for example, assume there is a high redshift galaxy called "A" with z=6, so how to calculate the distance between us and another galaxy B when the light left the galaxy A?

For example, say B has a redshift of 0.5, and let's assume when the light emitted from A, us and B both exist.

  • $\begingroup$ How far is B from us now? (for example) $\endgroup$ – James K Mar 22 '20 at 22:14
  • $\begingroup$ for example, B has redshift of 0.5, but lets assume when the light emitted from A, us and B both exists $\endgroup$ – Stargazer Mar 22 '20 at 22:35

I'm assuming you're talking about physical distances (as opposed to any of the other distance measures in cosmology).

The comoving distance to a galaxy at redshift $z$ is $$ d_C(z) = \frac{c}{H_0}\int_0^z \frac{dz}{\sqrt{ \Omega_r(1+z)^4 + \Omega_m(1+z)^3 + \Omega_k(1+z)^2 + \Omega_\Lambda }}, $$ where $c$ and $H_0$ are the speed of light and the Hubble parameter, and $\{\Omega_r,\Omega_m,\Omega_k,\Omega_\Lambda\}$ $\simeq$ $\{\lt 10^{-4},0.3,0,0.7\}$ are the radiation, matter, curvature, and dark energy density parameters, respectively.

By definition, the comoving and the physical distance $d_P$ coincide today, but at the time a galaxy at $z$ emitted the light we see today, it was at $$ d_P = \frac{d_C}{1+z}. $$

In your example, $z_A = 6$, and $z_B = 0.5$. The comoving distance to galaxy $B$ is, from the equation above and with a Planck Collaboration et al. (2016) cosmology, $1946\,\mathrm{Mpc}$ away. Hence, when $A$ emitted the light we see today, its physical distance was a factor $(1+6)$ smaller, i.e. $278\,\mathrm{Mpc}$.

In Python, with astropy:

from astropy.cosmology import Planck15
from astropy import units as u
zA,zB      = 6,.5
dC_B       = Planck15.comoving_distance(zB)
dP_B_at_zA = dC_B / (1+zA)
   -> 1945.5612969107992 Mpc
   -> 277.93732813011417 Mpc
print(dP_B_at_zA.to(u.Mlyr)) # convert to light-years
   -> 906.5103217447773 Mlyr
  • $\begingroup$ Thanks for your answer, I am really appercaite your help. However, i still have some quesitons, do I need to know the hubble constant at z=6? to solve this question, can i calculate the recession velocity of B galaxy at z=6, and then calculate the distance by divde the recession velocity of B by hubble constant at z=6? $\endgroup$ – Stargazer Mar 23 '20 at 2:34
  • $\begingroup$ @Stargazer You only need to know the Hubble parameter at $z_A = 6$ if you also want to know the distance to $A$. Your approach is sort of one step forward, and one step back: To calculate the recession velocity of $B$ at $z=6$, you must first calculate its physical distance $d_{P,B}(z=6)$, and then multiply by $H(z=6)$. Then you divide by $H$ again to get back to $d_{P,B}$ $\endgroup$ – pela Mar 23 '20 at 11:27
  • $\begingroup$ @Stargazer Sorry, I wrote a slight mistake/confusion: You don't really need to know the Hubble parameter at $z=6$ to know the distance. But in effect you do calculate it, since that is really what the denominator in the integral is (that square-root is sometimes written $\sqrt{\cdots} \equiv E(z)$, and $H(z) = H_0 E(z)$). $\endgroup$ – pela Mar 23 '20 at 22:15
  • $\begingroup$ Thank you very much, i know it now. $\endgroup$ – Stargazer Mar 24 '20 at 0:28

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