Blog #17: WS6.1, #2

2. The Hubble Classes have characteristic surface light distribution profiles. They are fairly well-described by parametrized equations in the form I(R) where I = intensity and R = distance from the center. Often each is scaled to their value at the effective radius, such that \(I_e=I(R_e)\). The most general profile is the Sérsic Profile, given by 
\(I(R)=I_eexp(-b_n[(\frac{R}{R_e})^{1/n}-1])\)
The constant \(b_n\) depends on the shape parameter n. n = 4 gives rise to the famous \(r^{1/4}\)-law, or the de Vaucoleur Profile, which approximates elliptical and the bulge of spiral galaxies. n = 1, on the other hand, is equivalent to a simple exponential profile, which often corresponds with the outskirts of spiral galaxies. The best fit is often obtained by a combination of the functional forms. 
a) Another way to write the exponential profile is \(I(R)=I_0exp(-R/b)\), where \(I_0\) is the central surface brightness and b is a characteristic lengthscale, a constant. 
i. Describe what b is, and suggest how one might measure it for a given spiral galaxy. 

b is the scale radius of the galaxy, which is defined as the radius that contains a specific proportion of the matter in the galaxy. To measure the scale radius, you could measure the fluxes in "bins" across the galaxy, then fit those measurements to the function \(I(R)=I_0exp(-R/b)\). Since you know how far each bin is from the center of the galaxy, you should be able to calculate b.

ii. The Milky Way has an estimated b = 3.5kpc. Plot the disk surface brightness profile relative to the central brightness and indicate the location of the characteristic lengthscale from the center. 

At the scale radius, b = R, so we can solve \(I(R)=I_0exp(-b/b)\). 

\(I(b)=I_0e^{-1}\)

\(I(b)=\frac{I_0}{e}\) 

iii. Assuming the entire Milky Way (including the central bulge) can be described by this profile, and that it is circularly symmetric. Approximately what fraction of all its stars are interior to the Sun (at 8kpc)? 

To find the fraction of the stars that are interior to the sun, we want to divide the total intensity in the sphere with radius = 8kpc by the total intensity of the Milky Way, regardless of radius. To find these total intensities, we can integrate the intensity function over the desired range. 

The intensity interior to the Sun: 

\(\int_{0}^{8}I(r)dr=I_0\int_{0}^{8}e^{-r/b}dr=-I_0be^{-r/b}|^8_0\)

The intensity of the entire Milky Way: 

\(\int_{0}^{\infty}I(r)dr=I_0\int_{0}^{\infty}e^{-r/b}dr=-I_0be^{-r/b}|^{\infty}_0\)

We can then take a ratio (thus canceling out the constants): 

\(\frac{e^{-r/b}|^8_0}{e^{-r/b}|^{\infty}_0}\)

Finally, we can evaluate: 

\(\frac{e^{-8/b}-e^0}{e^{-\infty/b}-e^0}=\frac{e^{-8/b}-1}{-1}=1-e^{-8/b}\) of the stars in the Milky Way are interior to the Sun.