A question often asked on forums such as Quora or Yahoo Answers or Physics Forums is what's the relationship between radiative flux and apparent magnitude (for the same spectral band). The correct answer is seldom forthcoming. But it isn't hard to figure out.

First, let's acknowledge that

1 parsec = 30856775814913673 meters

That'll be important down the page.

m : apparent magnitude

M : absolute magnitude

L : luminosity in watts

F : flux in watts per square meter

D : distance in parsecs

d : distance in meters

The distance modulus equation

m − M = 5 log D − 5

The inverse square law for radiative flux

F = L/(4πd²)

The magnitude scale definition

M = M☉ − 2.5 log (L/L☉)

M = M☉ − 2.5 log L + 2.5 log L☉

L = 4πd²F

M = M☉ + 2.5 log L☉ − 2.5 log(4πd²F)

M = M☉ + 2.5 log L☉ − 2.5 log(4π) − 5 log d − 2.5 log F

D = d / 30856775814913673

5 log D = 5 log(d) − 5 log(30856775814913673)

5 log D = 5 log(d) − 82.446752726110691733478528419098

m − M = 5 log D − 5

m − M = 5 log d − 87.446752726110691733478528419098

m − [M☉ + 2.5 log L☉ − 2.5 log(4π) − 5 log d − 2.5 log F] = 5 log d − 87.446752726110691733478528419098

m = M☉ + 2.5 log L☉ − 2.5 log(4π) − 2.5 log F − 87.446752726110691733478528419098

M☉ = 4.7554

L☉ = 3.827e26 watts

2.5 log(4π) = 2.7480246600552406119468651943497

m = 4.7554 + 66.457146155561248277821578807329

− 2.7480246600552406119468651943497

− 87.446752726110691733478528419098

− 2.5 log F

m = −2.5 log F − 18.982231230604684067603814806119

That's the equation that I derive, using my values for the sun's bolometric absolute magnitude, M☉, and bolometric luminosity, L☉. A possibly more authoritative source, whose name and publisher I've forgotten, gave this very similar equation, which agrees with mine to six significant figures. You should probably use his equation. Mine is just for showing how it can be found.

m = −2.5 log F − 18.982249379206

**Edited by David Sims, 16 September 2020 - 11:31 AM.**