general relativity - A Cosmological horizon at the Hubble radius?

I have calculated that if one extends a rigid ruler into space by a fixed proper distance $D$ then a clock at the end of the ruler, running on proper time $\tau$, will run more slowly than time $t$ at the origin by a time dilation factor:

$$\frac{dt}{d\tau} = \frac{1}{\sqrt{1 - H^2 D^2 / c^2}}$$

where $H$ is the Hubble parameter.

If one substitutes in Hubble's law, $v = H D$ (the theoretical law which is exact), one finds the following satisfying result that

$$\frac{dt}{d\tau} = \frac{1}{\sqrt{1 - v^2 / c^2}}.$$

Although this looks like a result from special relativity I derived it by combining the FRW metric line element and an equation for the path of the end of the ruler, $\chi = D / R(t)$, where $\chi$ is the radial comoving coordinate of the end of the ruler, $D$ is a fixed proper distance and $R(t)$ is the scalefactor.

Does this prove that there is a cosmological horizon at the Hubble radius $D=c/H$ where the proper time $\tau$ slows down to a stop compared to our time $t$?

This seems to be a general result which is true regardless of cosmological model.

Details of Calculation

The general FRW metric is given by:

$$ds^2 = -c^2 dt^2 + R(t)^2\left[d\chi^2+S^2_k(\chi)d\psi^2 \right]$$

where $d\psi^2 = d\theta^2 + \sin^2 \theta d\phi^2$ and $S_k(\chi)=\sin \chi$,$\chi$ or $\sinh \chi$ for closed ($k=+1$), flat ($k=0$) or open ($k=-1$) universes respectively. The scale factor $R(t)$ has units of length.

Consider a ruler of fixed proper length $D$ extending out radially from our position at the origin. The path of the far end of the ruler in comoving coordinates is

$$\chi = \frac{D}{R(t)}$$

Differentiating this equation by proper time $\tau$ gives us:

$$\frac{d\chi}{d\tau} = - \frac{D}{R^2} \frac{dR}{dt} \frac{dt}{d\tau}.$$

Using the FRW metric we can find a differential equation for the path of the far end of the ruler. We substitute in $ds^2=-c^2d\tau^2$ (end of ruler has a time-like path), $d\psi=0$ (the ruler is radial) and divide through by $d\tau^2$ to obtain:

$$c^2\left(\frac{dt}{d\tau}\right)^2 - R(t)^2 \left(\frac{d\chi}{d\tau}\right)^2 = c^2.$$

Substituting the expression for $d\chi/d\tau$ into the above equation we find:

$$c^2\left(\frac{dt}{d\tau}\right)^2 - D^2 \left(\frac{\dot R}{R}\right)^2 \left(\frac{dt}{d\tau}\right)^2 = c^2.$$

Using the definition of the Hubble parameter $H=\dot{R}/R$ we finally obtain:

$$\frac{dt}{d\tau} = \frac{1}{\sqrt{1 - H^2 D^2 / c^2}}.$$