Transcript 射电天文基础

射电天文基础
姜碧沩
北京师范大学天文系
2009/08/24-28日，贵州大学
大纲
1.
2.
3.
4.
5.
射电天文基础
射电望远镜
连续谱辐射机制
谱线辐射机制
星际分子
参考书：《射电天文工具》
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射电天文
• Radio
– 大气窗口
– 地面射电天文的频率上限和下限
– 空间
• Astronomy
– Astro-: star
• Radio astronomy
– 与其它波段的区别
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The waves used by optical
astronomers
•
•
•
•
•
•
•
Electromagnetic Spectrum
4000 to 8000 angstroms
7.51014Hz to 3.751014Hz
The Sun
The solar system
Stars
Galaxies
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The radio window
• Atmospheric Transmission
• From about 0.5mm to 20m
• 600GHz to 15MHz
–
–
–
–
Troposphere（对流层） to ionosphere
FM radio (and TV)
AM radio
Mobile phone…
• The solar system, stars, ISM, galaxies, cosmic
microwave background……..
– The Sun
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Some advantages of radio astronomy
• Transparent to terrestrial clouds: visible in
cloudy time
• The Sun is quiet: visible in day time
• Transparent to the vast clouds of
interstellar dust: able to see distant objects
• Different origin of radiation
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射电天文的辉煌
• 获得诺贝尔奖的发现
–
–
–
–
宇宙微波背景辐射的发现
脉冲星的发现：快速旋转的中子星
双星脉冲星的发现与引力波理论的验证
宇宙微波背景辐射的黑体形式以及非各向同性
• 其他重要贡献
–
–
–
–
–
星际分子
氢原子谱线
恒星形成区
磁场

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The world’s largest radio telescopes
• The Arecibo Telescope
Type: Fixed reflector, movable feeds
Diameter of reflector: 1000 ft (304.8 m)
Surface accuracy: 2.2 mm rms
Working wavelength: from cm to dm
• The Effelsberg Telescope
Type: Fully steerable
Diameter: 100-m
Working wavelength: up to 3mm, mainly cm
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Fundamentals of Radio Astronomy
• Some basic definitions
• Radiative transfer
• Blackbody radiation and brightness
temperature
• Nyquist theory and noise temperature
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I: specific intensity
•
dW  I cosddd
• dW＝infinitesimal power, in
watts，
• dσ＝infinitesimal area surface,
in cm2，
• dν＝infinitesimal bandwidth，in
Hz，
• θ＝angle between the normal to
dσand the direction to dΩ
• Iν＝brightness or specific
intensity, in Wm-2Hz-1sr-1。
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The total flux of a source
• Total flux of a source: integration over the total
solid angle of the source Ωs
S 
• Unit
 I ( ,  ) cosd
s
– W m-2Hz-1
– Jy
• 1Jy=10-26 W m-2Hz-1= 10-23 erg s-1 cm-2Hz-1
• A 1Jy source induces an signal of only 10-15W.
• Few sources are as bright as 1Jy
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Brightness is independent of the
distance
I 1 (r1)  I 2 (r 2)
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The total flux density depends on
distance as r-2
• Total flux received at an
point P from an uniformly
bright sphere
 c

S   I ( ,  )cos  d   I    sin  cos  d  d


s
0  c

2
R

R
sin  c 
S  I 2  I 
r
r
2
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Radiation energy density
• Energy density per solid angle: erg cm-3Hz-1
1
u  ()  I 
c
• Total energy density
1
u   u ( )d   I d
c ( 4 )
( 4 )
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Radiative transfer
• For radiation in free
space the specific
intensity is independent
of distance. But I
changes if radiation is
absorbed or emitted.
dI    I ds,
dI     ds,
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 I (s  ds)  I (s) dsd dd   I    dsd dd
dI
  I  
ds
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Limiting cases
• Emission only:   0
dI
  ,
ds
s
I ( s )  I ( s0 )    ( s )ds
s0
• Absorption only:   0
s


dI


  I , I ( s)  I ( s0 ) exp    ( s)ds 
ds


 s0

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Limiting cases (cont’d)
• Thermodynamic Equilibrium (TE): radiation is in
complete equilibrium with its surroundings, the
brightness distribution is described by the Planck
function, which depends only on the
thermodynamic temperature T of the surroundings
dI

 0, I  B (T ) 
ds

2h 3
1
B (T )  2
c e h / kT  1
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 ( s )
 cases (cont’d)

