<p>I think the main reason the answer doesn't come out correct is that the simple Hubble's law equation: </p>

<p><span class="math-container">$v = dH_{0}$</span></p>

<p>Is only applicable in the local universe (Hence <span class="math-container">$H_{0}$</span>, the value of the Hubble constant today). The rate of expansion of the Universe (the Hubble parameter) is determined by the composition of its energy density. Today, the composition is about 30% matter and 70% dark energy. In the recent past the energy density was dominated by matter. In the more distant past the energy density was dominated by relativistic particles (photons and neutrinos). The rate of expansion of the universe is different in each of these "epochs". In general the universe used to expand more rapidly (expansion has slowed thanks to the effects of matter), but distances also used to be smaller (the universe itself was smaller). Combine all of this together and you have a very complicated relationship between distance and the recessional velocity we observe today. This relationship does not have a simple functional form.</p>

<p>I think the easiest question to ask is: "How far has a photon traveled such that if it were emitted at the beginning of the universe it would reach me today?". The answer to this question is the "cosmological horizon", which we commonly talk about as being the size of the observable universe. To answer this question you need to use the spacetime metric which applies for our universe (Called the Friedmann-Robertson-Walker-Lemaitre metric). This is basically the relationship between speed, time, and distance in the realm of general relativity. If our universe is flat (which we think it must be) this equation is actually quite simple, but unfortunately it requires that you know how the universe has been expanding over its lifetime. As I mentioned this relationship has no closed form equation or number, which is why you will never see a simple equation for the solution you are looking for. You need to integrate the solution numerically. <a href="https://en.wikipedia.org/wiki/Particle_horizon" rel="nofollow noreferrer">Wikipedia</a> has a good little article on the horizon.</p>

<p>If you're interested, you'd write down the FLRW metric for a photon travelling in a "straight path":</p>

<p><span class="math-container">$ cdt = \frac{a(t) dr}{ \sqrt{1-kr^2} } $</span></p>

<p>Where <span class="math-container">$a(t)$</span> is the scale factor of the universe, which tells you about its expansion history. This is the complicated function. <span class="math-container">$k$</span> is the curvature constant, which for a flat universe is 0. You then get:</p>

<p><span class="math-container">$ \int^{t_{0}}_{0} \frac{c dt}{a(t)} = \int^{r_{H}}_{0} dr = r_{H} $</span></p>

<p>Here <span class="math-container">$t_{0}$</span> is the age of the universe and <span class="math-container">$r_{H}$</span> is the particle horizon you're interested in. To figure out <span class="math-container">$a(t)$</span> you need the <a href="https://en.wikipedia.org/wiki/Friedmann_equations" rel="nofollow noreferrer">Friedmann equations</a>, which are functions of the different energy densities (dark energy, matter, relativistic particles, curvature), and the rates at which they change in time. I used astropy to do the integration and got <span class="math-container">$\approx 3.35 c t_{0}$</span> which gives about 14 Mpc, so close enough to the answer that you could attribute any differences to a difference in measured parameters that go into solving the integral.</p>

<p>I'd recommend Introduction to Cosmology by Ryden as a textbook if you really want to get into it.</p>

CC BY-SA 4.0
2020-05-25T22:40:15.487Z
2020-05-25T22:40:15.487Z
astrolane
<p>I read this post</p>

<p><a href="https://astronomy.stackexchange.com/questions/26979/does-the-edge-of-the-universe-travel-faster-than-the-speed-of-light">Does the edge of the Universe travel faster than the speed of light?</a></p>

<p>but there was no calculation, so I tried it thusly:</p>

<p>The observable universe is 28.5 Gpc (gigaparsecs)
= 28500 Mpc (megaparsec)</p>

<p><a href="https://www.google.com/search?q=diameter+of+the+observable+universe&amp;oq=diameter+of+the+observable+universe&amp;aqs=chrome..69i57.6887j0j7&amp;sourceid=chrome&amp;ie=UTF-8" rel="nofollow noreferrer">https://www.google.com/search?q=diameter+of+the+observable+universe&amp;oq=diameter+of+the+observable+universe&amp;aqs=chrome..69i57.6887j0j7&amp;sourceid=chrome&amp;ie=UTF-8</a></p>

