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Lecture 6. Blackbody Radiation
Gustav Robert Kirchhoff
(12 March 1824 – 17 October 1887)
Review Particle physics
Fermion
mass
charge
spin
name
Boson
Radiation in Equilibrium with Matter
Typically, radiation emitted by a hot body, or from a laser is not in equilibrium: energy
is flowing outwards and must be replenished from some source. The first step towards
understanding of radiation being in equilibrium with matter was made by Kirchhoff,
who considered a cavity filled with radiation, the walls can be regarded as a heat
bath for radiation.
The walls emit and absorb e.-m. waves. In equilibrium, the walls and radiation must
have the same temperature T. The energy of radiation is spread over a range of
frequencies, and we define uS (,T) d as the energy density (per unit volume) of the
radiation with frequencies between  and +d. uS(,T) is the spectral energy
density. The internal energy of the photon gas:

u T    uS  , T  d
0
In equilibrium, uS (,T) is the same everywhere in the cavity, and is a function of
frequency and temperature only. If the cavity volume increases at T=const, the
internal energy U = u (T) V also increases. The essential difference between the
photon gas and the ideal gas of molecules: for an ideal gas, an isothermal
expansion would conserve the gas energy, whereas for the photon gas, it is the
energy density which is unchanged, the number of photons is not conserved, but
proportional to volume in an isothermal change.
A real surface absorbs only a fraction of the radiation falling on it. The absorptivity 
is a function of  and T; a surface for which ( ) =1 for all frequencies is called a
black body.
Photons
The electromagnetic field has an infinite number of modes (standing
waves) in the cavity. The black-body radiation field is a superposition of
plane waves of different frequencies. The characteristic feature of the
radiation is that a mode may be excited only in units of the quantum
of energy h (similar to a harmonic oscillators) :
T
  ni  1 / 2 h
i
This fact leads to the concept of photons as quanta of the electromagnetic field. The
state of the el.-mag. field is specified by the number n for each of the modes, or, in other
words, by enumerating the number of photons with each frequency.
According to the quantum theory of radiation, photons are massless
bosons of spin 1 (in units ħ). They move with the speed of light :
The linearity of Maxwell equations implies that the photons do not
interact with each other. (Non-linear optical phenomena are
observed when a large-intensity radiation interacts with matter).
E ph  h
E ph  cp ph
p ph 
E ph
c
h

The mechanism of establishing equilibrium in a photon gas is absorption and emission
of photons by matter. Presence of a small amount of matter is essential for establishing
equilibrium in the photon gas. We’ll treat a system of photons as an ideal photon gas,
and, in particular, we’ll apply the BE statistics to this system.
c
Chemical Potential of Photons = 0
The mechanism of establishing equilibrium in a photon gas is absorption
and emission of photons by matter. The textbook suggests that N can be
found from the equilibrium condition:
On the other hand,
F 

   ph

N

T ,V
F 

  0
  N T ,V
Thus, in equilibrium, the chemical
potential for a photon gas is zero:
 ph  0
However, we cannot use the usual expression for the chemical potential, because one
cannot increase N (i.e., add photons to the system) at constant volume and at the same
time keep the temperature constant:
F 


- does not exist for the photon gas
  N T ,V
G  N
Instead, we can use
G  F  PV
F T ,V 
 F 
P  
 
V
 V T
- by increasing the volume at T=const, we proportionally scale F
Thus,
GF
F
V 0
V
- the Gibbs free energy of an
equilibrium photon gas is 0 !
 ph 
G
0
N
For  = 0, the BE distribution reduces to the Planck’s distribution:
n ph  f ph  , T  
1
  
  1
exp 
k
T
 B 

1
Planck’s distribution provides the average
 h 
  1 number of photons in a single mode of
exp 
frequency  = /h.
 k BT 
The average energy in the mode:
  n h 
h
 h 
  1
exp 
k
T
 B 
  k BT
In the classical (h << kBT) limit:
In order to calculate the average number of photons per small energy interval d, the
average energy of photons per small energy interval d, etc., as well as the total
average number of photons in a photon gas and its total energy, we need to know the
density of states for photons as a function of photon energy.
Density of States for Photons
kz
1 4 / 3 k 3
k 3 volume
N k  

