interstellar extinction star colors

1

Interstellar Extinction&

Star Colors

James R. Graham

10/17/2006

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Visual Extinction

• Debate regarding presence of interstellar dust has a long history– Existence of absorbing interstellar dust remained controversial

• Star counts show few stars in some directions (W. Herschel1738-1822)– No stars, or something hiding them?

– No consensus until the 1930’s

• R. Trümpler (1930 PASP, 42, 214) at Berkeley conclusivelydemonstrated interstellar absorption– Compare distances from luminosity & angular diameter of open clusters

• Angular diameter distances are systematically smaller

• Discrepancy grows with distance

• Distant clusters are redder

– Extimated ~ 2 mag/kpc absorption• Attributed to Rayleigh scattering by gas

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Scattered Light in the Pleiades

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Dark Nebulae

B 33

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A Milky Way Panorama

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The Discovery of InterstellarExtinction by Dust

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Trumpler’s Argument

• Compare two distance estimates to starclusters– Angular diameter distance

– Luminosity distance

d

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• Angular diameter distance:

d! = D/!

!

d!

D

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• Luminosity distance:

dL = (L/4! F)1/2

d

L

F = L/4!d2

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• Assume all clusters have (on average) thesame diameter and luminosity

d!

dL

Expect d!

dL

Observe

Luminosity distances are systematically too large

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• This systematic trend implies– More distant clusters are progressively fainter

(anti Copernican)

– There is some absorption along the line of sight

• The correct formula for the brightness of adistant cluster is

Where A is the extinction [magnitudes]

�

F =L

4!d210

"0.4A

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The Equation of Radiative Transfer:F", I" & all that

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Equation of Radiative Transfer

• The transport of radiation along some path#s through a medium can be described interms of the– Specific intensity, Iv [W/m2/Hz/sr]

– Opacity, $v [m-1]– Emissivity, jv [W/m3/Hz/sr]

• In general these quantities depend onwavelength or frequency

�

dI!

ds= j

!"#vI!

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Radiative Transfer in a Vacuum

• In a vacuum there is no emission and noabsorption– Opacity, $v = 0

– Emissivity, jv = 0

• Specific intensity, Iv, is constant in a vacuum!– Flux varies as 1/r2, but specific intensity is constant

because Iv, is normalized to solid angle�

dI!

ds= 0

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Flux & Intensity

• Specific intensity is a nearly completedescription of the radiation field– Intensity measures the power along a ray. In

general Iv = Iv(!)– Intensity is distinct from flux because it

describes only those photons headed in aparticular direction, i.e. confined within a coneof solid angle, #%

• The flux [W/m2/Hz] is the total energycrossing a surface regardless of direction

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Flux & Intensity

#A

x

y

z

#&&

#% =sin! #! #&!

I"(!)

�

F! = I! "( )#

$ cos"d#

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Flux at the Surface of Star

• A star, radius R*, radiates out to dark space– I*v ! 0 in the outward direction only

• The star is a black body and I*v is the samein all outward pointing directions

– I*v (!, &) = I*v

�

F! R*( ) = I!

* "( )#$ cos"d#

= I!*d%

0

2&

$2&!

cos" sin"d"0

& 2

$1 2

" # $ $ % $ $

= &I!*

R*

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Flux Far from the Surface of Star

The inverse square lawapplies!– The star’s surface flux is "I*v

– The star’s luminosity is 4"R*2

"I*v

�

F! d( ) = I! "( )#

$ cos"d#

= 2% I!*cos" sin"d"

0

"max

$

= 2% I!*

xdx

0

sin"max

$

= % I!*R*

2

d2

R*

d !

#%=2" sin! d!

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Applications: Absorption in aCold Gas & Mean Free Path

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Radiative Transfer: The Black Body

• Consider a box with reflecting walls,containing an absorbing/emitting medium inthermal equilibrium– Iv = Bv : the Planck function which depends

only on T

– In equilibrium dIv /ds = 0

�

dI!

ds= j

!"# vI!

dI!

ds=dB

!

ds= 0 = j

!"# vB!

j!/# v = B

!

Shinywalls

Iv = Bv

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The Absorption Law a Cold Gas: I

• Consider absorption in cold (jv = 0) gas

�

dI!

ds= "#

vI! ,

dI!

I!

= "#vds $ d%!

dI!

I!

= " d%!

0

% 0

&I! (0)

I!

& , hence I! = I! (0) e"% 0

for uniformgas% 0 = s#v

Iv Iv (0)

s

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Optical Depth & Mean Free Path

• How do we relate optical depth to physicalconditions in the gas:– n, particle density?

– ', cross sectional area for absorption?

• Mean free path, (– Typical distance traveled by a photon before

begin absorbed

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Mean Free Path: 2-d

Two dimensionalexample*– Almost every

horizontal line of sightintersects a tree trunk

*In the approximation that trees are infinite cylinders!

