# What does cosmological redshift say about speed of separation and why?

I've been thinking about cosmological redshift lately and in particular its effect on the apparent speed of separation. In particular, let's imagine a source which emits light pulses with a period of $$\tau_e=1$$ to a source a distance $$D$$ away at present. If a pulse were emitted at some time $$T$$ then the distance between the source and the observer at the time this pulse is observed (let's call this time $$x_T$$) would be $$\int_{T}^{x_T} \frac{a(x_T)}{a(t)}cdt=Da(x_T)$$. Now, if we let $$a(t)=nt$$ (for simplicity, let's consider at least the case where the rate of change of the scale factor is constant), evaluate the integral, and solve for $$x_T$$, we get $$x_T=Te^\frac{Dn}{c}$$. The observed period of the pulses would be simply the time elapsed after the arrival of one pulse and before the arrival of the next one, so $$\tau_o=x_T-x_{T-1}=e^\frac{Dn}{c}$$, and hence $$\lambda_o=e^\frac{Dn}{c}$$c. As we can see, this expression depends only on the initial distance between the source and the observer and the rate of change of the scale factor - not on the speed of separation between the source and the observer.

Intuitively, then, it would appear to me that redshift can't be a measure of speed of separation; if anything, it might be a better measure of distance, as it is roughly indicative of the amount of time that it took the incoming light to reach the observer - in that its wavelength would have been stretching all this time.

The only relationship between $$z(t)$$ and $$d'(t_0)$$ that I can imagine is that (taking $$a(t_0)=1$$) $$\frac{da}{dt}≈\frac{\Delta a}{\Delta t}=\frac{z(t_0)}{(z(t_0)+1)(t_0-t_1)}$$, with $$z(t_0)+1=\frac{1}{a(t_1)}$$, but this only gives us an estimate for the Hubble constant and not speed of separation. Furthermore, astronomers don't seem to make use of this approximation when attempting to measure the Hubble constant, leading me to believe that it's not very useful (bonus question: why is such an approximation not useful?).

I've seen it being referenced that $$z(t)≈\frac{d'(t)}{c}$$ is a good approximation for small distances, but I fail to see why that must be the case.

Can anyone explain in what way cosmological distance is intuitively related to speed of separation and in particular where the approximation $$z(t)≈\frac{d'(t)}{c}$$ comes from?

• You should take a look at the classic paper by Tamara Davis and Charlie Lineweaver, Expanding Confusion: common misconceptions of cosmological horizons and the superluminal expansion of the Universe, there's also their related article from Scientific American. – PM 2Ring May 9 '20 at 6:08
• I've deleted comments that were not focused on clarifying the question or suggesting improvements. Please keep comments focused on improving the question. Anything else should be done in chat. Thanks! – tpg2114 May 9 '20 at 17:32
• @tpg2114 I think there was a comment by safesphere which pointed out an error in my question. In my opinion, you should've kept that one. – Max May 9 '20 at 20:55

## 2 Answers

Let us suppose there a celestial object with a proper distance $$D(z)$$ and have a measured velocity $$v(z)$$. By using Hubble's Law we can write

$$v(z) \equiv H_0D(z) = H_0\chi(z)~~(1)$$ for $$a(t_0) = 1$$

Now let me take the FLWR metric in the form of

$$ds^2 = -c^2dt^2 + a^2(t)[d\chi^2 + S_{\kappa}^2(\chi)d\Omega^2]$$

To calculate the comoving distance $$\chi$$ between two points, we can set $$d\Omega = ds = 0$$ for the path of a photon

So we have,

$$\int_{t_e}^{t_0}d\chi = c\int_{t_e}^{t_0}\frac{dt}{a(t)}$$

By using $$1+z = a(t)^{-1}$$ we can transform the above integral into

$$\chi(z) = \frac{c}{H_0}\int_{0}^{z}\frac{dz}{E(z)}~~(2)$$

For $$E(z) = \sqrt{\Omega_{r,0}(1+z)^4 + \Omega_{m,0}(1+z)^3 + \Omega_{\Lambda,0} + \Omega_{\kappa}(1+z)^2}$$

where $$\Omega_{\kappa} = 1 - \Omega_0 = 1 - \Omega_{r,0} - \Omega_{m,0} - \Omega_{\Lambda,0}$$

If you insert (2) into (1) we obtain

$$v(z) = c\int_{0}^{z}\frac{dz}{E(z)}~~(3)$$

Assume that z is small. This implies

$$\lim_{z \rightarrow 0} {E(z)} =\sqrt{\Omega_{r,0} + \Omega_{m,0} + \Omega_{\Lambda,0} + [1 - \Omega_{r,0} - \Omega_{m,0} - \Omega_{\Lambda,0}]} = 1$$

Thus we obtain $$v(z) = cz$$

But when $$z$$ is not small this approximation does not work. At this point, you can ask which $$z$$ is considered small.

