta }^{\mathrm{\prime }}}{c}(1-{\beta }^2)=\frac{{\gamma }^3\beta {\beta }^{\mathrm{\prime }}}{c}={\gamma }^3{q}_1^{\mathrm{\prime }}{q}_1^{\mathrm{\prime }\mathrm{\prime }}\mathrm{/}{c}^3$, where $1-{\beta }^2={\gamma }^{-2}$. Since $s$ is a rest vector ($\eta (n,s)=0$ by step 1.4), the projections act as the identity on $s$ and $\theta (s,s)=s^a{s}^b{\mathrm{\partial }}_a{n}_b$ in the first-order sense; all transverse components of ${\mathrm{\partial }}_a{n}_b$ vanish because $n$ depends only on $t$, so $\theta (s,s)$ is nonzero exactly when $q_1^{\mathrm{\prime }}(t)q_1^{\mathrm{\prime }\mathrm{\prime }}(t)\mathrm{\ne }0$, and then the first-order rigidity condition $\theta =0$ fails on a neighbourhood of $t$; the translated worldlines have, at every fixed chart time, the constant separation $\mathrm{\mid }{\xi }_1-{\xi }_2\mathrm{\mid }$, so this failure of Born rigidity is not in conflict with the constant labelling. No crack, fracture or elasticity conclusion is drawn.

Ring cylinder geometry (c), part 3: parameterize the cylinder $r=R$, $x^3=const$ by $(t,\phi )$ with $x^0=ct$ and $\phi =\varphi -\mathrm{\Omega }t$, so that the inertial angle is $\varphi =\phi +\mathrm{\Omega }t$ and the point is $R(\cos \varphi ,\sin \varphi )$. Substituting $d\varphi =d\phi +\mathrm{\Omega }dt$ into $d{s}^2=-c^{2}d{t}^2+R^{2}d{\varphi }^2$ gives the induced metric $-(c^2-{\mathrm{\Omega }}^2{R}^2)d{t}^2+2\mathrm{\Omega }{R}^{2}dtd\phi +R^{2}d{\phi }^2$, a nondegenerate Lorentzian metric on the cylinder since $\mathrm{\mid }\mathrm{\Omega }\mathrm{\mid }R<c$. The observer with label ${\varphi }_0$ has inertial angle $\varphi (t)={\varphi }_0+\mathrm{\Omega }t$, hence tangent $c{\mathrm{\partial }}_0+\mathrm{\Omega }R(-\sin \varphi {\mathrm{\partial }}_1+\cos \varphi {\mathrm{\partial }}_2)=cK$ evaluated at the ring, so its dimensionless four-velocity is $n=\gamma (R)K$ and its four-velocity is $cn$, whose components at the ring are $c\gamma (R)(1,-(\mathrm{\Omega }R\mathrm{/}c)\sin \varphi ,(\mathrm{\Omega }R\mathrm{/}c)\cos \varphi ,0)$ as stated. The covector $\kappa$ of step 2.2 restricts to the cylinder as $\kappa =-cdt+(\mathrm{\Omega }\mathrm{/}c)R^{2}d\varphi =-cdt+(\mathrm{\Omega }{R}^2\mathrm{/}c)(d\phi +\mathrm{\Omega }dt)=-\frac{c^2-{\mathrm{\Omega }}^2{R}^2}{c}dt+\frac{\mathrm{\Omega }{R}^2}{c}d\phi$, so the ambient rest covector $n^{\mathrm{♭}}=\gamma (R)\kappa$ is proportional to $-(c^2-{\mathrm{\Omega }}^2{R}^2)dt+\mathrm{\Omega }{R}^{2}d\phi$; its kernel on the cylinder is the line field generated by ${\mathrm{\partial }}_{\phi }+\frac{\mathrm{\Omega }{R}^2}{c^2-{\mathrm{\Omega }}^2{R}^2}{\mathrm{\partial }}_t$, since the kernel condition is $-(c^2-{\mathrm{\Omega }}^2{R}^2)dt(v)+\mathrm{\Omega }{R}^{2}d\phi (v)=0$ for $v=a{\mathrm{\partial }}_t+b{\mathrm{\partial }}_{\phi }$, that is $-(c^2-{\mathrm{\Omega }}^2{R}^2)a+\mathrm{\Omega }{R}^{2}b=0$, so $v$ is a multiple of ${\mathrm{\partial }}_{\phi }+\frac{\mathrm{\Omega }{R}^2}{c^2-{\mathrm{\Omega }}^2{R}^2}{\mathrm{\partial }}_t$; equivalently $dt\mathrm{/}d\phi =\mathrm{\Omega }{R}^2\mathrm{/}(c^2-{\mathrm{\Omega }}^2{R}^2)$ along rest lines.

