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Slide 1
Radiation Processes
High Energy Astrophysics
[email protected]
http://www.mssl.ucl.ac.uk/
Slide 2
Absorption Processes
So far, considered the production of X-rays.
Now, will consider X-ray absorption.
Emission processes
Recombination
Inverse Compton
e-/p+ annihilation
synchrotron emission
Absorption process
Photoionization
electron scattering
e-/p+ pair production
synchrotron self absorption
So far we have seen processes which lead
to the production of X-rays. Now we
consider processes which absorb
radiation, ie. lead to the removal of
photons from a beam. For example, one
these processes is scattering.
Every process responsible for the
emission of radiation has a corresponding
absorption process.
Photoionization leads to recombination of
the removed electron, thus
photoionization removes X-ray photons also extreme UV photons - and
dominates in these bands.
The two forms of electron scattering,
Thomson scattering and Compton
scattering, remove photons from the line
of sight. In Thomson scattering, the
photon energy is much lower than the rest
mass of the electron and is just deflected.
In Compton scattering, the photon
effectively transfers some of its energy to
the electron, thus light is absorbed.
At very high energies, photons can
collide with ambient photons and form an
electron-positron pair, ie photon energy is
lost and particles are formed. In the
inverse process, electrons and positrons
can annihilate each other, emitting
gamma-rays.
In synchtron self-absorption, the photons
produced by the spiralling of electrons in
the magnetic field, are then absorbed by
those same electrons.
Slide 3
Photoionization
e-
Atom absorbs photon
  EI 


h
3
Atom, ion or
molecule
h
Cross-section ()
characterized by edges
corresponding to
ionization edges.
Slide 4
Example of photoelectric absorption
eg. soft X-rays from a star absorbed by ISM
star
interstellar cloud
observer
I
I


Slide 5
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
 Z ( E )(m 2 )
E is given by:
dl (m)
dV
[High Energy Astrophysics, Longair, 2nd
Edition, Vol 1, p89-91]
An atom, ion or molecule absorbs a
photon and releases an electron with an
energy equivalent to the energy of the
incoming photon less the ionization
energy for the level that the electron
occupied.
The absorption cross-section is
characterized by edges which correspond
to the various ionization levels. The fall
in cross-section between the edges is
proportional to the cube-root of the
frequency. The position of the edges in
frequency space reveals which
atom/ion/molecule is involved and the
depth of the edge gives the amount of
material in the line of sight.
We will take the example of X-rays
leaving a star and passing through an
interstellar cloud, which is photoionized
by the star’s X-rays, absorbing photons
from out of the line of sight.
If the star has a spectrum of the form
shown on the left, the resultant spectrum
will be absorbed as shown on the right.
Slide 6
The fraction of volume dV which is blocked
by the presence of element Z is :
nZ  Z ( E )dl
Thus the fraction of flux lost in volume dV is:
dF   Fn Z  Z ( E ) dl
or :
dF
  nZ  Z ( E )dl
F
Slide 7
Integrating over length from source...
dF
  nZ Z ( E)dl   Z ( E) nZ dl
lnF + c 
F
 F  F0 exp( Z ( E )  nZ dl )
Including all elements in the line of sight:


n 
F  F0 exp    Z ( E )  nZ H dl 
nH  
 Z 
Slide 8
Optical depth
This becomes:  F0 exp   eff ( E ).N H 
This is ‘’, the optical depth,
which has no dimensions

