Astro 300B: Jan. 26, 2011
Thermal radiation and Thermal Equilibrium
Thermal Radiation, and
Thermodynamic Equilibrium
Thermal radiation is radiation emitted by matter in thermodynamic
equilibrium.
When radiation is in thermal equilibrium, Iν is a universal function of
frequency ν and temperature T – the Planck function, Bν.
I   B
Blackbody Radiation:
In a very optically thick media, recall the SOURCE FUNCTION
S 
So thermal radiation has
j

 I
S  B and
j    B
And the equation of radiative transfer becomes
dI 
dl
    I     B or
dI 
d
  I   B (T )
THERMODYNAMIC EQUILIBRIUM
When astronomers speak of thermodynamic equilibrium, they mean a lot
more than dT/dt = 0, i.e. temperature is a constant.
DETAILED BALANCE: rate of every reaction = rate of inverse reaction
on a microprocess level
If DETAILED BALANCE holds, then one can describe
(1) The radiation field by the Planck function
(2) The ionization of atoms by the SAHA equation
(3) The excitation of electroms in atoms by the Boltzman distribution
(4) The velocity distribution of particles by the Maxwell-Boltzman distribution
ALL WITH THE SAME TEMPERATURE, T
When (1)-(4) are described with a single temperature, T, then the system is
said to be in THERMODYNAMIC EQUILIBRIUM.
In thermodynamic equilibrium, the radiation and matter have the same
temperature,
i.e. there is a very high level of coupling between matter and radiation
 Very high optical depth
By contrast, a system can be in statistical equilibrium,
or in a steady state, but not be in thermodynamic equilibrium.
So it could be that measurable quantities are constant with time, but there
are 4 different temperatures:
T(ionization)
T(excitation)
T(radiation)
T(kinetic)
given by the Saha equation
given by the Boltzman equation
given by the Planck Function
given by the Maxwell-Boltzmann distribution
Where
T(ionization) ≠ T(excitation) ≠ T(radiation) ≠ T(kinetic)
LOCAL THERMODYNAMIC EQUILIBRIUM (LTE)
If locally, T(ion) = T(exc) = T(rad) = T(kinetic)
Then the system is in LOCAL THERMODYNAMIC EQUILIBRIUM,
or LTE
This can be a good approximation if the mean free path for
particle-photon interactions << scale upon which T changes
Example: H II Region (e.g. Orion Nebula, Eagle Nebula, etc)
Ionized region of interstellar gas around a very hot star
Radiation field is essentially a black-body at the temperature of the central
Star, T~50,000 – 100,000 K
However, the gas cools to Te ~ 10,000 K
(Te = kinetic temperature of electrons)
O star
H II
HI
Q.: Is this room in thermodynamic equilibrium?
FYI, we write down the following functions, without deriving them:
(1) The Boltzman Equation
Boltzman showed that the probability of finding an atom with an electron, e-,
in an excited state with energy χn above the ground state
decreases exponentially with χn
increases exponentially with temperature T
Nn
N1
and
 n 

exp  

g1
 kT 
gn
Where
Nn = # atoms in excited state n / volume
N1 = # atoms in ground state /volume
gn = 2n2 the statistical weight of level n
= number of different angular momentum quantum numbers
in energy level n
(2) The Planck Function
I   B 
2h
c
2
3
1
e
h  / kT
1
(3) The Maxwell-Boltzman distribution of speeds of electrons
 me
f ( v )  4  
 2  kT e




3/2
2
v e
 mev 2 


 2 kT 
e 

= fraction of electrons with velocity between v, v+dv
where me = mass of the electron
Te = temperature of the electrons
(4) The Saha Equation
ne
N m 1
Nm
Z m  1  2  m e kT 
2


3
Zm 
h

3/2

e
m
kT
Where ne = number density of free electrons
Nm = number density of atoms in the mth ionization state
Zm = partition function of the mth ionization state

Zm 

i 1

g ie
i
kT
Thermodynamics of Blackbody Radiation:
The Stefan-Boltzman Law
Consider a piston containing black-body radiation:
Inside the piston: T, v, p u
Move blue wall  extract or perform work
First Law of Thermodynamics:
dQ = dU + p dV
where dQ = change in heat
dU = total change in energy
p = pressure
dV = change in volume
Second Law of Thermodynamics:
dS = dQ/T
S = entropy
Recall,
U = uV
p = 1/3 u
u 
4
c
So…
u = energy density
energy/volume
p = radiation pressure in piston
 J d
dS 
J   B
and
dQ
T

