= Worked supernova distance comparison
{title2=$\Delta\mu=5\log_{10}(d_{L,1}/d_{L,2})$}
Stipulate $H_0=70\,\mathrm{km\,s^{-1}\,Mpc^{-1}}$ and $c=299792.458\,\mathrm{km\,s^{-1}}$, 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,
$$
E(z)=\frac{H(z)}{H_0}=\sqrt{\Omega_{m,0}(1+z)^3+\Omega_{k,0}(1+z)^2+\Omega_{\Lambda,0}},\qquad
\Omega_{k,0}=1-\Omega_{m,0}-\Omega_{\Lambda,0},
$$
$$
\chi(z)=\int_0^z\frac{dz'}{E(z')},\qquad
d_L=\frac{c}{H_0}(1+z)S_k(\chi),\qquad
S_k(\chi)=\begin{cases}
\sinh(\sqrt{\Omega_{k,0}}\chi)/\sqrt{\Omega_{k,0}},&\Omega_{k,0}>0,\\
\chi,&\Omega_{k,0}=0,\\
\sin(\sqrt{-\Omega_{k,0}}\chi)/\sqrt{-\Omega_{k,0}},&\Omega_{k,0}<0.
\end{cases}
$$
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. Thus
$$
\begin{array}{c|r|r}
(\Omega_{m,0},\Omega_{\Lambda,0})&d_L\ (\mathrm{Mpc})&\mu=5\log_{10}[d_L/(1\,\mathrm{Mpc})]+25\\\hline
(0.3,0.7)&2832.9381&42.26119\\
(0.3,0)&2565.1861&42.04559\\
(1,0)&2357.6975&41.86244
\end{array}
$$
The 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.
\Image[/supernova-flrw-distance-modulus-comparison.png]
{title=Stipulated FLRW distance moduli and residuals at a common Hubble constant}
{description=Calculated curves only, with no survey points. Residuals use the open matter-only model as reference. The coasting Milne model shows why curvature and the comparison family matter.}
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, $70\to77\,\mathrm{km\,s^{-1}\,Mpc^{-1}}$ 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.
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