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Mean Free Path: 2-dHow far, on average, does a photon travel in aforest before it intersects a tree trunk of width2r?

2r

(

Rectangle of area2r ( contains onetree trunk

)The corresponding number oftrees per unit area is * = 1/ (2r ()or

( = 1/ (2r *)View of the forest fromabove

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Mean Free Path: 3-d

s

2r ' = " r2

If a photon travels onaverage a distance (before absorption thevolume ( ' mustcontain one absorber

)n = 1/( '

Sphericalabsorberwith cross-sectionalarea '

(

Cylinder ofvolume = ( '

( = 1/n'=1/$

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Absorption in a Volume of Cold Gas: II

If the photon travels ( most of the light is absorbed• How much energy is absorbed over a small length #s << ( ?

– If the fractional area of the beam covered by absorbers is #A/A then

�

!I

I= "

!A

A, where !A = !V # n #$ and A = %R

2

= " !Vn$

%R2

= "!s #%R

2n$

%R2

= "n$!s

#s

2R

#V

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The Absorption Law in a Cold Gas: II

• How much energy is absorbed over a small length #s<< ( ?

�

!I

I= "

!A

A

= "n#!s Writing as an integral

dI

I=

I0

I x( )

$ " n#ds'0

s

$ = "% (s)

log I s( ) I0[ ] = "% (s)

I s( ) = I0 exp "% (s)[ ]

For a uniform medium % s( ) = n# s

#s

2R

#V

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Optical Thickness and Mean Free Path

• An optically thin medium, + <<1, causes only smallattenuation– The dimensions of an optically thin medium is small compared to

the mean free path

+ = n's = s/(

• When the optical depth is large, + >> 1, the intensity isstrongly attenuated by absorption– The medium is said to be optically thick

– An optically thick medium spans many mean free paths

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Application to the Colors of Stars:Reddening & Extinction

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Hertzsprung-Russell Diagram

• The Hertzsprung-Russell diagram is aplot of luminosity vs.temperature

• T originally deducedfrom spectral type– Color depends on T– Colors are easier to

measure thanspectral type

• Color is a differenceof two magnitudes– Ratio of fluxes

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Color Magnitude Diagram

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Interstellar Extinction

• Extinction is a strongfunction ofwavelength

• General (-1 trend inthe visible/near-IR

• Steep UV rise• Peak at ~ 800 Å

• Features• ( = 220 nm

• ( = 9.7 µm

UV Vis IR

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Barnard 68

B,V & I B , I & K

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Colors & Extinction

• Extinction is not“grey”

• Presence ofextinction modifiesthe apparent colorof a star 0.282J (1250)

0.023M (4800)

0.058L (3500)

0.112K (2200)

0.175H (1650)

0.482I (800)

0.748R (660)

1V (550)

1.324B (450)

1.531U (360)

A/AVBand (nm)

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Magnitudes & Extinction

• From the equation of radiative transfer

I = I0 exp(-+)• Expressed on the magnitude scale the change in

brightness due to extinction -+#m = -2.5 log10 (I/I0)

= -2.5 log10 (e) +

=1.0857 + =A

• Given extinction A (> 0) and the true magnitude,V0, the observed magnitude is

V = V0 + AV

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Color & Extinction

• Suppose we know the true color of a star

(B0-V0 )

• Observed color is

(B-V)

• What is the extinction to the star?

(B-V) = (B0+AB ) -(V0 + AV)

= (B0-V0 ) + (AB - AV)

= Intrinsic color + color excess

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Color Excess

• Observed color excess for a given star is

(B-V) - (B0-V0 ) = (AB - AV)

• Compare with the wavelength dependence of A

(AB - AV) = (AB /AV- 1) AV

= (1.324 - 1) AV

0.324 AV

• If you can measure the color excess and know the

extinction curve you can compute the extinction, A

AV = 3.086 [(B-V) - (B0-V0 )]

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Interpreting a Color-Magnitude Diagram

• Suppose you have made a color magnitudediagram from your photometry? What would youexpect?

• For example, the y-axis is absolute K magnitude– MK = K + (m-M)

– K = K-band mag.

– m-M is the distance modulus

– m-M = 5 log10 (d/pc) - 5

– You may have to search the literature to find d• For example, the Orion star forming region, including NGC

2024, is at 450 pc.

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Color Magnitude Diagram

• For this example, choose J-K for the x-axis– Since the wavelength difference between J (1.25 µm)

and K (2.2 µm) is larger than between H (1.65 µm) andK the difference in magnitude, or the color, should belarger and therefore easier to measure

• The color plotted on the x-axis is simply thedifference J-K– Since magnitudes are logarithmic the color is

proportional to the log of the flux ratio at these twowavelengths

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K vs. J-K Diagram

Fromtabular datain the Lab.handout

Main sequenced = 450 pcA = 10 mag.

Main sequenced = 450 pcA = 0 mag.

Reddening vectorAV = 10 mag.

interstellar extinction star colors

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