<p>The Hubble constant is 67.4 or 70 km/s/megaparsec.</p>

<p><a href="https://www.nasa.gov/feature/goddard/2019/new-hubble-constant-measurement-adds-to-mystery-of-universe-s-expansion-rate" rel="nofollow noreferrer">https://www.nasa.gov/feature/goddard/2019/new-hubble-constant-measurement-adds-to-mystery-of-universe-s-expansion-rate</a></p>

<p>I’ll use the older value 67.4</p>

<p>v=H x D = 28500 Mpc * 67.4 km/s Mpc = 1,920,000 km/s = 1.9 x 10^9 m/s</p>

<p>speed of light 2.998 x 10^8 m/s</p>

<p>So the velocity of expansion at the edge of the universe is >6x greater than the speed of light?</p>

<p>I read these posts</p>

<p><a href="https://physics.stackexchange.com/questions/107748/how-are-galaxies-receding-faster-than-light-visible-to-observers">https://physics.stackexchange.com/questions/107748/how-are-galaxies-receding-faster-than-light-visible-to-observers</a></p>

<p><a href="https://astronomy.stackexchange.com/questions/25954/so-where-are-these-measurements-of-galaxies-moving-faster-than-light">So where are these measurements of galaxies moving faster than light?</a></p>

<p>and this</p>

<p><a href="https://arxiv.org/pdf/astro-ph/0011070v2.pdf" rel="nofollow noreferrer">https://arxiv.org/pdf/astro-ph/0011070v2.pdf</a></p>

<p>From the arxiv paper, I get that galaxies can be receding at 3c based on GR, but my simple calculation comes up with 6c.</p>

<p>Can someone help me reconcile my curiosity with the facts?</p>

CC BY-SA 4.0
2020-05-20T16:03:19.347Z
Many mistakes. Please check your units. Please check the size of the observable universe.
ProfRob
CC BY-SA 4.0
2020-05-21T06:11:53.887Z
From your link, the Universe is 93 billion ly; this is 28500 Mpc, not 28.5 as you've started in your calculation.Try again.
Jim421616
CC BY-SA 4.0
2020-05-25T13:02:02.723Z
You're completely correct, except that you used the diameter rather than the radius of the observable Universe (we can always argue over the exact value — I think using the latest [Planck (2018)](https://ui.adsabs.harvard.edu/abs/2018arXiv180706205P) parameters, the radius would be some 14.4 Gpc.
pela
<observable-universe><velocity><size>
can we calculate the velocity of galaxies at the edge of the observable universe?
aquagremlin
# can we calculate the velocity of galaxies at the edge of the observable universe?

(Asked by: username_0 on 2020-05-20)

I read this post

[Does the edge of the Universe travel faster than the speed of light?](https://astronomy.stackexchange.com/questions/26979/does-the-edge-of-the-universe-travel-faster-than-the-speed-of-light)

but there was no calculation, so I tried it thusly:

The observable universe is 28.5 Gpc (gigaparsecs)
= 28500 Mpc (megaparsec)

<https://www.google.com/search?q=diameter+of+the+observable+universe&oq=diameter+of+the+observable+universe&aqs=chrome..69i57.6887j0j7&sourceid=chrome&ie=UTF-8>

The Hubble constant is 67.4 or 70 km/s/megaparsec.

<https://www.nasa.gov/feature/goddard/2019/new-hubble-constant-measurement-adds-to-mystery-of-universe-s-expansion-rate>

I’ll use the older value 67.4

v=H x D = 28500 Mpc \* 67.4 km/s Mpc = 1,920,000 km/s = 1.9 x 10^9 m/s

speed of light 2.998 x 10^8 m/s

So the velocity of expansion at the edge of the universe is >6x greater than the speed of light?

