PHOTONS:High Energy
Photon Flux Production and
Propagation
γ
γ
π0
γ
γ
γ
γ
γ
γ
γ
Andrew
Taylor
LECTURE PLAN:
1) COSMIC RAYS- proton interactions with photons,
composition, nuclei interactions with photons, different
photon targets
2) NEUTRINOS- presence of GZK-cutoff, photo-pion
production mechanism, interaction rate, cosmic ray
spectra, source distribution
3) PHOTONS- photon flux production, photon flux
attenuation, competition of rates, e/ cascades
4) MULTIMESSENGER
APPROACH
Andrew
Taylor
What Percentage of Cosmic
Rays are Expected to be
18
Photons? (>10 eV)
Dan Hooper
Subir Sarkar
Paolo Coppi
Andrew
Taylor
Aims
1) High Energy photon production through cosmic ray
interactions
2) Difficulties for a high energy photon in the Universe
3) A comparison of proton and photon propagation
through extragalactic space
4) The photon/proton ratio expected at Earth
5) What energy does the electron get?
EM Showers
6) What happens to the electron energy?
Andrew
Taylor
1) Photon Production
Andrew
Taylor
Why any Cosmic Rays should be
Photons?
Charged Pion production interaction with CMB 
For Eproton>1019.6 eV
(photon energy in proton frame)
γ
p
π+
n
Andrew
Taylor
Why any Cosmic Rays should be
Photons?
Neutral Pion production interaction with CMB 
For Eproton>1019.6 eV
(photon energy in proton frame)
γ
γ
γ
p
π0
p
From isospin considerations for the Deltaresonance decay, neutral pions are twice
as likely to be produced as charged pions
Andrew
Taylor
Why any Cosmic Rays should be
Photons?
Neutral Pion production interaction with CMB 
For Eproton>1019.6 eV
(photon energy in proton frame)
γ
γ
γ
p
π0
p
However, close to threshold, where direct
pion production dominates, charged pion
production is more likely
Andrew
Taylor
2) The Struggles of
the Photon
Andrew
Taylor
Why aren't most Cosmic Rays
19.6
Photons above 10 eV?
γ
γ
e+
e(and what happens to
the electrons energy?)
(photon energy in lab frame- for a GeV photon)
Andrew
Taylor
Uncertainties in the Radio Background
(the spanner in the works)
dust
stellar
Radio
Background
Andrew
Taylor
Uncertainties in the Radio Background
(the spanner in the works)
n ≈400 cm
n ≈1 cm
−3
−3
Andrew
Taylor
3) Photon and Proton
Attenuation Rates
Andrew
Taylor
Photon (Cosmic Ray)- Photon (CMB)
Interaction Lengths
Andrew
Taylor
Proton (Cosmic Ray)- Photon (CMB)
Loss Lengths
R=
m2p c 4
2E
2
∞
∫0 d 
n

2
2 E / m p c
∫0
2
d '  '  p  ' K p
(where R is the
energy loss rate)
Andrew
Taylor
A Comparison of the Photon
and Proton Loss Lengths
Andrew
Taylor
4) The Photon/Proton Ratio
at Earth
Andrew
Taylor
The Photon/Proton Loss Length
Ratio
Andrew
Taylor
Assuming.....
1) A large radio background
(UB = 10-8 eV cm-3)
2) A single pair production interaction occurs only
(synchrotron losses dominate electron cooling)
3) A uniform distribution of the cosmic ray sources locally
(in the < 300 Mpc region)
Andrew
Taylor
The Photon Flux
dip feature in photon fluxes
from more distant shells
Andrew
Taylor
The Photon Fraction
Andrew
Taylor
5) What energy does the electron
get?
Andrew
Taylor
Pair Production Physics
γ
γ
e+
e-
Pair Production:
γ
γ
e-
e+
In center-of-mass frame,
electron/positron pair are produced with
equal energy
Andrew
Taylor
Pair Production Physics (2)
But, boosting back from the center-of-mass frame
to lab frame, one of the electron's tends to take
nearly all the energy.
γ
γ
e-
Let,
s=
E  E bg

(the squared center-of-mass
energy of the collision)
2 2
m e c 
E e = 1 E cm
e =  s
E e = 1− E cm
e =
e+
E cm
e =
s
2
s
2
Andrew
Taylor
Pair Production Physics (2)
But, boosting back from the center-of-mass frame
to lab frame, one of the electron's tends to take
nearly all the energy.
γ
γ
ebg
Let,
s=
E E
(the squared center-of-mass
energy of the collision)
m e c 2 2
cm
e
E e = 1 E =  s
E e = 1− E cm
e =
e+
E
≈
s
s
2
Andrew
Taylor
Pair Production Physics (3)
So,
E e 4s−1
=
f  s
E
4s
At large center-of-mass energies (“s”), one of the
electrons produced via pair production carries most of
the energy
Andrew
Taylor
6) What happens to the electron
energy?
Andrew
Taylor
What Happens to the Electron
Energy?
e+
e-
Photon Background
γ
γ
γ
γ
Magnetic Field
γ
γ
γ
−3
U CMB
=0.25
eV
cm

