<p>All of these answers start with the formula and say therefore <span class="math-container">$H_0$</span> decreases, but I think I have a more intuitive explanation:</p>
<p>If you imagine the universe as an open straight rubber band, and the expansion of the universe stretching markings (galaxies) on the band, and say one end is being held stationary (this is us), then the rate at which each galaxy moves away from us is proportional to two things: how fast the rubber band is stretched, and the fraction of its distance over the entire length of the rubber band - why more distant galaxies move away faster.</p>
<p>Then you can imagine by the time a galaxy gets to be the same distance away as another galaxy was before, the expansion of the universe could be accelerating and the band is increasing in length faster, but now the distance to the galaxy is a much smaller fraction of the total length of the rubber band - and hence, depending on the function of time giving the length of the string, these two factors can multiply to give the speed of a galaxy at a certain distance to be decreasing.</p>
<p>Hence as v=<span class="math-container">$H_0$</span>d, <span class="math-container">$H_0$</span> decreases.</p>

CC BY-SA 4.0
2020-09-03T00:27:08.410Z
2020-09-03T00:27:08.410Z
user236529
<p>It really comes down to what you define as accelerated expansion.</p>

<p>Usually accelerated expansion is taken to mean that the first deriviative of the scale factor $a'(t)$ is increasing. However the Hubble parameter is given by:</p>

<p>$$H \equiv \frac{a'(t)}{a(t)}$$</p>

<p>As in expanding Universe $a(t)$ is increasing, it is possible for $a'(t)$ to be increasing, but $H$ to be decreasing.</p>

<p>I suppose it could be argued that an increasing $H$ (i.e. some kind of phantom energy scenario) is what we should truly reserve the term "accelerated expansion for. E.g. you could argue that a de Sitter Universe where $H$ is a constant, and therefore also any Universe where $H$ is decreasing, shouldn't be described as as "accelerated expansion", as there is effectively no difference between the expansion at some time $t$ and some later time. However we don't go with this definition.</p>

CC BY-SA 3.0
2016-10-31T19:33:09.433Z
2016-10-31T19:33:09.433Z
John Davis
<p>So The Hubble &quot;Constant&quot; has been and still is getting smaller since the beginning of time. It's at about 70 km/s/Mpc and will stop at about 60 or &quot;H=Λ/3&quot;. Its getting smaller at a slower rate originally because the matter density of the universe is decreasing more slowly but now the rate-of-H-slowing has decreased even more because the Universe is feeling the acceleration as Dark Energy dominates. &quot;Acceleration of the expansion today just means that the Hubble parameter is decreasing less rapidly than before. But it's still decreasing.&quot; Eventually the acceleration will reach a peak and H will be a constant. That is paraphrased from the physics stack, so I don't want credit, which has a much clearer explanation than here above but details above still educate.</p>
<p>Source <a href="https://www.physicsforums.com/threads/why-is-the-hubble-sphere-expanding.616551/" rel="nofollow noreferrer">https://www.physicsforums.com/threads/why-is-the-hubble-sphere-expanding.616551/</a></p>

CC BY-SA 4.0
2021-07-09T16:48:48.450Z
$H$ tends to $\sqrt{\Lambda/3} = \sqrt{\Omega_\Lambda}H_0$
ProfRob
CC BY-SA 4.0
2020-09-07T06:52:17.927Z
2020-09-07T06:52:17.927Z
ParityViolator
<p>The Hubble Law is just a approximation for "small" scales. When you look for a large scales the scale factor is like</p>

<p><span class="math-container">$$a(t) = a(t_0)[1+H_0(t-t_0) + ...+]$$</span></p>

<p>So, you have another contributions to the scale factor, which affects the value of the hubble's parameter.</p>

<p>So, this parameter is not a angular coefficient of a line. If it would were so the constant need to be increase in a accelerate universe. As this law is just a approximation, then this is not necessary.</p>

