Transcript The Continuous Spectrum of Light

Color/Temperature Relation
•Blackbody Radiation
•Planck’s law and Astrophysics
•Monochromatic Luminosity and Flux
•Bolometric Magnitude
•Filters, measured flux
•The Color Index
Color/Temperature Relation
Betelguese(3100-3900K)
What does the color of a
celestial object tell us?
Rigel (8000-13,000K)
Blackbody Radiation
•
•
•
Any object with temperature above
absolute zero 0K emits light of all
wavelengths with varying degrees of
efficiency.
An Ideal Emitter is an object that
absorbs all of the light energy incident
upon it and re-radiates this energy with
a characteristic spectrum.Because an
Ideal Emitter reflects no light it is known
as a blackbody.
Wien’s Law: Relationship between
wavelength of Peak Emission max and
temperature T.
Blackbody
lmax T = 0.002897755mK
•
Stefan-Boltzmann equation: (Sun example)
4
L = AsT
L:Luminosity A:area T:Temperature
Blackbody Radiation Spectrum
http://hyperphysics.phy-astr.gsu.edu/Hbase/hframe.html
http://hyperphysics.phy-astr.gsu.edu/Hbase/hframe.html
Blackbody Radiation
http://www.mhhe.com/physsci/astronomy/applets/Blackbody/frame.html
lmax T = 0.002897755mK
L = AsT 4
Blackbody Radiation
•
http://hyperphysics.phy-astr.gsu.edu/Hbase/bbcon.html#c1
Classical Spectral Energy Density
Rayleigh-Jeans/Wien’s
•
•
•
Energy per unit volume per unit
frequency within the blackbody cavity
u(f,T)
Power radiated J(f,T) is related to
u(f,T) by J(f,T)=u(f,T)c/4 (cavity
radiation is isotropic and
unpolarized)
How to calculate u(f,T)?
–
Wien’s Exponential Law (loosely
based on Maxwell’s velocity
distribution for molecules)
u( f ,T) = Af 3e- bf /T
–
•
Rayleigh-Jeans Law
8pf 2
u( f ,T)df = 3 kB Tdf
c
8p
u( l,T)dl = 4 kB Tdl
l
Consider EM radiation to be in
equilibrium with cavity walls
Not
Quite
Right
Ultraviolet
Catasrophe
Planck’s Law for Blackbody Radiation
•Planck used a mathematical “sleight of
hand” to solve the ultraviolet catastrophe.
•The energy of a charged oscillator of
frequency f is limited to discrete values
of Energy nhf.
•During emission or absorption of light
the change in energy of an oscillator is
hf.
•The mean energy at high frequencies
tends to zero because the first allowed
oscillator energy is so large compared
to the average thermal energy available
kBT that there is almost zero probability
that this state is occupied.
•Planck seemed to be an “Unwilling
revolutionary”.He viewed this “quantization”
merely as a calculation trick…Einstein
viewed it differently…light itself was
quantized.
u( l,T) =
8phc
l5 (e hc / lk T -1)
B
Planck’s Law and Astrophysics
Power radiated per unit wavelength per
unit area per unit time per steradian:
(units are W m -2 m-1 sr -1 )
2
5
Bl (T) =
c
2hc / l
u( l,T) = hc / lkBT
4p
(e
-1)
Note that:
ò
Power radiated per unit frequency per
unit area per unit time per steradian:
(units are W m -2 s sr -1 )
c
2hn 3 /c 2
Bn (T) =
u(n ,T) = hn / kB T
4p
(e
-1)
In spherical coordinates the amount of radiant
energy per unit time having wavelengths
between  and demitted by a blackbody
radiator of temperature T and surface area dA
into a solid angle
is given by:
In terms of B:
¥
0
sT 4
Bl (T)dl =
p
Consider a model star consisting of a
spherical blackbody of radius R and
temperature T. Assuming that each patch dA
emits isotropically over the outward
hemisphere, the energy per second having
wavelengths between  and demitted
by the star is:
Ll dl =
òf òq ò
2p
=0
p /2
=0
A
Bl dldAcosq sinqdqdf
Angular integration yields a factor of  , the
integration of dA over the surface of the
sphere yield 4R2.
