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

Physics Asked on February 14, 2021

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:
$$
r=cintfrac{dt}{a(t)}
$$

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

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:
$$
2(14.7cdot t_1)^{1/3}-t_1^{1/3} = t_3^{1/3}
$$

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

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

One Answer

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.

Answered by benrg on February 14, 2021

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