<p>The figure below shows the evolution of the Hubble parameter $H$ from 10 billion years (Gyr) ago, to 10 Gyr into the future:</p>

<p><a href="https://i.sstatic.net/ZexIv.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/ZexIv.png" alt="Hz"></a></p>

<p>As you can see, the change in $H$ is modest nowadays, compared to the past. The "acceleration" of the expansion at any point in time is given by the tangent to the curve at that time. Today, the change is $dH/dt \simeq -1.2\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}\,\mathrm{Gyr}^{-1}$, or roughly $-10^{-17}\,\mathrm{km}\,\mathrm{s}^{-2}\,\mathrm{Mpc}^{-1}$.</p>

<p>In other words, it will take roughly a billion years before the expansion rate has decreased by $1\,\mathrm{km}\,\mathrm{s}^{-1}$, and it asymptotically approaches some $57\,\mathrm{km}\,\mathrm{s}^{-1}$.</p>

<p>Since $H$ is defined as $(da/dt)\,/\,a$, where $a$ is the <em>scale factor</em> (the "size" of the Universe), a constant $H$ implies that $a\propto e^{Ht}$; that is, the size of the Universe increases exponentially.</p>

CC BY-SA 3.0
2017-09-11T17:09:12.103Z
Wait I thought this is backwards, and expansion is accelerating...
Евгений Артеменко
CC BY-SA 3.0
2017-09-11T18:45:08.370Z
@ЕвгенийАртеменко: The expansion rate at a given distance approaches a constant value, 55 km/s/Mpc. But the velocity with which a given galaxy recedes from us increases, because it increases its distance, and hence lies at a distance where the expansion rate is larger. A galaxy that is today 1 Mpc away, recedes at 68 km/s/Mpc. In a billion years, it will be at a distance of 1.1 Mpc, where the expansion rate is 73 km/s/Mpc, and so on.
pela
CC BY-SA 3.0
2017-09-11T20:10:06.170Z
Aw man, second time today! @RobJeffries.
pela
CC BY-SA 3.0
2017-09-11T20:25:15.993Z
@ЕвгенийАртеменко I added a paragraph on this.
pela
CC BY-SA 4.0
2020-04-19T17:27:56.527Z
The acceleration of the expansion is surely $d(Ha)/dt = H\dot{a} + a\dot{H}$?
ProfRob
CC BY-SA 4.0
2020-04-21T12:27:40.217Z
Yes, I seem to have answered what the OP asks in the _question_, not in the title.
pela
CC BY-SA 3.0
2017-09-11T14:41:43.757Z
2017-09-11T20:24:30.357Z
2017-09-11T20:24:30.357Z
pela
<p>Pela's answer gives the numerical value, I thought I would just explain the difference between $\dot{H}$, which is negative, and an acceleration, which is positive.</p>

<p>The <a href="https://en.wikipedia.org/wiki/Friedmann_equations#Equations" rel="nofollow noreferrer">Friedmann acceleration equation</a> is given by 
$$\frac{\ddot{a}}{a} =  -\frac{4 \pi G}{3}\left(\rho+\frac{3p}{c^2}\right) + \frac{\Lambda c^2}{3}, \tag*{(1)}$$
whilst the Hubble parameter is defined as
$$H = \frac{\dot{a}}{a} \tag*{(2)}$$</p>

<p>In equation (1) the first term gets smaller with time since both matter density $\rho$ and pressure $p$ due to matter and  radiation become smaller. The second term due to the cosmological constant $\Lambda$ is positive and eventually dominates and we are in that regime <em>now</em>. This is what is meant by an accelerating universe, where the second derivative of the scale factor $\ddot{a}$ is positive.</p>

<p>However, you have modified your question to ask what the time derivative of the Hubble parameter (not constant!) is. Differentiating equation (2) with respect to time, we have
$$\dot{H} = \frac{\ddot{a}}{a} - H^2$$
$$\dot{H} = -\frac{4 \pi G}{3}\left(\rho+\frac{3p}{c^2}\right) + \frac{\Lambda c^2}{3} - H^2 \tag*{(3)}\, .$$</p>

