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# The relationship between apparent magnitude and radiative flux for the same spectral band, a derivation

2 replies to this topic

### #1 David Sims

David Sims

Sputnik

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Posted 16 September 2020 - 11:24 AM

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.

### #2 TOMDEY

TOMDEY

Fly Me to the Moon

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Posted 16 September 2020 - 11:54 AM

Very good! An idealized bolometer is presumed to be a perfect lambertian black body absorber, 100% efficient. For practical applications, the actual spectrum of the stellar source vs that of the sun is by far the biggest driver, so the ratio of the integrations of those relative to the presumptive detector spectral responsivity vs plain vanilla (L/Lsun)... is what actual practitioners use. Otherwise, it's almost always idealized apple /vs/ real orange. And yeah, six significant figures is probably sufficient... thirty-two is a wee-bit too accurate.  Tom

[I worked quantitative radiometric metrology and calibrations for our imaging satellites... interesting stuff! The Gremlins are in the details.]

### #3 David Sims

David Sims

Sputnik

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• Loc: Hillsboro, West Virginia, USA

Posted 16 September 2020 - 04:25 PM

I kept all the digits returned by the "old" Windows calculator. Maybe I should have rounded, but I didn't bother.

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