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Lecture 13
• Today I plan to cover:
– The jansky.
– Photon noise at radio wavelengths?
– Flux calibration.
– A bit more about noise temperatures.
– Polarized radio signals.
– Radio spectroscopy.
NASSP Masters 5003S - Computational Astronomy - 2009
The jansky
• The standard unit of flux density in radio
astronomy.
• Symbol Jy.
• Named after a pioneer in the field.
• = 10-26 W m-2 Hz-1.
Photon noise..?
• High-energy astronomy usually is
dominated by this. But radio υ hugely less!
• Energy of a photon at 1.4 GHz is ~1e-24 J.
• Hence 10 mJy over 1 MHz at this freq
• ~ only 100 photons/sec. - a bit unexpected.
NASSP Masters 5003S - Computational Astronomy - 2009
Flux calibration
•
•
The bandwidth and gain of a radiometer tend
not to be very stable.
There are several methods of calibration. Eg:
1. Switching between the feed and a ‘load’ at a
temperature similar to the antenna temperature.
–
To detector
To detector
“Warm load” at T
“Warm load” at T
But, the required physical temperature of the load
resistor can be < 20 K...  need to cool with He.
NASSP Masters 5003F - Computational Astronomy - 2009
More widely used:
2. Periodic injection of a few % of noise into the feed.
Noise sources can be made much more stable than
noise detectors.
•
It is still good, while observing, to look
occasionally at an astronomical source
which has the following properties:
–
–
•
Known, stable flux density.
Should also be unresolved (compact).
These are difficult conditions to meet at
the same time! Compact sources tend to
vary with time.
NASSP Masters 5003S - Computational Astronomy - 2009
Typical noise temperatures
J D Kraus, “Radio Astronomy”
2nd ed., fig 8-6.(+ 7-25)
NASSP Masters 5003F - Computational Astronomy - 2009
Polarized EM waves – conventions:
y
x
Snapshot of a wave moving in the
positive z direction.
Left-hand circular polarization
according to IEEE convention.
(Physicists use the opposite
convention.)
z
Direction of rotation of
the field vector as seen
by an observer.
NASSP Masters 5003F - Computational Astronomy - 2009
Sources of polarized radio waves:
• Thermal? No
• Spectral line? No (unless in a strong B field)
• Synchrotron? YES.
– And this is the most common astrophysical emission
process.
• All jets emit synchrotron – and jets are everywhere.
Magnetic field B
Electron moving at
speed close to c
Linearly
polarized
emission.
• Scattering off dust? YES.
NASSP Masters 5003F - Computational Astronomy - 2009
How to describe a state of polarization?
Stokes parameters I, Q, U and V.
I = total intensity.
Q = intensity of horizontal pol.
U = intensity of pol. at 45°
V = intensity of left circular pol.
V axis
U axis
Q axis
Therefore need 4
measurements to
completely define
the radiation.
Polarization fraction d:
Visualize with the “Poincaré sphere.” of radius I.
Q2  U 2  V 2
d
I
NASSP Masters 5003F - Computational Astronomy - 2009
Antenna response, and coherency matrices.
• The antenna response is different for
different incoming polarization states.
• This may be quantified by 4 ‘Stokes
effective areas’ AI, AQ, AU, AV.
• But it is more convenient to express both
the radiation and the antenna response as
coherency matrices:
1
S
2I
 I Q
U  iV

U  iV 
I  Q 
and
1
A
2 Ae
 AI  AQ
 A  iA
V
 U
AU  iAV 
AI  AQ 
• Then the power spectral density detected is
w = AeI×Tr(AS) (‘Tr’ = the ‘trace’ of the matrix, ie the sum of all diagonal terms.)
NASSP Masters 5003F - Computational Astronomy - 2009
Depolarization due to finite resolution
Arrows show the polarization direction.
Half-power contour
of the beam.
Nett
polarization
observed.
Waves from different areas of the source add incoherently. Result: some degree
of depolarization. In general, the finer the resolution, the higher the polarization fraction.
NASSP Masters 5003F - Computational Astronomy - 2009
Faraday rotation.
• Any linear polarized wave can be decomposed
into a sum of left and right circularly polarized
waves.
• In a magnetized plasma, the LH and RH
components travel at slightly different speeds.
• Result:
– The plane of polarization rotates.
– The amount of rotation θ is proportional to distance
travelled x the field strength x the number density of
electrons.
– θ is also proportional to λ2.
• Most due to Milky Way, but the Earth’s
ionosphere also contributes – in a time-variable
fashion. The ionosphere is a great nuisance and
radio astronomers would abolish it if they could.
NASSP Masters 5003F - Computational Astronomy - 2009
Faraday rotation
J D Kraus, “Radio Astronomy”
2nd ed., fig 5-4
The slope of the line is called
the rotation measure.
Why is there progressive depolarization with increase in wavelength?
NASSP Masters 5003F - Computational Astronomy - 2009
Faraday rotation – another cause of
depolarization
Because the rotation measure is not uniform and may vary within the beam. Eg:
Half-power contour
of the beam.
NASSP Masters 5003F - Computational Astronomy - 2009
Radio spectroscopy
• The variation of flux with wavelength
contains a lot of information about the
source.
• We can pretty much divide sources into
– Broad-band emitters, eg
• Synchrotron emitters
• HII regions (ie ionized hydrogen)
• Thermal emitters
– Narrow-band emitters (or absorbers), eg
• HI (ie neutral hydrogen)
• Masers
• Neutral molecular clouds
NASSP Masters 5003F - Computational Astronomy - 2009
Broad-band emitters
• Most of these have spectra which, over
large ranges of wavelength, can be
described by a simple power law, ie

