Blog Post 17, Worksheet 6.1, Problem 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 centre. Often each is scaled to their value at the effective radius, such that \(I_e = I(R_e)\). The most general profile is the Sersic Profile, given by the equation \[I(R) = I_e \:\text{exp}(-b_n[(\frac{R}{R_e})^{1/n}-1])\] 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 = 4, 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: \[I(R)=I_0\:\text{exp}(-R/b)\] 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. Be as specific as possible about what kind of data and tools may be required. 

From the equation and given value for the Milky Way, we can infer b is the scale radius of the spiral galaxy, or the radius at which the galaxy's intensity has decreased by a factor of e. To measure the scale radius for a given spiral galaxy, one could use a telescope with a CCD to observe the brightness of the galaxy at different radii. From these data points, one could create the light curve for the galaxy across different radii, and find the radius at which the intensity has decreased by a factor of e.

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

The Milky Way's exponential profile is given by: \[I(R) = I_0\text{exp}(-R/3.5 \text{ kpc})\]The following is a plot showing the exponential profile of the Milky Way, given a central surface brightness of \(I_0 = 1.0\). (Not the actual value).

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 8 kpc)?

To determine the fraction of the Milky Way's stars that are interior of the Sun, we must find the ratio of the intensity at all radii interior the the Sun to the intensity at all radii, from zero to infinity. As we are assuming that the Milky Way is circularly symmetric, the total intensity between the center of the galaxy and an unspecified radius r is given by the intensity equation above integrated around the circumference of a circle at a given radius: \[\int_{0}^{r}(2πR)I_0e^{-R/b}dR,\] with b = 3.5 kpc, and \(I_0\) = the central surface brightness of the Milky Way. Thus, the ratio we are looking for is: \[\frac{\int_{0}^{8}(2πR)I_0e^{-R/b}dR}{\int_{0}^{\infty}(2πR)I_0e^{-R/b}dR}\] Using u-v integration by parts , with u = R and dv = \(e^{-R/b}dR\), we see that the general integral can be expressed as the following: \[\int_{0}^{r}(2πR)I_0e^{-R/b}dR\ = 2πI_0\int_{0}^{r}Re^{-R/b}dR\] \[=\:2πI_0[-Rbe^{-R/b}\rvert_{0}^{r} + b\int_{0}^{r}e^{-R/b}dR]\] \[= 2πbI_0(b-e^{-r/b}(r+b))\] The limit of this expression as r goes to zero is just \(2πI_0b^2\), so our fraction becomes: \[\frac{2πbI_0(b-e^{-r/b}(r+b))}{2πI_0b^2}\] \[\frac{b-e^{-r/b}(r+b)}{b}\] With b = 3.5 kpc and r = 8 kpc, our fraction is equal to 0.666, or approximately 2/3.