<p>$\dot{H}$ is the gradient of the plot shown in Pela's answer and it is <em>negative</em>. That is $H$ is getting smaller. As the universe gets bigger, $H$ reaches its asymptotic value when $\ddot{a}/a = H^2$. </p>

CC BY-SA 3.0
2017-09-11T20:04:02.873Z
2017-09-11T20:04:02.873Z
ProfRob
<p>The acceleration of the universe is written as $\dot{H}$, where $H$ is the Hubble Rate (today, the Hubble Rate is given the name the <em>Hubble Constant</em>, $H_0$, and is given the value of 70 km/s/10<sup>6</sup>pc). The <a href="https://en.wikipedia.org/wiki/Friedmann_equations#Equations" rel="nofollow noreferrer">Friedmann Equation</a> can be simplified to the form:</p>

<p>$$\dot{H} + H^2 = -\frac{4\pi G}{3}\left(\rho + \frac{3p}{c^2}\right)$$</p>

<p>Where $\rho$ is the density, and $p$ is the pressure. $G$ is the gravitational constant, and $c^2$ is the speed of light.</p>

<p>Now, to find the acceleration, simply substitute the relavent values into the equation below:</p>

<p>$$\dot{H_0} = -\frac{4\pi G}{3}\left(\rho_0 + \frac{3p_0}{c^2}\right) - H^2_0$$</p>

<p><sub>I will try to find the relavent values, but as the previous answer says, the value is roughly $-10^{-17}\,\mathrm{km}\,\mathrm{s}^{-2}\,\mathrm{Mpc}^{-1}$</sub></p>

CC BY-SA 4.0
2020-04-19T17:28:52.753Z
$\dot{H}$ is negative.
ProfRob
CC BY-SA 3.0
2017-09-11T17:04:59.803Z
2017-09-11T21:33:05.400Z
2017-09-11T21:33:05.400Z
Beta Decay
<p>You have 1pc=3.26ly=3.26*300000*365.25*24*3600 km = ...</p>

<p><a href="https://en.wikipedia.org/wiki/Parsec" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Parsec</a></p>

<p>So you could easily find H0 :)</p>

<p>EDIT :</p>

<p>Sorry I've read to quickly the question. The acceleration rate of the universe is given by second equation of Friedmann Lemaître.</p>

<p>$$\frac{\ddot{a}}{a} =  -\frac{4 \pi G}{3}\left(\rho+\frac{3p}{c^2}\right) + \frac{\Lambda c^2}{3}$$</p>

<p>with $\rho$ and p density of the fluids (matter, dark energy...) you consider in the cosmological (standard) model.</p>

<p><a href="https://en.wikipedia.org/wiki/Friedmann_equations" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Friedmann_equations</a></p>

<p><a href="https://en.wikipedia.org/wiki/Accelerating_expansion_of_the_universe" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Accelerating_expansion_of_the_universe</a></p>

CC BY-SA 3.0
2017-09-08T09:30:51.753Z
The OP is not asking about expansion rate $H_0$ (which he actually gives in the question), but about the _acceleration_ of this rate.
pela
CC BY-SA 3.0
2017-09-10T22:52:53.777Z
This still doesn't answer the question because you don't show how this equation relates to $dH_o/dt$ or what the actual numerical value works out to, which is what the OP was asking for.
zephyr
CC BY-SA 3.0
2017-09-11T09:16:40.270Z
@zephyr I started to answer the questions along the same line. It is not clear what is meant by the "acceleration rate of the Universe"; $\ddot{a}/a$ might be as good a definition as any, as might $\dot{H_0}$, but they are not the same thing.
ProfRob
CC BY-SA 3.0
2017-09-11T16:50:29.620Z
What the OP is wanting is $\dot{H_0}$. You can relate this to the equation you give by the fact that $$\dot{H} + H^2 = \frac{\ddot{a}}{a}$$
Beta Decay
CC BY-SA 3.0
2017-09-11T19:25:01.647Z
@BetaDecay well yes, that is *now* clear, though you might like to explain in your answer how a decreasing value of H corresponds to an "acceleration rate"...
ProfRob
CC BY-SA 3.0
2017-09-08T08:10:00.977Z
2017-09-08T14:19:15.397Z
2017-09-08T14:19:15.397Z
Richard
<p>I've found that the speed of expansion is about $70\frac{km}{s\cdot 10^{6}pc}$.</p>

