Download Astronomy 15 - Problem Set Number 4 1) Suppose one were to

Astronomy 15 - Problem Set Number 4
1) Suppose one were to observe a star for parallax with a telescope of 15 meters focal
length. Suppose the pixels of the CCD are 24 microns in size (1 micron = 10−6 meter).
A star’s image will be spread over several pixels. By carefully measuring this spread-out
image, it’s possible to estimate the center of a star’s image with a standard deviation σ of
0.03 pixels.
(A small, but very useful, statistical aside. The standard deviation, often denoted σ, is a
means of denoting the uncertainty in a measurement. In rough terms, it is a measure of
the width of the famous Gaussian “bell-shaped curve”. Its application in measurements
is as follows. Suppose, for example, one says that a distance is 10.0 ± 0.2 meters, where
the error quoted is supposed to be plus or minus one standard deviation. If this statement is accurate, and the relevant probabilities all follow the Gaussian distribution, the
mathematical properties of the Gaussian curve then dictate that there is a 68 per cent
probability that the true value of the distance is between 9.8 and 10.2 meters; that there
is a 95 per cent probability that the true length is within “2σ” of the quoted value, or
between 9.6 and 10.4 meters; and over a 99 per cent probability that the true value is
within “3σ” of the quoted value, or between 9.2 and 10.6 m. By the way, the ‘bell-shaped
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curve’ is proportional to e−x /2σ , where x is the distance from the mean value. There are
some general theorems in statistics which suggest that Gaussians should appear in many
different circumstances.)
a) At what distance is the parallax five times larger than the 1σ measurement error?
Express your result in parsecs, in light years, in AU, in kilometers, and in centimeters.
Along the way you will have to find the image scale of the telescope and figure out how
many arcseconds a pixel subtends.
b) A star is found to have a displacement on the plates of almost half a pixel - 10 microns due to parallax, and simultaneously has a yearly displacement of 20 microns due to proper
motion. Find its annual proper motion, its distance, and its transverse velocity.
c) For this same star, a heliocentric radial velocity of 10.0 km/s is found. What angle does
the star’s motion make to the line of sight?
d) This star has an atypically large proper motion since it is quite nearby. How long would
it take to move the angular diameter of the full moon (1/2 degree)?
e) There is a proposal to put an optical interferometer in space which would generate
positions good to the micro arcsecond level! If this came to pass, and the 1σ uncertainty
really were 1 micro-arcsec, how far away could one obtain 5σ parallaxes? Refer to your
notes and place this in context - to what objects could one reach?
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2) Astronomers build big telescopes because the objects they deal with are so faint. For
visible wavelengths and shorter, the amount of information an astronomer gets from an
object is ultimately limited by the inherent ‘graininess’ of the signal – by the fact that the
photon energy E = hν. This question, which is even more tutorial than most, explores
the issues involved in how accurately one can measure stars given this limited information.
It’s largely an exercise in statistics – despite its dreary reputation, statistical inference can
be very interesting, and it’s are fundamental to any kind of observational or experimental
science.
To measure a star’s brightness, the astronomer wants to measure the rate at which photons
come from the star. Let’s suppose the star’s true brightness is such that it should deliver
R photons per second to a detector on a telescope. This is the true rate, which you can’t
measure directly; all you can do is make a measurement, see how many photons you count
(call this N ), and divide the number of seconds t you counted; so your best guess is that
R = N/t. But this is only an estimate of R; if you tried the experiment again, random
chance would make N different, so you’d get a different guess for R.
There’s a well developed statistical theory for how much fluctuation you expect in N from
trial to trial. This ‘counting statistics’ is called more formally the theory of the Poisson
distribution, after the person who developed it. The basic result in Poisson or counting
statistics is very simple:
√
Size of average fluctuation in N counts = N .
Here the size of the average fluctuation is given more formally as the standard deviation,
which is defined as
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(standard deviation) = mean of (N − hN i)2 .
(1)
(Yes, this is the same as σ in the previous problem – in the limit of large N , the Poisson
distribution becomes indistinguishable from the Gaussian, which is a good thing, since the
Poisson distribution is difficult to compute with.) Since any one sample N will usually
be different from the mean value (denoted hN i; the angle brackets denote ‘take the mean
value of’), squaring the term in parentheses on the right will give a contribution always
greater than or equal to zero, so its mean is positive definite. It will be a measure of how
much the distribution of samples spreads out around its mean. A little algebra shows that
the standard deviation reduces to
standard deviation = [hN 2 i − hN i2 ]1/2 .
(2)
This is said out loud as the mean square minus the square mean. This sounds like doubletalk at first, but if you think about it, it makes sense.
Question: Show that the equation (1) above leads to equation (2). You can do this by
squaring the expression in the first one, then realizing that taking the mean is distributive
over a sum.
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Measuring a brightish star. Suppose a telescope is set up with a photometer, which (as
its name implies) is a device for measuring the exact amount of light coming from a star.