 eff ( E )    Z ( E )
Z

nZ 

nH 
This is the effective
cross-section,
weighted over the
abundance of
elements with respect to hydrogen
The fraction of volume, dV, which is
blocked by the presence of element Z can
also be thought of as the probability that
a photon from the star is absorbed in
volume dV.
Slide 9
Column density
The column density is given by :
N H   nH dl
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...
n_H is the interstellar hydrogen number
density, and N_H is the interstellar
hydrogen ‘column’ density and has units
of m^-2.
These derivations assume that the ratio of
n_Z to n_H is NOT a function of
distance, while n_H MAY BE a function
of distance.
For values of the effective cross-section,
see eg. Morrison and McCammon, 1983,
ApJ, 270, 119.
The column density of hydrogen in any
given direction is measured from the flux
in the 21cm atomic hydrogen line using
radio telescopes. Unfortunately though,
there is a factor of 2 uncertainty in
making these measurements - the 21cm
line is emitted during a transition in
which spins of the electron and proton
change from parallel to anti-parallel.
There are two possible spins of the
electron and proton, thus a factor of 2
uncertainty arises.
There are also problems due to the very
wide beams used (2x3 degrees for the
Stark et al all-sky survey) which give
poor spatial resolution… and there is an
unknown contribution to the column from
molecular hydrogen (although the
distribution of molecular hydrogen is
generally found to follow the atomic
hydrogen distribution).
Slide 10
Clumping of the ISM
Take an example at low energies, eg at ...
h  0.1keV ,  eff  10 24 m 2
Average ISM density
At a distance,
 H  106 m 3
d=100 pc
 31018 m
We are going to examine the effects of
clumping of the interstellar medium
(ISM) on the absorption of X-ray photons
by comparing how the light from a star
100 parsecs away is affected by passing
through a smooth medium with the light
passing across the same distance but
through a clumpy medium, with cold gas
clouds embedded in a hot gas.
Slide 11
Smooth versus clumpy
star
observer
smooth
10 6 / m 3
clumpy
Hot medium
0.1 10 6 / m 3
Cold dense
clouds
6
3
4  10 / m
Slide 12
Numerical example
• Through the smooth medium -
N H   H  d  3 10 24 / m 2
F
F  F0 exp  3  10 24  10  24   0  0.05 F0
20
• Through the clumpy medium -
N H  3  1018  0.1 10 6  0.3  10 24 / m 2


F  F0 exp  0.3  10 24  10 24  0.75 F0
One of the problems, as I mentioned
earlier, with measuring the amount of
absorbing gas along our line of sight to a
star, is that measurements of the 21cm
line emission are made using a very wide
beam, hence low spatial resolution. So it
is not known whether there is any fine
structure in the ISM or not. When we
measure the 21cm emission at a certain
position, we assume that this is for a
smooth distribution of cold hydrogen gas.
But in reality, we could be seeing the star
through a gap in cold clouds which are
suspended in a hot, low density medium.
This makes a dramatic difference to
amount of flux that is actually
transmitted. And if the amount of
absorption is overestimated because we
assume a smooth distribution of
hydrogen, then we calculate a much
higher intrinsic luminosity for the source
than is actually the case.
So we want to compare the amount of
lost through a smooth, relatively dense
medium with that lost through a clumpy
medium, where cold clouds are
suspended in a hot, low density gas, and
we are seeing the star through a gap in
the clouds.
Thus we need to calculate the flux
actually observed in each case, F.
Using
F = F_0 exp (-sigma_eff (E) x N_H)
we first calculate N_H in each case, from
the distance to the star and the density of
the interstellar medium. Then using this
value for N_H and the effective crosssection, sigma_eff, we can calculate F for
each scenario in terms of the intrinsic
flux, F_0.
If we assume the case of a smooth
medium, we calculate that we actually
observe only 1/20th of the intrinsic flux,
thus we would multiply the observed flux
by 20 to find the intrinsic luminosity.
If in fact the medium is clumpy as
described, and we see the star through a
gap, then the star has only lost about 75%
of its flux and we are vastly overcompensating for photoelectric
absorption along the line of sight.
Observations support the clumpy medium
model, with cold HI clouds (~80K)
suspended in hot HII gas (~9000K). The
clouds may contain 75% of the ISM mass
but only 10% of the volume.
Slide 13
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.
Slide 14
Thomson Scattering
low-E photon scattered by electron -
h
h
electron
Thomson cross-section is given by -
 