Ll dl =
òf òq ò
2p
=0
p /2
=0
A
Bl dldAcosq sinqdqdf
Monochromatic Luminosity and Flux
Monochromatic Luminosity
Ll dl = 4p 2 R2 Bl dl
8p 2 R 2 hc 2 / l5
Ll dl = hc / lkB T
dl
e
-1
Monochromatic Flux received at a
distance r from the model star is:
2
Ll
2phc 2 / l5 æ R ö
Fl dl =
dl = hc / lkB T ç ÷ dl
2
4p × r
e
-1 è r ø
Fd is the number of Joules of starlight energy
with wavelengths between  and dthat
arrive per second per one square meter of
detector aimed at the model star, assuming that
no light has been absorbed or scattered during
its journey from the star to the detector. Earth’s
atmosphere absorbs some starlight, but this can
be corrected. The values of these quantities
usually quoted for stars have been corrected and
would correspond to what would be measured
above Earth’s atmosphere.
Why do we keep the
wavelength dependence?
Filters!!!
Sf(
Color/Temperature Relation
Color Indices
•
•
•
A star’s absolute color magnitudes
can be determined from the
apprarent color magnitudes by using
eqn 3.6 if the distance is known.
U-B color index:difference between
its ultraviolet and blue magnitudes.
U - B = MU - MB
V-B color index:difference between
its blue and visual magnitudes.
B -V = MB - MV
Stellar magnitudes decrease with
increasing brightness. A star with a
smaller B-V index is bluer than a star
with a larger value of B-V!!!!
Because a star’s color index is a
difference in magnitudes it is
independent of the star’s distance
The Color Index
UVB Wavelength Filters
•
•
•
Bolometric Magnitude: measured over
all wavelengths.
UBV wavelength filters: The color of a
star may be precisely determined by using
filters that transmit light only through
certain narrow wavelength bands:
– U, the star’s ultraviolet magnitude.
Measured through filter centered at
365nm and effective bandwidth of
68nm.
– B,the star’s blue magnitude.
Measured through filter centered at
440nm and effective bandwidth of
98nm.
– V,the star’s visual magnitude.
Measured through filter centered at
550nm and effective bandwidth of
89nm
U,B,and V are apparent magnitudes
Sensitivity Function S()
Filter Response
U = -2.5log10
(ò
¥
0
)
Fl SU dl + CU
æ
U - B = -2.5log10 ç
ç
è
B = -2.5log10
ò
ò
Fl SU dl ö
0
÷+C
U -B
¥
Fl SB dl ÷ø
¥
0
(ò
¥
0
)
Fl SB dl + CB
æ
B - V = -2.5log10ç
ç
è
V = -2.5log10
ò
ò
¥
0
¥
0
Fl SB dl ö
÷+C
B-V
Fl SV dl ÷ø
http://astro.unl.edu/naap/blackbody/animations/blackbody.html
(ò
¥
0
)
Fl SV dl + CV
Apparent Magnitude and Radiant Flux
U = -2.5log10
(ò
¥
0
)
Fl SU dl + CU
æ
U - B = -2.5log10 ç
ç
è
B = -2.5log10
ò
ò
Fl SU dl ö
0
÷+C
U -B
¥
Fl SB dl ÷ø
¥
0
CU-B º CU - CB
(ò
¥
0
)
Fl SB dl + CB
æ
B - V = -2.5log10ç
ç
è
V = -2.5log10
ò
ò
¥
0
¥
0
(ò
Fl SB dl ö
÷+C
B-V
Fl SV dl ÷ø
¥
0
)
Fl SV dl + CV
CB-V º CB - CV
If one assumes that B is slowly varying across the bandwidth of the filter S can be
approximated by a step function S=1 inside the filter’s bandwidth and S=0 otherwise.
The integrals for U,V and B can be approximated by the value of the Planck function B
at the center of the filter bandwidth,multiplied by that bandwidth. Therefore for the
filters listed on p 75 of the text, we have
Look at example 3.6.2 T=42000K U-B=-1.19 and B-V=-0.33
The constants CU,CB and CV differ and are chosen such that the star Vega(T=9600K,
use applet) has a magnitude of zero as seen through each filter. This does not imply
that Vega would be equally bright when viewed through them.
