Transcript Lecture 3 By Tom Wilson Summary of Lecture 1 Noise in a Receiver  TRMS  TSYS   Analying bandwidth (for lines, need 3 resolution elements.

Lecture 3
By Tom Wilson
Summary of Lecture 1
Noise in a Receiver
 TRMS 
TSYS
 
Analying bandwidth (for lines, need 3 resolution
elements on the line above the ½ power point)
Receiver itself,
atmosphere,
ground
and source
time
on source
Temperatures from thermal hot and cold
load measurements using the receiver.
Hot-cold load measurements
(to determine receiver noise contribution)
Absorber at
a
given
temperature
OK for heterodyne receivers, but not for Bolometers
Input to
receiver
Current Receiver Noise Temperatures
Tmin=h/k
for coherent
receivers
TSYS=TRX e 
Atmospheric optical depth
Receivers
Heterodyne for spectral lines


High velocity resolution
flexibility, but not multi-pixel receivers in the mm/sub-mm
Bolometers for continuum

Multi-pixel cameras
TRMS
Noise
(about 1)
A  TSYS
A  NEP


  2  k    
Equivalent
power
Types of Receivers
Fractional Resolution
Lecture 3 page 7
BOLOMETERS VS COHERENT RECEIVERS
JCMT: 15 m sub mm dish, hA = 0.5 at l= 0.87 mm, = 345 GHz
With SCUBA, can detect a source with 0.16 Jy in 1 second. So RMS is ¼
of this peak value or S0.04 Jy in 1 sec.
Compare to a coherent receiver:
TSYS = 50 K, = 2 GHz, integration time= 1 sec  T
RMS 
In antenna temperature.
Or
S 
2  109
 12
.  103 K
TA
S  3520
ha Dm2
From Lecture 2:
For JCMT,
50 K
16
ha
 TA
 SRMS  4  102 Jy
So comparable for 1 beam, but SCUBA has 37 beams & MAMBO has 117
beams.
q(arc sec)=206 k l(mm)/D(m)
of order 1.2 for single dishes
S=3520 TA /(h A D2)
Summary of Lecture 2
TA
S  3520
ha D2
Rayleigh-Jeans
1
P  Ae  S    k  TA   
2
S  2.65
Or
In Jy

TMB  q0 (')
l (cm)
S  0.074
2

2
 2.65
Ae  ha  Ag
TMB  q0 2 (')
l2 cm

TMB  q0 ('')
l (mm)
2

2
(Show that these
are consistent)
True source size
Gaussian
beams:
apparent source size
and temperature
2
TMB  q obs
 Ts q s2
2
2
q obs
 q Beam
 q s2
TMB
q s2
 Ts  2
2
q s  q Beam
and temperature
Can make a relation for flux density similar to that for
Main Beam Brightness temperature:
S(total)=S(peak) . (q S 2 + q B 2 )/ q B 2
Example: Orion A is an HII region with a total flux
density of 380 Jy at 1.3 cm. The size is 2.5’ (FWHP). If the radio
telescope beam size is 40” (FWHP), what is the peak flux density?
Use the R-J relation to find the peak main beam brightness
temperature. Solution: peak Jy/beam=9.5; TA=8.8K, TB=24 K
Lecture 3 page 1
Far Field Diffraction and Fourier Transforms
(radiation passing through an opening)
(Exact calculations require programs such as GRASP)
Lecture 3 page 2
Grading Across the Aperture and Far E Field
Lecture 3, page 3
ALMA
Technical
Building
ANTENNA
Correlator
Front-End
Tunable Filter Bank
IF-Processing
(8 * 2-4GHz sub-bands)
Digitizer
8* 4Gs/s -3bit ADC
8* 250 MHz, 48bit out
Local
Oscillator
Digital De-Formatter
Digitizer
Clock
Optical De-Mux
& Amplifier
Data Encoder
12*10Gb/s
12 Optical Transmitters
12->1 DWD Optical Mux
Fiber
Fiber Patch-Panel
From 270 stations to
64 DTS Inputs
Sketch of 2 element
interferometer
Earth Rotation Aperture Synthesis
(u,v) plane and image plane
These are related by Fourier transforms
The distance between antennas varies, so
we sample different source structures
On the next overheads, we indicate how
structures are sampled. Following
tradition, u represents one dimension
distributions, with x as the separtion in
wavelenghts u=2px/l and v=2py/l
Above: the 2
antennas on
the earth’s
surface have a
different orientation
as a function
of time.
Below: the
ordering
of
correlated
data in
(u,v) plane.
Gridding and sampling in (u,v)
plane
Sensitivity: http://www.eso.org/projects/alma/science/bin/sensitivity.html
VLA uv plane response
Data
as
taken
Data
with
MEM
with
MEM
and
SelfCalibration
The
radio
galaxy
Cygnus A
as
measured
with all
configurations of
the
VLA
From W. D. Cotton (in ‘The Role of VLBI in Astrophysics, Astrometry,
And Geodesy, ed Mantovani & Kus, Kluwer 2004)
Lecture 3 page 16
BROADBAND RADIATION
•
•
•
•
•
Black body (Moon, planets, 3K background)
Dust thermal emission
Bremsstrahlung (free-free)
Synchrotron (relativistic electrons in magnetic fields)
Inverse Compton Scattering (S-Z)
Dust: Mostly carbon, silicon with ice mantles “ground up planets”
From Hildebrand (1983)
 DUST
 DUST : in mm
 z 
  b  N H  l2
 7  10 
 zSun 
21
b : factor to account to fit the data
N H : column density in cm 2
l2 : in m 2
Lecture 3 page 17
T  TDUST   DUST
For warm grains
Use R-J
get
S  2k  TDUST   DUST  l2   
4
S
(
mJy
)
l
mm
24 
N H  193
.  10
q 2 (")  z 

