1. High Energy Astrophysics emp@mssl.ucl.ac.uk http://www.mssl.ucl.ac.uk/
Radiation Processes
High Energy Astrophysics

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.

3. Photoionization e- Atom absorbs photon Atom, ion or molecule
[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.
Cross-section (s) characterized by edges corresponding to ionization edges.

4. Example of photoelectric absorption
eg. soft X-rays from a star absorbed by ISM
interstellar cloud
star
observer I I
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. n n

5. How much passes through?
Take a path of length dl (metres)
is the number density ( ) of element Z.
Cross-section offered by element Z at energy E is given by:
dl (m) dV

6. Thus the fraction of flux lost in volume dV is:
The fraction of volume dV which is blocked by the presence of element Z is :
Thus the fraction of flux lost in volume dV is:
or :
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.

7. Integrating over length from source...
Including all elements in the line of sight:

8. This is ‘t’, the optical depth, which has no dimensions
This becomes:
This is ‘t’, the optical depth, which has no dimensions
This is the effective cross-section, weighted over the abundance of
elements with respect to hydrogen

9. Column density The column density is given by :
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).

10. Clumping of the ISM Take an example at low energies, eg at ...
At a distance,
d=100 pc
Average ISM density
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.

11. Smooth versus clumpy star observer Hot medium Cold dense clouds smooth
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.
Hot medium
Cold dense clouds

12. Numerical example Through the smooth medium -
Through the clumpy medium -
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 cross-section, 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 over-compensating 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.

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.

14. Thomson Scattering low-E photon scattered by electron -
Thomson cross-section is given by -
electron
, where
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).

15. Thomson scattering cont.
If N = number of particles per
then fraction of area blocked by a square metre of path = 1m 1m
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.
If R is the extent of the absorbing region along the line of sight,
( = optical depth)
and

16. Compton scattering
In Compton scattering, the photon wavelength increases, ie its energy decreases.
electron q
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).
frequency change

17. Compton scattering cont.
On average,
This effect of cooling the radiation (ie because it loses energy) and transferring the energy to the electron is sometimes called the recoil effect.
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.
Compton scattering has the effect of broadening spectral lines and washing out edges.

18. Electron-positron pair production
g-ray y q x e+
e-/e+ photon
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.
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.

19. Minimum g-ray energy required
Must first demonstrate that is a relativistic invariant.
Rest energy of particle,

20. Thus, from and , And this is a relativistic invariant
Thus we demonstrate that E^2 - (pc)^2 is an invariant, because a particle’s rest-mass, 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.
And this is a relativistic invariant

21. Total initial momentum,
thus

22. But since ,
and -
So now we have a simple expression for the initial E^2-(pc)^2 of the gamma ray and the low energy photon.

23. Calculating the minimum energy
Assuming e+ and e- have no momentum…
and since ,
Which gives us this expression for the energy of the g-ray photon
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.

24. 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
e-/e+ photon
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.
The minimum energy is also inversely proportional to the energy of the photon with which it interacts.
And the minimum g-ray energy is given by:

25. Minimum energy for mm-wave photon
g-ray photon interacts with mm-wave
First converting to eV :
l=1.2mm corresponds to hn=10 eV -3
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!

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 cross-section, 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.
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 gamma-ray (about 1e-29 m^2). It is the number density of photons in space compared to the number density of particles which is important.

27. Photon versus particle density
eg., for 3K m-wave background photons - 9
Corresponding to about 10 photons / m 3
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. 6
No of nuclei in space is about 10 / m 3

28. Synchrotron Self-Absorption
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.
Relativistic electrons moving
in a magnetic field

29. Flux emitted as a function of frequency:
Synchrotron Spectrum
Flux emitted as a function of frequency:
We are now going to take a look at the synchrotron spectrum, and see how it is modified by the synchrotron self-absorption 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. E n
logF
logn

30. Blackbody turnover Assume power-law cut off, n , is given by:
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.
max
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.

31. On the Rayleigh-Jeans side...
impossible
logF
logn
synchrotron n
blackbody
R-J
Rayleigh-Jeans approximation to blackbody...
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.

32. So total energy flux at Earth is given by:
Total flux at Earth...
So total energy flux at Earth is given by:
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.

33. SSA spectrum SSA logF logn n Optically-thick regime Optically-thin
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.
n lies at the point where the observed synchrotron flux equals the blackbody limit. a

34. For d=source distance and R=source size,W R d

35. … and SSA frequency Substituting for W then: and
Substituting for Omega into the previous SSA equation in terms of F_nu, we derive the expression above. Then re-arranging 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 X-ray source?

36. SSA in Compact X-ray sources18
X-ray frequency, n= Hz
Assume F ~ J m s Hz
d = 10 kpc and B = 10 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 n to be observable in the X-ray band).
- but a neutron star diameter is 10 to 20km -
-29 -2 -1 n 8 a

37. Radiation processes (summary)
Thermal - Bremsstrahlung electron energies ~ photon energies to produce X-rays, b = v/c ~ 0.1
Non-thermal - Synchrotron and Inverse Compton
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.

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
from , => B=6x Tesla To produce X-rays,
First we are going to calculate the electron energies required to produce X-rays. 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
-10

39. Inverse Compton Scattering
Consider starlight: <hn> ~ 2eV (l~6000A)
or 3K background photons, <hn> ~3x eV
then
= for stars
= for the 3K background, to produce X-rays. We need cosmic rays!!! -4
Inverse Compton scattering requires a target radiation, ie a photon input is required to absorb energy from relativistic electrons and scatter up to X-ray 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.

40. Non-thermal process (cont.)
Energy distribution of cosmic ray particles within a unit volume has the form:
(over at least part of the energy range)
We use this to determine the relative importance of synchrotron and IC processes

41. Power radiated in the two processes is about equal in the case of equipartition of energy ie when ie an electron with a given g 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.

42. Ratio of IC to Synchrotron Xrays
For example:
Galactic X-rays require (stars)
(3K)
but for synchrotron

43. Ratio IC to Synchrotron (cont.)
Ratio = (no of electrons with )
(no of electrons with )
But:
The number density ratios (ie numbers per unit volume) are multiplied by the ratio of gamma^2 because energy is proportional to gamma^2 in both cases, and we want to find the ratio of energies produced by the IC and synchrotron processes.

44. Ratio IC to Synchrotron (cont.)
Thus:
So which is more important for producing
X-rays via IC; starlight or 3K background?

45. X-rays from IC scattering
(no. X-rays produced from starlight per )
(no. X-rays produced from 3K per )

46. IC - starlight versus 3K We know that and
Thus ie 3K photons more important!

47. IC or synchrotron for X-rays?
Remember
assuming for :
thus synchrotron dominates over IC in Galaxy

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 radiation depends on whether
there are enough electrons to produce IC.