Download Problem Set 2

Physics of Galaxies 2013 Assignment 2
Due: Friday, 24 May 2013
1. As we saw when we discussed the cosmic distance ladder, the surface brightness fluctuation
(SBF) method for distance determination depends on the weighted mean luminosity hLi,
where
Σi N̄i L2i
,
(1)
hLi =
Σi N̄i Li
and N̄i is the mean number of stars of luminosity Li in each resolution element (like a
pixel). Since hLi is a luminosity, if you know (or can calibrate) hLi for a galaxy, you can
determine its distance by measure the apparent “surface brightness fluctuation mangitude.”
Use the (total) luminosity function shown in Figure 1 (Figure 2.3 of SG, below) to compute
hLiV for the local Solar Neighborhood. The absolute magnitude of the Sun in the V -band
is MV = 4.78 mag. What absolute magnitude does hLiV correspond to? Is this a bright
star or a faint star? Describe why SBFs are difficult to use for young populations like those
in disk galaxies.
Figure 1: Fig 2.3 from SG
2. Recently, a number of galaxies with circumnuclear H2 O masers have been found, like the
famous “maser galaxy” NGC 4258. Like that galaxy, it is possible to measure the distances
to these galaxies and the masses of their black holes.
From Figure 2 (from Reid et al. 2009), we see that the masers in the distant galaxy UGC
3789 seem to follow a distribution in velocity versus radius that looks like a disk with a
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Figure 2. Map of the relative positions of individual maser spots toward
UGC 3789. High-velocity blueshifted (blue) and redshifted (red) masers straddle
the systemic masers (green and expanded view inset) and the linear arrangement
of spots suggests that we are viewing a nearly edge-on rotating disk, similar to
that seen in NGC 4258. The ≈2 mas extent of the maser spots in UGC 3789
is approximately seven times smaller than for NGC 4258, which is consistent
with UGC 3789 being at approximately seven times greater distance. Formal
fitting uncertainties are given in Tables 2, 3 and 4 and are typically less than
0.010 mas.
Figure 3. Position–velocity plot of the maser spots toward UGC 3789. The
high-velocity
blueshifted (blue) and redshifted (red) spots display a Keplerian
√
1/ R rotation curve, indicated by the curved dotted lines. The systemic masers
(green) are consistent with projected positions and velocities for gas in Keplerian
orbit at R ≈ 0.43 mas, indicated by the straight dotted line, but small deviations
from a linear distribution are apparent. Impact parameter is defined as distance
along a position angle of 41◦ east of north from an (east, north) offset of
(−0.4, −0.5) mas; the plus sign (+) at (−0.03 mas, 3265 km s−1 ) indicates the
assumed center of the SMBH.
∼1000 km s−1 at radii of ∼0.1 pc, presumably moving under
the influence of a ∼107 M& SMBH.
UGC 3789 has a recessional velocity of ≈3325 km s−1 and
is well into the Hubble flow. The VLBI results presented in this
paper will be followed by detailed spectral monitoring data
and disk modeling in Paper II to determine the distance to
UGC 3789. This angular-diameter distance, when combined
with its recessional velocity, should yield a direct and accurate
estimate of H0 .
Facilities: VLBA, GBT, Effelsberg
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no continuum
emission
was detected
at a 2σ limit
of less than
Figure 2: The postiion-velocity
diagram
of water
masers
in UGC
3789, from Reid et al. (2009).
0.14 mJy.
central hole. The hole appears to have an opening of angular size θin ≈ 0.43 milliarcseconds
(mas = 10−3 arcsec), and a peak velocity of vin ≈ 685 km s−1 . Furthermore, the masers
have been measured to have an acceleration of v̇ ≈ 4.75 km s−1 yr−1 .
(a) Assume the acceleration of the masers is entirely from centripedal acceleration. What
is the radius rin (in pc) of the central hole, where the velocity vin is maximal?
(b) Given the angular size of the inner hole θin and the physical radius rin you just
determined, what is the distance to UGC 3789?
(c) Correcting for the motion of the Cosmic Microwave Background and for the pull of
nearby galaxies, the radial velocity of UGC 3789 is v = 3481 km s−1 . What is the
value of the Hubble Constant H0 based on the distance you just determined? How
does it compare with the currently accepted value of H0 = 69 km s−1 Mpc−1 (is it
higher or lower)?
3. We want to measure the distance to the center of the Galaxy and the mass of its black
hole, so we spend a few decades observing a star near the Galactic Center. You notice that
one star moves back-and-forth across the sky in a straight line, meaning that its orbit is
edge-on. We take spectra of this star to measure its radial velocity vr from the wavelength
shift of its absorption lines, and we find that this repeats exactly each time the star is
at the same point in the sky. This means that most extreme points of the star’s motion
correspond to pericenter (when it’s closest to the black hole) and apocenter (when it’s
furthest from the black hole), as in Figure 3 (below). We measure the separation s of these
two points on the sky and find s = 0.00248, with an orbital period P = 15.24 yr.
Use Figure 2.1 and Equation 2.1 in SG to show that the orbit’s semi-major axis a is
s d s = 0.5 AU × 00
.
1
1 pc
What is a if the Galactic Center is at a distance of d = 8 kpc?
2
(2)
Figure 3: Fig. 2.6 from SG. Solid lines correspond to the case in Problem 3.
At the extreme points of the star’s motion, we observe the radial velocity to be v0 =
473 km s−1 an v1 = 7326 km s−1 . At which point is it closest to the black hole? Call this
velocity vp and the velocity at the farthest point va (for pericenter and apocenter velocity,
respectively).