8   
6 2
Lx Ly Lz
kx
dG 
g   
d
ky
extra factor of 2 due
to two polarizations:
  cp  ck G  
6  2 c 

d
h 
8 2
g    g  
h 2
 3
d
c
 c 3
2
3D
ph
3D
ph
3
3
k3
G k  
6 2
g
3D
ph
  
2
2  2 c 
8 2
g    3
c
3D
ph
3
h g   n   d  uS  , T d
8 2
3D
g ph    3
c
h g( )
The average energy of photons with frequency
between  and +d (per unit volume):

3

photon
energy
n( )
=


8 h
3
us  , T   h g   f    3
c exp  h   1
u as a function of the energy:
average number
of photons
h g( ) n( )
Spectrum of Blackbody Radiation

- the spectral density of the
black-body radiation
(the Planck’s radiation law)
u  , T d  u  , T d
u  , T   u  , T 
8
u  , T  
hc 3
d
 u h , T  h
d
3
  
  1
exp 
k
T
 B 
u(,T) - the energy density per unit photon
energy for a photon gas in equilibrium with
a blackbody at temperature T.
Classical Limit (small f, large ), Rayleigh-Jeans Law
At low frequencies:
 h  1 exp  h  1   h
8 h
3
8  2
us  , T   3
 3 k BT
c exp  h   1
c
- purely classical result (no h), can be
obtained directly from equipartition
Rayleigh-Jeans Law
This equation predicts the so-called
ultraviolet catastrophe – an infinite
amount of energy being radiated at
high
frequencies
or
short
wavelengths.
Planck curve
Rayleigh-Jeans Law (cont.)
u as a function of the wavelength:
3
u  , T d  u  , T d
 c
h 
hc 
8
1
 d
 hc  8 hc
 




u

,
T




 d
2 
hc 3 exp  hc   1  2  5 exp  hc   1
k T 
k T 
 B 
 B 
In the classical limit of large :
u  , T  large  
8 k BT

4
1
4
High  limit, Wien’s Displacement Law
At high frequencies/low temperatures:  h  1 exp  h  1  exp  h 
u s  , T  
8 h 3
 exp   h 
c3
Nobel 1911
The maximum of u() shifts toward higher frequencies with increasing temperature. The
3
position of maximum:


 h 

 

 3x 2
k
T
du
d
x 3e x 
  B 

 const 
 const   x

0
2
x
d
e

1
 h  
 h  
e  1 

  exp 
  1
d 
 k BT  
 k BT  

 max  2.8
k BT
h
3  x e x  3
 x  2.8
u(,T)
h max
 2.8
k BT


Wien’s displacement law
- discovered experimentally
by Wilhelm Wien
- the “most likely” frequency of a photon in a
blackbody radiation with temperature T
Numerous applications
(e.g., non-contact radiation thermometry)
max  max
u , T 
u , T 
 max
max
h max
 2.8 - does this mean that
k BT
hc
No!
 2.8 ?
k BT max
3
u  , T d  u  , T d
 c
h 
hc 
8
1
 d
 hc  8 hc
 




u

,
T




 d
2 
hc 3 exp  hc   1  2  5 exp  hc   1
k T 
k T 
 B 
 B 




du
d 
1
5
 x 2 exp 1 / x  
 const   5
 const   6
 5
0

2
df
dx  x exp 1 / x   1
 x exp 1 / x   1 x exp 1 / x   1 
5xexp 1/ x 1  exp 1/ x
T = 300 K
“night vision” devices


max 
hc
5 k BT
max  10 m
Solar Radiation
The surface temperature of the Sun - 5,800K.
max 
As a function of energy, the spectrum of
sunlight peaks at a photon energy of
hc
 0.5 m
5 k BT
umax  h max  2.8k BT  1.4 eV
- close to the energy gap in Si, ~1.1 eV, which has been so far the best material for solar cells
Spectral sensitivity of human eye:
Stefan-Boltzmann Law of Radiation
The total number of photons per unit volume:

N
8
n    n  g  d  3
V 0
c

2
 h 
8
d


0
c3
  1
exp 
 k BT 
The total energy of photons per unit volume :
(the energy density of a photon gas)
2 5 k B