For $$z<0.01$$ you can use $$v=cz$$ equation.

In general, astronomers expand the integral ($$3$$) and write

$$$$v(z) = \frac{cz}{1+z}[1+\frac{1}{2}(1-q_0)z - \frac{1}{6}(1-q_0-3q_0^2+j_0)z^2]$$$$

with $$q_0 = -0.55$$ and $$j_0 = 1.0$$ for the $$\Lambda CDM$$ model ($$\Omega_m = 0.3$$, $$\Omega_{\Lambda} = 0.7$$).

This approximation works for $$z < 0.3$$. When you have larger $$z$$ values its best to use the integral form of the equation.

• Okay, thanks for your answer, but unfortunately I am not familiar with Friedmann's equations, and so I still fail to see the intuition behind the formula that you derived. Could you explain the derivation a bit more intuitively, ideally without appealing to Friedmann's equations? – Max May 9 '20 at 22:00
• @Max Well you can think that in small distances (which corresponds to small $z$) the expansion of the universe becomes negligible (The case of $E(z) = 1$). This approximation allows you to write $v=cz$. However, if you go further distances the expansion of the universe becomes important and you can no longer assume $v=cz$ (because in this case $E(z) \ne 1$) – Layla May 10 '20 at 18:30
• @Max Well, there are two things. The first is the expansion of the universe. In this case, objects recede from us, also called "the recession velocity". This is via the Hubble expansion. There's also another type of velocity called "the peculiar velocity". In general, the objects total velocity is the combination of these two effects. Mathematically its written as, $$v_{tot} = v_{r} + v_{p}$$ – Layla May 10 '20 at 18:43
• Okay, thanks for the explanation. To be honest, I still don't quite follow your integral here. Based on my calculation, $\chi(z) = \int_{0}^{z} \frac{1+z}{\frac{dz}{dt}}dz$, but I don't see how we could calculate $\frac{dz}{dt}$ in a meaningful way. Even if expansion of the universe is negligible, $\frac{dz}{dt}$ still equals $a'(t)$, which is a dead end. Could you explain this bit to me? – Max May 10 '20 at 22:26
• Also, if that's possible, can you please explain how my calculation in op is wrong? I assumed that rate of change of the scale factor was constant and got that $z=ce^{Dn/c}-1$. Why is this result wrong? – Max May 10 '20 at 22:32

In your post, you ask a number of questions. Although your questions are valid, some of your questions seem to stem from a few misunderstandings. My response is not a direct answer to every single one of your questions, but it may clear things up for you.

If you don't mind, I'd like to tidy up your questions a bit:

1. What exactly is the relationship between a celestial object's redshift and it's distance from Earth?
2. While we're at it, how does a celestial object's redshift relate to the velocity with which it is moving away from us?
3. How does the "time of light travel" between the celestial object and the Earth figure in to all of this?

Before I answer these questions in earnest, it's important to know that the basic Hubble constant stuff assumes that the celestial objects aren't accelerating - they're moving at a constant rate. The question as to whether or not stars and such are accelerating away from us is an entirely different question. I'll assume that all of the "moving away from us" is due to ordinary expansion of the universe. Whether or not the expansion of the universe is accelerating (and the redshift increasing) is another topic for another day.

We can answer questions (1.) and (2.) with the basic Hubble equation which tells us that:

$$R = D \times H$$ where D is the distance the celestial object is from Earth, H is the hubble constant, and R is the amount of redshift. Based on what you stated in your question, I'm sure you have encountered this formula.

Simply put, the further a star is from the Earth, the greater it's redshift. It can be implied from all this that the stars further from Earth (provided they're outside our galaxy) are moving away from us faster, the further away from us they are in the first place.

Lastly, you are basically correct in your understanding that the further a star is from the Earth, the longer it takes the light to get here. The simplest response I can give to this is to say that astrophysicists are aware of this, and they go in and correct for it.

I suppose the distance traveled is again related to the redshift (see Hubble's law above) but if need be, astronomers can go in and correct for all this madness.

Sorry if I haven't addressed every facet of your question - this was just intended to clarify a few things so that you might be on your way to a better understanding.

• @RobJeffries "Cosmological redshift does not depend on the speed of an object" - Here you apparently imply the peculiar velocity. However, if the object is stationary relative to the Hubble flow. it still recedes away from us with a certain speed and its redshift certainly depends on this speed. – safesphere May 9 '20 at 8:09
• @safesphere and that is where many misconceptions start. e.g. faster than light travel? – ProfRob May 9 '20 at 8:14
• @RobJeffries Not faster than light, only faster than the local speed of light. If you look up at Alpha Centauri, you would see it rotating around the Earth at 10,000 times the speed of light, but of course its light still outruns it :) – safesphere May 9 '20 at 8:33
• @RobJeffries -- thank you, I made an appropriate correction to my answer. – the_photon May 9 '20 at 14:21