  b  TDUST
 zSun 
EXAMPLE: Dust emission from Orion KL
The Orion “hot core” has the following properties:
q  10"
TDUST  160 K
n( H2 )  107 cm3
Dist  15
.  1021 cm (  500 pc)
Lecture 3 page 18
Calculate the column density N(H2) = n L and the 1.3 mm dust flux density, S,
if z = zSun, b = 1.9
If the value of L = diameter, use L (diameter in cm) x (size in radians) =
= 7.5 1016 cm
Them N(cm-2) = 7.5 . 1016 cm x 107 cm-3 = 7.5 . 1023 cm-2 for H2
N(H) = 2 N(H2) = 1.5 . 1024 cm-2
So
4
S
(
mJy
)

(
1300
)
15
.  1024  193
.  1024
2
(10)  (19
. )  (160)
S (mJy )  10
4
S  10 Jy
At 4 mm, S is 81 times smaller or 120 mJy. At 0.39 mm, S is 81 times
larger or 810 Jy
Lecture 3 page 19
BREMSSTRAHLUNG (FREE-FREE)
Hydrogen is ionized by O, B, electron and protons interact, electrons
radiate. Classically:

velocity
2 2 v 2 (t )
P( t )  e
( Larmor )
3
c3
Power radiated during encounter:
Find
4 z 2  e6 Ni  N e
 
3c3
m2
2 e2
W
3 c3

 v
2
(t ) dt

‘ p’ is impact parameter
2m
p2
ln
p  k  Te p1
From the Kirchhoff relation, get


  c 2
 


2
k

T
B (T )
2 k  Te   2
e
l2
so
 
Ni  N e  L
3
2
Te   2
frequency
Lecture 3 page 20
TB  Te  (1  e   
When   0,  = 1:
 0  0.3045  (TB 
 0.643
 EM 


 pc cm6 
0.476
For Orion A, 0 = 1 GHz, or 30 cm.
   1GHz
TB  Te  8000 K
   1GHz
TB  Te    
at 5GHz
TB  400 K
at 23GHz
TB  24 K
Te
2
But
S 
2 kTB
l2
  
2 kTe
EM  const
l2   2
  8.235  10  (Te 
2
 1.35
What is the
 (GHz  EM
-2
Relation for
TB?
Lecture 3 page 21
Orion A HII Region
Te = 8500 K, q = 2.5’ (FWHP), so use
 is much less than 1, at  = 23 GHz so T = Te 
From 100-m, TMB (main beam) = 24 K in a 40” beam, so TMB (main beam) = T(true)
24 = (8500) . (8.235 10-2) . (8500)-1.35 . (23)-2.1 . EM
so EM = 4 . 106 cm-6 pc = Ne . Ni . L
If L = 25’ = 0.33 pc converted to radians @ 500 pc get Ne = Ni = 3.5 . 103 cm-3
This is the RMS density. Calculate the mass of ionized gas.
Rough number since know Orion A is not spherical. From spectral lines
know Ne = 104 cm-3, so L = 0.03 pc. Then M = 0.6 MSun in ionized gas
Lecture 3 page 22
Free-Free Intensity and Flux Density as function of Frequency
(Problem: Use the
example
of Orion to
check these
Curves)
Lecture 3 page 4
Free-Free Emission from Planetary Nebulae
NGC7027 (a PNe) has S = 5.4 Jy at 1.3 cm. What is the TMB (main beam
brightness temperature) if the 100-m FWHP beam size is 43”?
Use
S  2.65
TMB  q0 2
(Problem: Repeat for the
l (cm)
5.4  2.65
30-m, with beam
2
 43
TMB   
 60 
27’’, wavelength
2
3.5 mm, flux density
2
(13
. 
4.7 Jy)
TMB  6.7 K
Where q 0 is the telescope beam size in are min.
Suppose the “true” gaussian source size is 10”, what is TB (true brightness
temperature). Could use
S  5.4  2.65
 10 
TMB   
 60 
2
(13
. 
2
Lecture 3 page 5
Or
And get
 432  10 2 