dU
 p
T

d ( uV )
T
dV
(substitute dQ=dU+pdV)
T

1
3
u
dV
T
(substitute U=uV, p=1/3 u)
Vdu
dS 



dV 
T
T
V
dT
du
1
u
dV
3

4 u
T
dT
3T
V du
dT 
4 u
T dT
So...
u
T
dV
dV
3T
V du
 dS 

 
T dT
 dT  V
and
Differentiate these….
4u
 dS 

 
3T
 dV  T
2
d  V du  1 du



dTdV
dV  T dT  T dT
d S
(Eqn.1)
2
d  4u 
4 u
4 1 du




2
dVdT
dT  3T 
3T
3 T dT
d S
Combining (1) and (2) 
Multiply by T
1 du
 
4 u
T dT
3T
du
4 u
 
dT
du
dT
(Eqn. 2)
2
3T
4
u
T

4 1 du
3 T dT

4 du
3 dT
du
4
dT
u
T
log u  4 log T  log a
u (T )  aT
a=constant of integration
Energy density ~T4
4
u can be related to the Planck Function
u 
So…
u 
4
c
J
 u d 
For isotropic radiation,

4
c
 B (T ) d 

4
c
B (T )
I  J   B
Where B(T) = the integrated Planck function
 B d 


ac
4
T
4
For a uniform, isotropically emitting surface, we showed that the flux
F 
 F d 


ac
4
ac
T
T

 B d 
  B (T )
4
4
4
OR….
F  T
Where
 
4
ac
Stefan-Boltzmann Law
= 5.67x10-5 ergs cm-2 deg-4 sec-1
4
[flux] = ergs cm-2 sec-1
also
a
4
c
flux integrated over frequency,
per area per sec
= 7.56x10-15 ergs cm-3 deg-4
Blackbody Radiation; The Planck Spectrum
• The spectrum of thermal radiation, i.e. radiation in equilibrium with
material at temperature T, was known experimentally before Planck
• Rayleigh & Jeans derived their relation for the blackbody spectrum for
long wavelengths,
• Wien derived the spectrum at short wavelengths
• But, classical physics failed to explain the shape of the spectrum.
• Planck’s derivation involved the consideration of quantized
electromagnetic oscillators, which are in equilibrium with the radiation
field inside a cavity
 the derivation launched Quantum Mechanics
See Feynmann Lectures, Vol. III, Chapt.4; R&L pp. 20-21
Result:
B 
2h
c
3
3
1
e
h  / kT
1
ergs s-1 cm-2 Hz-1 ster-1
Or in terms of Bλ recall
I d   I  d 
d
  c so
B 
2 hc

5
d
2

c

2
1
e
hc /  kT
1
ergs s-1 cm-2 A-1 ster-1
The Cosmic Microwave Background
The most famous (and perfect) blackbody spectrum is the
“Cosmic Microwave Background.”
Until a few hundred thousand years after the Big Bang, the Universe was
extremely hot,
all hydrogen was ionized, and
because of Thomson scattering by free electrons,
the Universe was OPAQUE.
Then hydrogen recombined and the Universe became transparent.
The relict radiation, which was last in thermodynamic equilibrium
with matter at the “surface of last scattering” is the CMB.
Currently the CMB radiation has the spectrum of a blackbody with
T=2.73 K.
It is cooling as the Universe expands.
The first accurate measurement of the spectrum of the CMB
was obtained with the FIRAS instrument aboard the Cosmic
Background Explorer (COBE), from space:
See Mather + 1990 ApJLetters 354, L37
The smooth curve is the theoretical Planck Law. This plot was
made using the first year of data; in subsequent plots the error
bars are smaller than the width of the lines!
Properties of the Planck Law
Two limits simplify the Planck Law (and make it simpler to integrate):
Rayleigh-Jeans: hν << kT
Wien
hν >> kT
(Radio Astronomy)
Rayleigh-Jeans Law
h   kT
so
e
h  /kT
1 
h
kT
so
I  (T ) 
2h
c
3
2
1
e
h  / kT
1
becomes
I  (T ) 
RJ
2
c
2
2
kT
The Ultraviolet Catastrophe
If the Rayleigh-Jean’s form for the spectrum of a blackbody
held for all frequencies, then
 I d 
  as   
And the total energy
in the radiation field


Wien’s Law
h   kT
1
so
e
I  (T ) 
W
2h
c
2
h  / kT
1
1

e
h  / kT
3
e
 h  / kT
 Very steep decrease in brightness for
 
peak
Monotonicity with Temperature
If T1 > T2, then Bν(T1) > Bν(T2)
for all frequencies
Of 2 blackbody curves, the one with
higher temperature lies entirely
above the other.
dB  (T )
dT
d  2h


2
dT  c
2h 
2

2
c kT
4
2
3
e
e
e
1
h  / kT


 1 
h  / kT
h  / kT
1

2
>0 always
Wien Displacement Law
At what frequency does the Planck Law Bν(T) peak?
dB 
Bν(T) peaks at νmax, given by
d
d  2h