<p>But what is the acceleration of expansion in terms of $\frac{km}{s^{2}\cdot 10^{6}pc}$ ?</p>

<p>I want to know how fast this value of "70" is changing, ie what is $\mathrm{d}H_0/\mathrm{d}t$?</p>

CC BY-SA 3.0
2017-09-07T10:58:06.577Z
Did you read [the WIkipedia page on this subject](https://en.wikipedia.org/wiki/Accelerating_expansion_of_the_universe) and if so, can you be more specific about what you're having trouble with ?
StephenG - Help Ukraine
CC BY-SA 3.0
2017-09-07T12:00:50.563Z
I've read, I didn't found an answer, I've corrected my question to be a bit more specific
Евгений Артеменко
CC BY-SA 3.0
2017-09-08T13:58:21.100Z
Not clear to me we've had enough observation time to answer this question. Let's let the visible universe expand to double its current size, then remeasure the Hubble constant to see if it's still around 70, or gone up to 75. That'd allow your calculation, but I don't think 10 or 20 years worth of data is going to give much accuracy.
Wayfaring Stranger
CC BY-SA 3.0
2017-09-09T16:17:05.827Z
Double is too much and not necessary, obviously... So you've meant, we can not estimate acceleration rate with any reasonable error yet? Then I should learn how do we know, it it even exist :)
Евгений Артеменко
CC BY-SA 3.0
2017-09-10T22:54:30.290Z
@WayfaringStranger You don't have to wait for the universe to expand in the future to measure the rate of change of $H_o$. You have 13.8 billion years worth of measurements you can make to determine $dH_o/dt$.
zephyr
CC BY-SA 3.0
2017-09-10T23:23:48.073Z
@zephyr You may be right, however, that superluminal inflation bit back in the beginning gives me a slight case of the heebeegeebees over extrapolating back too far. They might be able to do the old supernova brightness vs increasing distance thing with increased precision, but I'm not sure how valid that'd be here.
Wayfaring Stranger
CC BY-SA 3.0
2017-09-11T12:32:37.843Z
@RobJeffries could you answer to both?
Евгений Артеменко
CC BY-SA 3.0
2017-09-11T12:35:14.323Z
Actually I just want to know how fast, this top thing 70 km/(s*Mpc) is growing... :)
Евгений Артеменко
CC BY-SA 3.0
2017-09-11T19:32:03.977Z
Please note that $H$ is getting smaller with time and that $\dot{H}$ is *negative*. Meanwhile, $\ddot{a}/a$ is positive.
ProfRob
<cosmological-inflation>
What is the acceleration for the expansion of the Universe?
Евгений Артеменко
# What is the acceleration for the expansion of the Universe?

(Asked by: username_0 on 2017-09-07)

I've found that the speed of expansion is about $70\frac{km}{s\cdot 10^{6}pc}$.

But what is the acceleration of expansion in terms of $\frac{km}{s^{2}\cdot 10^{6}pc}$ ?

I want to know how fast this value of "70" is changing, ie what is $\mathrm{d}H\_0/\mathrm{d}t$?

