astro 300b: jan. 26, 2011 thermal radiation and thermal equilibrium
TRANSCRIPT
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.
Blackbody Radiation: BI
In a very optically thick media, recall the SOURCE FUNCTION
Ij
S
So thermal radiation has BjBS and
And the equation of radiative transfer becomes
)(or TBId
dIBI
dl
dI
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) given by the Saha equationT(excitation) given by the Boltzman equationT(radiation) given by the Planck FunctionT(kinetic) given by the Maxwell-Boltzmann distribution
WhereT(ionization) ≠ T(excitation) ≠ T(radiation) ≠ T(kinetic)
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
LOCAL THERMODYNAMIC EQUILIBRIUM (LTE)
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 centralStar, T~50,000 – 100,000 K
However, the gas cools to Te ~ 10,000 K (Te = kinetic temperature of electrons)
H IH II
O star
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 and increases exponentially with temperature T
kTg
g
N
N nnn exp
11
WhereNn = # atoms in excited state n / volumeN1 = # 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
1
12/2
3
kThec
hBI
(3) The Maxwell-Boltzman distribution of speeds of electrons
e
e
kT
vm
e
e evkT
mvf
22
2/3 2
24)(
= fraction of electrons with velocity between v, v+dv
where me = mass of the electron Te = temperature of the electrons
(4) The Saha Equation
kTe
m
m
m
me
m
eh
kTm
Z
Z
N
Nn
2/3
311 2
2
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
1i
kTim
i
egZ
Thermodynamics of Blackbody Radiation: The Stefan-Boltzman Law
Consider a piston containing black-body radiation:
Inside the piston: T, v, p uMove 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 u = energy density energy/volumep = 1/3 u p = radiation pressure in piston
BJdJ
cu and
4
So…T
dQdS
T
dVp
T
dU (substitute dQ=dU+pdV)
T
dVu
T
uVd
3
1)( (substitute U=uV, p=1/3 u)
T
dVudV
T
u
T
VdudS
3
1
dVT
u
dT
dTdu
T
V
3
4
dVT
udT
dT
du
T
V
3
4
So...
T
u
dV
dS
dT
du
T
V
dT
dS
TV 3
4 and
Differentiate these….
(Eqn.1) 12
dT
du
TdT
du
T
V
dV
d
dTdV
Sd
2) (Eqn. 1
3
4
3
4
3
42
2
dT
du
TT
u
T
u
dT
d
dVdT
Sd
Combining (1) and (2) dT
du
TT
u
dT
du
T
1
3
4
3
412
Multiply by TdT
du
T
u
dT
du
3
4
3
4
T
u
dT
du4
T
dT
u
du4
aTu loglog4log a=constant of integration
4)( aTTu Energy density ~T4
u can be related to the Planck Function
J
cu
4 For isotropic radiation,
BJI
So… )(4
)(4
TBc
dTBc
duu
Where B(T) = the integrated Planck function
4
4T
acdB
For a uniform, isotropically emitting surface, we showed that the flux
)(TBdBdFF
4
4T
ac
4
4T
ac
OR….
4TF Stefan-Boltzmann Law
Where4
ac = 5.67x10-5 ergs cm-2 deg-4 sec-1
[flux] = ergs cm-2 sec-1 flux integrated over frequency, per area per sec
alsoc
a4
= 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:
Or in terms of Bλ recall
dIdI
1
12/3
3
kThec
hB
2 so
c
d
dc
1
12/5
2
kThce
hcB
ergs s-1 cm-2 Hz-1 ster-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 equilibriumwith matter at the “surface of last scattering” is the CMB.
Currently the CMB radiation has the spectrum of a blackbody withT=2.73 K.
It is cooling as the Universe expands.
The first accurate measurement of the spectrum of the CMBwas obtained with the FIRAS instrument aboard the CosmicBackground 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 (Radio Astronomy) Wien hν >> kT
Rayleigh-Jeans Law
kT
hkTh
1e so /kTh
so1
12)(
/2
3
kThec
hTI
becomes
kTc
TI RJ2
22)(
The Ultraviolet Catastrophe
If the Rayleigh-Jean’s form for the spectrum of a blackbody held for all frequencies, then
as dI
And the total energy in the radiation field
€
∞
Wien’s Law
kThkTh eekTh
//
1
1
1 so
kThW ec
hTI /
2
32)(
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 withhigher temperature lies entirely above the other.
1
12)(/2
3
kThec
h
dT
d
dT
TdB
2/
/
22
42
1
2
kTh
kTh
e
e
kTc
h
>0 always
Wien Displacement Law
At what frequency does the Planck Law Bν(T) peak?
