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,

I ν

frequency

ν

and temperature

T

is a universal function of – the Planck function,

B ν

. Blackbody Radiation:

I

 

B

 In a very optically thick media, recall the SOURCE FUNCTION So thermal radiation has

S

 

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

(3) The

excitation

of atoms by the SAHA equation 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 equation T(excitation) given by the Boltzman equation T(radiation) given by the Planck Function T(kinetic) 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 T e ~ 10,000 K (T e = kinetic temperature of electrons) O star H I H II

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 χ n above the ground state decreases exponentially with χ n increases exponentially with temperature T and

N n N

1 

g g

1

n

exp Where

N n N 1

= # atoms in excited state

n

/ volume = # atoms in ground state /volume 

n kT g n = 2n 2

the statistical weight of level

n

= number of different angular momentum quantum numbers in energy level

n

(2) The Planck Function

I

 

B

  2

h

 3

c

2 1

e h

 /

kT

 1

(3) The Maxwell-Boltzman distribution of speeds of electrons

f

(

v

)  4     2 

m e kT e

   3 / 2

v

2

e



m e v

2 2

kT e

= fraction of electrons with velocity between

v, v+dv

where

m e T e

= mass of the electron = temperature of the electrons

(4) The Saha Equation

n e N m

 1

N m

 2

Z m

 1

Z m

2 

m e kT h

3 3 / 2

e

 

m kT

Where n e N m = number density of free electrons = number density of atoms in the m th ionization state Z m = partition function of the m th ionization state

Z m



i

   1

g i e

 

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 u p = 1/3 u p

= energy density energy/volume = radiation pressure in piston

u



4



c



J



d



and

J

 

B

 So…

dS



dQ T



dU T



p dV T

(substitute dQ=dU+pdV) 

d

(

uV

)

T

 1 3

u dV T

(substitute U=uV, p=1/3 u)

dS



Vdu T



u T dV

 1 3

u dV T

So...



V T du dT dT

 4

u

3

T dV



V T du dT dT

 4

u

3

T dV dS dT



V



V T du dT

and  

dS dV



T

 4

u

3

T

Differentiate these….

d

2

S dTdV



d dV V du T dT



1

T du dT

(Eqn.1)

d

2

S dVdT



d dT

4

u

3

T

  4 3

u T

2  4 1 3

T du dT

(Eqn.

2) Combining (1) and (2)  1

du T dT

  4 3

u T

2  4 1 3

T du dT

Multiply by T

du dT

  4

u

3

T

 4 3

du dT du dT



u

4

T

du u

 4

dT T

log

u

 4 log

T

 log

a u

(

T

)



aT

4 a=constant of integration Energy density ~T 4

u

can be related to the Planck Function

u

  4 

c J

 For isotropic radiation, So…

u

 

u



d

  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

   

B



d

  

B

(

T

)  

ac

4 

T

4 

ac T

4 4 OR….

F

 

T

4 Stefan-Boltzmann Law Where  

ac

4 =

5.67x10

-5 ergs cm -2 deg -4 sec -1 [flux] = ergs cm -2 sec -1

flux integrated over frequency, per area per sec also

a

 4 

c = 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

  2

h

 3

c

3 1

e h

 /

kT

 1 Or in terms of B λ recall 

I

 

d

 

c

so

I



d



d



d

  

c

2

ergs s -1 cm -2 Hz -1 ster -1

B

  2

hc

2  5 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: h ν << kT (Radio Astronomy) Wien h ν >> kT

so Rayleigh-Jeans Law

h

 

kT

so e h  /kT  1 

h



kT I

 (

T

)  2

h

 3

c

2 1

e h

 /

kT

 1 becomes

I



RJ

(

T

)  2  2

kT c

2

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

 

1

kT

so

e h

 /

kT



1



e h



1

/

kT I



W

(

T

)



2

h

 3

e



h

 /

kT c

2  Very steep decrease in brightness for   

peak

Monotonicity with Temperature If T 1 > T 2 , then B ν (T 1 ) > B ν (T 2 ) for all frequencies Of 2 blackbody curves, the one with higher temperature lies entirely above the other.

dB

 (

T

)

dT



d dT

  2

h



c

2 3 1

e h

 /

kT

 1   

2

h

2 

c

2

kT

2 4 

e h



e h

 /

kT

/

kT



1

2  >0

always

Wien Displacement Law At what frequency does the Planck Law B ν (T) peak?

B ν (T) peaks at ν max , given by

dB



d

  max  0

d d

   2

h



c

2 3 1

e h

 /

kT

 1    0 

e h

 /

kT



1



d d

  

2

h

 3

c

2     

2

h

 3

c

2  

d d

 

e h

 /

kT



1

 

0

 

e h

 /

kT

 1   6

h

 2

c

2    2

h

 3

c

2  

h kT



e h

 /

kT

 Divide by exp(h ν/kT), cancel some terms Let 3  1 

e



h

 /

kT

 

h



kT x



h

 max

kT

Need to solve 3  1 

e



x

 

x

Solution is

x=2.82

. Need to solve graphically or iteratively.

2 .

82 

h

 max

kT

 max

T



5 .

88



10

10

Hz

deg

 1 Similarly, one can find the wavelength λ max at which B λ (T) peaks

dB



d

    max  0  max

T

 0 .