8
 r e 2, where re  2.82  10 15 m
3
29
2
e
   6.65  10
m
Slide 15
Thomson scattering cont.
If
N = number of particles per m3
then fraction of area
1m
blocked by a square
metre of path =
6.65  10
1m
If R is the extent of
the absorbing region
along the line of
sight,
29
N /m
  6.65  10 29 NR
( = optical depth)
and
F  F0 exp  
See Longair, HEA, 2nd Edition, Volume
1, p92-96
The full derivation of the cross-section
presented by an electron to an incoming
photon in the case of Thomson scattering
is given in Longair.
For this course, you will only need to
know that the cross-section, sigma_e, is
given by the above expressions (r_e is the
classical electron radius).
Thomson scattering is a very important
process and the Thomson cross-section
appears in many radiation formulae.
Some interesting facts about Thomson
scattering:
1. It is symmetric with respect to the
scattering angle, ie as much is scattered
forwards as backwards.
2. Electrons present the same Thomson
cross-section to 100% polarized light as
they do to unpolarized light.
3. The scattered radiation is polarized
however: 100% in the plane orthogonal
to the direction of travel of the incoming
photon. (0% in the direction of travel of
the incoming photon).
4. Thomson scattering is one of the most
important processes for impeding the
escape of photons through a medium. For
example, it is Thomson scattering which
builds up radiation pressure in very
luminous sources and defines the
Eddington limit.
Slide 16
Compton scattering
In Compton scattering, the photon wavelength
increases, ie its energy decreases.
h 0
frequency
change
h
electron
1


1
0


h
1  cos 
me c 2
Slide 17
This effect of cooling the radiation (ie
because it loses energy) and transferring
the energy to the electron is sometimes
called the recoil effect.
Compton scattering cont.
On average,
 0 
h

 0
me c 2


0

The frequency change of the photon in
the Compton scattering process is given
by the relationship shown, where m_e is
the mass of an electron, c is the speed of
light and h is Planck’s constant. nu_0 is
the initial frequency of the photon and nu
is the final energy (ie after being
scattered).
For very high energy electrons, the
proper quantum relativistic cross-section
for scattering must be used, rather than
the classical Thomson cross-section.
Also, if the electron is moving
ultrarelativistically, the quantum
relativistic cross-section must be used
and this is given by the Klein-Nishina
formula.
h
me c 2
Compton scattering has the effect of
broadening spectral lines and washing out
edges.
Slide 18
Electron-positron pair production
e-
-ray
y

x
e-/e+ photon
e+
Two photons, one of which must be a -ray,
collide and create an electron-positron (e-/e+)
pair. This is therefore a form of -ray absorption.
An electron-positron pair can be created
when a gamma-ray collides with another
photon. In the process, the gamma-ray is
absorbed.
In the coming viewgraphs, we will derive
the minimum energy of the gamma-ray
required for this process to occur. Theta
is the angle between the two colliding
photons - and we will be resolving
components of the momentum vectors
onto two mutually orthogonal vectors, x
and y, as shown.
Slide 19
Minimum -ray energy required
Must first demonstrate that
relativistic invariant.
E 2   pc  is a
2
E  mc 2
Rest energy of particle,
m  m 0
1

 v2 
1  2 
 c 
Slide 20
Thus, from E  mc

m c 
2 2
1  v
0

2
/c
2

and pc  mvc ,
2
m0vc 2
 1  v
2
/c




m02 c 2 c 2  v 2
1 v2 / c2




m 02 c 2 c 2  v 2
 m 02 c 4
c2  v2
And this is a
c2
relativistic invariant
Slide 21
 

p  p  p p
Total initial momentum,
thus
2
 pc 2
  pxc    p yc 
2
2
  p c  p p c cos     p p c sin  
2
2
 p2 c 2  p 2p c 2 cos 2  2
 2 p p p c 2 cos   p 2p c 2 sin 2 
 p2 c 2  p 2p c 2  2 p p p c 2 cos 
Thus we demonstrate that E^2 - (pc)^2 is
an invariant, because a particle’s restmass, m_0, is independent of its velocity,
v. Therefore in a collision, [E^2-(pc)^2]
before is the same as that afterwards.
This property will now be used to
examine the case of an interaction of a
gamma-ray with a low-energy photon.
Slide 22
p c  E
But since
So now we have a simple expression for
the initial E^2-(pc)^2 of the gamma ray
and the low energy photon.
,
 pc 2  E2  E p2  2 E E p cos
and -
[E2   pc ]initial  E  Ep 
2
2