I read these posts

<https://physics.stackexchange.com/questions/107748/how-are-galaxies-receding-faster-than-light-visible-to-observers>

[So where are these measurements of galaxies moving faster than light?](https://astronomy.stackexchange.com/questions/25954/so-where-are-these-measurements-of-galaxies-moving-faster-than-light)

and this

<https://arxiv.org/pdf/astro-ph/0011070v2.pdf>

From the arxiv paper, I get that galaxies can be receding at 3c based on GR, but my simple calculation comes up with 6c.

Can someone help me reconcile my curiosity with the facts?

**Tags:** observable-universe, velocity, size

> **Comments:**
> **username_1**: Many mistakes. Please check your units. Please check the size of the observable universe.  
> **username_2**: From your link, the Universe is 93 billion ly; this is 28500 Mpc, not 28.5 as you've started in your calculation.Try again.  
> **username_3**: You're completely correct, except that you used the diameter rather than the radius of the observable Universe (we can always argue over the exact value — I think using the latest [Planck (2018)](https://ui.adsabs.harvard.edu/abs/2018arXiv180706205P) parameters, the radius would be some 14.4 Gpc.  

---

## 1 answers

### Answer by username_4 on 2020-05-25. Score: 2 | Accepted answer

I think the main reason the answer doesn't come out correct is that the simple Hubble's law equation:

$v = dH\_{0}$

Is only applicable in the local universe (Hence $H\_{0}$, the value of the Hubble constant today). The rate of expansion of the Universe (the Hubble parameter) is determined by the composition of its energy density. Today, the composition is about 30% matter and 70% dark energy. In the recent past the energy density was dominated by matter. In the more distant past the energy density was dominated by relativistic particles (photons and neutrinos). The rate of expansion of the universe is different in each of these "epochs". In general the universe used to expand more rapidly (expansion has slowed thanks to the effects of matter), but distances also used to be smaller (the universe itself was smaller). Combine all of this together and you have a very complicated relationship between distance and the recessional velocity we observe today. This relationship does not have a simple functional form.

I think the easiest question to ask is: "How far has a photon traveled such that if it were emitted at the beginning of the universe it would reach me today?". The answer to this question is the "cosmological horizon", which we commonly talk about as being the size of the observable universe. To answer this question you need to use the spacetime metric which applies for our universe (Called the Friedmann-Robertson-Walker-Lemaitre metric). This is basically the relationship between speed, time, and distance in the realm of general relativity. If our universe is flat (which we think it must be) this equation is actually quite simple, but unfortunately it requires that you know how the universe has been expanding over its lifetime. As I mentioned this relationship has no closed form equation or number, which is why you will never see a simple equation for the solution you are looking for. You need to integrate the solution numerically. [Wikipedia](https://en.wikipedia.org/wiki/Particle_horizon) has a good little article on the horizon.

If you're interested, you'd write down the FLRW metric for a photon travelling in a "straight path":

$ cdt = \frac{a(t) dr}{ \sqrt{1-kr^2} } $

Where $a(t)$ is the scale factor of the universe, which tells you about its expansion history. This is the complicated function. $k$ is the curvature constant, which for a flat universe is 0. You then get:

$ \int^{t\_{0}}\_{0} \frac{c dt}{a(t)} = \int^{r\_{H}}\_{0} dr = r\_{H} $

Here $t\_{0}$ is the age of the universe and $r\_{H}$ is the particle horizon you're interested in. To figure out $a(t)$ you need the [Friedmann equations](https://en.wikipedia.org/wiki/Friedmann_equations), which are functions of the different energy densities (dark energy, matter, relativistic particles, curvature), and the rates at which they change in time. I used astropy to do the integration and got $\approx 3.35 c t\_{0}$ which gives about 14 Mpc, so close enough to the answer that you could attribute any differences to a difference in measured parameters that go into solving the integral.

I'd recommend Introduction to Cosmology by Ryden as a textbook if you really want to get into it.


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