PROBLEM SET 1
SOLUTIONS
Astronomy 208 -- Fall 2000

1.

Estimate the magnitude distribution of stars in a $1^{\circ}$ patch of sky To get the distribution of stars in the Galaxy, we can start with the IMF (Initial Mass Function, not International Monetary Fund). Note that the IMF is defined as the total number of stars ever formed per square parsec in the galaxy, which is different from the PDMF (Present Day Mass Function), which is the number of stars currently in existence per square parsec. The difference lies in the fact that some stars will have died since the formation of the galaxy, so using the IMF will overestimate the number of stars. For a constant birthrate of stars, the PDMF ($\phi$) is related to the IMF ($\xi$) by

$$\phi(\log M) = \left\{ \begin{array} {lc} \xi(\log M) T_{ms}... ...T_{ms}<T_0\ \xi(\log M), & T_{ms}\geq T_0\ \end{array}\right.$$ (1)

where the PMDF and IMF are both in terms of stars per square parsec per logarithmic mass interval, T_ms is the main sequence lifetime of a star, and T₀ is the age of the galaxy.

We can model the IMF as a broken power law as in equation (32) of Miller & Scalo (1979) [henceforth MS],  

$$\xi(\log M) \propto \left\{ \begin{array} {lc} M^{-0.4}, & 0... ... M \leq 10\ M^{-2.3}, & 10 \leq M \leq 62\ \end{array}\right.$$ (2)

where M is in units of $M_\odot$ and $\xi(\log M)$ is the number of stars per square parcecs per log mass interval. I have used the values given in Table 7 of MS, excluding the normalization constants, since I will renormalize this function later.

I will assume that T_ms=T₀ for $1\,M_\odot$, and that to get

 

T_ms/T₀ = M^-2.5.

Using equation 2, we find that the PDMF is  

$$\phi(\log M) \propto \left\{ \begin{array} {lc} M^{-0.4}, & ... ...eq M \leq 10\ M^{-4.8}, & 10 \leq M \leq 62.\end{array}\right.$$ (4)

Now we need to convert this into a magnitude distribution. I will assume that $L/L_\odot = (M/M_{\odot})^{3.5}$. Luminosity is related to absolute magnitude by so  

$$M_v = 4.72 - 8.75 \log (M/M_\odot).$$ (5)

or  

$$M/M_\odot = 10^{[(4.72-M_v)/8.75]}.$$ (6)

The conversion from $\phi(\log M)$ to $\phi(M_v)$ is

$$\phi(M_v) = \phi(\log M) \left\vert\frac{d\log M}{d M_v}\right\vert$$ (7)

Substituting in equation (4):  

$$\phi(M_v) \propto \left\{ \begin{array} {lrcl} 10^{0.4M_v/8.... ...4.8M_v/8.75}, & -4.03 &\geq M_v \geq& -10.96.\end{array}\right.$$ (8)

(Note: a shortcut is to take the luminosity function of stars per unit absolute magnitude per pc³ directly from Figure 1 of MS.)

If we normalize equation(8) so that and that the power-law segments are continuous at the boundaries, we find  

$$\phi(M_v) = \left\{ \begin{array} {lrcl} 0.040\times 10^{0.0... ...10^{0.55 M_v}, & -4.03 &\geq M_v \geq& -10.96\end{array}\right.$$ (9)

In order to convert this from a distribution in absolute magnitude (M_v) to a distribution in apparent magnitude (m_v) and distance (d) we need the relation

$$M_v = m_v - 5 \log(d/10\,\pc).$$

Since we eventually will be integrating this over the distance, I will rewrite the limits in terms of d:  

$$\phi(m_v,d) = \left\{ \begin{array} {lrcl} 0.040\times10^{0.... ...5} &\leq d \:(\pc) \leq& 1600\times10^{m_v/5}\end{array}\right.$$ (10)

Note that the ranges on d are dependent on m_v. The limits correspond to the minimum and maximum distances at which a star in a given mass range will have an apparent magnitude of m_v.

To get the total magnitude distribution, we need to integrate this over the volume of space seen in 1 square degree. This integral has the following form:  

$$\phi_{\mathrm{tot}}(m_v) = \int_V \phi(m_v,d) \: n_\star(d) \: dV$$ (11)

where $n_\star$ is the number density of stars and $\phi(m_v)\,dm_v$ is the number of stars with apparent magnitude between m_v and m_v+dm_v.

To determine $n_\star$, I will assume that the stellar density decreases exponentially as a function of Galactic radius, and also falls off exponentially with distance from the midplane of the disk, so  

$$n_{\star} = n_0 e^{-r/r_0}e^{-z/z_0}$$ (12)

where r₀ is the scale radius and z₀ is the half the scale height of the Galaxy. Using values from p. 680 of Binney & Merrifield (1998), r₀=3 kpc and z₀=180 pc. We will also assume that the sun is 8 kpc from the Galactic center. In the vicinity of the sun (r=8 kpc and z=0 pc), $n_\star \sim 0.5\,\pc^{-3}$. (I get this from the fact that the nearest star is about 1 parsec away, so the number density of stars should be bit less than 1 pc^-3.) Solving, I find .

