<p>What the Hubble constant really depends on is how old was the universe at the time, but if you have a dynamical model of the universe, you can map that into z and come up with a function H(z).  So in that sense, the answer is "yes," but be careful-- we also think of z as a measure of how far away the objects are, and H does not depend on location it depends on age. What's more, the z we get from a given measurement reflects all the expansion, so all the H's, since that light was emitted, not just the value of that H at that z.  It would be a bit like if you were using your height to talk about your age, and you look at a picture of yourself at 4 feet tall, and say you were growing two inches a year when you were that tall.  That would be like the H that applies in that picture, but the ratio of the height you are now to the height in that picture depends on more than just the rate you were growing in that picture.</p>

<p>(Also-- what do you mean the value of the Hubble constant is 0.7 now?  That sounds like the fraction of total energy that is "dark energy," so is more about the rate of acceleration of the expansion rather than the rate of expansion itself.  If you are asking about that, then this number has been rising with age of the universe, so can be mapped into a function of z if we are using z to talk about the age of the universe at the time.)</p>

CC BY-SA 3.0
2018-04-19T11:45:09.840Z
Great analogy! +1
pela
CC BY-SA 3.0
2018-04-19T11:45:47.157Z
I guess by "0.7" he means the reduced Hubble constant, i.e. $H_0$/100 km/s/Mpc.
pela
CC BY-SA 3.0
2018-04-19T15:19:56.250Z
Ah yes, you're probably right.
Ken G
CC BY-SA 3.0
2018-04-19T10:27:42.560Z
2018-04-19T10:35:05.360Z
2018-04-19T10:35:05.360Z
Ken G
<p>Yes, definitely.</p>

<p>The Hubble constant describes the expansion rate of the Universe, and the expansion may, in turn, may be <em>decelerated</em> by "regular" matter/energy, and <em>accelerated</em> by dark energy.</p>

<p>It's more or less the norm to use the term Hubble <em>constant</em> $H_0$ for the value today, and Hubble <em>parameter</em> $H(t)$ or $H(a)$ for the value at a time $t$ or, equivalently, a scale factor $a = 1/(1+z)$, where $z$ is the redshift.</p>

<p>The value is given by the Friedmann equation:
$$
\frac{H^2(a)}{H_0^2} = 
\frac{\Omega_\mathrm{r}}{a^4} +
\frac{\Omega_\mathrm{M}}{a^3} +
\frac{\Omega_k}{a^2} +
      \Omega_\Lambda,
$$
where 
$\{
\Omega_\mathrm{r},
\Omega_\mathrm{M},
\Omega_k,
\Omega_\Lambda
\} \simeq
\{
10^{-3},0.3,0,0.7
\}
$
are the fractional energy densities in radiation, matter, curvature, and dark energy, respectively.</p>

<p>For instance, you can solve the above equation at $z=0.1$ and find that the expansion rate was 5% higher than today.</p>

<p>Since everything but dark energy dilutes with increasing $a$, $H(a)$ will asymptotically converge to a value $H_0\sqrt{\Omega_\Lambda} \simeq 56\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$.</p>

<p>The figure below shows the evolution of the Hubble parameter with time:</p>

<p><a href="https://i.sstatic.net/Xn2nW.png" rel="noreferrer"><img src="https://i.sstatic.net/Xn2nW.png" alt="Ht"></a></p>

<p>As noted by KenG, the fact that $H$ <em>decreases</em> with time may seem at odds with the accelerated expansion of the Universe. But $H$ describes how fast a point in space at a given distance recedes. Later, that point will be farther away, and so will recede faster. From the definition of the Hubble parameter, $H\equiv\dot{a}/a$, multiplying by the scale factor shows the acceleration $da/dt$:</p>

<p><a href="https://i.sstatic.net/VbLK0.png" rel="noreferrer"><img src="https://i.sstatic.net/VbLK0.png" alt="aHt"></a></p>

CC BY-SA 3.0
2018-04-19T10:42:11.377Z
And just to stave off any possible confusion surrounding that wonderful answer, when people talk about the expansion "accelerating," they are talking about what is happening to the expansion speed H times a, not the expansion rate H itself.  So your result shows that while H is dropping with or without dark energy, dark energy makes H times a rise with a, whereas matter alone makes H times a drop with a.
Ken G
CC BY-SA 3.0
2018-04-19T11:44:41.590Z
@KenG Yes, that's an important point.
pela
CC BY-SA 3.0
2018-04-19T14:20:31.010Z
If an animal grew by 1% per year, a gedanken microbe on its skin might claim that the expansion was accelerating!
John Duffield
CC BY-SA 3.0
2018-04-19T15:38:44.020Z
@JohnDuffield Gedanken microbe  :D
pela
CC BY-SA 4.0
2019-04-14T14:31:26.860Z
Good to see this one is finally settled.
Wayfaring Stranger
CC BY-SA 3.0
2018-04-19T10:15:16.673Z
2018-04-22T21:11:28.147Z
2018-04-22T21:11:28.147Z
pela
<p>I know there are lot of questions on the Hubble constant already, but I am curious to know if it changes with redshift? If at current redshift, <span class="math-container">$z=0$</span>, we know its value to be 0.7, will it be different at higher redshift (<span class="math-container">$z=0.1$</span>)? If so, is there any relationship with redshift?</p>

CC BY-SA 3.0
2018-04-19T11:24:59.817Z
See also https://astronomy.stackexchange.com/questions/10585/evolution-of-the-hubble-parameter https://astronomy.stackexchange.com/questions/11408/hubble-law-cosmological-redshift-and-distance https://astronomy.stackexchange.com/questions/18880/how-is-the-universes-expansion-accelerating-if-the-hubble-constant-is-decreasin?rq=1
ProfRob
<hubble-constant>
Does the Hubble constant depend on redshift?
bhjghjh
# Does the Hubble constant depend on redshift?

