1
Probability
We start with a reminder of a few basic concepts in probability.
1.1
discrete random variables
Let x be a discrete random variable with some probability function p(x).
1. The Expectation value of x is the average value of x that we will measure
by doing many repetitions of the experiment described by the probability
function p(x):
X
µ = E(x) =
xi p(xi )
(1)
i
where the sum goes over all (perhaps infinite) possible values for x. This
is not necessarily the peak of p(x), nor does it have to be the median value
of the distribution (the value of x for which half the possible values are
above and half are below).
2. The Variance of x measures the spread of the distribution function:
X
X
X
V ar(x) = E[(x − µ)2 ] =
(xi − µ)2 p(xi ) =
x2i p(xi ) − 2µ
xi p(xi )
i
2
+µ
X
2
i
2
2
i
2
2
p(xi ) = E(x ) − 2E(x) + E(x) = E(x ) − E(x) .
i
(2)
The variance is of course non-negative.pA related term is the standard
deviation which is defined as: σ(x) = V ar(x).
3. The Cumulative distribution function describes the probability that
x will be found to have a value less than or equal to X:
X
F (X) = P (x ≤ X) =
p(xi ).
(3)
xi ≤X
This is a monotonically rising function, ranging between 0 and 1.
1.2
Continuous random variables
For continuous random variables, the probability function is replaced with a
probability density function (PDF). A PDF, is a function that describes the
relative likelihood for a random continuous variable to take on a given value.
Let f (x) be such a PDF. The chance of measuring x to lie in the range
a ≤ x ≤ b is given by integration over the PDF in the said range:
Z
P (a ≤ x ≤ b) =
f (x)dx.
a
1
b
(4)
Notice that the chance of measuring any specific value of x is zero (even if this
x is at the peak of the PDF), since the integral goes to 0 as a → b. In the case
of a continuous random variable, the sums in the expressions for the expectation value, variance, cumulative distribution functions are simply replaced by
integrals:
Z
E(x) = xf (x)dx
Z
V ar(x) = (x − µ)2 f (x)dx
(5)
Z X
F (X) =
f (x)dx
2
Black body radiation
A black body is a physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. A black body in thermal
equilibrium emits electromagnetic radiation called black-body radiation. The
radiation is emitted according to Planck’s law:
Rν =
1
2hν 3
,
hν
2
c e KB T − 1
(6)
where Rν is the power emitted per unit area of the body, per unit solid angle that
the radiation is measured over, per unit frequency. The spectrum is determined
by the temperature alone, not by the body’s shape or composition.
One can also write the spectrum per unit wavelength instead of per unit frequency. To do so, we must ensure that the emitted energy in a given wavelength
band is the same as that in the corresponding frequency band:
Rν dν = −Rλ dλ → Rλ = −
dν
ν2
Rν = − Rν
dλ
c
(7)
Dividing Rν by the energy carried by each photon (hν) we can obtain the
number of photons emitted per unit area of the body, per unit solid angle that
the radiation is measured over, per unit frequency:
Nν =
2ν 2
1
.
c2 e Khν
BT − 1
(8)
Nν can be thought of as a PDF describing the chance of the Black Body emitting
a photon with frequency ν (per unit area of the body, per unit solid angle that
the radiation is measured over).
Question: what is the average energy of photons emitted from a
black body with temperature T?
To answer this question we must find the expectation value of Nν according
to Eq. 5. Since the integral of Nν presented above is not normalized to 1, we
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need to divide by this integral to get the expectation value:
R∞
R∞
0
hνN
dν
ν
< E >= 0R ∞
=R∞
Nν dν
0
0
hν 3
hν
e KB T −1
ν2
hν
e KB T −1
dν
dν
=
4
KB
T4
h3
3 T3
KB
h3
y3
dy
0 ey−1
R ∞ y2
dy
0 ey−1
R∞
y3
dy
0 ey−1
R ∞ y2
dy
0 ey−1
R∞
= KB T
≈ 2.7KB T
(9)
where in the third equality we have switched variables in the integral to y =
hν/(KB T ). We see that the average energy is proportional to the temperature
of the black body. Notice that this energy is different than the peak of Rλ given
by Wien’s law, which is: EW ien = 4.97KB T .
2.1
An interesting mistake - why should we not expect
astrophysical X-ray sources?
Notice that Rν (T ) is always rising as a function of temperature, since:
hν
hν
KB T
2hν 3
dRν (T )
T2 e
= 2 KBhν
>0
dT
c (e KB T − 1)2
(10)
The implication is that black bodies with higher temperatures emit more then
those with lower temperatures in any frequency. Since in early astrophysical
observations, no such excess radiation was observed in optical telescopes, the
prevalent thought was that there could not be astrophysical objects which spectrum peaks at larger frequencies (hotter). For this reason it was argued that
there is no justification in building X-ray telescopes. This argument turned out
to be wrong since astrophysical objects are often not in thermal equilibrium and
do not emit a Planck spectrum. Indeed, today we know that the sky is full of
objects that peak in the X-rays and even γ-rays.