TB  TMB  
2
 10

TB  124K
We know that the electron temperature of NGC7027 is Te = 14000 K.
Use equation of radiative transfer:
(
TB  Te  1  e 
To get

  0.009
This is a source which is thermal, so the radiation is free-free or
Bremsstrahlung
Lecture 3 page 23
SYNCHROTON RADIATION (NON-THERMAL)
Highly relativistic electrons spiraling in a B field with a frequency
B 
e B
 m
  1
P: Power radiated by electron (lab)
P’: Power radiated by electron (rest frame)
d
P
dt
   '
Transformation of acceleration
d '
P' 
dt '
dt    dt '
a '    2  a
So
2e2 4
P  3   a 2
3c
so P = P’
a  B  v 
e B
v
 m 
Radiation patterns of an electron in a B field Lecture 3 page 24
B Field,
B Field,
V about 0
Velocity
V about 0.2 c
Lecture 3 page 25
2  e 2  v2  B 2 2
P
   E 2  B2
2
3
3 m  c
Then
E: Particle energy
Is difficult to separate energy of electron from B field strength
To get spectral distribution, use
dN
 k  E 
dE
Find a synchrotron spectrum :
1
  (  1
2
  0.7
  2.4
Synchrotron radiation is found to be linearly polarized
(power law
distribution of
Cosmic Rays)
Lecture 3 page 26
SINGLE ELECTRON SYNCHROTRON EMISSION
For relativistic electrons, the emitted pulse is 1/ shorter due to
relativistic beaming while the Doppler effect gives rise to a factor 1/2
B: Frequency of rotation
B 
So for B = 10 G,
17.6  B 


  G 
G
176
 B 
Hz


B is even lower when <1
3 2
   G  sin 
Thus in frequency reach a critical value
4p
So if B = 10 G, G = 176 Hz, to reach C = 10 GHz,  = 1.6 104
C 
In Synchrotron emission, we measure only the most relativistic
particles
Lecture 3 page 27
SYNCHROTRON ENERGY CONSIDERATIONS
Allow one to determine the minimum or equipartion energy
WTOT  (  S   R  V
n
4
7
8
7
3
7
Inverse Compton effect
When the radiation density is equal to magnetic energy density there
can be energy losses in addition to synchrotron energy losses. R & W
don’t do much, but Kellermann & Owen give:
LCompton
LSynch
5
 1  TMAX  5 
1  TMAX 
  12  max  1   2  
2  10 
 2  10  
This is the basis of the statement: “1012 K is the highest source
temperature possible”
Lecture 3 page 6
Non-thermal sources
Cas A: at 100 MHz, S= 3 104 Jy, qs=4’ (source size), l = 3 m = 300 cm
S  2.65

TMB  q0 (')
l (cm)
3  104  2.65  T

2
2
16
 9  10 
4
7.5  108 K  T ( source)
Thermal sources have limit T = 2 .104 K
Assume that for Cas A, T=7.5 108 (l(m / 3) -2.8
What is the source temperature at 3 mm?
Source
Specral
Index
Lecture 3 page 29
SUNYAEV-ZELDOVICH EFFECT
Clusters of galaxies are filled with hot diffuse gas. Photons from the 3 K
background are scattered in this cluster gas.
More photons are given energy than lose energy on the low frequency
side of the Planck curve.
On the high frequency side, some photons are shifted to lower energies,
but still a reduction in the 3 K background. At 160 GHz, have a cross over
from absorption at longer wavelengths to emission at shorter
wavelengths, so have zero absorption. The
 T
absorption is:

  2.24  1034  Te  N e  L
 T  SZ
When combined with X ray luminosity, which is Bremsstrahlung (freefree), proportional to Ne2 L, can solve for source distance. Given
velocity of source, get HUBBLE CONSTANT. However there can be
systematic effects such as clumping.
Lecture 3 page 30
EXAMPLE OF S-Z EFFECT
The cluster CL0016 +16 shows on S-Z absorption of –700  K at 1 cm
wavelength
Z = redshift of CL0016 +16 is 0.541
X ray data: Te = 1.6 108 K
Cluster size = 30“ to 19”
RMS Ne = 1.2 10-2 cm-3
So
 T 
34
8
2
24
4

  2.24 10  10  10  3 10 cm  7 10  700K
 T  SZ
( (
(
