The Hubble Constant: A Present Snapshot, Not a Cosmic Prophecy — Epoche C2
What the constant fixes, stated exactly A measurement of the Hubble constant $H_0$ — say the $73.04 \pm 1.04$ km s$^{-1}$ Mpc$^{-1}$ that the SH0ES team obtained in 2022 from Type Ia supernovae calibrated by Cepheid variables — fixes the vertical scale of the cosmic expansion history and, at fixed density parameters, nothing about its shape. That is the thesis of the compressed version of this note, and it is correct as far as it goes. What follows sets out exactly what it means, exhibits the shape-invariants that make it precise, and then identifies the place where it stops being true — because the density parameters are not independent of $H_0$ in any fit to real data, and forgetting this is what makes the tension between local and early-universe determinations look either more mysterious or more trivial than it is. Begin with the object. The Hubble parameter is $H(t) = \dot{a}(t)/a(t)$, where $a$ is the scale factor of the Friedmann–Lemaître–Robertson–Walker metric, normalised to $a = 1$ today; $H_0 = H(t_0)$ is its value now. In a homogeneous and isotropic spacetime the Einstein field equations reduce to the first Friedmann equation, which for total mass density $\rho$ reads $$H^2 = \left(\frac{\dot{a}}{a}\right)^{\!2} = \frac{8\pi G}{3}\rho - \frac{kc^2}{a^2} + \frac{\Lambda c^2}{3},$$ with $k$ the curvature constant in units of inverse length squared and $\Lambda$ the cosmological constant. The compressed version wrote this with a factor $8\pi G/3c^2$ multiplying a sum of energy densities and no separate $\Lambda$ term. That is the same equation in a different convention — energy density is $c^2$ times mass density, and $\Lambda$ can be absorbed as a component with energy density $\Lambda c^4/8\pi G$ — and nothing there was wrong. It is worth saying so explicitly, because the two conventions must not be mixed within a single expression, and a reader checking dimensions on the compressed form would have been left uncertain which was intended. The shape invariants, and why they contain no $H_0$ Divide the Friedmann equation by $H_0^2$ and use $a = 1/(1+z)$ to trade time for redshift. Defining the present-day density parameters $\Omega_{i,0} = \rho_{i,0}/\rho_{c,0}$ relative to the critical density $\rho_{c,0} = 3H_0^2/8\pi G$, and writing each component's density in terms of its scaling with redshift — matter dilutes as volume, so as $(1+z)^3$; radiation dilutes as volume and redshifts in energy, so as $(1+z)^4$; a cosmological constant does not dilute at all — one obtains $$E(z)^2 \equiv \frac{H(z)^2}{H_0^2} = \Omega_{m,0}(1+z)^3 + \Omega_{r,0}(1+z)^4 + \Omega_{k,0}(1+z)^2 + \Omega_{\Lambda,0}.$$ Setting $z=0$ forces $\sum_i \Omega_{i,0} = 1$, which is a constraint, not an assumption. The function $E(z)$ is where the note's thesis lives: it is dimensionless, it depends on the $\Omega_{i,0}$ alone, and $H_0$ has been divided out of it entirely. Multiply $E(z)$ by any $H_0$ you like and you get a legitimate expansion history with the same shape. "Shape" can be made quantitative rather than left as a metaphor, and two derived quantities do it. The deceleration parameter $q \equiv -\ddot{a}a/\dot{a}^2$ measures whether the expansion is speeding up or slowing down, and for a spatially flat universe with matter and a cosmological constant only. Dropping radiation is legitimate here and the size of the omission can be stated: the photon density implied by the measured background temperature of $2.7255$ K corresponds to $\Omega_{\gamma,0}h^2 = 2.47\times10^{-5}$, and adding the three species of relativistic neutrinos multiplies this by $1.69$ to give $\Omega_{r,0}h^2 = 4.17\times10^{-5}$, hence $\Omega_{r,0} = 4.17\times10^{-5}/0.4537 = 9.2\times10^{-5}$ — about one part in eleven thousand of the present total density, and about one part in $3.4\times10^3$ of the matter density. Differentiating the Friedmann equation then gives $$q(z) = \frac{\tfrac{1}{2}\Omega_{m,0}(1+z)^3 - \Omega_{\Lambda,0}}{\Omega_{m,0}(1+z)^3 + \Omega_{\Lambda,0}}.$$ With the Planck 2018 values $\Omega_{m,0} = 0.315$ and hence $\Omega_{\Lambda,0} = 0.685$, this gives $q_0 = \tfrac{1}{2}(0.315) - 0.685 = 0.158 - 0.685 = -0.527$. The redshift at which the expansion changed from decelerating to accelerating is where the numerator vanishes: $(1+z)^3 = 2\Omega_{\Lambda,0}/\Omega_{m,0} = 1.370/0.315 = 4.35$, so $1+z = 4.35^{1/3} = 1.63$ and $z_{\rm acc} = 0.63$. Neither number contains $H_0$. This is the compressed version's claim made checkable: two of the most physically consequential features of the expansion history — its present acceleration and the epoch at which acceleration began — are determined by the density parameters and are strictly independent of the Hubble constant. Where the claim breaks: the density parameters are not measured directly Now the turn, and it is the point the compressed version did not reach. The argument above holds $\Omega_{m,0}$ fixed while varying $H_0$. But $\Omega_{m,0}$ is a ratio to the critical density, and the critical density is proportional to $H_0^2$. The physical matter density — grams per cubic centimetre, the thing that actually governs how gravity behaves in the early universe — is $$\rho_{m,0} = \Omega_{m,0}\,\rho_{c,0} = \frac{3H_0^2}{8\pi G}\,\Omega_{m,0} \propto \Omega_{m,0}h^2,$$ where $h = H_0 / (100\ \text{km s}^{-1}\,\text{Mpc}^{-1})$. It is the combination $\omega_m \equiv \Omega_{m,0}h^2$, not $\Omega_{m,0}$, that has physical meaning independent of one's choice of units, and it is $\omega_m$ that observations of the early universe constrain. Planck 2018 gives $\omega_m = 0.1430$ and $\omega_b = 0.02237$ for the baryonic part. The consequence is immediate. Holding $\omega_m$ at its measured value and moving $H_0$ from the Planck value $67.36$ to the SH0ES value $73.04$ changes $h^2$ from $0.4537$ to $0.5335$, and therefore $$\Omega_{m,0} = \frac{0.1430}{0.5335} = 0.268 ,$$ down from $0.1430/0.4537 = 0.315$. The shape invariants move with it: $q_0$ becomes $\tfrac{1}