Ring holonomy (c), part 4: the vector field $Z:={\mathrm{\partial }}_{\phi }+\frac{\mathrm{\Omega }{R}^2}{c^2-{\mathrm{\Omega }}^2{R}^2}{\mathrm{\partial }}_t$ on the cylinder is smooth and nowhere zero (the coefficient of ${\mathrm{\partial }}_{\phi }$ is $1$), so by the flow box theorem [F9] applied on the two-dimensional cylinder there are local coordinates in which $Z=\mathrm{\partial }\mathrm{/}\mathrm{\partial }{u}^0$; in particular the rest line field is locally integrable, its integral curves satisfy $dt\mathrm{/}d\phi =\frac{\mathrm{\Omega }{R}^2}{c^2-{\mathrm{\Omega }}^2{R}^2}$ by the kernel computation of step 3.1, and integrating from $\phi =0$ to $\phi =2\pi$ at the initial label gives the signed offset $\mathrm{\Delta }t=\frac{2\pi \mathrm{\Omega }{R}^2}{c^2-{\mathrm{\Omega }}^2{R}^2}$, which returns to the same rotating label $\phi$ at a different chart time when $\mathrm{\Omega }\mathrm{\ne }0$. If $f$ were a globally single-valued real function on the whole cylinder with $df=\mu (dt-\frac{\mathrm{\Omega }{R}^2}{c^2-{\mathrm{\Omega }}^2{R}^2}d\phi )$ for a continuous multiplier $\mu >0$ (the time-orientation convention $df({\mathrm{\partial }}_t)>0$) (so that the kernel of $df$ is exactly the rest-line distribution, so the level sets of $f$ are tangent to those lines), then the angular lift $F(\phi ):=f(0,\phi )$ would have $F(0)=F(2\pi )$ by single-valuedness on the circle. Its derivative is $F^{\mathrm{\prime }}(\phi )=-\mu (0,\phi )\mathrm{\Omega }{R}^2\mathrm{/}(c^2-{\mathrm{\Omega }}^2{R}^2)$, with a fixed nonzero sign because the multiplier is time-oriented and $\mathrm{\Omega }\mathrm{\ne }0$, $R>0$. The mean value theorem The mean value theorem, as the case $g(x)=x$ of Cauchy's: for $f$ continuous on $[a,b]$ with $a<b$ and differentiable on $(a,b)$ there is $c\in (a,b)$ with $f(b)-f(a)=f^{\mathrm{\prime }}(c)(b-a)$ on $[0,2\pi ]$ therefore gives $F(2\pi )-F(0)=2\pi F^{\mathrm{\prime }}(\xi )\mathrm{\ne }0$ for some $\xi \in (0,2\pi )$; this contradiction shows that no such global synchronization function exists. The obstruction is the holonomy of the global ring and is distinct from the ambient pointwise nonintegrability of step 2.2, which is a local statement in the four-dimensional open set and does not apply to the one-dimensional rest line field on the cylinder. For $\mathrm{\Omega }=0$ the kernel condition reads $dt=0$ and both the offset and the obstruction vanish.

Conclusion: step 1.1 proves (a), including the preservation of norms and orthogonality and the observer-adapted tetrad; steps 1.2 and 2.1 prove (b), including the Rindler metric, the unit future field, the acceleration ${\rho }^{-1}{\mathrm{\partial }}_{\rho }$, Born rigidity, vanishing vorticity, the rest slices and the non-coordinate-frame bracket; steps 1.3, 2.2, 3.1 and 4.1 prove (c), including the Born-rigid rotating congruence, its nowhere-zero vorticity and ambient nonintegrability for $\mathrm{\Omega }\mathrm{\ne }0$, the inertial case $\mathrm{\Omega }=0$, the ring metric, the rest-line equation, the time offset $2\pi \mathrm{\Omega }{R}^2\mathrm{/}(c^2-{\mathrm{\Omega }}^2{R}^2)$ and the impossibility of a globally single-valued ring synchronization time; and steps 1.4 and 2.3 prove (d), with the first-order deformation $\theta (s,s)={\gamma }^3{q}_1^{\mathrm{\prime }}{q}_1^{\mathrm{\prime }\mathrm{\prime }}\mathrm{/}{c}^3$ and the exclusion of material fracture claims.