What is the correlation between the Hubble tension and Dark Energy?

Question

When Dark Energy was first discovered it was because we noticed that distance type 1A supernovae were dimmer given their perspective redshifts.

However, to determine the Hubble constant in the late universe, astronomers compare the true and apparent brightness of those same distant type 1A supernovae to measure out to their distance. They then compare those distance measurements with how the light from the supernovae is stretched to longer wavelengths by the expansion of space. They use these two values to calculate how fast the universe expands with time, called the Hubble constant.

But again, they also used these same supernovae to determine the existence of Dark Energy. So, is Dark Energy taken into account when they determined the value of the Hubble constant 74 km/s/Mpc which was obtained from the type 1A supernovae measurements?

And if Dark Energy is what's affecting those distant supernovae to make them appear dimmer, is it because Dark Energy has sped up their velocity to 74 km/s/Mpc over time so now they are farther away than excepted? If so, then why are we saying this obtained value of 74 km/s/Mpc, a value we have attributed to dark energy accelerating the expansion of the universe, maybe shouldn't be this value, and therefore, we have a tension?

Or does Dark Energy have nothing to do with the Hubble Constant and its measured value of 74 km/s/Mpc using type 1A supernovae?

Answer 1

Dark energy (and no doubt the OP read something like the NASA page) has a lot to do with the Hubble Tension.

The so-called local/late Universe measurement (i.e. SH$0$ES $H_0=73.04\pm{1.04}$) derived from supernova, does not match the remote/early Universe value predicted by $\Lambda$CDM (i.e. $Planck$ collaboration $H_0=67.36\pm0.54$). This tension, as of late 2023, is considered unlikely also to be due to some kind of systemic error.

The $Planck$ value is based on the cosmic microwave background, plus using our current best cosmological model of the Universe, $\Lambda$CDM, in which the density of dark energy $\Lambda$ is constant through time.

A great number of attempts have been made (In the Realm of the Hubble Tension etc, 2021 is an excellent summary) to solve the Hubble Tension (without success). A lot of recent effort has been focused on so-called Early Dark Energy (EDE), which essentially involves adding an "extra" dark energy component at early times that then dilutes away later. So, not just vanilla $\Lambda$. The Ups and Downs of Early Dark Energy etc, 2023 is a good summary. So, again, all about dark energy.

Unfortunately, circa 2023, it has also been realized that it is going to take something more than early dark energy to solve the Hubble Tension (and there are all sorts of attempts going on, like this, without much success, so far). For me, the kicker is that EDE makes the so-called sigma-eight or $S_8$ tension (you can read a non-technical summary of what the $S_8$ tension is here) worse, not better.

Answer 2

The calculation of the present-day value of the Hubble parameter $H_0$ can be done either by

1. looking at the local value of the Hubble parameter, which is done they the observation of type Ia supernovae.
2. looking at the value of the Hubble parameter at higher redshifts and then using the $\Lambda$CDM model to determine the present-day value of $H_0$.

When we use the latter approach, the calculated value of $H_0$ is not in agreement with the same value determined from the local observation of $H_0$. This can be seen as the cause of Hubble tension.

How does dark energy enter all this?

For case 1, $H_0$ can be determined observationally using Hubble's law (although you will find that there is already the need for dark energy to explain the observed accelerated expansion). However, for case 2, since the observation happens at a higher redshift, we need a cosmological model to calculate the present-day value of $H_0$. This is where the need to include dark energy becomes important. The mathematical equation can be determined from the Friedmann equations as

$$\frac{H^2}{H_0^2} = \Omega_{0,\mathrm R} a^{-4} + \Omega_{0,\mathrm M} a^{-3} + \Omega_{0,\Lambda} \,.$$

Here $\Omega_{0,\mathrm R}, \Omega_{0,\mathrm M}, \Omega_{0,\mathrm \Lambda}$ is today's value of the density parameter of radiation, matter, and cosmological constant or vacuum density (a.ka. dark energy) respectively, with.

$$ \Omega_{0,\mathrm R} + \Omega_{0,\mathrm M} + \Omega_{0,\mathrm \Lambda} = 1 .$$

Now, depending on your choice of $\Omega_{0,\Lambda}$, you will get a different value for $H_0$ given that you have a value of $H$ from high redshift observations.