<p>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.</p>

<p>As the universe expands, $\rho$ of course decreases, but $\Lambda$ remains constant.</p>

<p>Thus the Hubble "constant" actually decreases from its current value $H_0$ and asymptotically tends towards $ H = \sqrt{\Lambda/3}$ as time tends towards infinity.</p>

<p>As $\Lambda = 3H_0^{2} \Omega_\Lambda$, and measurements suggest that $\Omega_{\Lambda} \simeq 2/3$, then $\Lambda \simeq 2H_0^2$, and the Hubble parameter will therefore decrease to approximately $\sqrt{2/3}$ of its present value <em>if</em> the cosmological constant stays constant.</p>

<p>Of course if $\Lambda = \Lambda(t)$, (ie not the basic $\Lambda$-CDM model) then the behaviour will be different.</p>

<p>EDIT: Another useful form of the solution (for the case of a constant vacuum energy density) is</p>

<p>$$H^2 = H_0^2 \left(  \frac{\Omega_r}{a^4} + \frac{\Omega_M}{a^3} + \frac{\Omega_k}{a^2} + \Omega_{\Lambda}\right),$$
where $H_0$ is the Hubble parameter now, $a(t)$ is the scale factor of the universe, $\Omega_r$ is the current (i.e. $a=1$) ratio of the radiation density to the critical density and $\Omega_M$, $\Omega_k$ and $\Omega_{\Lambda}$ are the equivalent densities for the matter (baryonic and dark), curvature and (constant) vacuum energy densities.</p>

<p>As $a$ increases you can see that all three of the leading terms get smaller and the Hubble parameter decreases at all times. When $a$ is very large, $H$ approaches $\sqrt{\Omega_{\Lambda}} H_0$ as before.</p>

CC BY-SA 3.0
2015-05-05T20:10:53.447Z
Has the value of the Hubble parameter always been decreasing? From time=0 until now?
set5
CC BY-SA 3.0
2015-05-12T11:29:31.220Z
@mick: Yes it has, since from the Friedmann equation H decreases with density, which in turn decreases (for matter) or at least doesn't increase (for dark energy) with time (note though that things around the moment of the Big Bang after all still are quite uncertain and could have behaved differently from what our current knowledge suggests).
pela
CC BY-SA 3.0
2015-05-12T12:17:41.040Z
@mick I have added a more general form which perhaps makes it easier to see that as long as the scale factor increases, H decreases (and providing that the dark energy doesn't get bigger in the early universe). In fact at very early (inflationary) epochs this cannot be assumed and the Hubble parameter is thought to be approximately constant, leading to exponential growth.
ProfRob
CC BY-SA 4.0
2023-02-20T13:20:15.630Z
Unless $\Omega_k$ is negative.  But, it probably is very close to 0.
eshaya
CC BY-SA 4.0
2024-09-21T23:37:38.607Z
Hi, in the above equation, and pretending you don't know how to derive the Hubble constant from the local universe, how do you derive the value of $H_0^2$ to determine the value of $H(z)^2$ at any redshift?
Tivity
CC BY-SA 4.0
2024-09-22T01:15:48.960Z
@Tivity the present day Hubble parameter is measured.
ProfRob
CC BY-SA 4.0
2024-09-22T01:39:15.350Z
Measured from where, the local universe using the supernova data? If so, does that mean if we never measured the Hubble constant using the local universe, we couldn't use that derived value to then determine the Hubble parameter in the early universe?
Tivity
CC BY-SA 4.0
2024-09-22T08:23:57.973Z
@Tivity $H_0$ is a *parameter* that is obtained by fitting to data. That data can be at any redshift, from the CMB to the local universe. Of course $H$ changes, which is why you will need to know/also model/ the other parameters given in the equations above.
ProfRob
CC BY-SA 4.0
2024-09-22T08:27:52.437Z
@Tivity In other words, $H_0$ can be (in principle) be found by just itting $v = H_0 d$ at low redshifts or by fitting a more complicated function featuring $H_0$, $\Omega_m$, $\Omega_\Lambda$ etc. to data at any redshift.
ProfRob
CC BY-SA 3.0
2015-04-26T18:55:29.600Z
2015-05-12T15:28:36.130Z
2015-05-12T15:28:36.130Z
ProfRob
<p>In the lambda-CDM model describing an accelerating Universe, the Hubble parameter is currently decreasing with time. Will it continue to decrease forever?</p>