Bν(T) peaks at νmax, given by 0max
ddB
01
12/2
3
kThec
h
d
d
0122
1 /2
3
2
3/
kThkTh e
d
d
c
h
c
h
d
de
€
ehν / kT −1( )6hν 2
c 2
⎛
⎝ ⎜
⎞
⎠ ⎟=
2hν 3
c 2
⎛
⎝ ⎜
⎞
⎠ ⎟h
kTehν / kT
( )
Divide by exp(hν/kT), cancel some terms
kT
he kTh /13
Let kT
hx max
Need to solve xe x 13
Solution is x=2.82. Need to solve graphically or iteratively.
kT
h max82.2
110max deg 1088.5 HzT
Similarly, one can find the wavelength λmax at which Bλ(T) peaks
0max
ddB
deg cm 290.0max T
NOTE: cmaxmax
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
dc
d2
So equal intervals in wavlength correspond to very different intervals offrequency across the spectrum
With increasing l, constant dl (the Il case) corresponds to smaller and smaller dnso these smaller dn intervals contain smaller energy, comparedto constant dn intervals (the In case)
Radiation constants in terms of physical constants
Recall the Stefan-Boltzman law for flux of a black body
dB
TF
4
0 /
3
20 1
2 dec
hdB
kTh
then Let kT
hx
0
34
0 2 1
2dx
e
x
h
kT
c
hdB
x
0
432
444
0
3
15
2 so...
151T
hc
kdBdx
e
xx
32
45
15
2
hc
k So
Also, since
4
0
4aTdB
cu
33
45
15
8
hc
ka
As an example of the kind of things you can model with thePlanck 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 roughlyN = (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
)( BTBI
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 canassume that the Rayleigh-Jeans Law holds,
kTh so BkTcI
2
22
I
k
cTB 2
2
2or
2. The source need not be a blackbody, despite being describedas 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:
)(TBId
dI
Assume Rayleigh-Jeans, kTc
I2
22
BB kTc
TBI2
22)(
BkTcd
d
d
dI2
22
So the equation of radiative transfer becomes:TT
d
dTB
B
TTd
dTB
B materialtheofetemperaturT
IdescribingetemperaturbrightnessTB
eTeTT BB 1)0(
then,ith constant w is T If
BT then If The brightness temperature =The actual temperature at largeoptical depth
Otherwise, TTB
TTd
dTB
B materialtheofetemperaturT
IdescribingetemperaturbrightnessTB
eTeTT BB 1)0(
then,ith constant w is T If
BT then If The brightness temperature =The actual temperature at largeoptical depth
Otherwise, TTB
(2) Color Temperature, Tc
Often one can measure the spectrum of a source, and it is more orless 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 fromλ(max), the peak of the spectrum,or the ratio of the spectrumat 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
A
SBA TT
Where 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
4 effTF
The Einstein Coefficients
Einstein (1917) related αν and jν to microscopic processes,by considering how a photon interacts with a 2-level atom:
emission absorption
E2
E1
Level 2, statistical weight g2
Level 1, statistical weight g1
012 hEE
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
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
2
1
2. Absorption
An atom in level 1 absorbs a photon and ends up in level 2.
1
2
Due to the Heisenberg uncertainty principle, ΔE Δt > ħ, the energy levels are not precisely sharp
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
φ(ν)
0
1)( d
Einstein B-coefficients
B12 ≡ transition probability per unit time for absorption
Where
0
)( dJJ
J
Stimulated Emission
The presence of a radiation field will stimulate an atom to go from level 2 level 1
JB21Transition probability, per unit time for stimulated emission
Equation of Statistical Equilibrium
If detailed balance holds
Number of transitions/secfrom Level 1 Level 2
Number of transitions/secfrom Level 2 Level 1
Let n1 = # of atoms / volume in Level 1 n2 = # of atoms / volume in Level 2
Then:
JJ B n An Bn 212212121
Absorption Spontaneousemission
Stimulatedemission
=
hence
121
12
2
1
21
21
BB
nn
BA
J
In thermodynamic equilibrium, the Boltzman equation gives n1/n2
kT
ho
gg
nn exp
2
1
2
1
1exp 0
212
121
21
21
kTh
BgBg
BA
J
So (1)
In thermodynamic equilibrium, BJ
)( 0BJ
1exp
1
2
02
30
kT
hh
c
Since the Lorentzian is narrow, we can approximate
(2)
Comparing (1) and (2), we getthe EINSTEIN RELATIONS
BgBg 212121
cBA
h2
3
2121
2
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:
BgBg 212121
h
cBA 0
8 3
32121
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:
acmh
gg
v
v
222
22
2
1)(
See derivation in Osterbrock & Ferland