ASTR 534: Radio Astronomy
Final Exam
Due: 9 December, 2006
1. (15 points) The Ooty radio telescope in India is a cylindrical reflector operating at 327 MHz of dimension
30 m by 530 m. It is oriented with its long-axis in the North-South direction, and it can tip only in the
East-West direction (i.e. along its long axis).
(a) (4 points) Assuming the aperture is uniformly illuminated and 50% efficient, what is the telescope’s
“gain” in units of K/Jy (i.e. How many Kelvin in antenna temperature would we get per Jansky
of incident flux density.)
Solution: Wν = kTA = Ae Smatched = Ae Smatched /2, for an unpolarized source. The effective
area of the telescope is Ae = ηa Ag , where ηa is the aperture efficiency and Ag is the geometric
area. The “gain” equals
TA
Ae
ηa Ag
0.5 × 30 m × 530 m × 10−26 W m−2 Hz−1
=
=
=
= 2.9 K Jy−1
S
2k
2k
2 × 1.38 × 10−23 Jy K−1
(b) (7 points) What is the shape (and angular size) of the FWHM contour when the telescope is pointed
at the zenith? Sketch what it would look like on the plane of the sky (label the cardinal directions).
Solution: The beam will be approximately elliptical with the long axis in the E-W direction
and the short axis in the N-S direction. We can estimate the beamwidths in each case using
λ
(since this is a uniformly illuminated antenna).
θHPBW ∼ 0.89 D
λ=
c
3 × 108 m s−1
= 0.92 m
=
ν
327 × 106 Hz
Therefore,
θHPBW,EW ∼ 0.89
0.92 m
= 0.027 rad = 1.6 deg
30 m
and,
θHPBW,NS ∼ 0.89
0.92 m
= 0.0015 rad = 0.09 deg
530 m
(c) (4 points) Since the telescope cannot tip North-South, how does the shape of the beam and the
telescope “gain”vary as a function of the North-South zenith angle?
Solution: The projected North-South size is D = 530 cos ZANS m and therefore
θHPBW,NS ∼
0.10
deg.
cos ZANS
Similarly, Ag will change with the N-S zenith angle, and so
TA
= 2.9 cos ZANS K Jy−1 .
S
2. (17 points) Assume there is a P =2 ms pulsar with a dispersion measure (DM) of 100 pc cm−3 and
a gaussian pulse that has an intrinsic width of W = 0.1 P . You are searching for this pulsar using an
observing system with 100 MHz of bandwidth centered at 1.4 GHz, split into 1000 channels, and sampled
every 50 µs.
(a) (9 points) Sketch what a folded pulse profile would look like if the data were de-dispersed at DMs
of 100 pc cm−3 , 103 pc cm−3 , and 110 pc cm−3 .
Solution: De-dispersing the profile at incorrect DMs will cause smearing of the pulse profile
across the full band.
The amount of smearing (in seconds) is described by the relation we derived in problem set 9:
∆t = 8.3 × 103 DM BW ν −3 ,
although in this case, DM will be the error in the dedispersion.
The smearing caused by each channel is
∆tchan = 8.3 × 103 × 100 × (100/1000) × 1400−3 = 3 × 10−5 s,
which is less than the sampling time, and therefore doesn’t affect us.
For DM=100 pc cm−3 , there will be no extra smearing and so we’ll get a gaussian pulse profile
with a fractional width of 0.1.
For DM=103 pc cm−3 , the DM error is 3 pc cm−3 , and the smearing caused by the incorrect
dedispersion will be
∆terr = 8.3 × 103 × 3 × (100) × 1400−3 = 9 × 10−4 s,
which is almost 1 ms. This will smear the pulse profile by half a period, making it nearly
sinusoidal.
For DM=110 pc cm−3 , the DM error is 10 pc cm−3 , and the smearing caused by the incorrect
dedispersion will be
∆terr = 8.3 × 103 × 10 × (100) × 1400−3 = 0.003 s,
which is 1.5 times the pulse period. This will completely smear out the signal from the pulsar,
and so we will see only noise.
(b) (5 points) Sketch and describe what the Fourier power spectra of a 30 min observation (with plenty
of sensitivity to detect the pulsar) would look like if the data were dedispersed at the DMs listed
above. Make sure you label all the important parts of the power spectrum.
Solution: Using the information from above, for DM=100 pc cm−3 , there should be the fundamental harmonic at 500 Hz and then approximately 10 (i.e. ∼ P/W as per the problem
set) higher harmonics. The harmonics should have amplitudes that correspond to a gaussian
envelope (i.e. convolution of the Shah function with the gaussian profile shape).
For DM=103 pc cm−3 , there should only be a single harmonic visible at the fundamental frequency of 500 Hz. The signal is now practically a sine wave.
For DM=100 pc cm−3 we should see just white noise.
(c) (3 points) If during the 30 min observation the power levels from the system vary substantially and
randomly (including significant instantaneous step functions) on time scales ranging from about
one second to tens of minutes, what affect will that have on the power spectrum? Draw a sketch
for DM=100 pc cm−3 . How will it affect the detectability of the pulsar?
Solution: The power fluctuations will add a strong “rednoise”-like component to the power
spectrum. The component will be almost completely gone by ∼10 Hz, and so the pulsar harmonics will be unaffected (as will it’s detectability).