S 
• For thermal sources, the Rayleigh-Jeans
approximation to the black-body radiation
law gives a spectral index α = -2.
• Synchrotron sources have +ve α,
averaging around +0.8.
• HII regions exhibit a broken power law.
NASSP Masters 5003F - Computational Astronomy - 2009
Wavelength or frequency?
• Since λ=c/υ, a power-law spectrum
S    

implies
S    

• Ie, the sign of the spectral index depends whether
you give the spectrum as a function of wavelength
or frequency.
• Make sure you know which convention is intended
before using a spectral index (and always record
this convention in your own reports).
NASSP Masters 5003F - Computational Astronomy - 2009
Broad-band emitters
J D Kraus, “Radio Astronomy”
2nd ed., fig 8-9(a)
Mostly synchrotron.
Steep-spectrum
source
Reversed-spectrum
sources: mostly
thermal, ie slope = 2.
Log-log plot so all straight lines
are power-laws.
One spectrum, many spectra.
Note too that nearly all broadband spectra are quite smooth.
NASSP Masters 5003F - Computational Astronomy - 2009
HII regions
• The gas here is ionized and hot (10,000 K is
typical) – usually as a result of intense irradiation
from a massive young star.
• The radiation comes from electrons accelerated
(diverted) as they come close to a positive ion.
+
-e
• This radiation mechanism is called free-free,
because the electron being accelerated is not
bound to an atom either before its encounter or
after. But it is basically a thermal process.
• Otherwise known as bremsstrahlung (German for
“braking radiation”. Yes spellt with 2 esses.)
NASSP Masters 5003F - Computational Astronomy - 2009
Optical depth
• Whenever you have a combination of radio
waves and plasma, optical depth τ plays a role.
– High τ = opaque – behaves like a solid body.
– Low τ = transparent.
• τ for a plasma is proportional to λ2.
• Effective temperature Teff = T(1-e-τ).
– Long λ - high τ - Teff ~ T – thus α = -2.
– Short λ - low τ - Teff proportional to τ, thus λ2 means flux density S is constant, or α = 0.
•  Flat-spectrum source.
– The troposphere is a good example of a thermal
radiator which is optically thin at dm wavelengths.
NASSP Masters 5003F - Computational Astronomy - 2009
Some more about synchrotron
• Already covered the
basics in slide 7.
• Also subject to optical
depth effects:
J D Kraus, “Radio Astronomy”
2nd ed., fig 10-10
PKS 1934-63
– At low frequencies,
opacity is high, the
radiation is strongly
self-absorbed:
• α ~ -2.5.
• Effective temperature
limited to < 1012 K by
inverse Compton
scattering.
NASSP Masters 5003F - Computational Astronomy - 2009
Narrow-band spectra
• Molecular transitions:
– Hundreds now known.
– Interstellar chemistry.
– Tracers of star-forming regions.
– Doppler shift gives velocity information.
• Masers:
– Eg OH, H2O, NH3.
– Like a laser – a molecular energy transition
which happens more readily if another photon
of the same frequency happens to be passing
 radiation is amplified, coherent.
– Spatially localized, time-variable.
• Recombination lines.
NASSP Masters 5003F - Computational Astronomy - 2009
HI
• The I indicates the degree of ionization.
– Subtract 1 from this number to get the number of
electrons stripped from the atom.
• Eg FeVII, ‘iron seven’, means iron which has lost 6
electrons. Hence I means minus 0 electrons – just the
neutral atom. Hydrogen has only 1 electron so the
highest it can go is HII – which is just a bare proton.
• The neutral H atom has a very weak (lifetime ~
107 years!) transition between 2 closely
spaced energy levels, giving a photon of
wavelength 21 cm (1420 MHz).
• But because there is so much hydrogen, the
line is readily visible.
NASSP Masters 5003F - Computational Astronomy - 2009
HI
• Because the transition is so weak, and
also because of Doppler broadening,
hydrogen is practically always optically
thin (ie completely transparent).
• Thus the intensity of the radiation is
directly proportional to the number of
atoms.
– For a cloud of mass M solar masses at a
distance D megaparsecs, the measured flux φ
(in W m-2) is
 ~ 6 10
 23
M
2
Dc
(I think.)
NASSP Masters 5003F - Computational Astronomy - 2009
HI
• Concept of column density in atoms per
Distribution of atoms
square cm.
this way makes no difference.
So many square cm
Observer
• Hydrogen will be seen in emission if it is
warmer than the background, in
absorption otherwise.
NASSP Masters 5003F - Computational Astronomy - 2009
HI – Doppler information
• Hubble relation between distance and
recession velocity allows distance of far
galaxies to be estimated.
– Hence: 3D information about the large-scale
structure of the universe.
• Our Milky Way is transparent to HI – we
can see galaxies behind it at 21 cm,
whereas visible light is strongly absorbed.
• Cosmic Doppler red shift z is given by
obs  true
1 v c
v
z

