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Why isn't the 'Crisis in Cosmology' or 'Hubble Tension' explained by the Accelerating Universe?

Physics Asked on April 7, 2021

The value of the Hubble parameter determined locally from the Type Ia supernovae is found to be greater than that determined from the Cosmic Microwave Background (CMB) radiation. This is known as the ‘Hubble Tension‘ or even ‘Crisis in Cosmology‘.

Why is this a problem?

If we assume that the Universe is accelerating, the higher value of the Hubble parameter locally can simply be explained as a result of that acceleration.

Indeed, the higher near-value of the Hubble parameter compared to its farther-value was the sole basis for the proposal of Dark Energy and Accelerating Universe.

What am I missing?

2 Answers

Both experiments report $H_0$, the Hubble parameter at redshift $z=0$, now. The different experiments directly measure $H(z)$ at different $z$'s then use the standard model of cosmology to determine $H_0$.

Late time or local measurements, like those using Type Ia supernovae and a cosmic distance ladder, measure $H(z)$ for small $zsim 1$. Early time measurements, like the CMB observations, measure $H(z)$ for large $zsim 10^3$.

There are two likely ways to resolve the differences between the two classes of measurements. First, one or both measurements could have unaccounted for systematic errors. In that case each experiments' directly measured $H(z)$ could be incorrect. When they calculate the implied $H_0$ the input would be wrong, so the reported $H_0$ would be incorrect too. Second, the standard model of cosmology could be incomplete. In this case each experiment is correctly measuring their $H(z)$, but the comparison at $H_0$ is flawed.

Correct answer by Paul T. on April 7, 2021

Both sets of experiments are measuring the same quantity: $H_0$, the Hubble parameter today.

When people say that Type Ia measure H0 "locally" they mean that the observations are happening at a small redshift. On the other hand, the surface of last scattering was generated at a redshift of about 1100. Nevertheless, both sets of observations are being used to measure $H_0$, and not $H(z)$ at different redshifts.

Answered by Andrew on April 7, 2021

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