(Asked by: username_0 on 2018-04-19)

I know there are lot of questions on the Hubble constant already, but I am curious to know if it changes with redshift? If at current redshift, $z=0$, we know its value to be 0.7, will it be different at higher redshift ($z=0.1$)? If so, is there any relationship with redshift?

**Tags:** hubble-constant
> **Comments:**
> **username_1**: See also https://astronomy.stackexchange.com/questions/10585/evolution-of-the-hubble-parameter https://astronomy.stackexchange.com/questions/11408/hubble-law-cosmological-redshift-and-distance https://astronomy.stackexchange.com/questions/18880/how-is-the-universes-expansion-accelerating-if-the-hubble-constant-is-decreasin?rq=1  

---

## 2 answers

### Answer by username_3 on 2018-04-19. Score: 20 | Accepted answer

Yes, definitely.

The Hubble constant describes the expansion rate of the Universe, and the expansion may, in turn, may be *decelerated* by "regular" matter/energy, and *accelerated* by dark energy.

It's more or less the norm to use the term Hubble *constant* $H\_0$ for the value today, and Hubble *parameter* $H(t)$ or $H(a)$ for the value at a time $t$ or, equivalently, a scale factor $a = 1/(1+z)$, where $z$ is the redshift.

The value is given by the Friedmann equation:
$$
\frac{H^2(a)}{H\_0^2} =
\frac{\Omega\_\mathrm{r}}{a^4} +
\frac{\Omega\_\mathrm{M}}{a^3} +
\frac{\Omega\_k}{a^2} +
\Omega\_\Lambda,
$$
where
$\{
\Omega\_\mathrm{r},
\Omega\_\mathrm{M},
\Omega\_k,
\Omega\_\Lambda
\} \simeq
\{
10^{-3},0.3,0,0.7
\}
$
are the fractional energy densities in radiation, matter, curvature, and dark energy, respectively.

For instance, you can solve the above equation at $z=0.1$ and find that the expansion rate was 5% higher than today.

Since everything but dark energy dilutes with increasing $a$, $H(a)$ will asymptotically converge to a value $H\_0\sqrt{\Omega\_\Lambda} \simeq 56\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$.

The figure below shows the evolution of the Hubble parameter with time:

[![Ht](https://i.sstatic.net/Xn2nW.png)](https://i.sstatic.net/Xn2nW.png)

As noted by KenG, the fact that $H$ *decreases* with time may seem at odds with the accelerated expansion of the Universe. But $H$ describes how fast a point in space at a given distance recedes. Later, that point will be farther away, and so will recede faster. From the definition of the Hubble parameter, $H\equiv\dot{a}/a$, multiplying by the scale factor shows the acceleration $da/dt$:

[![aHt](https://i.sstatic.net/VbLK0.png)](https://i.sstatic.net/VbLK0.png)

> **Comments:**
> **username_2**: And just to stave off any possible confusion surrounding that wonderful answer, when people talk about the expansion "accelerating," they are talking about what is happening to the expansion speed H times a, not the expansion rate H itself.  So your result shows that while H is dropping with or without dark energy, dark energy makes H times a rise with a, whereas matter alone makes H times a drop with a.  
> **username_3**: @KenG Yes, that's an important point.  
> **username_4**: If an animal grew by 1% per year, a gedanken microbe on its skin might claim that the expansion was accelerating!  
> **username_3**: @JohnDuffield Gedanken microbe  :D  
> **username_5**: Good to see this one is finally settled.  

---

### Answer by username_2 on 2018-04-19. Score: 7

What the Hubble constant really depends on is how old was the universe at the time, but if you have a dynamical model of the universe, you can map that into z and come up with a function H(z). So in that sense, the answer is "yes," but be careful-- we also think of z as a measure of how far away the objects are, and H does not depend on location it depends on age. What's more, the z we get from a given measurement reflects all the expansion, so all the H's, since that light was emitted, not just the value of that H at that z. It would be a bit like if you were using your height to talk about your age, and you look at a picture of yourself at 4 feet tall, and say you were growing two inches a year when you were that tall. That would be like the H that applies in that picture, but the ratio of the height you are now to the height in that picture depends on more than just the rate you were growing in that picture.

(Also-- what do you mean the value of the Hubble constant is 0.7 now? That sounds like the fraction of total energy that is "dark energy," so is more about the rate of acceleration of the expansion rather than the rate of expansion itself. If you are asking about that, then this number has been rising with age of the universe, so can be mapped into a function of z if we are using z to talk about the age of the universe at the time.)

> **Comments:**
> **username_3**: Great analogy! +1  
> **username_3**: I guess by "0.7" he means the reduced Hubble constant, i.e. $H_0$/100 km/s/Mpc.  
> **username_2**: Ah yes, you're probably right.  

---