Bolometric Magnitude/Correction
Bolometric Magnitude mbol is a measurement of total
brightness integrated over all wavelengths.
Monochromatic Flux received at a
distance r from the model star is:
mbol = -2.5log10
2
Ll
2phc 2 / l5 æ R ö
Fl dl =
dl = hc / lkB T ç ÷ dl
4p × r 2
e
-1 è r ø
(ò
¥
0
)
Fl dl + Cbol
S
Bolometric correction BC=mbol-V=Mbol-MV is the difference
between the bolometric and visual magnitudes.
Cbol is a constant that allows the magnitude scale to be defined such that the bolometric
correction BC is negative for all stars while still being as close to zero as possible.
(However some supergiant stars have positive bolometric corrections!!) (see problem
3.17)
Color Index
•
A star’s absolute color magnitudes
can be determined from the apparent
color magnitudes by using the
following equation if the distance is
known.(example 3.2.2)
•
U-B color index:difference between
its ultraviolet and blue magnitudes.
•
B-V color index:difference between
its blue and visual magnitudes.
Stellar magnitudes decrease with
increasing brightness. A star with a
smaller B-V index is bluer than a star
with a larger value of B-V!!!!
d
M = m - 5log10 (
) = m - 5log10 (d) + 5
10 pc
Related to “color”/surface
temperature of star.
Because a star’s color index is
a difference in magnitudes it is
independent of the star’s
distance
Spectral Type, Color and Effective Temperature
for Main-Sequence Stars
Spectral Type
B-V
Te(K)
O5
-0.45
35,000
B0
-0.31
21,000
B5
-0.17
13,500
A0
0.00
9,700
A5
0.16
8,100
F0
0.30
7,200
F5
0.45
6,500
G0
0.57
6,000
G5
0.70
5,400
K0
0.84
4,700
K5
1.11
4,000
M0
1.24
3,300
M5
1.61
2,600
From Frank Shu, An Introduction to Astronomy(1982), Adapted from C.W. Allen, Astrophysical Quantities
Note that this table does not quite agree with our text!!!!
Spectral Type, Color and Effective Temperature
for Main-Sequence Stars
Spectral Type, Color and Effective Temperature
for Main-Sequence Stars (continued)
Color-Color Diagram
•
•
•
Relation between the U-B and B-V
color indices for main sequence
stars.
Would be a straight line if stars were
true black bodies. Some light is
absorbed as it travels through a
star’s atmosphere. The absorption
being wavelength dependent alters
the distribution of radiation from that
of a blackbody.
Best agreement to Blackbody
radiation for very hot stars….
Interstellar Reddening
One also needs to correct color indices for
interstellar
reddening. As the light
propagates through interstellar dust, the
blue light is scattered preferentially making
objects appear to be redder than they
actually are…
Hertzsprung-Russell Diagram
Spectroscopic Parallax
•Identify star as Main-Sequence
•Obtain Te from B-V measurement
•Use H-R diagram to read off absolute luminosity
•Measure apparent luminosity
•Calculate distance modulus to obtain distance to star
Rayleigh-Jeans Law
Lord Rayleigh reasoned that the
following assumptions could be used:
Electromagnetic Waves in Cavity
•Standing Waves
l = 2L, L, 2L / 3, 2L / 4, 2L / 5,....