At the heart of most photometers is a photomultiplier tube. These work by allowing light
to fall on a strip of metal; photons can free electrons from the metal by the photoelectric
effect. These electrons are then accelerated with an electric field arranged so that the
accelerated electrons strike another piece of metal. Because the electron is now much more
energetic than the original photon, it liberates more than one electron when it strikes the
metal.
Question: How many Volts of potential difference should exist between the first electrode
and the second if the final kinetic energy of the electron is to be 100 times the energy of the
photon that liberated it? Assume the photon is in the ‘V’ passband, say 5000 Ångstroms
wavelength.
These electrons are accelerated to strike yet another piece of metal, the process is repeated
until the original photoelectron has been multiplied by a million times or more. This
results in an easily detectable pulse of electric current corresponding to the original photon. Photomultipliers therefore make it practical to count individual photons. In practice
only about 20-30 per cent of the original photons create electrons (for visible wavelength
photons); the Si in CCDs has a higher quantum efficiency. But photomultipliers are still
useful for many specialized measurements, especially when studying objects which change
rapidly, because you can in principle time-tag every photon you detect, whereas a CCD
exposure takes a considerable amount of time to read out from the chip. The statistical
properties of a CCD signal are somewhat similar to those from a photomultiplier, but there
are several complicating factors, so a photomultiplier makes a simpler example.
Question: Suppose a star delivers about 1000 pulses per second in a photomultiplier
attached to a telescope. Assume for now that the background (discussed later) is negligible.
Suppose an astronomer wishes to measure the brightness of the star with a standard
deviation which is 1 per cent of the measurement. How many photons will the astronomer
have to count, and about how long will it take to count that many photons?
Measuring a way faint star. Measuring a faint star is harder. Not only are there fewer
photons to work with, but there is always ‘background’ light to contend with, which in this
case can be comparable to or stronger than the signal you’re trying to measure. The faint
‘background’ glow mostly arises from excited atoms and molecules in the earth’s atmosphere and the scattering of sunlight off interplanetary dust, so it’s literally ‘foreground’
light, but it’s an unwanted contribution to the signal. An astronomer who tries to measure
very faint stars or galaxies is faced with picking the signal from their faint star out of this
noise.
In a CCD picture, one usually can determine the average level of the background very
accurately, because much of the area of the picture is background. But in the small area
of the picture which contains the star of interest, most of the light is contributed by the
background, and the statistical fluctuation of this background determines the faintness of
the stars you can see. As it turns out, to be very confident of the detection of a star, one
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wants it to be 5 standard deviations above background. Since statistical fluctuations this
large are very rare, if you see such a fluctuation, you can be fairly confident that there is
really a star there, instead of just a large fluctuation in the background. (The statistics of
CCD measurement behave pretty much like ordinary counting statistics when the exposure
is reasonably strong, as it will be in the presence of all this background.)
Question: Suppose the light of a star is concentrated into a disk 1 arcsecond across on
your CCD. (It’s actually an approximately bell-shaped function, but its rough size might
be 1 arcsecond, so it’s good enough for an estimate.) Imagine you have exposed long
enough so that you have accumulated 105 photons from the background in each patch of
sky this large. What is the standard deviation of this number? Multiply by 5 to find how
many photons you need from your least detectable star. How small is the star’s signal
compared to the background (i.e., signal/background)?
Question: The night sky background in the ‘V’ passband in one square arcsecond amounts
to the light of a 21.5 magnitude star, more or less. How many times fainter is this than a
zero-magnitude star?
Question: A zero magnitude star delivers about 500,000 (5 × 105 ) photons per square
centimeter per second in the ‘V’ passband. Assuming the Keck telescope is a circle 10
meters in diameter, how many photons would it deliver into the Keck telescope per second?
You can assume all these are detected by the CCD – most will, but some will be lost.
Question: So how many photons per second does one get in the Keck telescope from each
square arcsecond of background?
Question: How long does it take to accumulate 105 photons in each square arcsecond of
background?
Question: Finally, how many times fainter is the 5-sigma star than the background? How
many magnitudes does this factor correspond to? And how faint is the faintest detectable
star, given this exposure time?
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3). HR Mania. Below is a gigantic table from C. W. Allen’s book Astrophysical Quantities. This is the ultimate astronomer’s “cheat sheet”, containing all manners of useful
tables. There are all sorts of atomic data, plus data on stars, compiled from various sources
over the years by Allen. It’s far from a definitive source, since it’s a secondary source and
since it’s getting to be a bit out of data (a revision is in the works), but it’s a nice place
to get a general idea of the sizes of quantities.
