Suppose that I prepare two weights with same mass and a incredibly strong rope, and tether those weights with proper distance about $1.8R_H$, so that each weight is distanced from their center of mass of $0.9R_H$. (Where $R_H$ is the Hubble radius.)
Also, the universe is de Sitter universe, so that the $H$ and $R_H$ are constant.
The rope is so strong, so that it endures the expansion of the universe, and keep the proper distance of two weights to be constant.
In this case, each weight will have peculiar velocity of $-0.9c\;\hat{r}$, while the Hubble flow is $0.9c\;\hat{r}$ in the perspective of the center of mass.
The question is, am I allowed to make such situations? I mean, it sounds terribly unrealistic, but at least there is no physics that prevents such situation, as long as I think.
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1 Answers
The answer to your question is yes.
In a De Sitter universe the rope diameter must be smaller than the Hubble diameter (in a universe with time dependend Hubble parameter it can be larger, as the event horizon in such universes can be larger than the Hubble radius, see here). The De Sitter metric in proper distance coordinates is
$$\rm g_{\mu\nu}= \left[ \begin{array}{cccc} \rm c^2-H^2 \ r^2 \ \ \ & \rm H \ r \\ \\ \rm H \ r & -1 \\ \end{array} \right]$$
or in static diagonal coordinates
$$\rm g_{\mu\nu}= \left[ \begin{array}{cccc} \rm c^2-H^2 \ r^2 & 0 \\ \\ 0 & -1/(\rm 1-H^2 \ r^2 /c^2) \\ \end{array} \right]$$
and the proper acceleration acting on a rope which is stationary with respect to the coordinate origin is
$$\rm |a|= \surd \ |\sum_{\alpha, \beta} g_{\alpha\beta} \ a^{\alpha} \ a^{\beta} \ |$$ where $\rm a^{\mu}= d^2 x^{\mu}/d \tau^2+\sum_{\alpha, \beta} \ \Gamma^{\mu}_{\alpha \beta} \ u^{\alpha} \ u^{\beta}$ and $\rm u^{\mu}=d x^{\mu}/d \tau=\{ c/ \surd g_{tt}, \ 0\}$, so for the 4-acceleration vector of a static $\rm d^2 x^{\mu}/d \tau^2=\{0, \ 0 \}$ rope we get
$$\rm a^{\mu}=\left\{\frac{H^3 \ r^2}{c^2-H^2 \ r^2}, -H^2 \ r\right\}$$ in proper distance coordinates, or in static diagonal coordinates $$\rm a^{\mu}=\left\{0, -H^2 \ r\right\}$$ which gives the proper acceleration (in both coordinates, since it is an invariant) $$\rm |a|= \frac{c \ H^2 \ r }{\sqrt{c^2-H^2 \ r^2}}$$ which gets infinite at the Hubble radius $\rm r=c/H$, so the length of the rope connected to a comoving observer at $\rm r=0$ must be shorter than that in both directions.
A different comoving observer at proper distance $\rm r>0$ along the rope will of course measure that the rope has a local peculiar velocity $\rm v=-H \ r$ relative to himself.
The static length of the rope is $\rm R=\rm \int_0^r dr/\surd | g^{rr} |$ by the way (the contravariant $\rm g^{rr}=-g_{tt}/c^2$ is the same in both coordinates), so in the limit of $\rm r \to c/H$ that would be $\rm R \to c/H \cdot \pi/2$.
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