 E2  E p2  2 E E p cos

 2 E E p 1  cos  
Slide 23
Calculating the minimum energy
Assuming e+ and e- have no momentum…


 [ E 2   pc  ] final  2me c 2
2
and since
 2 E E p 1  cos 
Which gives us
this expression
for the energy
of the -ray
photon
E 
2
,
2m c 
2 2
e
2 E p 1  cos 
Slide 24
And this is...
found by simply making the denominator as
large as possible, ie when cos()=-1, ie
when =180 degrees.
-ray
And the minimum
-ray energy is
given by:
If the electron and positron have no
momentum, then p=0 so the final energy
is simply (2E)^2 = (2mc^2)^2. Then
equating this to the expression to the
initial energy (because the quantity (E^2 (pc)^2) is a relativistic invariant), we
derive the expression for the gamma-ray
photon as shown.
So, at last, we derive an expression for
the minimum energy of the gamma-ray
photon. E_gamma is a minimum when
the denominator is a maximum, which is
when cos(theta)=-1, ie when theta=180
degrees.
e-/e+ photon
E min 
m c 
2 2
e
Ep
The minimum energy is also inversely
proportional to the energy of the photon
with which it interacts.
Slide 25
Minimum energy for mm-wave photon
-ray photon interacts with mm-wave
Taking the example of a very low-energy
mm-wave photon, the amount of energy
which the gamma-ray must have is at
least 2.5e14 eV, which is very high!
First converting to eV :
-3
=1.2mm corresponds to h=10 eV
E min 
m c   0.5 10 
2 2
6 2
e
Ep
10 3
 2.5  1014 eV
Slide 26
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 -ray does not differentiate
much between another photon or a nucleus.
• Then we must compare the photon density
with the particle density in space.
Slide 27
Photon versus particle density
eg., for 3K -wave background photons -
E  h  3  10 4 eV
U ph  5  10 14 Jm 3  3  10 5 eVm 3
Corresponding to about 109 photons / m 3
No of nuclei in space is about 106 / m3
When a gamma-ray interacts to produce
an electron-positron pair, it makes little
difference to the gamma-ray whether it
interacts with another photon or with a
nucleus. Indeed in a laboratory situation,
photon-nucleus pair-production is more
commonly considered.
In both types of interaction, a similar
cross-section is presented to the gammaray (about 1e-29 m^2). It is the number
density of photons in space compared to
the number density of particles which is
important.
So comparing the microwave background
photon density with the ISM proton
density, we see that photons are about
1000 times more numerous than particles,
thus photon-photon collisions are far
more important.
Slide 28
Synchrotron Self-Absorption
ee-
Relativistic electrons moving
in a magnetic field
Slide 29
Synchrotron Spectrum
Flux emitted as a function of frequency:
1
 2me c  2
2 1
E~
 me c .
c
 eB 
E
logF

log
Slide 30
Blackbody turnover
Assume power-law cut off, m , is given by:
 max 
E 2 eB
2 m e3 c 4
And assume each electron emits and absorbs
synchrotron radiation only at this peak
frequency. Then, we will replace this with
the mean energy per particle for a thermal
source, ~kT.
See Accretion Power in Astrophysics,
Frank, King and Raine, 2nd Edition p 224
Synchrotron emission is produced when
relativistic electrons spiral around the
force lines of a magnetic field. At
sufficiently low frequencies, relativistic
electrons in the same field can absorb the
photons produced earlier by other
electrons. This process is known as
synchrotron self absorption.
We are now going to take a look at the
synchrotron spectrum, and see how it is
modified by the synchrotron selfabsorption process.
The synchrotron flux emitted as a
function of frequency is shown. It states
that the flux is proportional to the square
root of frequency, producing the
spectrum shown (a straight line with a
gradient of -1/2 in F_nu versus nu).
But this spectrum cannot continue rising
forever as the source would then have an
infinite luminosity. It must cut-off at
some point. And of course the spectrum
cannot exceed its blackbody intensity,
because this is the maximum intensity
that any source can reach.
We want to work out what the
synchrotron self-absorption spectrum
looks like and how it behaves. We will do
this by taking a simplistic approach to the
problem.
We assume that each electron emits and
absorbs at the peak frequency shown in
the equation above. Then the emitted
intensity can be obtained by replacing the
mean energy per particle for a thermal
source (~kT) with the energy given by
the equation above.
Slide 31
On the Rayleigh-Jeans side...
impossible
logF
blackbody
synchrotron
log
R-J
Rayleigh-Jeans approximation to blackbody...
I  d  
2 kT 2
 d
c2
Slide 32
Total flux at Earth...
So total energy flux at Earth is given by:
F  I   
 8  m e3
 
