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.



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 Sep 18 '20 at 14:04
  • $\begingroup$ Can you clarify how you substituted for redshift after you integrated? That isn't obvious to me, thanks. $\endgroup$ – Astroturf Sep 18 '20 at 14:07
  • 1
    $\begingroup$ @Astroturf Your derivation looks fine. I added some clarifying text. $\endgroup$ – benrg Sep 18 '20 at 14:16
  • $\begingroup$ Apparently my answer was wrong according to the professor. So I guess we are both wrong... $\endgroup$ – Astroturf Sep 23 '20 at 2:01
  • $\begingroup$ @Astroturf So what's the correct answer? $\endgroup$ – benrg Sep 23 '20 at 2:06

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.