Talk:Surface brightness

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I think the equation is not correct due to Dimensions of the parameters,

[edit] equation derivation

This derivation isn't very easy to follow. Among other things ,some of the terms aren't defined. Privong (talk) 19:20, 22 February 2008 (UTC)

[edit] Examples

The examples also seem quite pointless... Stating "plug into the equation and get the answer" isn't very informative. Privong (talk) 19:21, 22 February 2008 (UTC)

I removed the textbook-like derivation and examples (per WP:NOT). For reference, here is the text I pulled. Ashill (talk) 19:28, 10 March 2008 (UTC)

Converting Surface Brightness in magnitude per square arcseconds to Solar Luminosity per square parsec

Formula

$S(mag/arcsec^2)=M_{V,\odot}+21.57-2.5\log S (L_{\odot}/pc^2)$

Where $M_{V,\odot}=$ Absolute Magnitude of the sun in V band (in our example)

NOTE: Other absolute magnitudes of the sun can be got from Galactic Astronomy or Absolute Magnitude of the Sun in Several Bands

Derivation

Assume that we are interested in finding the surface brightness of a galaxy at a distance d Mpc.

The luminosity enclosed within a box of 1 arcsec on each side is: $L_{V} \sim d^2\times a^2 \times S(L_{\odot}/pc^2)$

where $a=\frac{4.848 \times d}{d} \frac{pc}{Mpc} = 4.848\times 10^{-6}$ for a galaxy (show it)

$S(mag/arcsec^2)-M_{V,\odot}=-2.5\log L_{V}+5\log(d/10)$

substituting

$S(mag/arcsec^2)-M_{V,\odot}=-2.5\log(d^2\times a^2\times S(L_{\odot}/pc^2))+5\log d +5$

expanding and simplifying

$S(mag/arcsec^2)=M_{V,\odot}-2.5\log(S(L_{\odot}/pc^2))-5\log a-5$

Substituting for a, we get the required result.

Examples

(Problem 1.10 of Galaxies in the Universe: An Introduction): Show that the surface brightness $I_{B}=27 mag/arcsec^{2} \sim 1 L_{\odot}/pc^{2}$

Ans:
It is known that . 
Substituting into the formula and re-arranging, we get $\log S(L_{\odot}/pc^2)\sim 0$ which gives the required answer.

(Problem 5.2 of Galaxies in the Universe: An Introduction):

Show that the central surface brightness of 15 $m a g / a r c s e c 2$ in the I band corresponds to $18000 L_{\odot}/pc^2$

Ans:

$M_{I,\odot}=4.08$

Substituting in to the formula the answer is arrived at.