Worked Rindler wedge congruence: chart, metric, acceleration, rigidity and non-coordinate frame

Bijection and inverse: $\mathrm{\Phi }$ is smooth because its components are sums of smooth functions of $(\theta ,\rho ,y,z)$ by [L1]. For $(\theta ,\rho ,y,z)$ in the domain one has $x^2-(x^0{)}^2={\rho }^2({\cosh }^2\theta -{\sinh }^2\theta )={\rho }^2$ and $x=\rho \cosh \theta >0$, so $\mathrm{\Phi }$ maps into $W$. Conversely, given $(x^0,x,y,z)$ with $x>\mathrm{\mid }{x}^0\mathrm{\mid }$, put $\rho :=\sqrt{x^2-(x^0{)}^2}>0$ and $\theta :=\frac{1}{2}\log ((x+x^0)\mathrm{/}(x-x^0))\in \mathbb{R}$ (the argument of the logarithm being positive because $x\pm x^0>0$); then $e^{\theta }=\sqrt{(x+x^0)\mathrm{/}(x-x^0)}$ and $e^{-\theta }=\sqrt{(x-x^0)\mathrm{/}(x+x^0)}$, so $\rho \sinh \theta =\rho (e^{\theta }-e^{-\theta })\mathrm{/}2=\frac{(x+x^0)-(x-x^0)}{2}=x^0$ and $\rho \cosh \theta =\rho (e^{\theta }+e^{-\theta })\mathrm{/}2=\frac{(x+x^0)+(x-x^0)}{2}=x$ by [L1], which gives a two-sided inverse. The $(\theta ,\rho )$ derivative block has determinant $\rho ({\cosh }^2\theta -{\sinh }^2\theta )=\rho >0$, with identity transverse block, so [L5] makes this inverse smooth locally everywhere. Hence $\mathrm{\Phi }$ is a smooth bijection onto $W$ with smooth inverse, that is, a diffeomorphism by [L5].

Pullback metric: differentiating the four components of $\mathrm{\Phi }$ and using [L1] gives $d{x}^0=\sinh \theta d\rho +\rho \cosh \theta d\theta$, $dx=\cosh \theta d\rho +\rho \sinh \theta d\theta$, $dy=dy$, $dz=dz$. Substituting into $-(d{x}^0{)}^2+(dx{)}^2+d{y}^2+d{z}^2$: the cross terms are $-2\rho \sinh \theta \cosh \theta +2\rho \cosh \theta \sinh \theta =0$, the $d{\theta }^2$ coefficient is $-{\rho }^2{\cosh }^2\theta +{\rho }^2{\sinh }^2\theta =-{\rho }^2$ and the $d{\rho }^2$ coefficient is $-{\sinh }^2\theta +{\cosh }^2\theta =1$; hence the pullback metric is $-{\rho }^{2}d{\theta }^2+d{\rho }^2+d{y}^2+d{z}^2$.

Unit future field: in these coordinates $g({\mathrm{\partial }}_{\theta },{\mathrm{\partial }}_{\theta })=-{\rho }^2$, $g({\mathrm{\partial }}_{\rho },{\mathrm{\partial }}_{\rho })=g({\mathrm{\partial }}_y,{\mathrm{\partial }}_y)=g({\mathrm{\partial }}_z,{\mathrm{\partial }}_z)=1$ and all mixed components vanish, so $n={\rho }^{-1}{\mathrm{\partial }}_{\theta }$ satisfies $g(n,n)=-{\rho }^2\cdot {\rho }^{-2}=-1$. In the inertial chart, ${\mathrm{\partial }}_{\theta }=\rho (\cosh \theta {\mathrm{\partial }}_0+\sinh \theta {\mathrm{\partial }}_1)$ by step 2.1, hence $n=\cosh \theta {\mathrm{\partial }}_0+\sinh \theta {\mathrm{\partial }}_1$, whose time component is $\cosh \theta >0$ and whose $\eta$-square is $-{\cosh }^2\theta +{\sinh }^2\theta =-1$; by [L2] $n$ is future timelike, so $n$ is a congruence on $W$.

