\[I(R)=I_e exp\left(-b_n \left(\left(\frac{R}{R_e}\right)^{1/n} -1 \right) \right)\]
The constant bn 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.
Another way to write the exponential profile is:
\[I(R)=I_0 exp(-R/b)\]
where I0 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.
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.
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)?
Since b is in the denominator of a term in the arguments of an exponential, when R=b, the coefficient of the exponential is multiplied by 1/e (e=2.718...). Therefore, b is a scale length, such that every length of b that one moves away from the center of the galaxy, the surface brightness decreases by a factor of e.
\[I(R)=I_0 exp(-R/b)\]
\[I(b)=I_0 exp(-b/b)=I_0 exp(-1)=\frac{I_0}{e}\]
To determine e for a galaxy, one would first need a telescope large enough to resolve the galaxy. By taking pictures of the galaxy with a CCD, one could use the image light counts at different radii of the galaxy to construct a surface brightness profile for the galaxy. One could then compare their profile to the exponential profile and solve for b.
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.
On this plot, radius b is at the dark vertical line, and the y-axis measures the fraction of the central brightness.
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)?
This is going to involve some calculus. First, let's use this profile to get the total brightness of the galaxy. Since we are using polar coordinates, we will need to introduce a factor of R into our integration.
\[I_T=\int^{\infty}_0 \int^{2\pi}_0 I_0 Re^{-R/b} dR d\theta\]
\[I_T=2\pi I_0 \int^{\infty}_0 Re^{-R/b} dR\]
This involves integration by parts, so we'll let Wolfram Alpha take care of that for us.
\[I_T=2\pi I_0 b^2\]
Now that we know the total brightness of the galaxy, let's solve for the part inside the sun's galactic radius.
\[I_{r_*}=2\pi I_0 \int^{r_*}_0 Re^{-R/b} dR\]
\[I_{r_*}=2\pi I_0 b(b-(r_*+b)e^{-r_*/b})\]
Taking the ratio of this with the total brightness gives us an equation for brightness fraction as a function of radius.
\[\frac{I_{r_*}}{I_T}=\frac{2\pi I_0 b(b-(r_*+b)e^{-r_*/b})}{2\pi I_0 b^2}\]
\[\frac{I_{r_*}}{I_T}=\frac{b-(r_*+b)e^{-r_*/b}}{b}\]
Now let's plug in our numbers, R=8 kpc and b=3.5 kpc.
\[\frac{I_{r_*}}{I_T}=\frac{3.5-(8+3.5)e^{-8/3.5}}{3.5}\]
\[\frac{I_{r_*}}{I_T}=.67\]
We know that brightness directly scales to number of stars, so 67% of the stars in the Milky Way are inside the sun's galactic radius.
I worked with B. Brzycki, G. Grell, and N. James on this problem.
Very nice.
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