In measurements concerning distances, luminosity, etc. to evaluate the Hubble's parameter, do scientists take into considerations the fact that the light, emitted from a star, and received by a detector, is actually not representing the actual distance of the star? Thus when we say the current value of the Hubble's parameter is wrong, in a sense, it reflects measurements obtained from past events.

For example, if we denote (D), (O) for detector and object respectively,

(D)----$d_0$-----(O) -perceived

(D)-----$d$---------(O) -actual

then the actual distance $d=d_0(1+v/c)$.


The answer is yes. Hubble's law relates the "velocity of recession" to the proper distance - which is the distance to the other galaxy now.

That is not usually the distance we measure, as you have identified.

The difference between the two is very small at low redshifts, but becomes larger at higher redshifts, and depends on the expansion history of the universe and hence the cosmological parameters like $\Omega_M$, $\Omega_\Lambda$ as well as $H_0$.

| cite | improve this answer | |
  • $\begingroup$ So, In a perfect theoretical cosmological model, if we use $t_{age}$ (the age of the universe), then $H(t_{age})$ should be different than $H_0 \approx 70$. $\endgroup$ – PseudoYousef Jun 22 at 12:44
  • $\begingroup$ Yes @PseudoYousef $H_0$ means the Hubble parameter now. It varies with cosmic epoch. $\endgroup$ – Rob Jeffries Jun 22 at 13:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.