The volume integral is over the volume seen in one square degree. As we go further out in d, the cross-sectional area increases, as shown in Figure 1. The volume element dV can be rewritten as

$$dV = A\,dd = (d\theta)^2\,dd = \theta^2 d^2 \, dd.$$ (13)

where $\theta = 1^{\circ}= \pi/180\,\mathrm{radians} = 0.0175$.

Some if you integrated with and .This is incorrect. See the Appendix for the correct way to do it.

 

Figure 1:  

Thus, equation (11) becomes

where $n_\star$ depends on the line of sight.

 

Figure 2:  

(a)

Toward the Galactic center: If the distance from the earth/sun is d and the distance from the Galactic center is r, then as we move toward the Galactic center, r=8-d kpc. Past the Galactic center, r=d-8 kpc. Substituting into the equation for $n_\star$ and taking z=0,

Note: d needs to be converted to parsecs to be consistent with the units used in the integral.

The resulting distribution (via Mathematica) is plotted in Fig. 2 as stars.

(b)

Toward the Galactic anticenter:

$180^{\circ}$ away from Galactic center, d=r+8 kpc, so

$$n_\star(d) = 7.2\,\pc^{-3} \, e^{-(d+8000)/3000}$$ (14)

where d is in pc.

The resulting distribution (via Mathematica) is plotted in Fig. 2 as squares.

(c)

Directly overhead:

Here, r=8 kpc is constant and z changes:

$$n_\star = n_0\,e^{-0.8} e^{-z/z_0} = 0.5 \mathrm{cm}^{-3}\,e^{-z/z_0}$$ (15)

where z₀ is 180 pc.

The resulting distribution (via Mathematica) is plotted in Fig. 2 as triangles.

2.

Estimate the number of stars per square degree to a limiting magnitude of $m_v=14\,\mathrm{mag}$

For this problem, we just integrate the functions calculated in Problem 1 for $m_v \leq 14,\mathrm{mag}$. I will also choose an arbitrary minimum magnitude of $m_v \geq -1$, since the brightest star visible from earth has a visual magnitude of -1. (Really, if your lower limit is between -1 and $-\infty$, your answer shouldn't be affected too much.)

(a)

Toward the Galactic center:

$$\fbox {$2.1\times10^5$\space stars per square degree.}$$

(b)

$180^{\circ}$ away from Galactic center:

(c)

Directly overhead:

3.

Estimate the HI column density

Let's assume that the average HI number density in the solar neighborhood is $1\,\mathrm{cm}^{-3}$. From §9.2 of Binney & Merrifield, the distribution of HI is approximately uniform out to 16 kpc from the center of the Galaxy. So, let's assume that the gas in uniform in the radial direction, and falls off exponentially with radius with a scale height of 200 pc. (See Fig. 3 for illustration.) Then,

$$n_H = (1\,\mathrm{cm}^{-3}) \times e^{-z/200\,\pc}.$$ (16)

 

Figure 3:  

The column density along a given line of sight is

$$N_H = \int_\ell n_H \, d\ell$$ (17)

where $\ell$ is the length of the column.

(a)

Toward the Galactic center In the Galactic plane, n_H is constant. Thus,  

$$N_H = n_H\ell = 1\,\mathrm{cm}^{-3} \times \ell \times \fra... ...m}}{\pc} = (3.1 \times 10^{18} \mathrm{cm}^{-2})\: \ell [\pc]$$ (18)

The total column length toward the Galactic center is . So the total column density is

$$N_H = (3.1 \times 10^{18} \mathrm{cm}^{-2})\times(24,000) = \fbox {$7.4 \times 10^{22} \mathrm{cm}^{-2}$.}$$

(b)

$180^{\circ}$ away from Galactic center

This line of sight is also in the Galactic plane so we can use Equation (20). Away from the Galactic center, $\ell = 8\,\mathrm{kpc}$ so

$$N_H = (3.1 \times 10^{18} \mathrm{cm}^{-2})\times(8000) = \fbox {$2.5 \times 10^{22} \mathrm{cm}^{-2}.$}$$

(c)

Directly overhead

Out of the plane of the Galaxy, $\ell = 200\,\pc$ so

$$N_H = \int_\ell n_H\, d\ell = \int_0^{\infty} (1\,\mathrm{c... ... (200\,\pc) = \fbox {$6.2 \times 10^{20} \mathrm{cm}^{-2}.$}$$

4.

Estimate the extinction, using  

$$A_v[\mathrm{mag}] = N_H \times 10^{-21} [\mathrm{cm}^{-2}]$$ (19)

(a)

Per kpc:

i. & ii.

