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Let say we measure the superficial brightness of an object in the $B$ band, $I_{B}$ measured in $\displaystyle\frac{L_{*}}{\mbox{pc}^2}$ where $L_{*}$ is solar luminosity and $\mbox{pc}$ is parsec. I have to show that, in magnitudes per arc sec

$$\mu_{B}=27.05-2.5\log(I_{B}) $$

My guess is that the 27.05 comes from conversion factors but I don't see it actually. It is not stated but I think that $\mu_{B}$ is the apparent magnitude in the $B$ band.

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up vote 1 down vote accepted

$\mu_{B}$ is surface brightness of an extended source in the B-band. This is given as (apparent) magnitude per square arc seconds.

You need to calculate the area you observe at the distance R through an angle of 1 X 1 arc seconds. With some trigonometry applied you will get; $$ A = 4 R^{2} \frac{1 - \cos \varphi}{1 + \cos \varphi} $$ Where the angle is 1''.

The luminosity of this area is; $$ L_{\Box} = I_{B} A $$ The radiation from the area you observe is spread over a sphere with the radius R. So you calculate the flux at a distance of R. This will eliminate the distance from the equations; $$ F_{\Box} = \frac{L_{\Box}}{4 \pi R^{2}} = \frac{I_{B}}{\pi} \frac{1 - \cos \varphi}{1 + \cos \varphi} $$ Furthermore you need the flux of the Sun. You probably had the Suns absolute magnitude in the B-band stated in the original exercise ($M_{B,\odot} = 5.48$). Absolute magitudes are calculated at the distance of 10pc. Using units of solar luminosity and parsec, the flux is; $$ F_{\odot} = \frac{L_{\odot}}{A_{Sphere, 10 pc}} = \frac{1}{4 \pi 10^{2}} = \frac{1}{400 \pi} $$ For the final step, we need an equation to calculate magnitudes. This one for instance; $$ m_{1} - m_{2} = -2,5 \log_{10} \left( \frac{F_{1}}{F_{2}} \right) $$ By assigning $m_{1} = \mu_{B}$, $m_{2} = M_{B,\odot}$, $F_{1} = F_{\Box}$ and $F_{2} = F_{\odot}$ and by recalling that $\log (a * b) = log (a) + log (b)$ you should get the final answer.

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