CC BY-SA 4.0
2020-03-19T18:10:16.017Z
Welcome on the Astronomy SE! Look, how wonderful Latex support do we have. :-) Just write `$H_0$` and you will get $H_0$.
peterh
CC BY-SA 4.0
2020-03-20T02:50:35.900Z
[MathJax basic tutorial and quick reference](https://math.meta.stackexchange.com/q/5020/284619)
uhoh
CC BY-SA 4.0
2020-03-19T17:01:50.877Z
2020-03-19T18:09:35.370Z
2020-03-19T18:09:35.370Z
user32135
<p>Note that if all velocities remained constant, then as distances increased, the expansion RATE, <span class="math-container">$H_0$</span>, would decrease because it is defined as velocity divided by distance. If <span class="math-container">$H_0$</span> is constant over time, then as each galaxy's distance increases, its velocity increases.  So, while the expansion rate is asymptotically decreasing to a constant, the galaxies are accelerating.  The expansion rate is not increasing, the expansion is.</p>

CC BY-SA 4.0
2024-01-30T17:51:04.457Z
If $H_0$ were constant the expansion rate would increase exponentially with time.
ProfRob
CC BY-SA 4.0
2024-01-30T19:52:17.067Z
If $\dot{a} = H_0 a$, where $H_0$ is a fixed constant, then $a = a_0 \exp(H_0 t)$. Exponential growth in $a$ and $\dot{a}$.
ProfRob
CC BY-SA 4.0
2024-01-31T16:02:31.193Z
I agree with CuriousCat, astronomers consistently call $H_0$ the expansion rate of the universe.  Calling $\dot{a}$ the expansion rate also leads to much confusion. The word rate does not necessarily mean a derivative with respect to time.  Here it means the derivative of velocity with distance.
eshaya
CC BY-SA 4.0
2024-01-30T17:23:35.593Z
2024-02-15T15:45:46.627Z
2024-02-15T15:45:46.627Z
eshaya
<p>The Hubble parameter is defined as the rate of change of the distance between two points in the universe, divided by the distance between those two points.</p>
<p>Both the numerator and denominator get bigget in this ratio for an accelerating, expanding universe.
The Hubble parameter is getting smaller because the denominator is getting bigger more quickly than the numerator.</p>
<p>In the future, the cosmological constant, <span class="math-container">$\Lambda$</span> could result in an exponential expansion with time. A simple piece of maths shows you that the Hubble parameter is a Hubble constant only for an exponential expansion.</p>
<p>For example, suppose <span class="math-container">$H(t) = (da/dt)/a$</span>, where <span class="math-container">$a=a(t)$</span> is the time-dependent distance between two arbitrary points in the universe at epoch <span class="math-container">$t$</span>. Now let's imagine an expansion rate <span class="math-container">$da/dt \propto t$</span> (i.e. accelerating). But in this case <span class="math-container">$a(t) \propto t^2$</span> and <span class="math-container">$H(t) \propto t^{-1}$</span> (i.e. decreasing with time).</p>
<p><strong>Some extra details:</strong></p>
<p>The solution to the Friedmann equation in a flat universe is
<span class="math-container">$$H^2 = \frac{8\pi G}{3}\rho + \frac{\Lambda}{3},$$</span>
where <span class="math-container">$\rho$</span> is the matter density (including dark matter) and <span class="math-container">$\Lambda$</span> is the cosmological constant.</p>
<p>As the universe expands, <span class="math-container">$\rho$</span> of course decreases as <span class="math-container">$a(t)^{-3}$</span>, but <span class="math-container">$\Lambda$</span> remains constant. Thus the first term on the RHS becomes unimportant if indeed <span class="math-container">$\Lambda$</span> is a cosmological constant.</p>
<p>Thus the Hubble &quot;constant&quot; actually decreases from its current value <span class="math-container">$H_0$</span> and asymptotically trends towards <span class="math-container">$ H = \sqrt{\Lambda/3}$</span> as time tends towards infinity.</p>

CC BY-SA 4.0
2016-10-28T06:35:37.870Z
2024-11-12T07:30:03.660Z
2024-11-12T07:30:03.660Z
ProfRob
<p>I am just getting into the field of cosmology and was wondering what it means practically to say that the Hubble Constant is decreasing while the expansion rate of the Universe is accelerating. I am not sure how this phenomenon is possible. How do we observe this? </p>

