Question

You observe a galaxy face-on. The galaxy has a surface luminosity density of S solar luminosities per square parsec (in some filter band). Prove that the observed surface brightness μ in magnitudes per square arcsecond (in that same filter band) is given by μ=M⊙ +21.572−2.5log10S where M⊙ is the absolute magnitude of the Sun in this filter band.

Answer 1

Answer:

given that

An object surface brightness depends on its surface luminosity density, i.e , its luminosity emitted per unit surface area.

in visible and infrared astronomy, surface brightness is often quoted on a magnitude scale, in magnitudes per square arcsecond in a particular filter band or photometric system.

The galaxy has a surface luminosity density of s solo luminositie per square parse.c. since we know the total luminosity of a galaxy is given by Lot = '21 2(R) RdR I(R) is Radially averaged function of Intensity and we also known that sustace brightness (M) in unit of magnitude per square arcsec we can relate surface brightness as Uz -3.510910 [+c I is measured in Lopet Now determine the amount of light in a square arcsec for a galary observed at distance (d) L-Id802 60 - Yodian 206265 Then the distance modulus formula become mam +50ggiodicipe)s where, m is Absolule mass solar absolute Now in terms of magnitude mo

Mz-2.5logio 440 + MO Now if we plug in L for the paten of tright in Isquare arsec in the above formula recognised that um then inz -25.logio Cidazoy+5logo-s+ mo a= -25 logic In slog, (so) -5+Mo Nes - .Stoglo II (Lopen) +M@ +21572 we know I | (LOP (²) = 5 surface luminosity Density) le= NO +21.512 - 3. stogos