High Energy Astrophysics
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Transcript High Energy Astrophysics
Radiation Processes
High Energy Astrophysics
[email protected]
http://www.mssl.ucl.ac.uk/
Absorption Processes
So far, considered the production of X-rays.
Now, will consider X-ray absorption.
Emission processes
Recombination
Absorption process
Photoionization
Inverse Compton
electron scattering
e-/p+ annihilation
e-/p+ pair production
synchrotron emission
synchrotron self absorption
Photoionization
e-
Atom absorbs photon
EI
h
3
Atom, ion or
molecule
h
Cross-section ()
characterized by edges
corresponding to
ionization edges.
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
Example of photoelectric absorption
eg. soft X-rays from a star absorbed by ISM
star
interstellar cloud
observer
I
I
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
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
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
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
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...
Clumping of the ISM
Take an example at low energies, e.g. at
...
24
2
h 0.1keV, eff 10 m
Average ISM density
H 10 m
6
3
At a distance,
d=100 pc
310 m
18
Smooth versus clumpy
star
observer
smooth
6
3
10 / m
clumpy
Hot medium
0.110 / m
6
3
Cold dense clouds
4 10 / m
6
3
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 310 0.110 0.310 / m
18
6
24
24
F F0 exp 0.310 10
24
2
0.75F
0
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.
Thomson Scattering
low-E photon scattered by electron -
h
electron
h
Thomson cross-section is given by -
8
2
re , where re 2.821015 m
3
29
2
e
6.6510 m
Thomson scattering cont.
If
3
N = number of particles per m
then fraction of area
1m
blocked by a square
metre of path =
29
6.6510
1m
If R is the extent of
the absorbing region
along the line of
sight,
N /m
t 6.6510 NR
29
( = optical depth)
and
F F0 exp t
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
Compton scattering (cont.)
On average,
0
h
2
0
mo c
0
h
2
mo c
Electron-positron pair production
e
g-ray
y
q
x
photon
e+
Two photons, one of which must be a g-ray, collide and
create an electron-positron (e-/e+) pair. This is therefore
a form of g-ray absorption
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
1
v
1 2
c
2
2
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
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
2
pg c p c 2 pg p pc cosq
2 2
2 2
p
2
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
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
And this is...
found by simply making the denominator as
large as possible, ie when cos(q)=-1, ie
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
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.
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 310 eVm
5
Corresponding to about 109 photons / m 3
6
No of nuclei in space is about 10 / m3
Synchrotron Self-Absorption
e-
e-
Relativistic electrons moving
in a magnetic field
Synchrotron Emission
Electrons, mainly responsible for emission at frequency ,
have energy, E, given by:
1
2
2mo c
2 1
E~
mo c .
c
eB
and for a power law electron spectrum
logF
log
Blackbody turnover
Assume Synchrotron power-law cut off, max, is
given by:
max
2
E eB
3 4
2mo c
And assume each electron emits & absorbs only at
this peak frequency. Then, we will replace this with
the mean energy per particle for a thermal source, ~ kT.
On the Rayleigh-Jeans side...
impossible
logF
R-J
blackbody
synchrotron
log
Rayleigh-Jeans approximation to blackbody...
2kT 2
I d 2 d
c
Source distance
For d=source distance and R=source size,
R
d
2
R
2
d
Total flux at Earth...
So total energy flux at Earth is given by:
2E 2
F I 2
c
8m
Be
3
o
5
1
2
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.
… and SSA frequency
Substituting for then:
8m
F
Be
3
o
5
1/ 2
2
R
2
d
and
R 310 F B d
17
1/ 2
1/ 4
5 / 4
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
Radiation processes (summary)
• Thermal - Bremsstrahlung
electron energies ~ photon energies
to produce X-rays, b = v/c ~ 0.1
• Non-thermal - Synchrotron and Inverse
Compton
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
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
Inverse Compton Scattering
For a relativistic electron colliding with a low
energy photon, gIC2 ≈ hfinal/hinitial
For X-ray production consider:
- starlight: <h> ~ 2eV (l~6000A)
-4
- 3K background: <h> ~3x10 eV
then
3
g
2
IC
8keV
h
= 410 for stars
7
= 310 for the 3K background
We need cosmic rays!!!
RADIATION PROCESSES
END OF TOPIC