# derivation of reduced magnitude of asteroids

I've noticed that there is this parameter called reduced absolute magnitude $$V(\alpha)$$ and it is defined as below according to the paper by Mahlke et al. 2021 https://doi.org/10.1016/j.icarus.2020.114094

$$V(\alpha) = m + 5 log (r\Delta)$$

, where where $$r$$ is the distance between the asteroid and the Sun at the epoch of observation and $$Δ$$ the respective distance between the asteroid and Earth, $$m$$ is the observed apparent magnitude, according to the authors. $$V(\alpha)$$ is referred to as the phase curve, and, by definition, $$H = V(0)$$, observed at a phase angle of 0° at 1 AU from both Earth and Sun.

I've tried to find the the derivation of this equation, and haven't succeeded yet. I've also referred the absolute magnitude of solar system bodies given in the the wikipedia page https://en.wikipedia.org/wiki/Absolute_magnitude and could not find the derivation of the above equation.

Can anyone point me to the right direction please? Thanks !

Muinonen et al. 2010, before presenting their $$H G_1 G_2$$ model, express the older $$H G$$ model as $$V(\alpha) = H - 2.5 \log_{10}[(1 - G) \phi_1(\alpha) + G \phi_2(\alpha)].$$
If we substitute $$d_{BS} = r,\ d_{BO} = \Delta,\ d_0 = 1$$ in the formula from the Wikipedia article $$m = H + 5 \log_{10} \left( \frac{d_{BS}\ d_{BO}}{{d_0}^2} \right) - 2.5 \log_{10} q(\alpha),$$ where $$q(\alpha) = (1 - G) \phi_1(\alpha) + G \phi_2(\alpha)$$, then we have $$m = V(\alpha) + 5 \log_{10}(r \Delta),$$ so $$V(\alpha) = m - 5 \log_{10}(r \Delta).$$
The $$5 \log_{10}(r\Delta)$$ term comes from the inverse square law; increasing either $$r$$ or $$\Delta$$ by a factor of 10 decreases the observed flux by a factor of 100 and increases the apparent magnitude by 5.