•Considered at temperature T,
each mode an equal amount of
energy kT
•Analogous to 1-d Harmonic
Oscillators
From Boltzmann’s distribution law
the average oscillator energy is
kBT. The probability of finding an
individual system (oscillator) with
energy above E0 with an ensemble
of sytems at temperature T is:
P(E) = P0e-(E-E0 )/ kB T
For a discrete set of energies the
varage would be:
E=
Energy Density
•Product of number of standing
waves(oscillators)in the
frequency range f - f+df and the
oscillators average energy
u( f ,T)df = EN( f )df
å E × P(E)
å P(E)
In the classical case considered
by Rayleigh, an oscillator can have
any energy in a continuous range
from 0 to infinity. Leading to:
ò Ee
ò e
¥
E classical =
0
-E / kB T
¥ -E / k T
B
0
dE
dE
= kB T
Density of Modes
•
Calculation of the number of modes
in a cavity
–
–
–
Consider cubical cavity of side L
Consider standing waves that have
frequencies between f and f+df
Use Maxwell’s Equations and
boundary condition that the wave
vanishes at the boundaries.Each
component of the electric field
satisfies an equation of the type:
¶2Ex ¶2Ex ¶2Ex
+
+ 2 + k 2Ex = 0
2
2
¶x
¶y
¶z
E x = E x (x, y,z)
2p
k=
l
–
Solutions of the form:
E x (x, y,z) = Asin(kx x)sin(ky y)sin(kz z)
Where
k2 =
p2
L2
(n x2 + n y2 + nz2 )
•Number of standing waves
between k and k+dk is:
1 2
pk dk Vk 2 dk
N(k)dk = 2
=
(p /L) 3
2p 2
•Accounting for 2 polarizations
•Using
N(k)dk k 2 dk
= 2
V
p
k=
2pf
c
•We obtain the number of modes
between f and f+df
8pf 2
N( f )df = 3 df
c
•And the number of modes between
 and +d
N( l)dl =
8p
l
4
dl
Rayleigh-Jeans Law
•
Putting everything together we obtain
the Rayleigh-Jean’s expression for
spectral density.
8pf 2
u( f ,T)df = 3 kB Tdf
c
u( l,T)dl =
8p
l
4
kB Tdl
Ultraviolet
Catasrophe
Derivation of Planck’s Law
•
Planck solved the “catasrophe” by
“quantizing” the energy of the
standing waves E = nhf in the
cavity to obtain a mean energy:
¥
å (nhf )P e
-nhf / kB T
n= 0
¥
åP e
-na
=-
dæ 1 ö
e-a
=
ç
÷
da è1- e-a ø (1- e-a ) 2
å ne
-na
=-
dæ 1 ö
e-a
=
ç
÷
da è1- e-a ø (1- e-a ) 2
å ne-nhf / kBT =
0
E=
å ne
e-hf / kB T
(1- e-hf / kBT ) 2
-nhf / kB T
0
n= 0
•
•We obtain the mean oscillator energy:
¥
åe
¥
1
= å rn
1- r n= 0
•
hfe-hf / kB T
hf
E=
=
1- e-hf / kB T e hf / kBT -1
Solving this by using
-nhf / kB T
n= 0
=
1
(1- e
-hf / kB T
And the steps
E = hf (1- e
-hf / kB T
¥
)å ne-nhf / kB T
n= 0
a = hf /kB T
å ne-na = -
¥
d
åe-na
da n= 0
)
•Multiplying by the density of modes
8pf 2
N( f )df = 3 df
c
•We obtain the spectral energy distribution
ö
8pf 2 æ hf
u( f ,T)df = 3 ç hf / k T ÷df
c èe
-1ø
8phc
u( l,T)dl = 5 hc / lk T
dl
l (e
-1)
B
B
Derivation of Stefan-Boltzmann Law
•Stefan-Boltzmann’s Law can be
obtained from Planck’s Law by simply
integrating the spectral density function
over all wavelengths
¥
c ¥
2phc 2
etotal = ò u(l,T)dl = ò 5 hc / lk T
dl
0 l (e
4 l= 0
-1)
B
•Subsituting and evaluating
ò
x = hc / lkB T
2pkB T 4
=
c 2h 3
4
etotal
¥
0
ò
x3
p4
dx =
(e x -1)
15
¥
0
x3
dx
x
e
-1
( )
•We obtain:
2p 5 kB 4
4
=
T
=
s
T
15c 2 h 3
4
etotal
The Stefan-Boltzmann constant is a
derived constant depending on kB, h
and c !!!!!
(2)(3.141) 5 (1.381 ´10-23 J /K) 4
s=
(15)(2.998 ´10 8 m /s) 2 (6.64 ´10-34 J × s) 3
s = 5.67 ´10-8 W × m-2 × K -4
Derivation of Wien’s Law
http://en.wikipedia.org/wiki/Wien%27s_displacement_law#Derivation_from_Planck.27s_Law