15h 3c 2
4
the Stefan-Boltzmann
constant
 k BT 


 h 
3
- increases as T 3
U
  g  
8 5 k BT 
uT    
d 
3
V 0 exp    1
15hc 

u T  
4
4 4
T
c
the StefanBoltzmann Law
uT 
8 5 k BT  hc 
4



k BT  2.7 k BT
3
3
N
15hc  8 k BT   2.4 15  2.4
4
The average energy per photon:
3
x 2 dx
 kB  3

8

0 e x  1  hc  T  2.4
3
(just slightly less than the “most” probable energy)
Some numbers:

8
4 2
The value of the Stefan-Boltzmann constant:   5.76 10 W / K m
Consider a black body at 310K. The power emitted by the body:
 T 4  500 W / m2
While the emissivity of skin is considerably less than 1, it still emits a considerable
power in the infrared range. For example, this radiation is easily detectable by modern
techniques (night vision).

Temperature of the Sun
• Stefan also determined the temperature of the Sun's surface. He learned from
the data of Charles Soret (1854–1904) that the energy flux density from the
Sun is 29 times greater than the energy flux density of a warmed metal
lamella. A round lamella was placed at such a distance from the measuring
device that it would be seen at the same angle as the Sun. Soret estimated the
temperature of the lamella to be approximately 1900 °C to 2000 °C. Stefan
surmised that ⅓ of the energy flux from the Sun is absorbed by the Earth's
atmosphere, so he took for the correct Sun's energy flux a value 3/2 times
greater, namely 29 × 3/2 = 43.5.
• Precise measurements of atmospheric absorption were not made until 1888
and 1904. The temperature Stefan obtained was a median value of previous
ones, 1950 °C and the absolute thermodynamic one 2200 K. As 2.574 = 43.5,
it follows from the law that the temperature of the Sun is 2.57 times greater
than the temperature of a lamella, so Stefan got a value of 5430 °C or 5700 K
(modern value is 5780 K). This was the first sensible value for the temperature
of the Sun. Before this, values ranging from as low as 1800 °C to as high as
13,000,000 °C were claimed. The lower value of 1800 °C was determined by
Claude Servais Mathias Pouillet (1790-1868) in 1838 using the Dulong-Petit
law. Pouilett also took just half the value of the Sun's correct energy flux.
Perhaps this result reminded Stefan that the Dulong-Petit law could break
down at large temperatures.
Power Emitted by a Black Body
For the “uni-directional” motion, the flux of energy per unit area  c u
T
energy density u
1m2
c  1s
Integration over all angles
provides a factor of ¼:
power emitted by unit area 
1
cu
4
(the hole size must be >> the wavelength)
Thus, the power emitted by a unit-area
surface at temperature T in all directions:
power 
c
c 4 4
u T   
T  T4
4
4 c
The total power emitted by a black-body sphere of radius R:
 4R 2 T 4
Sun’s Mass Loss
The spectrum of the Sun radiation is close to the black body spectrum with the
maximum at a wavelength  = 0.5 m. Find the mass loss for the Sun in one second.
How long it takes for the Sun to loose 1% of its mass due to radiation? Radius of the
Sun: 7·108 m, mass - 2 ·1030 kg.
max = 0.5 m 
max
hc

5 k BT
 T
hc
5 k B max


6.6 1034  3 108
 

K

5
,
740
K
 23
6
 5 1.38 10  0.5 10

2 5 k B
W
8


5
.
7

10
15h 3c 2
m2K 4
4
P power emitted by a sphere   4R 2 T 4
This result is consistent with the flux of the solar radiation energy received by the Earth
(1370 W/m2) being multiplied by the area of a sphere with radius 1.5·1011 m (Sun-Earth
distance).

hc
2
P  4 RSun   
 2.8 k B max
4

2
W
  4 7 108 m  5.7 10 8 2 4  5,740K 4  3.8 10 26 W
mK


the mass loss per one second
1% of Sun’s mass will be lost in

dm P 3.8 1026 W
9
 2 

4
.
2

10
kg/s
2
8
dt c
3 10 m


0.01M
2 1028 kg
t 

 4.7 1018 s  1.5 1011 yr
9
dm / dt 4.2 10 kg/s
Dewar
Radiative Energy Transfer
Liquid nitrogen and helium are stored in a vacuum or Dewar flask, a
container surrounded by a thin evacuated jacket. While the thermal
conductivity of gas at very low pressure is small, energy can still be
transferred by radiation. Both surfaces, cold and warm, radiate at a rate:
J rad  1  r  Ti
4
W / m2
i=a for the outer (hot) wall, i=b for the inner (cold) wall,
r – the coefficient of reflection, (1-r) – the coefficient of emission
Let the total ingoing flux be J,
and the total outgoing flux be J’:
The net ingoing flux:
J  1  r  Ta4  rJ  J   1  r  Tb4  rJ
J  J 
1 r
 Ta4  Tb4
1 r