2
d   c
e
h  / kT
3
1
e
h  / kT
0
 max

0
 1 
d  2h

 1
2
d   c
3
  2h

  c2
 
3
 d
h  / kT

 1  0
 d  e

e
h  / kT
6 h  2  2 h  3  h
h  / kT
 1 2    2 
e

 c   c  kT

Divide by exp(hν/kT), cancel some terms

3 1 e
Let
x
 h  / kT

h
kT
h  max
kT
Need to solve
3 1  e
Solution is x=2.82.
x
 x
Need to solve graphically or iteratively.
2 . 82 
h  max
kT
 max
 5 . 88  10
10
Hz deg
1
T
Similarly, one can find the wavelength λmax at which Bλ(T) peaks
 dB  
0


 d      max
 max T  0 . 290 cm deg
 max  max  c
NOTE:
That is to say, Bν and Bλ don’t peak at the same wavelength, or
frequency.
For the Sun’s spectrum,
λmax for Iλ is at about 4500 Å whereas λmax for Iν is at about 8000 Å
Why?
recall
d 
c

2
d
So equal intervals in wavlength
correspond to very different intervals of
frequency across the spectrum
With increasing l, constant dl (the Il case)
corresponds to smaller and smaller dn
so these smaller dn intervals contain smaller energy, compared
to constant dn intervals (the In case)
Radiation constants in terms of physical constants
Recall the Stefan-Boltzman law for flux of a black body
F  T

Let x 
h
4

B d 

0
x
e 1
x
So 

0
3
dx 
B d  
0

then
kT




c
2
 2 h   kT 
B d    2  

 c  h 
4

so...
15
 
2h

0
2 k
5
4
2
3
15 c h



e
0
4

h  / kT

0
x
1
3
e 1
x
2 k
4
B d  
3
2
15 c h
dx
4
3
T
4
d
Also, since
u 
4
c


0
B d   aT
8 k
5
a
3
4
15 c h
3
4
As an example of the kind of things you can model with the
Planck radiation formulae, consider the following:
(see http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/nickel.html)
(1) How much radiant energy comes from a nickel at room
temperature per second?
Measured properties of the nickel are diameter =
2.14 cm, thickness 0.2 cm, mass 5.1 grams.
This gives a volume of 0.719 cm3 and a surface
area of 8.54 cm 2.
The radiation from the nickel's surface can be calculated from the
Stefan-Boltzman Law
F= σT4
The room temperature will be taken to be 22°C = 295 K.
Assuming an ideal radiator for this estimate, the radiated power is
P = σAT4
A=surface area of nickel
= (5.67 x 10-8 W/m2K4)x(8.54 x 10-4 m2)x(295 K)4
= 0.367 watts.
So the radiated power from a nickel at room temperature is about 0.37 watts
2. How many photons per second leave the nickel?
Since we know the energy, we can divide it by the average photon energy.
We don't know a true average, but the wavelength of the peak of the blackbody
radiation curve is a representative value which can be used as an estimate.
This may be obtained from the Wien displacement law.
lpeak = 0.0029 m K/295 K = 9.83 x 10-6 m = 9830 nm, in the infrared.
The energy per photon at this peak can be obtained from the Planck
relationship.
Ephoton = hν = hc/λ = 1240 eV nm/ 9830 nm = 0.126 eV
Then the number of photons per second is very roughly
N = (0.367 J)/(0.126 eV x 1.6 x 10-19 J/eV) = 1.82 x 1019 photons
Characteristic Temperatures for Blackbodies
1. BRIGHTNESS TEMPERATURE, Tb
Instead of stating Iν, one can state Tb, where
I   B (T B )
i.e. Tb is the temperature of the blackbody having the same
specific intensity as the source, at a particular frequency.
Notes:
1. TB is often used in radio astronomy, and so you can
assume that the Rayleigh-Jeans Law holds,
h   kT
or T B 
I 
so
c
2
c
2
2
kT B
2
2 k
2
I
2. The source need not be a blackbody, despite being described
as a source with brightness temperature TB.
3. Units of TB are easier to remember than units of Iν
TB and the equation of Radiative Transfer:
dI 
d
  I   B (T )
I 
Assume Rayleigh-Jeans,
2
c
2
2
kT
I   B (T B ) 
dI 
d
2
c
2
2
kT B
2

d  2
 2 kT B 


d    c

So the equation of radiative transfer becomes:
dT B
d
 TB  T
dT B
d
 TB  T
T B  brightness
T  temperatur e of the material
If T is constant w
TB  TB (0)e
 