290 cm deg

NOTE :  max  max 

c

That is to say, B ν frequency.

and B λ don’t peak at the same wavelength, or For the Sun’s spectrum, λ max for I λ is at about 4500 Å whereas λ max for I ν 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 d l ( the I l case) corresponds to smaller and smaller d n so these smaller d n intervals contain smaller energy, compared to constant d n intervals (the I n case)

Radiation constants in terms of physical constants Recall the Stefan-Boltzman law for flux of a black body

F

 

T

4   

B



d

  0 

B



d

  2

h c

2  0 

e h

  /

kT

3  1

d

 Let

x



h



kT

then  0 

B



d

  2

h



c

2  

kT h

4  0 

e x x

3  1

dx

 0 

e x x

3 

1

dx

  4

15 so...

 0 

B



d

 

2

 4

k

4

15

c

2

h

3

T

4 So    2  5

k

4 15

c

2

h

3

Also, since

u



4



c

0  

B



d

 

aT

4

a

 8  5

k

4 15

c

3

h

3

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 cm 3 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= σ T 4 The room temperature will be taken to be 22 °C = 295 K. Assuming an ideal radiator for this estimate, the radiated power is P = σAT 4 A=surface area of nickel = (5.67 x 10-8 W/m 2 K 4 )x(8.54 x 10 -4 m 2 )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

.

l peak = 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 .

E photon = h ν = 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 10 19 photons

Characteristic Temperatures for Blackbodies 1. BRIGHTNESS TEMPERATURE,

T b

Instead of stating

I ν

, one can state

T b

, where

I

 

B

 (

T B

) i.e. T b is the temperature of the blackbody having the

specific intensity same

as the source, at a particular frequency.

Notes: 1. T B is often used in radio astronomy, and so you can assume that the Rayleigh-Jeans Law holds,

h

 

kT

so

I

  2  2

c

2

kT B

or

T B

 2 

c

2 2

k I

 2. The source need not be a blackbody, despite being described as a source with brightness temperature T B .

3. Units of T B are easier to remember than units of I ν

T B and the equation of Radiative Transfer:

dI



d

   

I

 

B

 (

T

) Assume Rayleigh-Jeans,

I



I

  2  2

kT c

2 

B

 (

T B

)  2  2

c

2

kT B dI



d

  

d d

    2 

c

2 2

kT B

  So the equation of radiative transfer becomes:

dT B d

   

T B



T

dT B d

   

T B



T T B T

 

brightness temperatur temperatur e of the e describing material I



If

T B

T is



constant w

T B

( 0 )

e

   

T

ith



1

  

e

 

, then

  If     then

T B

  The brightness temperature = The actual temperature at large optical depth Otherwise,

T B



T

dT B d

   

T B



T T B T

 

brightness temperatur temperatur e of the e describing material I



If

T B

T is



constant w

T B

( 0 )

e

   

T

ith



1

  

e

 

, then

  If     then

T B

  The brightness temperature = The actual temperature at large optical depth Otherwise,

T B



T

(2) Color Temperature, T c 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 I ν , but only F ν , if for example the source is unresolved.

T c 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, T A A radio telescope mearures the brightness of a source, Often described by

T A

 

T B



S



A

Where η = the beam efficiency of the telescope, typically ~0.4-0.8

Ω s = solid angle subtended by the source Ω A = solid angle from which the antenna receives radiation (“beam”)

(4) Effective Temperature, T eff If a source has total flux F, integrated over all frequencies we can define T eff such that

F

 

T eff

4

The Einstein Coefficients

Einstein (1917) related α ν and j ν to microscopic processes, by considering how a photon interacts with a 2-level atom:

E

2 

E

1 

h

 0 E 2 emission E 1 Level 2, statistical weight g 2 absorption Level 1, statistical weight g 1 Absorption: system goes from Level 1 to Level 2 by absorbing a photon with energy hν 0 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 A 21 ≡ transition probability per unit time for spontaneous emission [A 21 ]= sec -1 Examples: permitted, dipole transitions A 21 ~ 10 8 magnetic dipole, forbidden transitions A 21 ~10 3 sec sec electric quadrupole, forbidden transitions A 21 ~1 sec -1 -1 -1 2 1

2

2. Absorption

An atom in level 1 absorbs a photon and ends up in level 2.

1 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 ν o .

We define φ(ν) so that   0  (  )

d

  1 φ(ν)

Einstein B-coefficients B

J

≡ transition probability per unit time for absorption Where

J

   0

J

  (  )

d



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

n 1 n 2 = # of atoms / volume in Level 1 = # of atoms / volume in Level 2 Then: n 1 B 12

J

 n 2 A 21  n 2 B 21

J

Absorption Spontaneous emission Stimulated emission

hence

J



n

1

n

2

A

21

B

21

B B

12 21  1 In thermodynamic equilibrium, the Boltzman equation gives n 1 /n 2

n

1

n

2    

g

1

g

2    exp  

h



kT o

  So 

J

   

g

1

g

2

B

12

B

21   

A

21 exp

B

21  

h



kT

0    1 (1)

In thermodynamic equilibrium,

J

 

B



Since the Lorentzian is narrow, we can approximate

J



B

 (  0 )  2

h

 0 3

c

2 exp    1

h

 0

kT

    1 (2)

Comparing (1) and (2), we get the EINSTEIN RELATIONS

g

1

B

1 2 

g

2

B

2 1

A

21 

B

21  

2

h

 3

c

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, u ν instead of Jnu, so there’s an extra factor of 4π/c:

g

1

B

1 2 

g

2

B

2 1

A

21 

B

21   8 

c

3  

h

 0 3

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:  (

v

) 

g

1

g

2

m h

2 2 

c

2 2 v 2

a

 See derivation in Osterbrock & Ferland