<p>It is my understanding that the Hubble Constant is the rate at which the Universe is expanding, and it is decreasing with time. (Although it is asymptotic)... Is this conception flawed?</p>

CC BY-SA 3.0
2016-10-28T13:04:25.347Z
This is a great question, but I did want to point out that there are indications that the original research which showed the universe's expansion was accelerating may have been wrong or flawed. See for example [this article](https://www.sciencedaily.com/releases/2016/10/161021123238.htm). Essentially it all comes down to how well we think we understand supernovas (and some people argue we don't really understand them well enough to say the universe's expansion is accelerating).
zephyr
CC BY-SA 3.0
2016-10-29T11:58:14.623Z
@zephyr - fascinating news, did not know that
Fattie
CC BY-SA 3.0
2017-05-12T12:24:40.670Z
@zephyr But with little support in the literature - e.g. see https://arxiv.org/pdf/1702.08244.pdf and https://arxiv.org/abs/1610.08972
ProfRob
<expansion><dark-energy><hubble-constant>
How is the Universe's Expansion Accelerating if the Hubble Constant is Decreasing?
nosn
# How is the Universe's Expansion Accelerating if the Hubble Constant is Decreasing?

(Asked by: username_0 on 2016-10-28)

I am just getting into the field of cosmology and was wondering what it means practically to say that the Hubble Constant is decreasing while the expansion rate of the Universe is accelerating. I am not sure how this phenomenon is possible. How do we observe this?

It is my understanding that the Hubble Constant is the rate at which the Universe is expanding, and it is decreasing with time. (Although it is asymptotic)... Is this conception flawed?

**Tags:** expansion, dark-energy, hubble-constant
> **Comments:**
> **username_1**: This is a great question, but I did want to point out that there are indications that the original research which showed the universe's expansion was accelerating may have been wrong or flawed. See for example [this article](https://www.sciencedaily.com/releases/2016/10/161021123238.htm). Essentially it all comes down to how well we think we understand supernovas (and some people argue we don't really understand them well enough to say the universe's expansion is accelerating).  
> **username_2**: @zephyr - fascinating news, did not know that  
> **username_3**: @zephyr But with little support in the literature - e.g. see https://arxiv.org/pdf/1702.08244.pdf and https://arxiv.org/abs/1610.08972  

---

## 6 answers

### Answer by username_3 on 2016-10-28. Score: 13 | Accepted answer

The Hubble parameter is defined as the rate of change of the distance between two points in the universe, divided by the distance between those two points.

Both the numerator and denominator get bigget in this ratio for an accelerating, expanding universe.
The Hubble parameter is getting smaller because the denominator is getting bigger more quickly than the numerator.

In the future, the cosmological constant, $\Lambda$ could result in an exponential expansion with time. A simple piece of maths shows you that the Hubble parameter is a Hubble constant only for an exponential expansion.

For example, suppose $H(t) = (da/dt)/a$, where $a=a(t)$ is the time-dependent distance between two arbitrary points in the universe at epoch $t$. Now let's imagine an expansion rate $da/dt \propto t$ (i.e. accelerating). But in this case $a(t) \propto t^2$ and $H(t) \propto t^{-1}$ (i.e. decreasing with time).

**Some extra details:**

The solution to the Friedmann equation in a flat universe is
$$H^2 = \frac{8\pi G}{3}\rho + \frac{\Lambda}{3},$$
where $\rho$ is the matter density (including dark matter) and $\Lambda$ is the cosmological constant.

As the universe expands, $\rho$ of course decreases as $a(t)^{-3}$, but $\Lambda$ remains constant. Thus the first term on the RHS becomes unimportant if indeed $\Lambda$ is a cosmological constant.

Thus the Hubble "constant" actually decreases from its current value $H\_0$ and asymptotically trends towards $ H = \sqrt{\Lambda/3}$ as time tends towards infinity.


---

### Answer by username_4 on 2020-09-03. Score: 5

All of these answers start with the formula and say therefore $H\_0$ decreases, but I think I have a more intuitive explanation:

If you imagine the universe as an open straight rubber band, and the expansion of the universe stretching markings (galaxies) on the band, and say one end is being held stationary (this is us), then the rate at which each galaxy moves away from us is proportional to two things: how fast the rubber band is stretched, and the fraction of its distance over the entire length of the rubber band - why more distant galaxies move away faster.

