In an E0 (round) elliptical galaxy, the surface brightness is given by where r i

Question

In an E0 (round) elliptical galaxy, the surface brightness is given by

where r is the radius out from the center of the galaxy, I0 is the central surface brightness, and re is the eective radius of the galaxy. The constant A = 7.67. I(r) is given in units of ergs/s/cm2/(arcsec)2, and the radii r and re are both given in arcsec.

i) What is the total ux of the galaxy F in units of ergs/s/cm2?
ii) What is the ux Fe within a projected radius of re?
iii) Show that Fe = F/2, so that the eective radius contains one half of the light in projection.

Explanation / Answer

(a) Since I = I0 e-A(r / re)1/4, therefore we have to first find r, which is

- A(r / re)1/4 = log[I/I0]

r / re = -(1/ A x log[I/I0])4 , therefore,

r = -re x (1/A x log[I/I0])4 , therefore, the total flux density will be

F (directly proportional to) r-2 . therefore,

F = k/r2 , where k is a constant,

F = -k/[re2 x (1/A x log[I/I0])6]

(b)  Taking similar calculations of the part(a) we get,

Fe = -k/[ r2/(1/A log[I/I0])6]

Fe = -k/[A6 r2 /(log[I/I0])6]

(c) If re = r/2, therefore,

F = Fe

-k/r2 = -k/re2 , since re = r/2, therefore, dividing first equation by second, we get,

r2 = re2 , taking underoot both sides, we get,

r = re

F/-k/r2 = Fe/-k/re2 , by cross multiplication we get

F x - k/re2 = Fe x - k/r2 ,

F/re2 = Fe/r2

F/2 = Fe.

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