**Tags:** cosmological-inflation
> **Comments:**
> **username_1**: Did you read [the WIkipedia page on this subject](https://en.wikipedia.org/wiki/Accelerating_expansion_of_the_universe) and if so, can you be more specific about what you're having trouble with ?  
> **username_0**: I've read, I didn't found an answer, I've corrected my question to be a bit more specific  
> **username_2**: Not clear to me we've had enough observation time to answer this question. Let's let the visible universe expand to double its current size, then remeasure the Hubble constant to see if it's still around 70, or gone up to 75. That'd allow your calculation, but I don't think 10 or 20 years worth of data is going to give much accuracy.  
> **username_0**: Double is too much and not necessary, obviously... So you've meant, we can not estimate acceleration rate with any reasonable error yet? Then I should learn how do we know, it it even exist :)  
> **username_3**: @WayfaringStranger You don't have to wait for the universe to expand in the future to measure the rate of change of $H_o$. You have 13.8 billion years worth of measurements you can make to determine $dH_o/dt$.  
> **username_2**: @zephyr You may be right, however, that superluminal inflation bit back in the beginning gives me a slight case of the heebeegeebees over extrapolating back too far. They might be able to do the old supernova brightness vs increasing distance thing with increased precision, but I'm not sure how valid that'd be here.  
> **username_0**: @RobJeffries could you answer to both?  
> **username_0**: Actually I just want to know how fast, this top thing 70 km/(s*Mpc) is growing... :)  
> **username_4**: Please note that $H$ is getting smaller with time and that $\dot{H}$ is *negative*. Meanwhile, $\ddot{a}/a$ is positive.  

---

## 4 answers

### Answer by username_5 on 2017-09-11. Score: 4 | Accepted answer

The figure below shows the evolution of the Hubble parameter $H$ from 10 billion years (Gyr) ago, to 10 Gyr into the future:

[![Hz](https://i.sstatic.net/ZexIv.png)](https://i.sstatic.net/ZexIv.png)

As you can see, the change in $H$ is modest nowadays, compared to the past. The "acceleration" of the expansion at any point in time is given by the tangent to the curve at that time. Today, the change is $dH/dt \simeq -1.2\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}\,\mathrm{Gyr}^{-1}$, or roughly $-10^{-17}\,\mathrm{km}\,\mathrm{s}^{-2}\,\mathrm{Mpc}^{-1}$.

In other words, it will take roughly a billion years before the expansion rate has decreased by $1\,\mathrm{km}\,\mathrm{s}^{-1}$, and it asymptotically approaches some $57\,\mathrm{km}\,\mathrm{s}^{-1}$.

Since $H$ is defined as $(da/dt)\,/\,a$, where $a$ is the *scale factor* (the "size" of the Universe), a constant $H$ implies that $a\propto e^{Ht}$; that is, the size of the Universe increases exponentially.

> **Comments:**
> **username_0**: Wait I thought this is backwards, and expansion is accelerating...  
> **username_5**: @ЕвгенийАртеменко: The expansion rate at a given distance approaches a constant value, 55 km/s/Mpc. But the velocity with which a given galaxy recedes from us increases, because it increases its distance, and hence lies at a distance where the expansion rate is larger. A galaxy that is today 1 Mpc away, recedes at 68 km/s/Mpc. In a billion years, it will be at a distance of 1.1 Mpc, where the expansion rate is 73 km/s/Mpc, and so on.  
> **username_5**: Aw man, second time today! @RobJeffries.  
> **username_5**: @ЕвгенийАртеменко I added a paragraph on this.  
> **username_4**: The acceleration of the expansion is surely $d(Ha)/dt = H\dot{a} + a\dot{H}$?  
> **username_5**: Yes, I seem to have answered what the OP asks in the _question_, not in the title.  

---

### Answer by username_4 on 2017-09-11. Score: 1

Pela's answer gives the numerical value, I thought I would just explain the difference between $\dot{H}$, which is negative, and an acceleration, which is positive.

