Rindler and rigid rotating bath domains

Rindler right wedge. Let $a_0>0$ be an acceleration in $\mathrm{m}{\mathrm{s}}^{-2}$, use coordinates $(x^0,x,y,z)$ with $x^0=ct$, and take the open domain $x>0$ with metric

$d{s}^2=-{\left(\frac{a_{0}x}{c^2}\right)}^2(d{x}^0{)}^2+d{x}^2+d{y}^2+d{z}^2\mathrm{.}$

The selected Killing field ${\xi }_R={\mathrm{\partial }}_{x^0}$ has lapse $N_R=a_{0}x\mathrm{/}{c}^2$ and observer acceleration $a_x=1\mathrm{/}x$, so the physical proper-acceleration magnitude is $c^2\mathrm{/}x$. The selected stationary-bath model gives

$T_R(x)=\frac{T_K{c}^2}{a_{0}x}\mathrm{.}$

At the reference point $x_{\mathrm{r}\mathrm{e}\mathrm{f}}=c^2\mathrm{/}{a}_0$, $N_R=1$ and $T_R=T_K$. Any $T_K>0$ is allowed by this classical geometric model; the metric and temperature relation do not select the Unruh value.

Rigid rotation in Minkowski spacetime. In cylindrical coordinates let

$d{s}^2=-(d{x}^0{)}^2+d{r}^2+r^{2}d{\varphi }^2+d{z}^2,{\xi }_{\mathrm{\Omega }}={\mathrm{\partial }}_{x^0}+\frac{\mathrm{\Omega }}{c}{\mathrm{\partial }}_{\varphi },$

where $\mathrm{\Omega }$ is a constant angular velocity in ${\mathrm{s}}^{-1}$. For $\mathrm{\Omega }\mathrm{\ne }0$, use the annular domain $0<r<c\mathrm{/}\mathrm{\mid }\mathrm{\Omega }\mathrm{\mid }$; for $\mathrm{\Omega }=0$, ${\xi }_{\mathrm{\Omega }}={\mathrm{\partial }}_{x^0}$ is timelike throughout Minkowski space. On the rotating domain,

$N_{\mathrm{\Omega }}=\sqrt{1-\frac{{\mathrm{\Omega }}^2{r}^2}{c^2}},T_{\mathrm{\Omega }}(r)=\frac{T_K}{N_{\mathrm{\Omega }}}\mathrm{.}$

The radial acceleration component is $a_r=-{\mathrm{\Omega }}^{2}r\mathrm{/}[c^2(1-{\mathrm{\Omega }}^2{r}^2\mathrm{/}{c}^2)]$, directed inward, with physical magnitude ${\gamma }^2{\mathrm{\Omega }}^{2}r$, where $\gamma =(1-{\mathrm{\Omega }}^2{r}^2\mathrm{/}{c}^2{)}^{-1\mathrm{/}2}$. At the light cylinder $r=c\mathrm{/}\mathrm{\mid }\mathrm{\Omega }\mathrm{\mid }$, ${\xi }_{\mathrm{\Omega }}$ becomes null and the normalized stationary observer does not exist. A differential-rotation field ${\mathrm{\partial }}_{x^0}+[\mathrm{\Omega }(r)\mathrm{/}c]{\mathrm{\partial }}_{\varphi }$ is not Killing when ${\mathrm{\Omega }}^{\mathrm{\prime }}(r)\mathrm{\ne }0$: its metric Lie derivative has $({\mathcal{L}}_{\xi }g{)}_{r\varphi }=r^2{\mathrm{\Omega }}^{\mathrm{\prime }}(r)\mathrm{/}c$.

One-mode trace check. If an additional model specifies a single bosonic mode with lab Hamiltonian $H=ϵ\mathcal{N}$ and angular momentum $J=\mathrm{\hslash }j\mathcal{N}$, then its rotating generator is $H-\mathrm{\Omega }J$. Its Gibbs trace is finite exactly when $ϵ-\mathrm{\hslash }j\mathrm{\Omega }>0$. This is the separate one-mode spectral condition of Rotating mode Gibbs obstruction; it does not prove a continuum-field trace obstruction or describe every rotating equilibrium state.

The left panel uses dimensionless Minkowski coordinates $(X\mathrm{/}\mathrm{\ell },X^0\mathrm{/}\mathrm{\ell })$ for a fixed length $\mathrm{\ell }>0$ and the exact Rindler embedding

$X=x\cosh (a_0{x}^0\mathrm{/}{c}^2),X^0=x\sinh (a_0{x}^0\mathrm{/}{c}^2)\mathrm{.}$

It maps $x>0$ onto the right wedge $X>\mathrm{\mid }{X}^0\mathrm{\mid }$; the red curve is the constant-Rindler-position orbit $x=\mathrm{\ell }$, which obeys $X^2-(X^0{)}^2={\mathrm{\ell }}^2$. The right panel is a fixed-$z$ cross-section of the $\mathrm{\Omega }\mathrm{\ne }0$ rotating timelike domain. Its dashed circle is the excluded null boundary $r=c\mathrm{/}\mathrm{\mid }\mathrm{\Omega }\mathrm{\mid }$, and the shaded open disk is where $g({\xi }_{\mathrm{\Omega }},{\xi }_{\mathrm{\Omega }})<0$. The sketches show domains only; every lapse, acceleration, and temperature claim is established by the coordinate calculations in Verification.