 1  for v  c
2
true
c
1  v c 
NASSP Masters 5003F - Computational Astronomy - 2009
HI – Doppler information
• Within galaxies:
– Doppler broadening tells about the distribution of
velocities within a cloud of hydrogen.
– the Doppler shift of the HI line maps the rotation curve
of the galaxy, eg:
NGC 2403
Credit: F Walter et al (2008).
(Courtesy Erwin de Blok.)
NASSP Masters 5003F - Computational Astronomy - 2009
Model of galactic hydrogen
• A good model for many galaxies is a disk of
uniform surface density ρ within a radius R,
rotating with a uniform speed |v|.
– Such decoupling of speed from radius is not what is
observed in eg planetary systems and is one of the
indicators for large dominance of dark matter.
• With distance D and inclination angle α this
leads to a predicted spectral shape of
R2
S f  
D
2
f
2
0
 v c  sin
2
2
   f  f centre 2
– Here I have used f instead of υ to avoid confusion
with velocity v. (Caveat! This is my own derivation.)
NASSP Masters 5003F - Computational Astronomy - 2009
Diagram of HI spectrum
NASSP Masters 5003F - Computational Astronomy - 2009
An example exercise
• You have an antenna of diameter 12 m,
efficiency η=70%. The feed is linearly
polarized and the detection chain has a noise
temperature of 45 kelvin. How long do you
have to observe to see an unresolved HI
source of average flux density 2 Jy in a
channel width of 15 kHz?
– The relation between source flux density S and
equivalent noise temperature T, for an unresolved
source, is
kT
S
pAeffective
– where p is the fraction of signal the polarised
detector is picking up.
NASSP Masters 5003F - Computational Astronomy - 2009
An example exercise
– HI emission is unpolarised, so p is 0.5 in the present
case.
– The effective area is the true area times the efficiency;
= πr2η = in the present problem.
– Hence
2
Tsource 
r S
2k
• We need to observe long enough that the
uncertainty in the total noise temperature (source
plus instrumental) is equal to the source noise
temperature.
– A common trick: assume Tsource<<Tinst.
– This is valid because we don’t need to know our
observing time to more than a couple of significant
figures!
NASSP Masters 5003F - Computational Astronomy - 2009
An example exercise
• We can therefore write, to a reasonable
approximation:
Ttotal
Tinst
T 
~
t
t
• Setting this equal to the source
temperature allows us to derive:
4
t 

 Tinstk 


2 
 r S 
2
• = 41 s. (Don’t forget to convert Jy to MKS!)
NASSP Masters 5003F - Computational Astronomy - 2009
Spectra - baselines
• The spectrum of interest
sits on top of a high
‘mesa’ due to system and
background ‘temperature’
(ie noise).
• We usually want to
subtract the mesa and
just leave the spectrum.
• We could do that by
alternating between onand off-source
observations, and
subtracting the two:
– But this needs 4 times as
much observing time to
reach the same SNR!
NASSP Masters 5003F - Computational Astronomy - 2009
Spectra - baselines
• More commonly, the
mesa is slowly varying
compared to the
spectrum, so one can fit
some fairly smooth
function to the mesa,
then subtract it.
• The examples (for which
this has already been
done) show it is not
always so simple!
• These show the
infamous ‘Parkes ripple.’
NASSP Masters 5003F - Computational Astronomy - 2009
The ‘Parkes ripple’
A weak Fabry-Perot resonance occurs between the dish and the feed.
2D = nλ = (n+1)(λ-Δλ)
=> Δν ~ 5.5 MHz.
D ~ 26 m
NASSP Masters 5003F - Computational Astronomy - 2009