Transcript High Energy Astrophysics - Mullard Space Science Laboratory

Radiation Processes
High Energy Astrophysics
[email protected]
http://www.mssl.ucl.ac.uk/
1. Interaction of radiation with matter:
Photoelectric absorption and the ISM;
Thomson and Compton scattering; Pair
production; Synchrotron self-absorption;
Inverse Compton scattering
[2]
2
Absorption Processes
Photon emission processes have
corresponding absorption processes
We will consider
X-ray absorption.
Emission processes
Recombination
Absorption process
Photoionization
Inverse Compton
electron scattering
e-/e+ annihilation
e-/e+ pair production
synchrotron emission
synchrotron self absorption
3
Photon Absorption Process Cross Sections
• Absorption coefficients are plotted
against photon energy for the three
processes:
- Photoelectric absorption
- Compton effect
- Pair production
• Absorber is lead – plots shift
up and down in energy with
Z increasing or decreasing
• Photoelectric absorption is
dominant at low energies
and pair production at high
energies (E > 2moc2) while
the Compton effect is
dominant at intermediate
energies
Z
E (MeV) →
4
Photoionization
e-
Atom absorbs photon
(  EI )


h
3.5
Atom, ion or
molecule
h
Cross-section ()
characterized by edges
corresponding to
ionization edges.
5
Photoelectric Absorption Cross-section
The photoelectric absorption cross-section for
photons with E > EI and h << mec2 is given by K = 4√2 T a4 Z5 (moc2/)7/2
where EI is the electron binding energy, a is the
fine structure constant and T is the Thomson
cross-section
Note dependence on Z5 and on -7/2
6
Example of photoelectric absorption
eg. soft X-rays from a star absorbed by ISM
star
interstellar cloud
observer
I
I


7
How much passes through?
Take a path of length dl (metres)
nZ is the number density (m 3 ) of element Z.
Cross-section offered by element Z at energy
2
 Z ( E)(m )
E is given by:
dl (m)
dV
8
The fraction of volume dV which is blocked
by the presence of element Z is :
nZ Z ( E)dl
Thus fraction of flux F lost in volume dV is:
dF  FnZ Z ( E)dl
or :
dF
 nZ  Z ( E )dl
F
9
Integrating over length from source...
dF
 F   nZ Z ( E )dl   Z ( E ) nZ dl
 F  F0 exp(  Z ( E )  nZ dl )
Including all elements in the line of sight:




n
H

F  F0 exp    Z ( E )  nZ
dl 
n
Z
H



10
Optical depth
This becomes:
 F0 exp(  eff ( E).N H )
This is ‘t’, the optical depth,
which has no dimensions

nZ 
 eff ( E )    Z ( E ) 
nH 
Z 
This is the effective
cross-section,
weighted over the
abundance of
elements with respect to hydrogen
11
Interstellar Medium Absorption Cross-section
The effective photoelectric
absorption cross-section, eff,
is plotted against wavelength
in Å for the interstellar medium
for an assumed set of interstellar
element abundances (Morison and
McCammon, 1983, Ap.J., 270, 119)
12
Column density
The column density given by :
N H   nH dl
is the number of H – atoms per m2 column
Column density is measured from the 21cm
atomic hydrogen line - but not foolproof.
There is a factor of 2 uncertainty, wide beams,
molecular hydrogen contamination...
13
Clumping of the ISM
Take an example at low energies, e.g. at
h  0.1keV , eff  10 m
24
Average ISM density
 H  10 m
6
3
At a distance,
d=100 pc
 310 m
18
14
2
Smooth versus clumpy
star
observer
smooth
6
3
10 / m
clumpy
Hot medium
0.110 / m
6
3
Cold dense clouds
4 10 / m
6
3
15
Numerical example
• Through the smooth medium -
N H   H  d  3 10 / m
24
(
F  F0 exp  3 10 10
24
 24
)
2
F0

 0.05 F0
20
• Through the clumpy medium -
N H  310  0.110  0.310 / m
18
(
6
24
24
F  F0 exp  0.310 10
24
2
)  0.75F
0
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Electron scattering
• Thomson scattering
- the scattering of a photon by an electron
where the photon energy is much less than
the rest mass of the electron.
• Compton scattering
- photons have a much higher energy in this
case and lose some of their energy in the
scattering process.
17
Thomson Scattering
low-E photon scattered by electron -
h
electron
h
Thomson cross-section is given by -
8
2
  re , where re  2.821015 m
3
29
2
e
   6.6510 m
18
Thomson scattering cont.
If
3
N = number of particles per m
then fraction of area
1m
blocked by a square
metre of path =
29
6.6510
1m
If R is the extent of
the absorbing region
along the line of
sight,
N /m
t  6.6510 NR
29
( = optical depth)
and
F  F0 exp( t )
19
Compton scattering
In Compton scattering, wavelength increases
and frequency decreases i.e. photon energy
decreases
electron
h 0
frequency
change
h
q
h
(1  cosq )
 
2
  0 mo c
1
1
20
Compton scattering (cont.)
On average,
 0 
h

2
 0
mo c


0
h

2
mo c
21
Electron-positron pair production
e-
g-ray
y
q
x
photon
e+
Two photons, one of which must be a g-ray with E > 2mec2,
collide and create an electron-positron (e-/e+) pair. This is
therefore a form of g-ray absorption
22
Minimum g-ray energy required
Must first demonstrate that
relativistic invariant.
E  ( pc) is a
Rest energy of particle,
m  gm0
g 
2
2
E  mo c
2
1
 v 
1  2 
 c 
2
23
Thus, from E  m c and pc  mvc ,
2
(m c )
2 2
(m0vc)
(
m c c v