γ
γ
γ
γ
γ
γ
γ
U B=10−8 eV cm−3 Andrew
Taylor
What Happens to the Electron
Energy?
e+
e-
Photon Background
γ
γ
γ
γ
Magnetic Field
γ
γ
γ
−3
U CMB
=0.25
eV
cm

γ
γ
γ
γ
γ
γ
γ
U B=10−8 eV cm−3 Andrew
−10
Taylor
B=3 x 10 G
What Happens to the Electron
Energy?
2 options for electron
interactions:
Photon Background
Inverse Compton
Scattering:
eγ
eγ
Magnetic Field
Synchrotron:
e-
e-
γ
γ
Andrew
Taylor
Competition of Processes
Both interactions are described by similar physics:
If Thomson scattering
Since
dE 4
= T 2 U
dt 3
−3
U CMB
=0.25
eV
cm

energy density of
scattering field
U B=10
−8
−3
eV cm
Naively IC should win! ....but is the scattering in the
Thomson regime?
Andrew
Taylor
Thomson Regime
For Thomson scattering, assumption is that the photon in the
electron's frame has it's momentum reversed- ie. it's the
center-of-mass frame!
Applies when,
bg
 E  m e c
Thompson scattering
2
(underlying assumption)
bg
bg
 E
 E
e-
e-
4
E =  2 E bg

3
(in electron rest-frame)
Andrew
Taylor
Synchrotron Photons
Treating magnetic field as a virtual photon field,
 
B
2
13
E =
m e c , where Bcrit ≈4x10 G
B crit
bg

−10
If,
B=10
bg
−18
E  =10
If,
eV
E e =10
Since,
G
19
eV , =2x10
4 2 bg
E =  E 
3
The synchrotron
photons get
energy
13
bg
−6
 E  =2x10
eV (Thomson)
Andrew
4
26
−18
6
E = x 4 x 10 x 10 =5x10
eV
Taylor
3
Inverse Compton Photons
−3
E bg
=10
eV

If,
Then,
E e =10
(CMB background)
19
bg
eV , =2x10
10
 E  =2x10 eV
13
Not Thomson
Hence, our naïve conclusion that IC out-competes
synchrotron was wrong
Andrew
Taylor
When Thomson Scattering Doesn't
Apply- the Klein Kishina regime
When
bg
 E  ≥m e c
2
bg
Let,
b=
Can re-write,
as
4 Ee E 
m e c 2 2
(center of mass frame is very
different to electron rest frame)
(which physically represents the squared
center-of-mass energy in the collision)
4 2 bg
E =  E 
3
1
E = b E e
3
Andrew
Taylor
When Thomson Scattering Doesn't
Apply- the Klein Nishina Regime
So the energy exchange is,
E b
=
Ee 3
Thomson:
b1
E
But
shouldn't become larger than 1....what went wrong?
Ee
Complete description is actually,
E
b
=
f b
E e 1b
Andrew
Taylor
Klein-Nishina Description
Andrew
Taylor
e/ Cascades
The repetition of 2 processes:
1) Pair Production
γ
γ
e+
e-
2) Inverse Compton
-
e
e-
γ
γ
Andrew
Taylor
Leading Particle Description
19
E =10 eV
e+
ee+
5 Mpc
e+
(ignore secondary
particles)
e-
5 Mpc
5 Mpc
5 Mpc
−3
E =10 19 eV, E bg
=10
eV

Ee
 s=4x10 ,
≈0.99
E
4
γ
γ
19
bg
−3
E e =10 eV, E  =10
e+
γ
eV
E
b=1x10 ,
≈0.99
Ee
5
So the high energy particle simply changes from a
Andrew
neutral to charged state and back spending
Taylor
roughly equal times in each state.
The Spanner in the Works
However, with the radio background with bg
−8
E  =10
If,
Then,
−8
−3
U CRB
=10
eV
cm

eV
E e =10
19
eV , =2x10
13
5
 E bg
=2x10
eV

~Thomson
e-+
e-+
e+
e
-
e
e+
e-+
new particles cannot be
ignored- particle
multiplication headache!
e+
ee+
Andrew
Taylor
The Photon Flux- with cascading
no cascading
with cascading
Andrew
Taylor
Conclusion
• High energy photon production is an inevitable
consequence of the GZK cut-off's existence
• The present uncertainty in the radio background and
extragalactic magnetic field strengths lead to various
possibilities for the propagation of the electromagnetic
energy through the system
• If the radio background and extragalactic magnetic field
are low, a simple leading particle description with the
particle alternating between neutral and charged states
may be used
• The radio background provides low energy photons,
prematurely putting the cascade into the ThomsonAndrew
regime
Taylor
The Photon Fraction- data
Auger data
Andrew
Taylor