If r=0.98 (walls are covered with silver mirror), the net flux is reduced to 1% of the
value it would have if the surfaces were black bodies (r=0).
Superinsulation
Two parallel black planes are at the temperatures T1 and T2 respectively. The
energy flux between these planes in vacuum is due to the blackbody
radiation. A third black plane is inserted between the other two and is allowed
to come to an equilibrium temperature T3. Find T3 , and show that the energy
flux between planes 1 and 2 is cut in half because of the presence of the third
plane.

T1
J 0   T1  T2
Without the third plane, the energy flux per unit area is:
4
4
T3 T2

The equilibrium temperature of the third plane can be found from the energy balance:

 
 T14  T3 4   T3 4  T2 4

1/ 4
T1  T2  2T3
4
4
 T14  T2 4 

T3  

2


4
The energy flux between the 1st and 2nd planes in the presence of the third plane:

J   T1  T3
4
4



 4 T14  T2 4  1
1
   T14  T2 4  J 0
   T1 
 2
2
2


Superinsulation: many layers of aluminized Mylar foil loosely
wrapped around the helium bath (in a vacuum space between the
walls of a LHe cryostat). The energy flux reduction for N heat
shields:
J0
JN 
N 1
- cut in half
The Greenhouse Effect
Absorption:



2
4 R
Power in    RE  TSun   Sun 
 Rorbit 
2
the flux of the solar radiation energy
received by the Earth ~ 1370 W/m2
Emission:
Power out  4 RE  TE
2
4
1/ 4
  R  2 
TE    Sun   TSun
 4  Rorbit  
Rorbit = 1.5·1011 m
Transmittance of the Earth atmosphere
 = 1 – TEarth = 280K
In reality
 = 0.7 – TEarth = 256K
To maintain a comfortable temperature on the
Earth, we need the Greenhouse Effect !
However, too much of the greenhouse effect
leads to global warming:
RSun = 7·108 m
Thermodynamic Functions of Blackbody Radiation
The heat capacity of a photon gas at constant volume:
This equation holds for all T (it agrees with
the Nernst theorem), and we can integrate it
to get the entropy of a photon gas:
16
 u T 
cV  

VT 3

c
 T V
c T 
16V
16
2
3


S T    V
dT  
T
d
T

VT
T
c 0
3c
0
T
T
Now we can derive all thermodynamic functions of blackbody radiation:
the Helmholtz free energy:
F  U  TS 
4
16
4
VT 4 
VT 4  
VT 4
c
3c
3c
G  U  TS  PV  F  PV  0
the Gibbs free energy:
the pressure of a photon gas
(radiation pressure)
4 4 1
 F 
P  

