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Choose $D=7.5\mathrm{kpc}$, $\alpha =0.110$ arcseconds, $P=18$ years, $e=0.70$ and $i=55$ degrees. The small-angle conversion gives $a=825\mathrm{AU}$. Taking one model year as $365.25$ days gives $\mu \simeq 2.30\times 10^{26}{\mathrm{m}}^3{\mathrm{s}}^{-2}$ and $K\simeq 1566km/s$.

Dividing by the nominal solar mass parameter, $(GM{)}_{\odot }^{\mathrm{N}}=1.3271244\times 10^{20}{\mathrm{m}}^3{\mathrm{s}}^{-2}$, gives $1.73\times 10^6$ nominal solar masses. The nominal parameter is an exact conversion convention, rather than an assertion that the Sun's measured mass is exact.^[ref]

The chosen periapsis radius is $r_p=a(1-e)=247.5\mathrm{AU}$. Its corresponding Schwarzschild radius, $r_s=2\mu /c^2$, is about $0.0342\mathrm{AU}$. Thus this illustrative orbit remains roughly $7200$ Schwarzschild radii from the centre at closest approach. Inferring a very concentrated mass from such an orbit is different from directly resolving its event horizon. Every number here follows from the chosen orbit, not from a fit to the actual Galactic-centre measurements.

The stretch method rescales the light curve's time axis by $s(1+z)$ and corrects peak apparent magnitude by $\alpha (s-1)$, fitting $\alpha$ jointly with cosmology.^[ref]

Here is a deliberately stipulated local model that exposes the signs. Let $t_{\mathrm{obs}}$ be observer time in days, $t_{\max }$ the observed peak time and $T(u)$ a dimensionless template with $T(0)=1$. In a single photometric passband, writeHere $u$ is rest-frame time in days; the template's argument carries those units. For illustration, assign the intrinsic relation $M_B(s)=M_{B,0}-\alpha (s-1)$ with $\alpha >0$: broader events are intrinsically brighter. Define the sign of the K correction by $m_{\mathrm{obs}}=M_B(s)+\mu +A_B+K$, where $A_B$ is interstellar extinction in magnitudes. ThenRemoving attenuation makes the magnitude smaller; standardizing a broad, intrinsically bright event makes it larger. The fitted cosmological time dilation must be removed before comparing intrinsic durations. This schematic factorization explains the bookkeeping; an actual multicolour fit must also model changing intrinsic colours and correlated uncertainties.

For two equally standardized sources, cancellation of their common absolute magnitude gives an especially clean result:A stipulated residual of $+0.20$ mag therefore means a distance ratio $10^{0.04}=1.09648$ and a radiative flux ratio $10^{-0.08}=0.83176$ relative to the comparison prediction, provided luminosity and attenuation have been correctly standardized. This is a conversion of a hypothetical residual, not a measured survey value.