Limiting
I ( s)  I (0)e
  B (T ( ))e d


(s)

0
• Local Thermodynamic Equilibrium (LTE)
– Kirchhoff’s Law  
 B (T )
s

    ds
– Optical depth
d    ds
0
– Equation of transfer
– Solution
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dI
1 dI


 I  B (T )
 ds d 
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LTE (cont’d)
• The medium is isothermal
– T(τ)＝T(s)＝T=const.
I ( s)  I (0)e  ( s )  B (T )(1  e  ( s ) )
• Optical depth is very large
– τ(0) 
I  B (T )
– Difference with the intensity in the absence of
an intervening medium
I ( s)  I ( s)  I (0)  ( B (T )  I (0))(1  e  )
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Blackbody radiation
• Planck law
2h 3
1
B (T )  2
c e h / kT  1
B (T ) 
2hc 2
1
5 e hc / kT  1
• Total brightness of a blackbody
2 k
5
 2 1
4
B(T )  T ,  

1
.
8047

10
erg
cm
s
K
15c 2 h 3
4
4
4
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Wien’s displacement law
• Maxima of B (T) and Bλ(T)
– Bν/ν=0 and Bλ/λ＝0
– νmax
 max
T 
– λmax
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 58.789 
GHz
K
 max

 cm
 T 
   0.28978
 K 
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Rayleigh-Jeans Law
• Rayleigen-Jeans Law
h  kT
T 
 20.84 
GHz
K
e
c
h
1
J (T ) 
I
2
h / kT
k
2k
e
1
2

h / kT
• Radiation
temperature
h
 1

kT
2
BRJ ( , T )  2 kT
c
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2
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Wien’s Law
h  kT
e
h
kT
 1
2h h / kT
BW ( , T )  2 e
c
3
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Brightness temperature Tb
• One of the important features of the Rayleigh-Jeans
law is the implication that the brightness and the
thermodynamic temperature of the blackbody that
emits the radiation is strictly proportional.
• In radio astronomy, the brightness of the extended
source is measured by its brightness temperature
which would result in the given brightness if inserted
into the Rayleigh-Jeans law
c 1

Tb 
B 
B
2
2k 
2k
2
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2
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Transfer equation of Tb
• Transfer equation
dTb (s)
 Tb (s)  T (s)
d 
• General solution
Tb ( s)  Tb (0)e  ( s ) 
 ( s )

T
(
s
)
e
d

0
• Two limiting cases when Tb(0)=0
– Optically thin, τ<<1
Tb    T
– Optically thick, τ>>1
Tb  T
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The Nyquist Theorem
• Johnson noise
– The thermal motion of the electrons in a resistor will
produce a noise power which is the noise determined
by the temperature of the resistor
• The average noise power per unit bandwidth
produced by a resistor R is proportional to the its
temperature, i.e. the noise temperature, and
independent of its resistance
P=kTN
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Electromagnetic wave propagation
fundamentals
•
•
•
•
•
•
•
•
Maxwell’s equations
Energy conservation and the Poynting vector
Complex field vectors
The wave equation
Plane waves in nonconducting media
Wave packets and the group velocity
Plane waves in dissipative media
The dispersion measure of a tenuous plasma
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Maxwell’s equations
• Material equations
J  E
D  E
B  H
• Maxwell’s equations
  D  4
B  0
1
  E   B
c
4
1
 H 
J  D
c
c
• Continuity equation of charge density and current
  J    0
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Energy conservation and the
Poynting vector
• Energy density of an electromagnetic field

1
1
E  D  B  H    E 2   H 2
u
8
8
• Poynting vector
c
S
E H
4π
• Equation of continuity for S
u
   S  E  J射电天文暑期学校
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t

Complex field vectors
• Complex field vectors
E  E1  iE 2 e it
H  H 1  iH 2 e it
• The Poynting vector
c
S
ReE ReH
4

c
*
S 
Re E  H
4
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
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The wave equation
  4 
 H 2 H 2 H
2
c
c
  4 
 E  2 E 2 E
2
c
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c
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Plane waves in nonconducting media
• Nonconducting media
– σ＝0
– The wave equation
1
 u  2 u  0
v
2
– Velocity of the wave
v
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c