T  u

3
 V T , N 3c
 N 
1
PV  U
3
For comparison, for a non-relativistic monatomic gas – PV = (2/3)U. The difference
– because the energy-momentum relationship for photons is ultra-relativistic, and
the number of photon depends on T.
In terms of the average
density of phonons:
1
1
1
PV  U  V  n    V  n  2.7k BT  0.9 N k BT
3
3
3
Radiation in the Universe
Approximately 98% of all the photons
emitted since the Big Bang are observed
now in the submillimeter/THz range.
In the spectrum of the Milky Way galaxy,
at least one-half of the luminous power is
emitted at sub-mm wavelengths
The dependence of the radiated energy
versus wavelength illustrates the main
sources of the THz radiation: the
interstellar dust, emission from light and
heavy molecules, and the 2.7-K cosmic
background radiation.
Cosmic Microwave Background
A. Penzias
R. Wilson
Nobel 1978
In the standard Big Bang model, the radiation
is decoupled from the matter in the Universe
about 300,000 years after the Big Bang,
when the temperature dropped to the point
where neutral atoms form (T~3000K). At this
moment, the Universe became transparent
for the “primordial” photons. The further
expansion of the Universe
can be
considered
as
quasistatic
adiabatic
(isentropic) for the radiation:
S T  
16
VT 3  const
3c
Since V  R3, the isentropic expansion
leads to :
T  R 1
CMBR (cont.)
“…
for
their
discovery of the
blackbody form and
anisotropy of the
CMBR”.
Mather, Smoot, Nobel 2006
At present, the temperature of the Planck’s
distribution for the CMBR photons is 2.735
K. The radiation is coming from all
directions and is quite distinct from the
radiation from stars and galaxies.
Alternatively, the later evolution of the radiation temperature may be considered as a
result of the red (Doppler) shift (z). Since the CMBR photons were radiated at T~3000K,
the red shift z~1000.
Results: WMAP 3 years 3/16/06
Power Spectrum of temperature fluctuations
Correlation
Power
Spectra of
temperature
fluctuations
and E/B
polarization
Fitted Cosmological Parameters
WMAP results
• Big Bang=> very high temperatures!
– When T≈3000K, p+e recombine =>H and universe
becomes transparent
– Redshifted by expansion of the universe: 2.73K
• Very efficient thermalization
– In particular no late release of energy
• No significant ionization of universe since!
– e.g., photon-electron interactions= Sunyaev-Zel'dovich
effect
• Fluctuations of T => density fluctuations
Problem
The frequency peak in the spectral density of radiation for a certain distant star
is at 1.7 x 1014 Hz. The star is at a distance of 1.9 x 1017 m away from the Earth
and the energy flux of its radiation as measured on Earth is 3.5x10-5 W/m2.
a)
b)
c)
d)
(5) What is the surface temperature of the star?
(5) What is the total power emitted by 1 m2 of the surface of the star?
(5) What is the total electromagnetic power emitted by the star?
(5) What is the radius of the star?
h
6.6 10 34 1.7 1014
T

 3000 K
 23
2.7 k B
1.38 10  2.7
(a)


(b)
J   T 4  5.7 10 8 W / K 4  m 2  3000 K 4  4.6 106 W / m 2
(c)
power  4 r 2 J r   4  1.9 1017 m2  3.5 105W / m2  1.6 1031W
(d)

 
4 R J RS  4 r J r 
2
S
2
4

J r 
3.5 10 5
17
11
RS  r
 1.9 10 m 

5
.
2

10
m
6
J RS
4.6 10
 
Problem
The cosmic microwave background radiation (CMBR) has a temperature of
approximately 2.7 K.
(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density
u(λ,T) of the cosmic background radiation?
(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density
u(,T) of the cosmic background radiation?
(c) (5) Do the maxima u(λ,T) and u(,T) correspond to the same photon energy? If
not, why?
(d) (15) What is approximately the number of CMBR photons hitting the earth per
second per square meter [i.e. photons/(s·m2)]?
(a)
(b)
(c)
max
hc
6.6 1034  3 108
3



1
.
1

10
m  1.1mm
 23
5 k BT 5 1.38 10  2.7
 max
k BT 2.8 1.38 10 23  2.7
11
 2.8


1
.
58

10
Hz
34
h
6.6 10
hc
max
 1.8 10  22 J
h max  1.04 10 22 J
The maxima u(λ,T) and u(,T) do not correspond to the same photon energy. As 
increases, the frequency range included in unit wavelength interval increases as 2,
moving the peak to shorter wavelengths:
u  , T d  u  , T d
c
8 hc
 d




u

,
T

 d
2 
5
1
 hc 
  1
exp 

k
T
 B 
Problem cont.
(d)


J   TCMBR  5.7 108 W / K 4  m2  2.7 K 4  3 106W / m2
4
4
uT 
8 5 k BT  hc 
4



k BT  2.7 k BT
3
3
N
15hc  8 k BT   2.4 15  2.4
4
3
W 
J 2 
3 106
m 
 photons 

16 photons
N



3

10

2
 J 
2.7 1.38 1023  2.7
s  m2
 sm 
Problem
The planet Jupiter is a distance of 7.78 x 1011 meters from the Sun and is of radius
7.15 x 107 meters. Assume that Jupiter is a blackbody even though this is not
entirely correct. Recall that the solar power output of Sun is 4 x 1026 W.
a) What is the energy flux of the Sun's radiation at the distance of Jupiter's orbit?
PSun
4 1026W
J

 52.6W / m2
2
2
4Rorbit 4 7.78 1011 m


b) Recognizing that Sun's energy falls on the geometric disk presented to the Sun
by the planet, what is the total power incident on Jupiter from the Sun?


