 Worked supernova distance comparison

Stipulate $H_0=70\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$ and $c=299792.458\mathrm{km}{\mathrm{s}}^{-1}$, so $c/H_0=4282.7494\mathrm{Mpc}$. A megaparsec is $10^6$ parsecs. Neglect radiation in cosmology. Compare an accelerating flat model $({\Omega }_{m,0},{\Omega }_{\Lambda ,0})=(0.3,0.7)$, an open matter-only model $(0.3,0)$ and the Einstein-de Sitter universe $(1,0)$. Their present deceleration parameters are respectively $-0.55$, $0.15$ and $0.5$. For these models,Here $\chi$ is dimensionless; the radial null geodesic in FLRW spacetime supplies its integral and spatial curvature of an FLRW universe supplies $S_k$. The observer-to-source comoving radial distance is $(c/H_0)\chi$, whereas the transverse distance uses $S_k$. Omitting the latter distinction would bias the open-model comparison.

At $z=0.5$, numerical integration gives $\chi =0.440984335$ for the flat accelerating model and $\chi =0.392227378$ for the open matter-only model. ThusThe accelerating model is $0.21559$ mag fainter than the open matter-only prediction at the same redshift and standardized absolute magnitude, corresponding to $1.10438$ times the luminosity distance and $0.81991$ times the radiative flux. These are model calculations, not observations from either collaboration.

The Milne model has $q=0$, ${\Omega }_{m,0}={\Omega }_{\Lambda ,0}=0$ and ${\Omega }_{k,0}=1$, giving $\chi =\log (1+z)$ and $d_L=(c/H_0)(z+z^2/2)$. The Einstein-de Sitter luminosity-distance relation independently checks the flat matter-only integral. At $z=0.5$, the accelerating model exceeds the Milne model by only $0.12317$ mag, illustrating the dependence of a residual on the chosen reference. Over a longer redshift interval, luminosity distance integrates past deceleration as well as recent acceleration; a residual need not increase forever.

Multiplying $H_0$ by $f>0$ at fixed density fractions divides every $d_L$ by $f$ and shifts every distance modulus by $-5{\log }_{10}f$. The predicted apparent magnitude is unchanged if $M_{B,0}$ simultaneously increases by $5{\log }_{10}f$. This is the absolute magnitude–Hubble constant degeneracy. For example, $70\to 77\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$ shifts $\mu$ by $-0.20696$ mag and $M_{B,0}$ by $+0.20696$ mag. Relative curves at the same $H_0$ retain their separation; determining $H_0$ requires an external absolute magnitude calibration.