Be

Slide 33
2E 2
 
c2
1
5
2
 

The diagram illustrates the problem - the
synchrotron emission cannot continue to
rise towards lower frequencies (ie longer
wavelengths) because it will exceed the
blackbody intensity. The part of the
blackbody spectrum which applies in this
case is described by the Rayleigh-Jeans
approximation as shown. Omega is the
angle subtended by the source at the
Earth.
So to calculate the spectrum in the regime
where the synchrotron emission would
exceed the Rayleigh-Jeans flux, we
assume that it can be described as a
thermal source and replace the mean
energy per particle (~kT) with the value
of E at the peak frequency nu_max. Then
to calculate the flux, the intensity has to
be multiplied by the angle subtended by
the source at the Earth, Omega.
Note that the resulting emission will not
be blackbody though, because the
electrons do not have a thermal
distribution.
At an observed frequency nu_a, the
observed synchrotron flux equals the blac
body limit. Below nu_a then, the source
becomes optically-thick (ie blackbody
like) and the spectrum turns over,
following the Rayleigh-Jeans tail.
Slide 34
Source distance
For d=source distance and R=source size,

R
d

R2
d2
Slide 35
… and SSA frequency
Substituting for  then:
 8 m e3 5
F  
 Be



1/ 2
R2
d2
and
R  3  1017 F1/ 2 B1/ 4 d 5 / 4
Slide 36
SSA in Compact X-ray sources
X-ray frequency, =1018 Hz
Assume F ~ 10 -29 J m-2 s -1 Hz
d = 10 kpc and B = 10 8 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 -
Substituting for Omega into the previous
SSA equation in terms of F_nu, we
derive the expression above. Then rearranging and inserting values for
numerical constants we find the
expression for R shown.
This expression for R shows that R is
proportional to nu^(-5/4), ie as the source
size increases, the frequency at which the
self-absorption occurs decreases (moves
to longer wavelengths).
Thus if a source is large and emits
strongly in the radio, SSA is an important
process (eg. in extended, lobe-dominated
radio galaxies).
But how important is it in a compact Xray source?
Slide 37
Radiation processes (summary)
• Thermal - Bremsstrahlung
electron energies ~ photon energies
to produce X-rays,  = v/c ~ 0.1
• Non-thermal - Synchrotron and Inverse
Compton
Slide 38
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
B2
 U PH
• from 2  0
, => B=6x10 -10 Tesla
2
16
To produce X-rays,  S ~ 5  10
We are going to consider the background
emission from ‘Space’ in the Galaxy and
determine the relative contribution of
various processes, ie the diffuse
continuum emission.
There are two dominant processes 1. thermal bremsstrahlung emission,
where electrons are decelerated in the
electrostatic fields of ions and atomic
nuclei. The electron energies are
approximately equal to the photon
energies, thus taking a photon energy of
2.5 keV, beta=v/c (where v is the velocity
of the electron and c is the speed of
light), =0.1.
2. Non-thermal emission, ie synchrotron
and inverse Compton emission. In both
cases, the processes depend on other
properties of the source.
We are going to determine the electron
energies required in the non-thermal
processes to produce X-rays. Then we
will compare the relative contributions of
the non-thermal processes to see which
one dominates.
First we are going to calculate the
electron energies required to produce Xrays. To do this, we first assume an
equipartition of energy between starlight,
cosmic rays and magnetic fields - ie all
three have the same energy density in the
Galaxy.
Energy density in starlight in the galaxy,
U_ph = 10^6 eV /m^3
Thus B = 6 x 10^-10 Tesla.
To produce X-rays, gamma_s^2 = (3 x
10^7) / B
So substituting for B, in X-rays ie for
nu_max = 10^18 Hz,
gamma_s^2 ~ 5 x 10^16
In comparison, we take the case of
1. a supernova remnant has B~1e-7
Tesla, so gamma_s^2 ~ 3e14
2. A neutron star surface has B~1e8
Tesla, so gamma_s^2 ~ 0.3
Slide 39
Inverse Compton Scattering
Consider starlight: <h> ~ 2eV (~6000A)
or 3K background photons, <h> ~3x10 -4 eV
8keV
then  IC 
 h 
= 4103 for stars
7
3

10
=
for the 3K background, to
produce X-rays. We need cosmic rays!!!
2
Slide 40
Non-thermal process (cont.)
Energy distribution of cosmic ray particles
within a unit volume has the form:
N (E)  E

