# What's the redshift of a return signal in a flat universe between two observers?

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.

Thanks

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.

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