2
2
2
2
2
2
1 v / c
1 v / c
1 v / c
(
0
2
) (
(
2 2
0
) (
2
2
)
)
)
m c c v
2 4


m
c
0
2
2
c v
2
And
this
is
a
c
relativistic invariant
2
0
2
2
2
24
 

p  pg  p p
Total initial momentum,
( pc)
2
thus
 ( px c )  ( p y c )
2
2
 ( pg c  p p c cosq )  ( p p c sin q )
2
2
 pg c  p c cos q
2
2 2
2
 2 pg p pc cosq  p pc sin q
2 2
2 2
p
2
 pg c  p c  2 pg p pc cosq
2 2
2 2
p
2
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pg c  Eg ,
But since
( pc)
2
 Eg  E  2Eg E p cosq
2
2
p
and -
[ E  ( pc) ]initial  (Eg  E p )
2
2
2
(
 Eg  E  2Eg Ep cosq
2
2
p
)
 2Eg E p (1  cosq )
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Calculating the minimum energy
Assuming e+ and e- have no momentum…
(
 [ E  ( pc ) ] final  2mo c
2
2
and since
 2Eg E p (1  cosq )
Which gives us
this expression
for the energy
of the g-ray
photon
Eg 
)
2 2
,
(2m c )
2 2
o
2 E p (1  cos q )
27
And this is...
found by simply making the denominator as
large as possible, ie when cos(q)= -1, or when
q=180 degrees.
g-ray
And the minimum
g-ray energy is
given by:
e-/e+ photon
(
mc )

2 2
Eg min
o
Ep
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Photon-nucleus pair production
• In the laboratory, it is more usual to
consider photon-nucleus production.
So why do we ignore it in space?
• Photons and nuclei have a similar crosssection, and the g-ray does not differentiate
much between another photon or a nucleus.
• Then we must compare the photon density
with the particle density in space.
29
Photon versus particle density
e.g. for 3 K m-wave background photons -
E  h  3 10 eV
4
14
3
3
U ph  5 10 Jm  310 eVm
5
Corresponding to about 109 photons / m 3
6
No of nuclei in space is about 10 / m3
30
Synchrotron Self-Absorption
e-
e-
Relativistic electrons moving
in a magnetic field
31
Synchrotron Emission
Electrons, mainly responsible for emission at frequency ,
have energy, E, given by:
1
2
 2mo c 
2 1
E~
 mo c .
c
 eB 
and for a power law electron spectrum
logF
log
32
Blackbody turnover
Assume Synchrotron power-law cut off, max, is
given by:
max
2
E eB

3 4
2mo c
and assume each electron emits and absorbs only at
this peak frequency. Then, we will replace this with
the mean energy per particle for a thermal source or
E ~ kT
33
On the Rayleigh-Jeans side...
impossible
logF
R-J
blackbody
synchrotron
log
Rayleigh-Jeans approximation to blackbody...
2kT 2
I ( )d  2  d
c
34
Source distance
For d=source distance and R=source size,

R
d
2
R
 2
d
35
Total flux at Earth...
So total energy flux at Earth is given by:
2E 2
F  I ( )  2  
c
 8m 


Be

3
o
5
1
2





36
SSA spectrum
SSA
log F
Optically-thick
regime
a
log 
Optically-thin
 a lies at the point where the observed
synchrotron flux equals the blackbody limit.
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… and SSA frequency
Substituting for  then:
 8m 
F  
 Be
3
o
5
1/ 2



2
R
2
d
and
R  310 F B d
17
1/ 2
1/ 4
5 / 4
38
SSA in Compact X-ray sources
X-ray frequency, =1018 Hz
-29
If F ~ 10 J m-2 s-1 Hz - typical X-ray source value
d = 10 kpc and B = 108 Tesla
(the field for a neutron star)
This gives a maximum for R of ~1 km for SSA of
X-rays to occur (ie for a to be observable in the
X-ray band).
but a neutron star diameter is 10 to 20km
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Radiation processes (summary)
• Thermal - Bremsstrahlung
electron energies ~ photon energies
to produce X-rays, b = v/c ~ 0.1
• Non-thermal - Synchrotron and Inverse
Compton
40
Synchrotron Emission
For an electron spiralling in a magnetic field B with
energy E, the peak radiated frequency, m is
m = g2 B e/2  mo
= E2 B e/2  mo3 c4
But
Hence
E = g mo c2 - for a relativistic electron
g2 = 2  mo m/B e
41
Electron energies required
• Synchrotron emission
depends on the magnetic field strength.
Assuming equipartition of energy - starlight,
cosmic rays + magnetic fields have all the
same energy density in Galaxy
2
B
-10
 U PH , => B=6x10 Tesla
and from
2m 0
To produce X-rays of m ~ 1018 Hz, we need
g ~ 5 10
2
S
16
42
Inverse Compton Scattering
For a relativistic electron colliding with a low
energy photon, gIC2 ≈ hfinal/hinitial
For X-ray production consider:
- starlight: <h> ~ 2eV (l~6000A)
-4
- 3K background: <h> ~3x10 eV
then
3
g
2
IC
8keV

 h 
= 410 for stars
7
= 310 for the 3K background
We need cosmic rays!!! 43
RADIATION PROCESSES
END OF TOPIC
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