INTRODUCTION
So far, the only information we have been able to get about the universe
beyond the solar system is from the electromagnetic radiation that reaches us
(and a few cosmic rays). So doing Astrophysics is like a big puzzle: look at the
radiation and use it to ¤gure outw well, everything! Where is the object, what
is it, when and how did it form, how has it evolved since then, how does it ¤t
in the "astrophysical zoo"?. So our objective in this course is to learn some of
the tricks for solving the puzzle.
We start by looking at the way that radiation is modi¤ed as it travels from
its source to the observer (us).
Radiative transfer
Vocabulary
Intensity
You can think of intensity as measuring the energy traveling along a light
ray. Of course a single ray carries no energy, so we look at a bundle of rays
within a solid angle d- of the central ray. The energy crossing area dA normal
to the ray in time dt and frequency range dº is
dE = Iº dAdtdº d-
(1)
The intensity Iº =energy per unit time per unit frequency per unit area per
unit solid angle units (astronomers like cgs units)
sec
erg
¢ Hz ¢ ster
¢cm2
Mean intensity Jn (ergs/cm2 ¢s¢Hz) is obtained by averaging over angles:
Z
1
Jº =
Iº d(2)
4¼
An isotropic radiation ¤eld has J º = Iº :
Flux Fº measures the energy crossing a surface, in erg/cm2 ¢s¢Hz. To get
the energy crossing the surface, we include only the component of the direction
vector along the ray that is normal to the surface. That introduces a factor of
cosµ:
1
Fº =
Z
Iº cos µ d-
(3)
Flux is usually the easiest quantity to measure for a typical astrophysical
source, such as a star. We can only measure intensity if the source is resolved,
like the sun or the moon. The ¢ux at Earth due to an unresolved source with
luminosity L erg/s a distance D away is
F =
L
4¼D 2
Momentum ¢ux and radiation pressure. The momentum of a photon is its
energy/c, so we can get momentum ¢ux from energy ¢ux. Pressure is normal
force/unit area=rate of change of momentum normal to the surface/area. So
we compute the total momentum crosing area dA in time dt due to a bundle of
rays at µ; Á as
d~p (µ; Á) =
dFº
Iº cos µ
^{ dAdt =
d-^{ dAdt
c
c
where ^{ is a unit vector along the ray. The normal component is
dpn or ma l (µ; Á) = dp cos µ =
dFº
Iº cos2 µ
dAdt cos µ =
d-dAdt
c
c
and integrating over the whole radiation ¤eld, we have
Z
Iº cos2 µ
dp n orm al =
d- dAdt
c
2
and so the pressure on the surface patch is
Z
dpn or ma l
1
Pº =
=
Iº cos 2 µ ddAdt
c
(4)
Finally the total radiation energy in a volume dV = dAd` contributed by radiation moving in direction µ; Á is
dE = du º (µ; Á) dAd`dº
All the energy passes out of the volume in a time dt = d`=c; so from (1),
d`
dºdc
and comparing the two expressions we ¤nd
dE = Iº dA
Iº
dc
To get the total energy density we sum the contributions from rays in all directions, to get
Z
Z
Iº
4¼
uº =
duº (µ; Á) =
d- =
Jº
(5)
c
c
Letzs use these de¤nitions in a few simple examples.
Radiation pressure in a perfectly re¢ecting enclosure with an isotropic radiation ¤eld.
At each re¢ection the momentum tranferred is twice the incident momentum,
so
Z
2
Pº =
Iº cos2 µdc
But an isotropic radiation ¤eld means that Iº is independent of direction, so
Iº = Jº and it comes out of the integral. Radiation only travels over half the
total solid angle since no radiation is coming in from outside the enclosure:
¯+ 1
Z +1
Z 2¼
2
4¼ Jº ¹3 ¯¯
4¼
1
2
P º = Jº
¹ d¹
dÁ =
¯ = 3c J º = 3 uº
c
c
3
0
0
0
du º (µ; Á) =
Flux from a uniformly bright sphere:
3
Rays reaching P leave the sphere at angles from µ = 0 to µ = µ ma x ; as shown.
The intensity is the same along all rays. The ¢ux at P is
F =
Z
I cos µd- = 2¼I
Z
1
cos µm ax
¡
2
¢
2
¹d¹ = ¼I 1 ¡ cos µ m ax = ¼I sin µm ax = ¼I
µ
where in the last step we used the fact that sin µ ' tan µ for small angles.
Notice that we cannot use the result for r = R (as is done in the book)
because that violates the small angle approximation we made. But at the surface
the integral becomes
Z
+1
F = 2¼I
¹d¹ = ¼I
0
So the textbook result is correct even though their method of getting it is not.
Radiative Transfer.
So letzs start with the intensity and look at the ray crossing a cylinder of
area dA and thickness ds: As the ray crosses the cylinder, the intensity will
increase due to emission in the cylinder, and decrease due to absorption. (We
will omit scattering for the moment, and come back to that later.) The emission
coe¢cient is jº (erg/cm3 ¢s¢Hz)
and
¡
¢ the absorption may be written as the density
of particles in the slab n cm¡3 times the absorption cross section per particle
¡
¢
¾ º (cm/Hz) ; or just as ® º = n¾ º cm¡ 1 Hz¡1 :
There are ndAds particles in the slab with a total area Ap art = n¾dAds:
The photons hitting that area will be absorbed, so
Energy out = energy in + emission - absorption
Dividing a factor of d-dtdº from each term, we have
dAIº (s + ds) = dAIº (s) + j º dAds ¡ ®º Iv dAds
or
dIº
= jº ¡ ® º Iº
(6)
ds
This is the equation of radiative transfer. To simplify, we de¤ne the optical
depth ¿ º by
d¿ º = ®º ds
so the radiative transfer equation may be written as
dIº
jº
=
¡ I º = Sº ¡ I º
d¿
®º
where
Sº ´
is the source function.
4
jº
®º
(7)
R
r
¶2
We may solve the equation formally as follows.
e ¿ v and rearrange.
Multiply the equation by
d
dIº
(e¿ v Iº ) = e¿ v Iº + e¿ v
= e¿ º Sº
d¿ º
d¿ º
Thus
Iº (¿ º ) = e¡ ¿ º Iº (0) + e¡ ¿ º
5
Z
¿º
0
0
e ¿ º Sº (¿ 0º ) d¿ 0º
(8)