# Is Hubble's Law Only Valid Nearby?

I am looking this paper by LIGO, and it claims at the very beginning,

At nearby distances $$(d \lesssim 50 \text{ Mpc})$$ [the Hubble constant] is well approximated by the expression $$v_H = H_0d, \tag{1}$$ where $$v_H$$ is the local “Hubble flow” velocity of a source, and $$d$$ is the distance to the source.

Are they implying that Eq. (1) is not valid at larger distances, and if so, why is this true? My cosmology book (Ryden) defines $$H_0$$ as $$\dot{a}/a$$, where $$a$$ is the scale factor, so it seems like Eq. (1) above must be true at all distances by that definition.

• Ryden is wrong. $H_0 = \dot{a}_0/a_0$. – Rob Jeffries May 8 at 7:03

You need to think carefully about what is meant by $$v$$ and $$d$$ in your equation.
There are various ways of measuring and expressing distances in cosmology. The gravitational wave "standard siren" technique described in the paper referred to actually gives a "luminosity distance" to the source, which in an expanding universe will be larger than the proper distance by a factor of $$\sim (1+z)$$.
For the paper in question, they argue that given the other uncertainties (the GW standard-siren distance is uncertain by $$\sim 10$$%) that this difference is unimportant for $$d<50$$ Mpc because $$z < 0.012$$. At larger distances and higher redshifts then a plot of $$v$$ vs luminosity distance would be a curve, rather than a straight line, that depended on the cosmological parameters $$\Omega_m$$, $$\Lambda$$ etc.