<cosmology><expansion>
Evolution of the Hubble parameter
set5
# Evolution of the Hubble parameter

(Asked by: username_0 on 2015-04-26)

In the lambda-CDM model describing an accelerating Universe, the Hubble parameter is currently decreasing with time. Will it continue to decrease forever?

**Tags:** cosmology, expansion


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## 1 answers

### Answer by username_1 on 2015-04-26. Score: 10 | Accepted answer

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, but $\Lambda$ remains constant.

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

As $\Lambda = 3H\_0^{2} \Omega\_\Lambda$, and measurements suggest that $\Omega\_{\Lambda} \simeq 2/3$, then $\Lambda \simeq 2H\_0^2$, and the Hubble parameter will therefore decrease to approximately $\sqrt{2/3}$ of its present value *if* the cosmological constant stays constant.

Of course if $\Lambda = \Lambda(t)$, (ie not the basic $\Lambda$-CDM model) then the behaviour will be different.

EDIT: Another useful form of the solution (for the case of a constant vacuum energy density) is

$$H^2 = H\_0^2 \left( \frac{\Omega\_r}{a^4} + \frac{\Omega\_M}{a^3} + \frac{\Omega\_k}{a^2} + \Omega\_{\Lambda}\right),$$
where $H\_0$ is the Hubble parameter now, $a(t)$ is the scale factor of the universe, $\Omega\_r$ is the current (i.e. $a=1$) ratio of the radiation density to the critical density and $\Omega\_M$, $\Omega\_k$ and $\Omega\_{\Lambda}$ are the equivalent densities for the matter (baryonic and dark), curvature and (constant) vacuum energy densities.

As $a$ increases you can see that all three of the leading terms get smaller and the Hubble parameter decreases at all times. When $a$ is very large, $H$ approaches $\sqrt{\Omega\_{\Lambda}} H\_0$ as before.

> **Comments:**
> **username_0**: Has the value of the Hubble parameter always been decreasing? From time=0 until now?  
> **username_2**: @mick: Yes it has, since from the Friedmann equation H decreases with density, which in turn decreases (for matter) or at least doesn't increase (for dark energy) with time (note though that things around the moment of the Big Bang after all still are quite uncertain and could have behaved differently from what our current knowledge suggests).  
> **username_1**: @mick I have added a more general form which perhaps makes it easier to see that as long as the scale factor increases, H decreases (and providing that the dark energy doesn't get bigger in the early universe). In fact at very early (inflationary) epochs this cannot be assumed and the Hubble parameter is thought to be approximately constant, leading to exponential growth.  
> **username_3**: Unless $\Omega_k$ is negative.  But, it probably is very close to 0.  
> **username_4**: Hi, in the above equation, and pretending you don't know how to derive the Hubble constant from the local universe, how do you derive the value of $H_0^2$ to determine the value of $H(z)^2$ at any redshift?  
> **username_1**: @username_4 the present day Hubble parameter is measured.  
> **username_4**: Measured from where, the local universe using the supernova data? If so, does that mean if we never measured the Hubble constant using the local universe, we couldn't use that derived value to then determine the Hubble parameter in the early universe?  
> **username_1**: @username_4 $H_0$ is a *parameter* that is obtained by fitting to data. That data can be at any redshift, from the CMB to the local universe. Of course $H$ changes, which is why you will need to know/also model/ the other parameters given in the equations above.  
> **username_1**: @username_4 In other words, $H_0$ can be (in principle) be found by just itting $v = H_0 d$ at low redshifts or by fitting a more complicated function featuring $H_0$, $\Omega_m$, $\Omega_\Lambda$ etc. to data at any redshift.  

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