3
2
(over at least part of the energy range)
We use this to determine the relative
importance of synchrotron and IC processes
Slide 41
Power radiated in the two processes is about
equal in the case of equipartition of energy
ie when
B2
U
20
ph
ie an electron with a given  loses energy
equally rapidly by the two processes
However, it does not mean that X-rays are
produced at the same rate in the two cases.
Inverse Compton scattering requires a
target radiation, ie a photon input is
required to absorb energy from
relativistic electrons and scatter up to Xray energies. We calculate the gamma
factors required for the electrons to
produce X-rays from stars and the 3K
microwave background.
Because, in IC scattering,
gamma^2 ~ (final photon energy / initial
photon energy)
we calculate the gamma factors for stars
and the 3K microwave background as
shown.
Thus electrons with cosmic ray energies
are required to upscatter photons with
these energies.
Slide 42
Ratio of IC to Synchrotron Xrays
For example:
Galactic X-rays require  IC
but
2
 S2  51016
 4103 (stars)
 3107 (3K)
for synchrotron
Slide 43
Ratio IC to Synchrotron (cont.)
Ratio = (no of electrons with IC )
(no of electrons with S )

But:
 IC2 N IC
 S2 N S
N IC  E IC
 
NS
 ES




3
2
 IC2
 S2

  IC
 S




3
2
The number density ratios (ie numbers
per unit volume) are multiplied by the
ratio of gamma^2 because the power
produced is proportional to gamma^2 in
both cases, and we want to find the ratio
of power produced by the IC and
synchrotron processes.
Since we are assuming that the energy
distribution of cosmic ray particles within
a unit volume has the form N(E) which is
proportional to E^-(3/2), the ratio of IC to
synchrotron produced X-rays reduces to
the equation shown.
[Recall that cosmic rays are required to
produce X-rays by the IC and
synchrotron processes].
Slide 44
Ratio IC to Synchrotron (cont.)
Thus:

R   IC
 S



2
3
2

  IC
 S
1
2


So which is more important for producing
X-rays via IC; starlight or 3K background?
So the ratio reduces to the square root of
the ratios of the gamma factors.
The next task, before we can work out
whether Inverse Compton or Synchrotron
is more important to the production of Xrays in the Galaxy, is to establish which
is more important for the production of
Inverse Compton emission, starlight or
the 3K microwave background.
Slide 45
X-rays from IC scattering
3
(no. X-rays produced from starlight perm )
(no. X-rays produced from 3K perm3 )
2

  N  OPT 
U

 OPT  OPT  
U 3 K   3 K   N  3 K 

U OPT
U 3K
  OPT

  3K



2
3
2

U OPT
U 3K
  OPT

  3K
1
2


Slide 46
IC - starlight versus 3K
We know that U OPT  106 eVm3
U 3 K  3 105 eVm3
and
 OPT
4 103

 10  2
 3K
3 107
1
Thus R '
ie 3K photons more important!
3
Slide 47
IC or synchrotron
for X-rays?
1
  IC  2

 S 
Remember R  
assuming
3K
for IC :
1
 3 10 7  2
  5 10 3
R
 5 1016 


thus synchrotron dominates over IC in Galaxy
So we want to find out which is more
important to the production of X-rays via
the Inverse Compton process, starlight or
the 3K microwave background.
This is done by finding the ratio of the
number of X-rays produced by the two
sources of photons, and this is given by
the energy density (the amount of energy
per cubic metre) multiplied by the energy
distribution of cosmic ray particles per
unit volume, multiplied by the square of
the gamma factor (again, because this is
proportional to the power emitted in the
IC process).
This reduces to the ratio of the energy
densities multiplied by the square root of
the ratio of the gamma factors.
Substituting the numerical values given
and those we calculated on slide 42, we
find that this ratio is about 1/100, ie the
3K photons are more important in
producing X-rays via the Inverse
Compton process than starlight.
This is especially true when we consider
that the 3K microwave background is
also external to our Galaxy.
Slide 48
Synchrotron emission
Synchrotron emission requires very high
energy particles however - and electron
energy distribution may well have tailed
off if there is no continuous re-supply.
Also 3K radiation extends outside our
Galaxy. Extragalactic 3K radiation
depends on whether there are enough
electrons to produce IC.