If     then T B  
Otherwise,
temperatur e describing
TB  T
ith   , then

 T 1 e
 

The brightness temperature =
The actual temperature at large
optical depth
I
dT B
d
 TB  T
T B  brightness
T  temperatur e of the material
If T is constant w
TB  TB (0)e
 
If     then T B  
Otherwise,
temperatur e describing
TB  T
ith   , then

 T 1 e
 

The brightness temperature =
The actual temperature at large
optical depth
I
(2) Color Temperature, Tc
Often one can measure the spectrum of a source, and it is more or
less a blackbody of some temperature, Tc.
We may not know Iν, but only Fν, if for example the source is
unresolved.
Tc can be estimated from
λ(max), the peak of the spectrum,
or the ratio of the spectrum
at 2 wavelengths.
e.g. B-V colors of stars
The solar spectrum vs. blackbody – from Caroll & Ostlie
(3) Antenna Temperature, TA
A radio telescope mearures the brightness of a source,
Often described by
T A   TB
Where
S
A
η = the beam efficiency of the telescope, typically ~0.4-0.8
Ωs= solid angle subtended by the source
ΩA= solid angle from which the antenna receives radiation
(“beam”)
(4) Effective Temperature, Teff
If a source has total flux F, integrated over all frequencies
we can define Teff such that
F  T
4
eff
The Einstein Coefficients
Einstein (1917) related αν and jν to microscopic processes,
by considering how a photon interacts with a 2-level atom:
E 2  E1  h  0
E2
emission
E1
Level 2, statistical weight g2
absorption
Level 1, statistical weight g1
Absorption: system goes from Level 1 to Level 2 by absorbing a photon with
energy hν0
Emission: system goes from Level 2 to Level 1 and a photon is emitted.
Three processes can occur:
1. Spontaneous Emission
2. Absorption
3. Stimulated Emission
1. Spontaneous Emission
2
An atom in Level 2 drops to Level 1,
emitting a photon,
even in the absence of a radiation field
Einstein A coefficient
A21 ≡ transition probability per unit time for
spontaneous emission
[A21]= sec -1
Examples:
permitted, dipole transitions
A21 ~ 108 sec-1
magnetic dipole, forbidden transitions
A21~103 sec-1
electric quadrupole, forbidden transitions A21~1 sec-1
1
2. Absorption
An atom in level 1 absorbs a photon and ends up in level 2.
2
Due to the Heisenberg uncertainty principle,
ΔE Δt > ħ, the energy levels are not precisely sharp
1
Each level has a “spread” in energy, called the “natural”
Line width, a Lorentzian.
So let’s parameterize the line profile as φ(ν),
Centered on frequency νo.
We define φ(ν) so that

  ( ) d 
0
1
φ(ν)
Einstein B-coefficients
B12 J ≡ transition probability per unit time
for absorption
Where

J 
 J   ( ) d 
0
Stimulated Emission
The presence of a radiation field will stimulate
an atom to go from level 2  level 1
B 21 J 
Transition probability, per unit time
for stimulated emission
Equation of Statistical Equilibrium
If detailed balance holds
Number of transitions/sec
from Level 1  Level 2
=
Number of transitions/sec
from Level 2  Level 1
Let n1 = # of atoms / volume in Level 1
n2 = # of atoms / volume in Level 2
Then:
n 1 B 12 J
Absorption
 n 2 A 21  n 2 B 21 J
Spontaneous
emission
Stimulated
emission
hence
A
J 




n
n
1
2
21




B
B
B
21
12
21

 1


In thermodynamic equilibrium, the Boltzman equation gives n1/n2
n
n
1
2





g
g

1 
exp

2 
 h o 


 kT 


A
So 
J 
21
B




g 1 B 12 
exp

g 2 B 21 




21
h
kT

0 
1


(1)
In thermodynamic equilibrium,
J   B
Since the Lorentzian is narrow, we can approximate
J  B ( 0 )

2h 0
c
2
3
1
 h 0 
exp 
 1
 kT 
(2)
Comparing (1) and (2), we get
the EINSTEIN RELATIONS
g 1 B 12
A
21

g 2 B 21
 2h 3
 B 21 
2

 c




Comments:
• There’s no “T” in the Einstein Relations, they relate atomic constants
only. Hence, they must be true even if T.E. doesn’t hold.
• Sometimes people derive the Einstein relations in terms of energy
density, uν instead of Jnu, so there’s an extra factor
of 4π/c:
g 1 B 12
A
21

g 2 B 21
 8
 B 21  3

c

3

 h 0

The Milne Relation
Another example of using detailed balance to derive
relations which are independent of the LTE assumption
Relate photo-ionization cross-section at frequency nu,
with cross-section for recombination for electrom with
velocity v:
g h
a
g m c v
2
 (v ) 
2
1
2
2
See derivation in Osterbrock & Ferland
2
2