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Compton Scattering
Compton scattering is the inelastic scattering of a photon by a free charged
particle, usually an electron. It results in a decrease in energy (and hence
frequency) of the photon, called the Compton effect. The frequency of the
scattered photon is given by:
ν0 =
ν
ν(1 − cos θ)λc /c + 1
(11)
h
where λc = mc
is the Compton length scale. We see that the frequency of
the outgoing photon changes depending on the scattering angle. Since for the
electron λc = 2.4 · 10−2 Å you need x-ray photons to see the effect.
Example - The maximum energy loss of a photon due to Compton
A photon of energy E = 500MeV strikes a proton at rest. What is the
maximal energy gain of the proton?
3
The maximum energy transfer to the photon occurs for a head-on collision.
In this case:
hν
hν 0 =
.
(12)
1 + 2 mhν
2
pc
The energy gain of the proton is the energy loss of the photon:
0
∆E = −(hν − hν) =
2hν mhν
2
pc
1 + 2 mhν
2
pc
=
1
2E 2
mp c2
+ 2 mE
2
pc
.
(13)
With E = 500MeV and mp c2 = 940MeV we obtain ∆E = 260MeV.
3.1
Inverse Compton
If the electron is moving relativistically it can increase rather than decrease the
energy of the photon. In many cases in astrophysics you see two similar spectra
but one is shifted to a higher frequency than the other. This is due to some
photons being up-scattered by relativistic electrons. Assume that we have an
electron moving at a velocity βc and a Lorentz factor γ which hits a photon. A
simple Lorentz transformation to the electron rest frame will give us the same
formula for the scattering as above except that the original frequency will now
be blueshifted by the factor γ(1 + β) since the transformation is
γ βγ
E/c
Ẽ/c
=
(14)
βγ γ
p
p̃
In the electron’s rest frame the frequency after scattering will be
ν0 =
νγ(1 + β)
ν(1 − cos θ)γ(1 + β)λc /c + 1
(15)
For simplicity let us assume θ = π so that the photon is involved in a head-on
collision with the electron and changes its direction by 180 degrees. Transforming back into the lab frame will result in an additional factor of γ(1 + β) so the
final frequency will be
νγ 2 (1 + β)2
ν0 =
(16)
2νγ(1 + β)λc /c + 1
In the relativistic limit (β → 1) and for hνγ mc2 the electron’s frequency is
boosted by a factor of 4γ 2 . We also get that there is a maximum boost which
occurs for hνγ mc2 that gives us a maximum frequency of γme c2 /h.
4
Pair production
The Compton effect is not the only way in which photons and massive particles
can interact. Another result achieved by combining the particle nature of light
with special relativity, is that a collision between two photons in a vacuum may
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lead to the production of a pair of particle and anti-particle (an anti-particle
has the same mass as the particle but opposite charge). The opposite effect,
annihilation of matter and anti-matter, resulting in two or more photons, can
also occur. In what follows, we aim to find the condition for production of an
electron-positron pair.
Denote by P1 , P2 the four-momenta of the incoming photons, and by P3 , P4
the four-momenta of the outgoing electron and positron. P1 , P2 can be written
using the directions of motion of the incident photons, î1 , î2 :
P1 = [E1 /c, (E1 /c)î1 ];
P2 = [E2 /c, (E2 /c)î2 ];
(17)
Conservation of four-momenta requires:
P1 + P2 = P3 + P4
(18)
To find the threshold for creation we assume the pair is created without kinetic
energy (and momentum), such that the only energy resides in the mass of the
newly created particles:
P3 = P4 = [me c, 0, 0, 0].
(19)
Squaring both sides of Eq. 18 and remembering that P1 · P1 = P2 · P2 = 0 (because photons are mass-less particles) and that P3 · P3 = P3 · P4 = P4 · P4 =
m2e c2 we obtain:
E1 E2
2m2e c4
E1 E2
2 2
−
cosθ)
=
4m
c
→
E
=
2
e
c2
c2
E1 (1 − cosθ)
(20)
where θ is the incident angle between the direction of the photons. Thus, if
electron–positron pairs are created, the threshold for the process occurs for
head-on collisions, θ = π, and hence:
2P1 · P2 = 4m2e c2 → 2(
E2 =
m2e c4
2.6 × 1011
=
eV
E1
E1
(21)
where E1 is in electron Volts. This process thus provides not only a means
for creating electron–positron pairs but also results in an important source of
opacity for high-energy γ-rays. For instance, the energy of optical light (such
as produced by the sun, for example) is of order ∼eV. Plugging this number
to E1 shows that photons with energies above ∼300 GeV can be absorbed by
starlight.
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