I know that the equation for the difference in apparent magnitudes can be represented as: $\Delta m=-2.5\log\left(\dfrac{I_{1}}{I_{2}}\right)$
and the apparent brightness of a celestial object is shown as:
$\ B=\dfrac{L}{4\pi\ D^2}$

Some people in my class told me that$\ B$ and$\ I$ were the same thing but used in different cases. This explanation didn't help me though. While playing around with the equations, I found this nice relation:
$\ \dfrac{B_{1}}{B_{2}}=\left[\dfrac {\left(\dfrac{L}{4\pi\ D_{2}^2}\right)}{\left(\dfrac{L}{4\pi\ D_{1}^2}\right)} \right]$
$\ \space\ \space\ \space\ ={\dfrac{4\pi\ D_{1}L}{4\pi\ D_{2}^2L}}$
$\ \space\ \space\ \space\ ={\dfrac{D_{1}^2}{D_{2}^2}}$

So if$\ {B_{1}}={I_{1}}$ and$\ B_{2}=I_{2}$holds true, is this following equation also true?

$\ {\dfrac{D_{1}^2}{D_{2}^2}} \space\ = 10^\left(-{\dfrac{\Delta {m}}{2.5}}\right)$
$\ \space\ \space\ \space\ \space\ =\dfrac{1}{10^\left({\dfrac{\Delta {m}}{2.5}}\right)}$

However, if$\ B$ and$\ I$ are not the same, what is the difference between the two measurements? And can either one be used to accurately measure how bright a non-luminous celestial body (ie. a planet) appears from Earth?

Also... does $\ m$ have an SI unit? I've never used any units when I solve for it


1 Answer 1


Yes, $B$ and $I$ are the same things in this context. The $\Delta m$ you derived goes by the name "distance modulus" when one of the distances is set to $10\operatorname{pc}$. The thing to be careful of is that $I$ is also sometimes used for spectral radiance (or surface brightness) when working with radiative transfer. Spectral radiance doesn't obey the inverse square law, though, so there's no risk that we're confusing them here.

  • $\begingroup$ Ok thank you. So since one distance has to be set to 10pc, does this mean that I'm not able to use this to calculate the apparent brightness of a planet within our solar system? $\endgroup$ Oct 19, 2016 at 22:39
  • 1
    $\begingroup$ Not at all. The formula works whether one distance is set to 10 parsecs or not, it just has a special name when one of them is set there (I don't recall which one, because I never use magnitudes, myself). $\endgroup$
    – Sean Lake
    Oct 19, 2016 at 22:41
  • 4
    $\begingroup$ Absolute magnitude $=$ the magnitude at 10 parsecs. $\endgroup$
    – ProfRob
    Oct 19, 2016 at 22:49

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