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If the total mass of the universe is smaller than estimated by neglecting the gravitational pull of dark matter, the estimated expansion rate should be greater.

Does this consideration in the CMB estimation of the Hubble constant agree with local distance ladder measurements?

Edit: I want to discard the energy density of dark matter so that the energy density of matter is only baryonic in the estimation of the Hubble constant.

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You would have to specify exactly what you meant by "ignoring dark matter" (for example, how would your proposed alternative model make up the missing energy density, or would you propose a Universe with spatial curvature?), but regardless of the details, unless you replace "dark matter" with "something that acts exactly like dark matter from the point of view of observations done to date" (eg, modified gravity, maybe), you will quickly run into some problem with observations. For example, dark matter explains the relative heights of first three peaks of the CMB in a way that can't be reproduced by baryonic matter.

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  • $\begingroup$ I want to ignore the gravitational effect of dark matter in cosmic scale. I know this contradicts galactic rotation curves and measurements of the CMB. But the question is still valid out of curiosity. $\endgroup$
    – Independent Physics
    Commented Apr 14 at 12:32
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    $\begingroup$ @Manuel The difficulty is that the same CMB power spectrum that demands dark matter is what also drives the $H_0$ inference. Without dark matter, you can't fit the CMB power spectrum, so there's no meaningful CMB inference of $H_0$ in the first place. $\endgroup$
    – Sten
    Commented May 26 at 17:17
  • $\begingroup$ @Manuel I don't understand the sense in which you can say that the premise of your question contradicts known observations but is still a valid question. (a) Would you be happy with a solution that resolved the Hubble tension but created a worse tension in some other parameter or observation? (b) In order to say how the Hubble tension is affected, you have to explain how you will measure the Hubble constant using Supernova and using CMB observations. Since dark matter is currently needed to fit to the CMB, how you are planning to fit the CMB to measure the Hubble constant? $\endgroup$
    – Andrew
    Commented May 26 at 18:37
  • $\begingroup$ @Andrew As I already said, it is just out of curiosity. I am not trying to solve the dark energy problem here. $\endgroup$
    – Independent Physics
    Commented May 28 at 18:00
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    $\begingroup$ @Manuel But the premise of your question would contradict observations, as you've said. So it's a little bit like asking a question that starts, "Supposing $1+1\neq 2$, then...". In some sense you can get any answer since you start with a contradiction. In this case, we wouldn't be able to fit the CMB well enough to measure $H_0$ if we forced $\Omega_{\rm DM}=0$, so then there's no tension. I'm all for "in principle" questions and to be clear I'm not trying to attack you, it's just that I genuinely don't understand what you are curious about. What would a "yes" or "no" answer tell you? $\endgroup$
    – Andrew
    Commented May 28 at 18:55
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Let us assume (as you want) that no dark matter exists. We will ignore its impact on structure formation and assume that the Universe can still be treated as spatially homogenous and isotropic on $\approx 100 $Mpc scale.

Firstly, the local value of the Hubble constant $ H_0$ can be directly measured using Type Ia supernovae. At higher redshifts, however, we can only infer $H$ and then use the Friedmann equations as

$$\frac{H^2}{H_0^2} = \Omega_{0,\mathrm R} a^{-4} + \Omega_{0,\mathrm M} a^{-3} + \Omega_{0,\Lambda} \,. \tag{1}$$

to determine the local value $H_0$. If I understand the question correctly, if dark matter is absent, then $\Omega_{0,\mathrm M}$ will be lower than what we think. In particular, $\Omega_{0,\mathrm M} \approx 0.05$ in contrast to what we expect it to be today, i.e., $\Omega_{0,\mathrm M} \approx 0.3$. So you will have to make $\Omega_{0,\mathrm \Lambda} \approx 0.95$ for the total sum of $\Omega_{0,\mathrm R} + \Omega_{0,\mathrm M} + \Omega_{0,\mathrm \Lambda} = 1$. Thus, post-radiation-dominated era, we may approximate Eq. (1) as

$$ H_0 \approx \frac{H}{\Omega_{0,\mathrm \Lambda}}.$$

For a given value of $H$ say inferred around redshift $z \approx 1100$ (CMB), implying that $a \approx 10^{-3}$, I think this will result in $H_0$ that is much bigger than what we get from local measurements.

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    $\begingroup$ Nice answer, +1. Also just pointing out another option would be to add $\Omega_{\rm curvature}$ into the Friedmann equation. $\endgroup$
    – Andrew
    Commented May 29 at 1:27
  • $\begingroup$ Nice point, agreed. $\endgroup$
    – S.G
    Commented May 29 at 4:46
  • $\begingroup$ Would the measurement of the universe being flat (and critical density being almost equal to the sum of all energy densities) be dependent of the LCDM model and thus, on dark matter? In other words, could it be that the sum of energy densities is not 1? $\endgroup$
    – Independent Physics
    Commented Jun 1 at 11:22
  • $\begingroup$ LCDM does not require the universe to be flat; this is a free parameter in the theory. On the other hand, the sum of all energy densities being equal to one is just another reformulation of the Friedmann equation $\left(\frac{\dot{a}}{a}\right)^{2}=\frac{8 \pi G_{N}}{3} \rho-\frac{\kappa}{a^{2}}$ so it has to be one. $\endgroup$
    – S.G
    Commented Jun 1 at 23:03

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