Now we can determine the eccentricity e of the star’s orbit. Use conservation of angular
momentum to show that
vp (1 − e) = va (1 + e)
(3)
and there that e = 0.879 for this star’s orbit.
Now we can solve for the black hole’s mass and the distance to the Galactic Center. At a
distance r from the black hole, if the star (which has mass m) is moving with a velocity v,
the star has kinetic energy KE = mv 2 /2 and potential energy PE = −GmMBH /r. Since
the total energy KE + PE cannot change during the orbit, show that
GMBH
4e
2
2
vp − va =
.
(4)
a
1 − e2
If we measure v in km s−1 , MBH in M , and a in pc, then G = 4.3 × 10−3 . Convert a into
AU to show that MBH /M = 3906(a/1 AU).
Now substitute this last equation into Kepler’s Third Law,
P 2 (yr2 ) =
a3 (AU3 )
,
MBH (M )
(5)
to show that a = 952 AU. Finally, compute the distance d to the Galactic Center and the
mass MBH of the Galaxy’s black hole.
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4. Here we’ll compute the distance and absolute magnitude of SN1987A, the Type IIc supernova that exploded in the Large Magellanic Cloud. The ring around SN1987A measures
1.0062 × 1.0018 on the sky. Assume that the ring is circular; what is its inclination i? Now,
the first signal is seen at t0 = 83 d and the second signal is seen at t1 = 395 d. What is the
ring’s radius in parcsec? Recall from class that the ring’s angular radius is the projected
ellipse’s semi-major axis. What is the distance d to the supernova? Finally, at its brightest, SN1987A had an apparent magnitude of mV ≈ 3 mag. What was its peak absolute
magnitude? Given that the Sun has an absolute magnitude of MV = 4.78 mag in the V
band, what was its peak luminosity (in solar units, L )?
5. In cylindrical coordinates centered on the Galactic Center (R, φ, z), the double-exponential
profile for the distribution of stars in the disk is written (SG Eq. 2.8)
n(R, z, S) = n(0, 0, S) exp[−R/hR (S)] exp[−|z|/hz (S)],
where S represents some population of stars with scale length hR (S) and scale height hz (S).
Show that at radius R, the number of stars per unit area – the surface density – of type S
is Σ(R, S) = 2n(0, 0, S)hz (S) exp[−R/hR (S)]. If each star has lmuinosity L(S), the surface
brightness I(R, S) = L(S)Σ(R, S). Assuming that hR and hz are the same for all S, show
that the disk’s total luminosity is LD = 2πh2R I(R = 0).
For the Milky Way, LD ≈ 1.5 × 1010 L in the V band and hR ≈ 4 mathrmkpc. Show
that the disk’s surface brightness at the Sun’s position, 8 kpc from the Galactic Center, is
∼ 20 L pc−2 . The mass density in the disk is 40 − 60 M pc−2 , so M/LV ∼ 2 − 3 (in solar
units). Why is this larger than M/LV ≈ 0.67 for stars within 100 pc of the Sun?
6. This problem is designed to make you understand Malmquist bias, which affects all measurements (not just distance measurements!) when using samples that are defined in terms
of a flux limit (i.e., an apparent magnitude limit). If we observe stars (or anything else)
down to a fixed apparent brightness, we do not get a fair mixture of all of the stars in the
sky (or observed volume), but we include more of the most luminous stars, which creep
into the sample because of the uncertainty in the magnitude determinations at the faint
limit of the sample.
(a) Consider a model of sky with G stars (i.e., like the Sun) in three regions: A (70 <
d (pc) < 90), B (90 < d (pc) < 110), and C (110 < d (pc) < 130). If the density is
uniform, and you have ten stars in region B, how many stars are there in regions A
and B (round to the nearest integer)?
Not all G stars have the same luminosity: the variation corresponds to about 0.3 mag
around the solar value (MV = 4.78 mag). What fractional change in luminosity is
this?
Now, for each star in each region, compute its absolute magnitude: roll a die, note
the number N1 on the upturned face, and give your star an absolute magnitude
MV = MV + 0.3 × (N1 − 3.5). If you like to program, write a program to place the
stars randomly in the volume and assign an absolute magnitude with mean MV and
standard deviation 0.3 mag. What is the mean absolute magnitude of your stars?
Now compute the apparent magnitude for each star. For simplicity, you can assume
that all the stars in region A are at a distance of 80 pc, those in B at 100 pc, and
those in C at 120 pc.
(b) Now observe your sky. Assume that you can only see stars with apparent magnitude
mV ≤ 10 mag. These stars are your sample. What is the mean absolute magnitude
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of your sample? How does this compare to the mean absolute magnitude of all the
stars? Why?
What is the average distance of the stars in your sample? Suppose you assumed that
your sample stars stars had the same average absolute magnitude as the full set of
stars in part (a). Now calculate the distances from their apparent magnitudes: what
is their average distance? Why is this different from the true average distance of the
sample?
(c) As we’ve discussed in class, metal-poor main sequence stars are fainter at fixed color
and therefore fixed spectral class. If the star’s metallicity as a fraction of the Sun’s
metallicity is Z/Z , then the star’s magnitude changes by ∆MV ≈ −0.87 log10 (Z/Z ).
Now determine Z/Z for each of your stars (in the full model sky): roll the die again,
note the number N2 on the upturned face, set Z/Z = (N2 + 0.5)/6, and change the
absolute magnitude by ∆MV .
Once again, observe the sky with your (apparent) magnitude limit of mV ≤ 10 mag.
What is the average metallicity Z/Z for those sample stars in region B and C? Are
they more or less metal-rich than all of the stars in your model sky? Why?
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