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Plane waves (cont’d)
• Harmonic wave solution of the wave equation
u  u 0 e i kx t 
• Wave number
• Phase velocity
k 
2
v

c2

k

2
c

• Index of refraction
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c
c
n     k
v

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Plane waves (cont’d)
• A wave that propagates in the positive z
direction is considered to be plane if the
surfaces of constant phase forms planes
z=const.
E z  0, H z  0
E H  0


H

E
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c
S 
4
 2
E

Group velocity
• Dispersion equation
   (k )
d
 (k )   0 
(k  k 0 )
dk
• Group velocity
0
d
vg 
dk
• Energy and information are usually
propagated with the group velocity
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Plane waves in dissipative media
• Dissipative media  0
• Harmonic waves propagating in the direction of
increasing x E( x, t )  E 0 e i ( kx t )
• Wave equations
 2   2
4   E 
k   c 2   i c 2  H＝0

  

• Dispersion equation
k 
2
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 2 
c
2
4 
1  i

 

射电天文暑期学校
Cont’d
• Wave number k  a  ib
2


 1
4




a  
1 

1



c 2
 



2


 1
 4 

b  
1 

1


c 2
 



• Field
E( x, t )  E 0 e
 bx i ( axt )
e
 n
 


E( x, t )  E 0 exp   nx  exp i  x  t 
 c


 c
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Cont’d
• Index of refraction and absorption coefficient
2


1
4




n  
1 

1



2
 



2


1
4



n  
1 
 1


2
 



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Dispersion measure of a tenuous
plasma
• Plasma: free electrons and ions are uniformly
distributed so that the total space charge
density is zero
• Tenuous plasma
– Interstellar medium
– dissipative medium
• Equation of motion of free electrons
me v  mer  -eE 0 e it
– Solution
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e
e
it
v
E0e
 i
E
im e
me 
射电天文暑期学校
Cont’d
• Conductivity of the plasma
Ne 2
 i
me ω
• Wave number for a thin medium with ε≈1
andμ≈1
2



 
p
2
k  2 1 2 
c   
2
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2
4

Ne
p2 
me
射电天文暑期学校
Cont’d
• Phase velocity and group velocity
– For ω>ωp, k is real, v>c, vg<c
c
v
1
vvg  c
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2

2
p
2

vg  c 1  2

2
p
n  1
射电天文暑期学校

2
p

2
Dispersion measure of pulsars
• A pulse emitted by a pulsar at a distance L
will be received after a delay
2


dl 1  1 p
 D     1
2

v
c
2

0 g
0
L
L
L
2



1
e
1
dl  1 
N (l ) dl
2



c 0  2me 


• The difference between the pulse arrival time
measured at two frequencies
2
e
 D 
2cme
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1
1
 2  2   N (l )dl
 1  2  0
L
射电天文暑期学校
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射电天文暑期学校
Cont’d
• Dispersion Measure

 N   l 
DM    3 d  
  pc 
0  cm




DM
1
1
 4   D  




2
.
410

10

3
2
2 



cm pc
 μs    1 
 2 



 
  MHz 
 MHz  
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1
Dispersion Measure, DM, for pulsars at different Galactic latitudes
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Faraday rotation
• In 1845, Faraday detected that the polarization angle of
dielectric material will rotate if a magnetic field is applied
to the material in the direction of the light propagation
• The rotation of the plane of polarization of an EM wave
as it passes through a region containing free electrons
and a magnetic field, also known as Faraday effect. The
amount of rotation, in radians, is given by RMλ2, where
RM is the rotation measure of the source and λ is the
wavelength. Observation of the Faraday rotation in
pulsars is the most important means of determining the
magnetic field of the Galaxy. It is named after the
English physicist Michael Faraday.
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射电天文暑期学校
Equation of motion for an electron
in the presence of a magnetic field
mv  mr  -eE  r  B
If the magnetic field B is oriented in the z direction
rx
ry
e
e
 Bry   E x
m
m
e
e
 Brx   E y
m
m
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e
e
r  i Br   E 
m
m
r  rx  ir y
E   E x  iE y
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Solution
Linearly polarized wave can be regarded as
the superposition of circularly polarized waves
1
1
E x  ( E   E  ), E y  ( E   E  )
2
2i
Solution in the form of harmonic waves
E  Aei ( k x t )
r  r0 e
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i ( k  x t )
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Parameters of the material
2
Ne
Conductivity: purely imaginary   i