In the plane of the Galaxy, we derive the extinction per kpc from Equation (20). In 1 kpc, the column density is

$$N_H (1\,\mathrm{kpc}) = (3.1 \times 10^{18} \mathrm{cm}^{-2})\times1000 = 3.1 \times 10^{21}\,\mathrm{cm}^{-2}$$

so the extinction per kpc is

$$\frac{A_v}{1\,\mathrm{kpc}} = \frac{N_H}{1\,\mathrm{kpc}}\times10^{-21} = \fbox {3.1 mag/kpc}$$

toward both the Galactic center and anticenter.

iii.

Out of the plane of the Galaxy, we can define the extinction per kpc as where $\ell$ is in units of kpc.

(b)

in total:

For this part, we can simply take the total column densities calculated in Problem 3 and plug into Equation (21).

i.

Toward the Galactic center so

$$A_v = 7.4 \times 10^{22} \times 10^{-21} = \fbox {74 mag}$$

ii.

$180^{\circ}$ away from Galactic center

so

$$A_v = 2.5 \times 10^{22} \times 10^{-21} = \fbox {25 mag}$$

iii.

Directly overhead

so

$$A_v = 6.2 \times 10^{20} \times 10^{-21} = \fbox {0.62 mag}$$

5.

Suppose Barnard photgraphed the Taurus dark clouds with a limiting magnitude of $m_v=14\,\mathrm{mag}$.

First, we should find the position of the Taurus clouds in terms of galactic coordinates. Taurus is at and the conversion to galactic coordinates is

where (l,b) are Galactic longitude and latitude, respectively, $(\alpha_{GP},\delta_{GP})=(192.9^{\circ}, 27.1^{\circ})$ is the RA and declination of the North Galactic Pole and $l_{CP}=123.932^{\circ}$ is the Galactic longitude of the North Celestial Pole. Solving, we find that the Taurus clouds are at $l=173^{\circ}, b= -17^{\circ}$, indicating that Taurus is away from the Galactic center ($l=0^{\circ}$ points toward the Galactic center).

(a)

How many stars per square degree could he expect without the ``extra'' extinction due to his ``dark nebulae?'' This calculation is similar to Problem 2, but we have to take two things into account: (1) the line of sight, which is non-trivial, and (2) extinction as described in Problem 4.

 

Figure 4:  

The line of sight is $17^{\circ}$ out of the plane of the Galaxy, and $173^{\circ}$ offset from the Galactic center (see Fig. 4). We can decompose d into its projections onto the plane of the galaxy and the z-axis. From Fig. 4, $d_z = d\sin 17^{\circ}$ and $d_r = d\cos 17^{\circ}$. In particular,

$$z = d_z = d\sin 17^{\circ}.$$

To calculate r, you can do some more trigonometry, saying that r is the third side of a triangle with sides 8 kpc and d_r with an interior angle of $173^{\circ}$, and get

$$r = \sqrt{(8\,\mathrm{kpc})^2+d_r^2-2(8\,\mathrm{kpc})d_r\cos173^{\circ}}$$

or you can say that $173^{\circ}$ is awfully close to $180^{\circ}$, and

$$r = d_r+8\,\mathrm{kpc}= d\cos 17^{\circ}+ 8\,\mathrm{kpc}.$$

I will opt for the latter. Thus $n_\star$ is

$$n_\star(d) = n_0 e^{-r/r_0-z/z_0} = (7.2\,\pc^{-3})\exp\lef... ...pc}}{3\,\mathrm{kpc}} -\frac{d\sin17^{\circ}}{180\,\pc}\right]$$

The increase in apparent magnitude due to extinction is simply the A_v calculated in Problem 4. So,

$$m_v = M_v + 5 \log(d/10\,\pc) + 3.1(d/1\,\mathrm{kpc}).$$ (20)

The integral we need to solve looks like  

$$N_\star = \int_{-1}^{14} \phi_(m_v) \:dm_v= \int_{-1}^{14} \int_d \phi(m_v,d) \: n_\star(d) \,\theta^2d^2\:dd\,dm_v.$$ (21)

After plugging into my numerical integrator (Mathematica) I find

with no extinction
and
including extinction.

(b)

How much extinction would be associated with ``almost black'' and ``totally black'' regions?

I have plotted the cumulative number of stars visible out to a a magnitude m_v in Figure 5. Additional extinction due to the Taurus clouds has the effect of shifting this curve to the right, or to higher magnitude.

 

Figure 5:  

i.

``almost black'' regions: Barnard's ``almost black'' have approximately 30 stars per square degree. Looking at Fig. 5, we see 30 stars per square degree up to a limiting magnitude of about 10.4. Thus the extra extinction is

$$A_v = 14-10.4 = \fbox {3.6 mag.}$$

ii.

``totally black'' regions:

Let's say the cutoff for a ``totally black'' region is 1 star per square degree. Then Fig. 5 indicates that a limiting magnitude of about 7 is appropriate. Thus

$$A_v = 14-7 = \fbox {7 mag.}$$