The [Friedmann acceleration equation](https://en.wikipedia.org/wiki/Friedmann_equations#Equations) is given by
$$\frac{\ddot{a}}{a} = -\frac{4 \pi G}{3}\left(\rho+\frac{3p}{c^2}\right) + \frac{\Lambda c^2}{3}, \tag\*{(1)}$$
whilst the Hubble parameter is defined as
$$H = \frac{\dot{a}}{a} \tag\*{(2)}$$

In equation (1) the first term gets smaller with time since both matter density $\rho$ and pressure $p$ due to matter and radiation become smaller. The second term due to the cosmological constant $\Lambda$ is positive and eventually dominates and we are in that regime *now*. This is what is meant by an accelerating universe, where the second derivative of the scale factor $\ddot{a}$ is positive.

However, you have modified your question to ask what the time derivative of the Hubble parameter (not constant!) is. Differentiating equation (2) with respect to time, we have
$$\dot{H} = \frac{\ddot{a}}{a} - H^2$$
$$\dot{H} = -\frac{4 \pi G}{3}\left(\rho+\frac{3p}{c^2}\right) + \frac{\Lambda c^2}{3} - H^2 \tag\*{(3)}\, .$$

$\dot{H}$ is the gradient of the plot shown in Pela's answer and it is *negative*. That is $H$ is getting smaller. As the universe gets bigger, $H$ reaches its asymptotic value when $\ddot{a}/a = H^2$.


---

### Answer by username_6 on 2017-09-11. Score: 0

The acceleration of the universe is written as $\dot{H}$, where $H$ is the Hubble Rate (today, the Hubble Rate is given the name the *Hubble Constant*, $H\_0$, and is given the value of 70 km/s/106pc). The [Friedmann Equation](https://en.wikipedia.org/wiki/Friedmann_equations#Equations) can be simplified to the form:

$$\dot{H} + H^2 = -\frac{4\pi G}{3}\left(\rho + \frac{3p}{c^2}\right)$$

Where $\rho$ is the density, and $p$ is the pressure. $G$ is the gravitational constant, and $c^2$ is the speed of light.

Now, to find the acceleration, simply substitute the relavent values into the equation below:

$$\dot{H\_0} = -\frac{4\pi G}{3}\left(\rho\_0 + \frac{3p\_0}{c^2}\right) - H^2\_0$$

I will try to find the relavent values, but as the previous answer says, the value is roughly $-10^{-17}\,\mathrm{km}\,\mathrm{s}^{-2}\,\mathrm{Mpc}^{-1}$

> **Comments:**
> **username_4**: $\dot{H}$ is negative.  

---

### Answer by username_7 on 2017-09-08. Score: 0

You have 1pc=3.26ly=3.26\*300000\*365.25\*24\*3600 km = ...

<https://en.wikipedia.org/wiki/Parsec>

So you could easily find H0 :)

EDIT :

Sorry I've read to quickly the question. The acceleration rate of the universe is given by second equation of Friedmann Lemaître.

$$\frac{\ddot{a}}{a} = -\frac{4 \pi G}{3}\left(\rho+\frac{3p}{c^2}\right) + \frac{\Lambda c^2}{3}$$

with $\rho$ and p density of the fluids (matter, dark energy...) you consider in the cosmological (standard) model.

<https://en.wikipedia.org/wiki/Friedmann_equations>

<https://en.wikipedia.org/wiki/Accelerating_expansion_of_the_universe>

> **Comments:**
> **username_5**: The OP is not asking about expansion rate $H_0$ (which he actually gives in the question), but about the _acceleration_ of this rate.  
> **username_3**: This still doesn't answer the question because you don't show how this equation relates to $dH_o/dt$ or what the actual numerical value works out to, which is what the OP was asking for.  
> **username_4**: @zephyr I started to answer the questions along the same line. It is not clear what is meant by the "acceleration rate of the Universe"; $\ddot{a}/a$ might be as good a definition as any, as might $\dot{H_0}$, but they are not the same thing.  
> **username_6**: What the OP is wanting is $\dot{H_0}$. You can relate this to the equation you give by the fact that $$\dot{H} + H^2 = \frac{\ddot{a}}{a}$$  
> **username_4**: @BetaDecay well yes, that is *now* clear, though you might like to explain in your answer how a decreasing value of H corresponds to an "acceleration rate"...  

---