e 

m   B 
m 

e
c  B
Cyclotron frequency which is in resonance
m
with the gyration frequency of the electrons in
e
the magnetic field
c 
B
2m
2
2 


 
p
2

k

1


Wave number
c 2   (   c ) 
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Phase propagation velocity
Index of refraction
2

p
2
n  1 
 (   c )
Phase propagation velocity
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v  c / n
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Relative phase difference
Two circularly polarized waves will have a relative phase
difference after a propagation distance due to the slightly different
phase velocity
2  (k   k  )z
3
p2c
2 Ne3 B
 
z  2 2 z
2
2c
m c
L
e
1
 
B N (z )dz
2
2  //
2 m c  0

5  
 8.1  10  
rad
m
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2 L / pc

0
 B//  N   z 

 3  d  
 Gauss  cm   pc 
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Rotation Measure
RM
5
 8.110
-2
rad m
L / pc

0
 B//  N   z 

 3  d  
 Gauss  cm   pc 
  1    2 



rad   rad 


2
2
 1   2 
   
m  m
B//
6 RM
 1.23  10
Gauss
DM
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Magnetic field parallel to
the line of sight
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Example
• Determine the upper limit of the angle
through which a linearly polarized EM wave is
rotated when it traverses the ionosphere.
Take the following parameters: an
ionospheric depth of 20km, an average
electron density of 105cm-3 and a magnetic
field strength (assumed to be parallel to the
direction of wave propagation) of 1G.
– Find RM
– Carry out the calculation for the Faraday rotation,
Δψ for frequencies of 100MHz, 1GHz and 10GHz,
if the rotation is Δψ/rad=(λ/m)2RM
– What is the effect if the magnetic field direction is
perpendicular to the direction of propagation?
What is the effect on
circularly polarized EM
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waves?
• Repeat previous problem for the
conditions which hold in the solar
system: the average charged particle
density in the solar system is 5 cm-3, the
magnetic field 5μG, and the average
path 10AU. What is the maximum
amount of Faraday rotation of an EM
wave of frequency 100MHz, 1GHz?
Must radio astronomical results correct
for this?
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Example
• A source is 100% linearly polarized in the
north-south direction. Express this in terms of
Stokes parameters.
• Intense spectral line emission at 18cm
wavelength is caused by maser action of the
OH molecule. At certain frequencies, such
emission shows nearly 100% circular
polarization, but little or no linear polarization.
Express this in terms of Stokes parameters.
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examples
• If the DM for a given pulsar is 50, and the
value of RM is 1.2×102, what is the value
of the line-of-sight magnetic field? If the
magnetic field perpendicular to the line of
sight has the same strength, what is the
total magnetic field?
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Homework
• A plane electromagnetic wave perpendicularly
approaches a surface with conductivityσ. The wave
penetrates to a depth of δ. Apply equation (2.25),
taking σ>>ε/4π, so 2 E  (4 / c 2 ) E
The solution
to this equation is an exponentially decaying wave.
Use this to estimate the 1/e penetration depth δ.
Estimate the value of
  c / 4
for copper, which has (in CGS units) σ=1017s-1 and μ≈1
for =1010Hz.
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Cont’d
• Assume that pulsars emit narrow periodic
pulses at all frequencies simultaneously. Use
eq. (2.83) to show that a narrow pulse (width of
order 10-6s) will traverse the radio spectrum at
a rate, in MHz s-1, of
v  1.2 104 ( DM ) 1[ / MHz ]3
• Show that a receiver bandwidth will lead to the
smearing of a very narrow pulse which passes
through the ISM with dispersion measure DM,
to a width
t  8.3 10 DM [ / MHz ] B s
3
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3
Examples
• In the near future there may be an anticollision radar installed on automobiles. This
will operate at ~70GHz. The bandwidth is
proposed to be 100MHz, and at a distance of
3m, the power per area is 10-9Wm-2. Assume
the power level is uniform over the entire
bandwidth of 100MHz. What is the flux
density of this radar at 1km distance? A
typical radio telescope can measure to the
mJy level. At what distance will such radars
disturb such radio astronomy measurements?
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Examples
• A signal passes through two cables with
the same optical depth τ. They have
temperatures T1 and T2, with T1>T2.
Which should be connected first to
obtain the lowest output power from this
arrangement?
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Examples
• The 2.73K microwave background is one of the
most important pieces of evidence in support of
the big bang theory. The expansion of the
universe is characterized by the redshift z. The
