Does the expansion of the universe stretch distances in non-radial/angular directions?

Question

At time $t = t_0$ a galaxy is situated at a proper distance $r_0$ and has a proper diameter $D_0$. Setting $a(t_0) = 1$ the proper distance to the galaxy will then of course evolve with the expansion of the universe as:

$$r = a\chi = \frac{\chi}{1+z} = \frac{r_0}{1+z}$$

What happens to the proper diameter of the galaxy as the universe expands? Will it also scale as:

$$D = \frac{D_0}{1+z}$$

like the proper radial distance does?

And if this is true does this mean that the angle subtended by the galaxy as seen by an observer at $r = 0$ is constant as:

$$\Delta\theta = \frac{D_{em}}{r_{em}} = \frac{D_0}{r_0}$$

where 'em' designates distances at the time when the photons were emitted that are received by the observer at time $t$.

Answer 1

FLRW metric can be written as,

$$ds^2 = -c^2dt^2 + a(t)^2[dr^2 + S_{\kappa}(r)^2d\Omega^2]$$

In the calculations of the angular diameter distance, we set $dt = dr = d\phi =0$ which leads to

$$ds = a(t_e)S_{\kappa}(r)d\theta$$

If the object has a diameter $D$ then we can write. $$D = a(t_e)S_{\kappa}(r)d\theta$$

or

$$D = \frac{S_{\kappa}(r)d\theta} {1+z}$$.

I think it's not right to say the diameter of the object decreases. I think only angular size ($d\theta$) changes.

If I put a $1m$ ruler in some distance and measure its angular size, and If I move it further away, Its length will not change. However, the angular size will get smaller since its getting further away.