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3. (13 points) Suppose you observed a very weak spectral line from a source near the zenith (zenith angle
= 0 deg) while the zenith system temperature was T0 = 40 K, the zenith atmospheric opacity was
τ0 = 0.07, and the air temperature was Tair = 273 K, and it took t0 = 20 minutes of integration to detect
the source with a signal-to-noise ratio of 5. A few hours later, you reobserve the same spectral line
when that source is 60 deg from the zenith. If the zenith atmospheric opacity and the air temperature
are unchanged, how long must you integrate to reach the same signal-to-noise ratio? You may ignore
receiver gain fluctuations.
Solution: The atmosphere increases the system noise temperature and absorbs part of the signal
from the source. The ideal radiometer equation implies that the integration time t needed to reach
a given signal-to-noise ratio is proportional to the square of the noise and inversely proportional to
the square of the signal strength:
t60
∝
t0
T60
T0
2 S60
S0
−2
,
where the subscripts indicate zenith angles in degrees. At the zenith, the contribution of atmospheric
emission to the system temperature is
Tatm = Tair (1 − e−τ0 ) = 273 K(1 − e−0.07 ) = 18.5 K ,
so the non-atmospheric contribution to the system temperature is 21.5 K. The fraction of the source
flux reaching the ground (S0 ) is e−τ = e−0.07 = 0.932 at the zenith.
At 60◦ zenith angle, the atmospheric opacity is multiplied by sec(60◦ ) = 2.00 so the atmospheric
contribution to the system temperature increases to 273 K(1−e−0.14 ) = 35.7 K and the total system
temperature becomes T60 = 21.5 + 35.7 = 57.2 K. The fraction of the source flux received on the
ground (S60 ) decreases to e−0.14 = 0.869 at z = 60◦ . Thus
t60 = t0
57.2
40
2 0.869
0.932
−2
= 20 min × 2.04 × 1.15 ≈ 47 min
This example shows that the noise produced by atmospheric emission is a bigger problem than the
signal attenuation produced by atmospheric absorption when the system noise temperature is much
smaller than the air temperature.
4. (7 points) The J = 1 − 0 spontaneous emission rate for the common 12 C16 O form of carbon monoxide
is A10 ≈ 7.4 × 10−8 sec−1 . The rare heavy form 13 C18 O obtained by adding one neutron to the carbon
atom and two neutrons to the oxygen atom is chemically very similar (same size molecule, same charge
distribution). Estimate A10 for 13 C18 O.
Solution: The molecular-line emission coefficient is proportional to the cube of the line frequency
ν10 :
64π 4 3
ν |µ10 |2 .
A10 =
3hc3 10
The line frequencies of two otherwise-similar diatomic molecules are inversely proportional to their
reduced masses
mA mB
m≡
,
m A + mB
so
−1 13 × 18
12 × 16
ν(13 C18 O)
=
≈ 0.90842
ν(12 C16 O)
13 + 18
12 + 16
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and
ν(13 C18 O)
A10 ( C O) = A10 ( C O) ×
ν(12 C16 O)
13
18
12
16
3
≈ 7.4 × 10−8 s−1 × 0.908423 ≈ 5.55 × 10−8 s−1
5. (12 points) The quantum-mechanical explanation for “spontaneous” emission is that there is no truly
spontaneous emission, and what appears to be spontaneous emission is just stimulated emission by
“virtual” photons in the quantum vacuum. These photons imply that the quantum vacuum has some
minimum energy density U ∗ .
(a) (9 points) Show that
4
2πhνmax
,
3
c
is the maximum frequency of these photons (a cutoff needed lest U ∗ become infinite).
U∗ =
where νmax
Solution: If “spontaneous” emission is just stimulated emission by photons in the quantum
vacuum, then the energy density Uν∗ per unit frequency of these photons needed to produce the
observed “spontaneous” emission rate AUL must satisfy:
AUL = BUL Uν∗ .
Einstein showed that
AUL
8πhν 3
=
BUL
c3
so
Uν∗ =
8πhν 3
AUL
=
BUL
c3
is the vacuum energy density contributed by photons of frequency ν. Integrating over frequencies up to νmax gives the total energy density
4
Z νmax
Z νmax
4
8πhν 3
8πh νmax
2πhνmax
∗
∗
U =
Uν dν =
dν
=
=
c3
c3
4
c3
0
0
(b) (3 points) If the maximum frequency corresponds to a photon whose energy is equal to the Planck
energy, then νmax ≈ 1042 Hz. Calculate U ∗ for this value of νmax and the corresponding mass
density ρ∗ = U ∗ /c2 of the quantum vacuum. The quantum vacuum should act as a “dark energy.”
Compare the calculated ρ∗ with the observed density of dark energy in the universe, ρ∗ ≈ 10−29 g
cm−3 . This enormous discrepancy is a major unsolved problem for physics.
Solution: Inserting νmax ≈ 1042 Hz gives
U∗ ≈
2π × 6.63 × 10−27 erg s × (1042 Hz)4
≈ 1.5 × 10111 erg cm−3
(3 × 1010 cm s−1 )3
and
ρ∗ = U ∗ /c2 ≈
4 × 10111 erg cm−3
≈ 1.7 × 1090 g cm−3
(3 × 1010 cm s−1 )2
This is about 120 orders-of-magnitude higher than the observed density of dark energy!
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