ratio of the observed wavelength λo to the
(laboratory) rest wavelength λr is related to z by
z=(λo / λr)-1. The dependence of the temperature
of the 2.73K microwave background on z is
T=2.73(1+z). What is the value of T at z=2.28?
What is the value at z=5 and z=1000?
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Examples
• The pulsar in the Crab nebula has a
dispersion measure DM=57 cm-3pc, and a
period of 0.0333s. Staelin and Reifenstein
(1969 Science 162, 1481) discovered this
pulsar at ν=110MHz, using a 1MHz-wide
receiver bandwidth. Someone tells you that
“this pulsar would not have been found at
110MHz if the pulses all had the same
amplitude.” Do you believe this? Use the
following relation to support your decision:
the smearing Δt of a short pulse is
(202/νMHz)3DM ms per MHz of receiver
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bandwidth.
Homework
• A cable has an optical depth τof 0.1 and a
temperature of 300K. A signal of peak
temperature 1K is connected to the input of this
cable. Use equation (1.34) in the textbook with
T being the temperature of the cable and T (0)
the temperature of the input signal. What is the
temperature of the output of the signal? Would
cooling the cable help to improve the
detectability of the input signal?
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Homework (cont’d)
• A signal passes through two cables with
the same optical depth, t. These have
temperatures T1 and T2, with T1>T2.
Which cable should be connected first to
obtain the lowest output power from this
arrangement?
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Homework (cont’d)
• Apply the Stefan-Boltzman relation to the Sun and the
planets to estimate the surface temperature if each
planet is assumed to absorb all of the radiation it
receives (this is an albedo of zero – this is the upper limit
the planet can absorb since in reality some radiation is
reflected). As a first approximation, assume that the
planets have no atmosphere and no internal heating
sources and that the rapid rotation equalizes the surface
temperatures. The distances for assumed circular orbits
(in AU) are: Mercury (0.39AU), Venus (0.72AU), Earth (1
AU) , Mars (1.5AU), Jupiter (5.2AU). At a wavelength of
68cm, Jupiter was found to have a brightness
temperature of more than 500K. Could the temperature
of Jupiter be caused by solar heating?
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Telescopes
• The Green Bank Telescope
Type: off-axis, fully steerable
Diameter: 100 by 110 meters
Surface accuracy: 1.2mm--0.3mm
Working wavelengths: cm to mm
• The Parkes Telescope
Diameter: 64-m, in the southern sky
Working wavelength: cm
• The Nobeyama 45-m
• JCMT JCMT with no membrane
15-m, sub-mm(surface accuracy 14-18 m), Mauna
Kea
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Telescopes
• Interferometers
VLBA: 10 radio telescopes across USA
VLA: 27 25-m antennas, Y-shape, largest separation of
antenna 36km (0.04 arcsecond at 43GHz)
The VLA looking south
MERLIN: an array of radio telescopes in UK, with
separation up to 217km (0.05 arcsecond at 5GHz)
• List of radio telescopes
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Radio astronomy in China
•
Telescopes
•
•
•
•
•
Miyun Synthesis Radio Telescope: linear array of 28
9-m antennas working at 232MHz
Shanghai: 25-m
Urumuqi: 25-m
Qinghai Delingha: 13.7-m
Projects
•
•
•
FAST: Five hundred meter Aperture Spherical
Telescope 30 elements, Guizhou
Large radio telescope: 50-m
MSRT FAST DLH Urumqi Sheshan
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The future of radio astronomy
• Bigger telescopes
– Atacama Large Millimeter Array(ALMA)
• ESO,IRAM,OSO,NFRA,NRAO,NAOJ……
• 64 12-m antennas, 10mm-0.35mm, 150m-10km
• Year 2012
– VSOP-2
• Research
Fainter objects, finer structure
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Homework
• If the average electron density in the interstellar
medium is 0.03 cm-3, what is the lowest
frequency of electromagnetic radiation which one
can receive due to the plasma cutoff? Compare
this to the ionospheric cutoff frequency if the
electron density, Ne, in the ionosphere is ~105cm3. Use
νp
N
kHz
 8.97
e
3
cm
Where p is the plasma cutoff frequency.
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JCMT
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JCMT without membrane
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Parkes
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佘山
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乌鲁木齐
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VLA
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FAST
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Nobeyama
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White light, radio and X-ray Sun
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德令哈
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