Acceleration and connection coefficients: by the coordinate map, ${\mathrm{\partial }}_{\theta }=\rho (\cosh \theta e_0+\sinh \theta e_1)$ and ${\mathrm{\partial }}_{\rho }=\sinh \theta e_0+\cosh \theta e_1$. Differentiating these inertial components with [L6] gives ${\mathrm{\nabla }}_{{\mathrm{\partial }}_{\theta }}{\mathrm{\partial }}_{\theta }=\rho {\mathrm{\partial }}_{\rho }$, ${\mathrm{\nabla }}_{{\mathrm{\partial }}_{\theta }}{\mathrm{\partial }}_{\rho }={\mathrm{\nabla }}_{{\mathrm{\partial }}_{\rho }}{\mathrm{\partial }}_{\theta }={\rho }^{-1}{\mathrm{\partial }}_{\theta }$ and ${\mathrm{\nabla }}_{{\mathrm{\partial }}_{\rho }}{\mathrm{\partial }}_{\rho }=0$; all derivatives involving the constant transverse basis fields vanish. Hence the only nonzero coordinate connection coefficients are ${\mathrm{\Gamma }}_{\theta \theta }^{\rho }=\rho$ and ${\mathrm{\Gamma }}_{\theta \rho }^{\theta }={\mathrm{\Gamma }}_{\rho \theta }^{\theta }={\rho }^{-1}$, as claimed. Since ${\mathrm{\partial }}_{\theta }\rho =0$, $a={\mathrm{\nabla }}_{{\rho }^{-1}{\mathrm{\partial }}_{\theta }}({\rho }^{-1}{\mathrm{\partial }}_{\theta })={\rho }^{-2}{\mathrm{\nabla }}_{{\mathrm{\partial }}_{\theta }}{\mathrm{\partial }}_{\theta }={\rho }^{-1}{\mathrm{\partial }}_{\rho }$, and $\eta (a,a)={\rho }^{-2}$ by step 2.1. At $\rho =2$, this is $1\mathrm{/}4$ and $\mathrm{\mid }a\mathrm{\mid }=1\mathrm{/}2$.

Rigidity, irrotationality and the coframe: the $\theta =const$ slices are the portions $\{x^0=\tanh \theta x\}\cap W$ of hyperplanes through the origin, and their tangent space is spanned by ${\mathrm{\partial }}_{\rho },{\mathrm{\partial }}_y,{\mathrm{\partial }}_z$, which are orthogonal to $n$ by the metric of step 2.1; their induced metric is $d{\rho }^2+d{y}^2+d{z}^2$, independent of $\theta$. The rest distribution is therefore integrable, being the tangent distribution of the foliation by these slices; by [L3] its vorticity vanishes and the congruence is irrotational. Moreover $h(n,\cdot )=0$, the rest-basis pairings $h({\mathrm{\partial }}_{\rho },{\mathrm{\partial }}_{\rho })=h({\mathrm{\partial }}_y,{\mathrm{\partial }}_y)=h({\mathrm{\partial }}_z,{\mathrm{\partial }}_z)=1$ are constant, and $[n,{\mathrm{\partial }}_{\rho }]=(1\mathrm{/}\rho )n$ while $[n,{\mathrm{\partial }}_y]=[n,{\mathrm{\partial }}_z]=0$. Thus ${\mathcal{L}}_{n}h(X,Y)=n(h(X,Y))-h([n,X],Y)-h(X,[n,Y])=0$ on that basis and on $n$, so the congruence is Born rigid; equivalently, the metric components on the rest slices are constant along ${\mathrm{\partial }}_{\theta }$, which is the content of the supplier's rigidity computation (Fermi-Walker transport preserves the metric, and explicit rigid and non-rigid observer congruences clause (b)).

Non-coordinate frame: by [L4], $[n,{\mathrm{\partial }}_{\rho }]=[{\rho }^{-1}{\mathrm{\partial }}_{\theta },{\mathrm{\partial }}_{\rho }]={\rho }^{-1}[{\mathrm{\partial }}_{\theta },{\mathrm{\partial }}_{\rho }]-{\mathrm{\partial }}_{\rho }({\rho }^{-1}){\mathrm{\partial }}_{\theta }=0+{\rho }^{-2}{\mathrm{\partial }}_{\theta }={\rho }^{-1}n$, which is nonzero since $n\mathrm{\ne }0$; here $[{\mathrm{\partial }}_{\theta },{\mathrm{\partial }}_{\rho }]=0$ because $\theta ,\rho$ are coordinates. Hence the members of the tetrad field $(n,{\mathrm{\partial }}_{\rho },{\mathrm{\partial }}_y,{\mathrm{\partial }}_z)$ do not commute pairwise, while the members of any coordinate frame do; so this tetrad field is not the coordinate frame of any chart.

Conclusion: steps 1.1-2.1 verify the diffeomorphism and the exact pullback metric; step 3.1 checks that $n$ is a future unit field; step 4.1 computes the acceleration $a={\rho }^{-1}{\mathrm{\partial }}_{\rho }$ with $\eta (a,a)={\rho }^{-2}$; step 4.2 verifies Born rigidity and vanishing vorticity with the explicit rest slices; and step 4.3 exhibits the nonzero commutator showing that the frame is not a coordinate frame. All identities are exact symbolic computations in the hyperbolic functions.