I'm trying to determine the redshift observed for a light signal between 2 observers (let's call them observers one and two) in a flat matter-dominated universe. Observer one sends a signal ($t_1$) at light speed to observer 2 and she records it as a z of 5 ($t_2$). Observer two then sends a signal directly back. At what redshift does observer one see this incoming signal ($t_3$)?

We know the scale factor for a matter-dominated universe is as follows: $$ \frac{a}{a_0}=\frac{{t^{2/3}}}{t_0} $$

and $a_0$ and $t_0$ are $1$ for the present time

Additionally, the comoving coordinate will stay the same for both time intervals by definintion and can be described as follows: $$ r=c\int\frac{dt}{a(t)} $$

Using the relation between the scale factor and z, we can find the following relationship: $$ a(t_2) = 6\cdot a(t_1) $$

Describe the comoving coordinates for the time intervals and set them equal to each other: $$ \int_{t_1}^{t_2}\frac{dt}{{t}^{\frac{2}{3}}} = \int_{t_2}^{t_3}\frac{dt}{{t}^{\frac{2}{3}}} $$

After integration, substitute the relation found above in terms of t to get the ratio between $t_1$ and $t_3$: $$ t_2 = 6^{\frac{3}{2}}\cdot t_1 $$ ...to get: $$ 2(14.7\cdot t_1)^{1/3}-t_1^{1/3} = t_3^{1/3} $$

Lastly, now that we have a relationship between the initial signal emitted ($t_1$) and the final signal received ($t_3$), we can determine $z$: $$ \frac{a(t_3)}{a(t_1)}=1+z=\frac{t_3^{\frac{2}{3}}}{\left [ \frac{t_3}{59} \right ]} \therefore z\approx 14 $$

Above is the work I did but I'm not confident my answer is correct although I don't see anything glaring wrong with what I did. I expect the redshift to be greater which is consistent with the redshift calculated, but I don't have much greater intuition that this. Perhaps there is a more elegant way to solve it. I'd appreciate if someone could let me know if the solution seems sound.



1 Answer 1


Your derivation looks fine. I don't have any substantially easier way to solve the problem but maybe this will provide some insight. Your experiment is equivalent to one in which 2 sends the signal to 3 who is the same comoving distance away as 1 but in the opposite direction. That in turn is equivalent to 2 not receiving or sending a signal at all, but simply letting it pass from 1 to 3. So the problem boils down to finding the redshift of the signal as a function of comoving distance traveled, and looking at what happens to the redshift when you double the distance.

You have $$Δr_{ab} = c \int_{t_a}^{t_b} \frac{dt}{a(t)} = K \, (t_b^{1/3} - t_a^{1/3}) = K \, t_a^{1/3} \, (\sqrt{1{+}z_{ab}}-1)$$ (using the fact that $1{+}z_{ab} = a(t_b)/a(t_a) = (t_b/t_a)^{2/3}$), for some constant $K$. Therefore, for a fixed starting time, $$Δr_{ab} \propto \sqrt{1{+}z_{ab}}-1.$$

Then you solve $\sqrt{1{+}z'}-1 = 2(\sqrt{1{+}z} - 1)$ for $z'$ and plug in $z=5$. I get $z'\approx 14$.

Note that you're calculating the redshift from 1 to 3 here, or in the original problem the redshift from 1 back to 1 under the assumption that 2 sends a signal of the same frequency she received (or just uses a mirror). If you calculate the redshift from 2 to 1 instead, it will be smaller than the redshift from 1 to 2, because the expansion has slowed down and less relative speed means less redshift.

  • $\begingroup$ Thanks for sharing your solution. Glad we got a similar answer. To clarify, you don't see anything wrong with the steps I've taken? $\endgroup$
    – Astroturf
    Commented Sep 18, 2020 at 14:04
  • $\begingroup$ Can you clarify how you substituted for redshift after you integrated? That isn't obvious to me, thanks. $\endgroup$
    – Astroturf
    Commented Sep 18, 2020 at 14:07
  • 1
    $\begingroup$ @Astroturf Your derivation looks fine. I added some clarifying text. $\endgroup$
    – benrg
    Commented Sep 18, 2020 at 14:16
  • $\begingroup$ Apparently my answer was wrong according to the professor. So I guess we are both wrong... $\endgroup$
    – Astroturf
    Commented Sep 23, 2020 at 2:01
  • $\begingroup$ @Astroturf So what's the correct answer? $\endgroup$
    – benrg
    Commented Sep 23, 2020 at 2:06

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