2
PJupiter  J  RJupiter
 52.6W / m2   7.15 107 m  8.4 1017W
2
c) Jupiter rotates at a rather rapid rate (one revolution per 0.4 earth days) and
therefore all portions of the planet absorb energy from the Sun. Hence all
portions of the surface of this planet radiate energy outward. On the basis
of this information find the surface temperature of Jupiter.
2
4
PJupiter  4RJupiter
TJupiter
 PJupiter
TJupiter  
 4R 2 
Jupiter

1/ 4




1/ 4


8.4 1017W


2
7
8
4 2
 4 7.15 10 m 5.76 10 W / K m 


 123K
Problem (blackbody radiation)
The black body temperature is 3000K. Find the power emitted by this black body
within the wavelength interval  = 1 nm near the maximum of the spectrum of
blackbody radiation.
c
Jd  u  , T d
4
hc
8 hc
1
max 
u  , T d  5
d
5 k BT

 hc 
  1
exp 
  k BT 
the power emitted by a unit area in all directions:
near the maximum:
c
c 8 hc5k BT 
1
2 c 5k BT 
1
J d    u  , T d 
d


d
5
4




4
4
exp
5

1
exp
5

1
hc 
hc 
5
J d  

2  3 108  5 1.38 10  23  3000
6.6 10
34
 3 108

4
5

5

1
110 9  3.1103 W/m 2
exp 5  1
Problem (radiation pressure)
(a)
(b)
(c)
(d)
At what temperature will the pressure of the photon gas be equal to 105 Pa (=
one bar )?
At what temperature will the pressure of the photon gas be equal to 10-5 Pa?
The temperature at the Sun’s center is 107 K. What is the pressure of the
radiation? Compare it to the pressure of the gas at the center of the Sun, which
is 1011 bar.
Calculate the pressure of the Sun’s radiation on the Earth’s surface, given that
the power of the radiation reaching earth from the Sun is 1400 W/m2.
4 4
P
T
3c
(b)
(c)
1/ 4
 3cP 
T 

 4 
P=10-5
Pa
 3  3 108  1 105
(a) P=105 Pa T  
8
 4  5.7 10
 3  3 108  1 10 5
T  
8
 4  5.7 10
 
4  5.7 108  107
P
3  3 108
1/ 4



 1.4  105 K
1/ 4



 450 K
4
 2.5 1012 Pa  2.5 107 bar
- at this T, the pressure of radiation is still negligible in the balance for mechanical equilibrium
(d)
P
1
4
u T  u T   I
3
c
P
4I
4 1400

 6 10 6 Pa
8
3c 3  3 10
Problem
Tungsten has an emissivity of 0.3 at high temperatures. Tungsten
filaments operate at a temperature of 4800 K.
a) At what frequency does a tungsten filament radiate the most
energy?
b) What is the power/unit surface emitted by the tungsten filament?
c) If the surface area of a tungsten filament is 0.01 cm2 what is the
power output of the bulb?
d) What is the energy flux of radiation emitted by the filament 2
meters from the bulb?
e) What fraction of radiation incident on a tungsten filament is
reflected?
[Answers: a) 2.88 x 1014 Hz, b) 9.03 x106 W/m2, c) 9.03 watts, d) 0.180 watts, and e)
70%.]
Problem Quiz
The spectrum of Sun, plotted as a function of energy,
peaks at a photon energy of 1.4 eV. The spectrum for
Sirius A, plotted as a function of energy, peaks at a
photon energy of 2.4 eV. The luminosity of Sirius A
(the total power emitted by its surface) is by a factor of
24 greater than the luminosity of the Sun. How does
the radius of Sirius A compare to the Sun’s radius?
L  4R  T
2
4
R
L
T
2
The temperature, according to Wien’s
law, is proportional to the energy that
corresponds to the peak of the photon
distribution.
LSirius / LSun
RSirius
24


 1.67
2
2
RSun
TSirius / TSun  2.4 / 1.4