Then you can imagine by the time a galaxy gets to be the same distance away as another galaxy was before, the expansion of the universe could be accelerating and the band is increasing in length faster, but now the distance to the galaxy is a much smaller fraction of the total length of the rubber band - and hence, depending on the function of time giving the length of the string, these two factors can multiply to give the speed of a galaxy at a certain distance to be decreasing.

Hence as v=$H\_0$d, $H\_0$ decreases.


---

### Answer by username_5 on 2016-10-31. Score: 4

It really comes down to what you define as accelerated expansion.

Usually accelerated expansion is taken to mean that the first deriviative of the scale factor $a'(t)$ is increasing. However the Hubble parameter is given by:

$$H \equiv \frac{a'(t)}{a(t)}$$

As in expanding Universe $a(t)$ is increasing, it is possible for $a'(t)$ to be increasing, but $H$ to be decreasing.

I suppose it could be argued that an increasing $H$ (i.e. some kind of phantom energy scenario) is what we should truly reserve the term "accelerated expansion for. E.g. you could argue that a de Sitter Universe where $H$ is a constant, and therefore also any Universe where $H$ is decreasing, shouldn't be described as as "accelerated expansion", as there is effectively no difference between the expansion at some time $t$ and some later time. However we don't go with this definition.


---

### Answer by username_6 on 2020-09-07. Score: 1

So The Hubble "Constant" has been and still is getting smaller since the beginning of time. It's at about 70 km/s/Mpc and will stop at about 60 or "H=Λ/3". Its getting smaller at a slower rate originally because the matter density of the universe is decreasing more slowly but now the rate-of-H-slowing has decreased even more because the Universe is feeling the acceleration as Dark Energy dominates. "Acceleration of the expansion today just means that the Hubble parameter is decreasing less rapidly than before. But it's still decreasing." Eventually the acceleration will reach a peak and H will be a constant. That is paraphrased from the physics stack, so I don't want credit, which has a much clearer explanation than here above but details above still educate.

Source <https://www.physicsforums.com/threads/why-is-the-hubble-sphere-expanding.616551/>

> **Comments:**
> **username_3**: $H$ tends to $\sqrt{\Lambda/3} = \sqrt{\Omega_\Lambda}H_0$  

---

### Answer by username_7 on 2020-03-19. Score: 1

The Hubble Law is just a approximation for "small" scales. When you look for a large scales the scale factor is like

$$a(t) = a(t\_0)[1+H\_0(t-t\_0) + ...+]$$

So, you have another contributions to the scale factor, which affects the value of the hubble's parameter.

So, this parameter is not a angular coefficient of a line. If it would were so the constant need to be increase in a accelerate universe. As this law is just a approximation, then this is not necessary.

> **Comments:**
> **username_8**: Welcome on the Astronomy SE! Look, how wonderful Latex support do we have. :-) Just write `$H_0$` and you will get $H_0$.  
> **username_9**: [MathJax basic tutorial and quick reference](https://math.meta.stackexchange.com/q/5020/284619)  

---

### Answer by username_10 on 2024-01-30. Score: -1

Note that if all velocities remained constant, then as distances increased, the expansion RATE, $H\_0$, would decrease because it is defined as velocity divided by distance. If $H\_0$ is constant over time, then as each galaxy's distance increases, its velocity increases. So, while the expansion rate is asymptotically decreasing to a constant, the galaxies are accelerating. The expansion rate is not increasing, the expansion is.

> **Comments:**
> **username_3**: If $H_0$ were constant the expansion rate would increase exponentially with time.  
> **username_3**: If $\dot{a} = H_0 a$, where $H_0$ is a fixed constant, then $a = a_0 \exp(H_0 t)$. Exponential growth in $a$ and $\dot{a}$.  
> **username_10**: I agree with CuriousCat, astronomers consistently call $H_0$ the expansion rate of the universe.  Calling $\dot{a}$ the expansion rate also leads to much confusion. The word rate does not necessarily mean a derivative with respect to time.  Here it means the derivative of velocity with distance.  

---