The Main Sequence
Type
MV
B−V
L/L
Teff
O5V
B0V
−5.7
−4.1
−0.35
−0.31
7.9 × 105
4.17 × 104
40000
28000
A5V
F0V
+2.0
+2.6
+0.13
+0.27
8500
7400
F5V
G0V
G5V
+3.4
+4.4
+5.1
+0.42
+0.58
+0.70
1.66 × 101
7.24 × 100
K0V
K5V
+5.9
+7.3
+0.89
+1.18
4900
4130
M0V
M5V
+9.0
+11.8
+1.45
+1.63
3.80 × 10−1
1.66 × 10−1
M8V
+16
+1.8
B5V
A0V
−1.1
+0.7
−0.16
+0.00
8.71 × 102
7.24 × 101
15500
9900
3.47 × 100
1.51 × 100
7.94 × 10−1
6580
6030
5520
6.03 × 10−2
9.55 × 10−3
3480
2800
2400
Note that these values are averages for the various spectral types; real main-sequence
stars show some variation around these average values. Sometimes there are also systematic differences from these tables caused primarily by modest effects of age and chemical
composition.
Get a piece of ordinary graph paper (or a graphics program on a computer). Graph the
absolute visual magnitude on the vertical axis and the B − V color on the horizontal. Note
that since more negative MV means a brighter star, you want MV to increase downwards
so that brighter stars are at the top. You’ll want to scale things so that one horizontal
unit takes up the same space as four or five vertical units, or even more. Now plot points
on this graph for each of the spectral types, and label each point with its spectral type.
Note that this diagram is the companion to the one you drew last time – here the horizontal
axis is a ‘surface-temperature-like’ quantity, and the vertical axis is a ‘brightness-like’
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quantity. Given the appropriate transformations between B − V and Teff , and between
MV and L for a given surface temperature, it would be possible to ‘warp’ this diagram
into the one you drew last time.
4) A-type stars, with effective temperatures of about 10 000 kelvin, have the strongest
visible-wavelength hydrogen lines, despite the fact that hydrogen is the main constituent
of the outer layers of all main-sequence stars. Let’s try to understand this quantitatively.
First, recall that only absorptions from the n = 2 state give rise to visible-wavelength
hydrogen lines. So our task is, for each temperature, to compute what portion of the
hydrogen is (1) still in atomic form (that is, not ionized) and (2) excited to the second
state.
The first thing we need is a way to compute how much hydrogen remains in atomic form
at each temperature. Let
Ni = number density of ions (protons)
N0 = number density of neutral atoms
Ne = number density of electrons
Ei = ionization energy (13.6 eV in this case)
Then the Saha Ionization Formula states that
(2πmkT )3/2 −Ei /kT
Ni Ne
= 2α
e
,
N0
h3
where most of the symbols have their usual meanings. Here α is a factor of order unity,
the “ratio of statistical weights”; we’ll ignore it.
This may be a scary-looking equation, but you’ve undoubtedly seen similar things in
chemistry. Look at the left side; it’s a lot like one of those solubility constants. In fact,
it’s similar in principle. The right side is less familiar, but expresses the tendency for the
ionization to go up as the temperature goes up. The factor (2πmkT )3/2 arises from the fact
that the electrons have lots of room to “spread out” in velocity space as the temperature
goes up; the fact that there is an h3 in the denominator arises from the profound fact that
the states available to free electrons are enumerable, by the laws of quantum statistical
mechanics. The last factor on the right is the by now slightly familiar Boltzmann factor.
It turns out that one can crank this around to a more convenient form for computation.
To do this, we multiply each side by kT , which leaves us with a factor of Ne kT on the left;
this is just proportional to the pressure of the electrons, Pe . After much numerical horsing
around, one finds that
log
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Ni Pe
= −Ei
+ log T − 0.48,
N0
T
2
where I’ve dropped the dependence on the small factor α. In order for this equation to
work, T must be in kelvin, Pe must be in dynes per square centimeter, and the energies
must be in electron Volts. The logarithms are all base 10.
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Now, it turns out that for most main sequence stars, Pe in the atmosphere is about 100
dynes per square centimeter. Let’s assume this is the case.
a) Compute Ni /N0 at three temperatures: 3000, 10000, and 50000 degrees; noting that the
total amount of hydrogen is Ni +N0 , compute the fraction of hydrogen (N0 /[Ni +N0 ])which
is neutral at the three temperatures.
b) All that remains is to find the fraction of the neutral atoms which are in the second
state. This is at first a formidable proposition, since there are an infinite number of states!
Let’s simplify this by just assuming (once again) that hydrogen has only two states, of equal
a priori probability; a ground state and a state 10.2 eV above it (the n = 2 state). Then
we can simply use the Boltzmann factor – refer back to your earlier HW – to calculate the
fraction of the neutral atoms in the n = 2 state at the same three temperatures; remember
that the fraction in n = 2 is given by
N2
.
N1 + N2
c) Find the fraction which is neutral and in the n = 2 state at the three temperatures by
multiplying the factors found in (a) and (b). Which one is highest?
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