Question from Introduction to Modern Cosmology by Andrew Liddle

Question

The exact question goes like this: In the real Universe the expansion is not completely uniform. Rather, galaxies exhibit some random motion relative to the overall Hubble expansion, known as their peculiar velocity and caused by the gravitational pull of their near neighbours. Supposing that a typical (e.g. root mean square) galaxy peculiar velocity is 600 km s-1, how far away would a galaxy have to be before it could be used to determine the Hubble constant to ten per cent accuracy, supposing

(a) The true value of the Hubble constant is 100 km s-1 Mpc-1?

(b) The true value of the Hubble constant is 50 km s-1 Mpc-1

Assume in your calculation that the galaxy distance and redshift could be measured exactly. Unfortunately, that is not true of real observations.

I started off by writing the hubbles equation and as they mentioned it should be accurate to 10% so its peculiar velocity (given as about 600 km/s) is less than or equal to 10% of its expansion velocity. But this gives me different answer.

The actual answer to the problem as stated in the book is Slightly different answers are possible depending how you deal with the rms velocity. You should get something like r > 35 Mpc for H0 = 100km s-1 and r > 70Mpc for Ho = 50kms-1

What am I missing here? Has it got to do with the RMS velocity as mentioned? Please explain.

Answer 1

From the definition of the rms (e.g. here), $$ \mathrm{rms}^2(x) = \langle x \rangle^2 + \sigma_x^2, $$ where $\langle x \rangle$ is the mean value, and $\sigma_x$ is the dispersion.

For galaxies with random velocities, the mean velocity should be $\langle x \rangle = 0$, unless they drift in some direction. Hence, the velocity dispersion should be $600\,\mathrm{km}\,\mathrm{s}^{-1}$.

For randomly distributed velocities, the speed $V_{||}$ you measure along your line of sight (LOS) is a factor $\sqrt{3}$ smaller, i.e. $\sim350\,\mathrm{km}\,\mathrm{s}^{-1}$.

Thus, to get a 10% accuracy, the mean value should be 10 times larger, i.e. $\sim 3500\,\mathrm{km}\,\mathrm{s}^{-1}$.

For a Hubble constant of $H_0 = 100\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$, that corresponds to a distance of $$ d = \frac{v_{||}}{H_0} \simeq 35\,\mathrm{Mpc}, $$ while for $50 \,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$ you get $70\,\mathrm{Mpc}$.