Unformatted text preview: NASA, NOAO, ESAand The Hubble Heritage Team (STScl/
AURA) Angles and Solid Angles Solid angles describe a fraction of a
sphere. If an emitting source covers an area
A on a spherical surface of
radius 7“, then the solid angle is Q:A/r2 Solid angles are measured in
steradians How many steradians in a
hemisphere? AST 203 (Spring 2011) Intensity I(u)du = energy/unit time/unit surface area in the frequency
range 1/ to V + dz/ being emitted into a cone of solid angle d9 Radiation moves through a small area dA into the cone described
by d9 Energy moving through this area
into the solid angle dB is dEzIV COSQdAdI/dﬂdt Intensity is measured in units of erg s"1 cm"2 Hz"1 ster'1 AST 203 (Spring 2011) Intensity (.059 {Actor [Vt lVl‘l‘eL/lsl‘ly ciefmd‘wh AST 203 (Spring 2011) Intensity In terms of wavelength, I(A)d)\ = energy/unit time/unit area in
the wavelength range A to A + dA emitted into solid angle d9 Then
dE : I,\ cos QdA dA d9 dt It is important to remember that intensity has a direction. Note—your book is a bit sloppy here. It leaves out the solid
angle part. AsT 203 (spring 2011) Blackbody  We won't go deeply into what intensity means (AST 341 topic) — We'll use intensity to derive some important results.
 Important concept: blackbody (thermal radiator) — Object in thermodynamic equilibrium
— Emission = absorption
— Emitted light has wellknown intensity spectrum  Function ofT only
 Isotropic, unpolarized.  Blackbody emission spectrum can be very different than that it
absorbs — Only requirement is that the net energy gain is zero, is it is in
equilibrium. AST 203 (Spring 2011) Blackbody A blackbody absorbs ALL incident
radiation —> black. If emission ¢ absorption, then T
would change —> no equilibrium. Stellar interior: atoms opaque to
radiation—absorb and reemit it. We see the stellar surface —>
stars act as good blackbodies. (Fimoozwmema) AST 203 (Spring 2011) Thermal Radiation 1433.2?
mm
.123 .100 uﬂr Seeing in IR maritime 3520 NA Elﬁn"I P A C Ashttpwmnlopsmos.ipac.caltech .ed u/cosmic_kids/learn_ir/ind ex. html Blackbodies infrared visibleUV Ximy Gammuimy 1015 As T increases,
the peak of the
emission moves to
higher frequencies
(shorter
wavelengths) 10m 105 Hotter objects
appear bluer. 107m “3715 105 10‘“ 10‘5 102” Notice that the higher temperature blackbody emits more radiation than the lower temperature blackbodies at all frequencies.
AST 203 (Spring 2011) Wien's Law Position of the peak emission moves to shorter A (higher u) with
increasing T. Wien's law:
AmaXT = 0.29 cm K = 2.9 X 106 nm K For a star: peak of its spectrum —> surface temperature Note, T of a star increases toward the center. AST 203 (Spl inc; 2011 i Wien's Law Consider the constellation Orion. Betelgeuse's spectrum peaks at 800 nm
(what color does it appear?) What is its temperature? 2£3X UT3HH11{
T:—:3600K
800nH1 Rigel's spectrum peaks at 220 nm, what
is its temperature? X 11m K (Mouservwniams)
T = — = 13000 K
220nH1 AST 203 (Spring 2011) Wien's Law
The average T of the earth is 300 K. What wavelength does it radiate in? Why don't we see it glowing? 2£lx UTSHH1I{ cr'
Amax : W : nm (NASA/Apollo11) What part of the electromagnetic spectrum is this? AST 203 (Spring 2011) RayleighJeans Law What's a first principles explanation of the blackbody spectrum? Classical physics: 2kTV2
2 I(V,T) : C Works well at low V, but does not work at higher frequencies. As we see, as V —> 00 this predicts an infinite amount of energy. ultraviolet catastrophe AST 203 (Spl ing 2011 ) Blackbod ies
10‘5— I Rludiol I I I lnfrhredlvisibleuvl >l<emyI I Gangtmuilroy I NOtice the
' RayleighJeans
law gets the slope right for low
frequencies. AST 203 (Spl ing 2011 i Planck's Law Planck: emperical formula that fits the entire blackbody spectrum 2hV3/62
[(y’ : Bhu/kT _ 1 For low frequencies, hy << kT, and we can use expand: 3132 m3 (B— _ _
e —1+£L‘+2! +3! —l—... and write
ehll/kZT _ 1 % h_y
kT resulting in [(y, T) m ZkTZ/2/c2 —RayleighJeans expression. Development required quantization of energy of photon:
E = nhv = nhc/A AST 203 (Spring 2011) Planck's Law Wavelength instead of frequency: ‘
I T _ 2hV3/02
(y’ )— Bhu/kT _ 1 Simply replace V with C/A—incorrect. Look at the definition of I so dE : L, cos 0dA d9 dt dz/ : I,\ cos QdA d9 dt dA Judy : IAdA AST 203 (Spring 2011) Planck's Law Since V = C/A , we have C
dl/ : —Ed/\ Minus sign: frequency increases = wavelength decrease. Ignoring this sign, we have I(/\, T) : [(u, T)%
or
2h02/A5
[(A’T) : ehc/AkT _ 1 Wien's law: define cc : hc/AkT and find maximum of 1(95) AsT 203 (Spring 2011) Planck's Law Note, to differentiate between a general intensity, [(y, T) , and a
blackbody, most books write 2hz/3/c2 B04 : Bhv/kT _ 1 AsT 203 (Spring 2011) Photons Photons are particles which are the “units” of light
Countable
Described by a frequency and wavelength Carries momentum and energy hc
E=h :—
V /\ AST 203 (Spring 2011) Intensity and Flux (Zeilik and Smith) Intensity has a direction, i.e. it is the energy/time/area/frequency
emitted per unit solid angle in a specific direction. Detectors measure the energy flux (erg s‘1 cm'2 Hz'1) hitting the
detector. Records energy hitting the detector area from all directions.
Frequency dependent—monochromatic flux. Integrate over all frequencies —> total flux (erg s‘1 cm'z). AST 203 (Spring 2011) Flux (Karttunen et al.) If we start with the definition of intensity,
dE : IV cos 6dA d1/ (19 (175 then we can write the fluX as: dE
f_/—dAdt _/IV cosddﬂdu Flux from a star: intensity = Planck function. We are interested in the outward emitted fluX from a patch dA, so f /dE fr /7r/2/OOB 0 ' 0d0d¢d
: — : V cos sm V
dAdt ¢=0 9:0 u=0 Note: we will not be tested on the details of this derivation,
but we will need to understand the end result. AST 203 (Spring 2011) Flux
The angular integration gives f—7r/ Budu =0 Integral over the frequency (we won't do this integral here) gives:
f : 0T4 This is called the StefanBoltzmann law. Here,
a = 5.67 X 10—5 erg 01111—2 K_4 53—1 (see Karttunen et al. for a detailed derivation). We will use this result often. AST 203 (Spring 2011) Luminosity and Flux
Note the very strong T dependence. If we double T, the radiated flux increases by 16x! Integrate over the stellar surface —> luminosity (erg/s) L : 47rR20T4 We measure this in erg 3'1. ln SI units, this is measured in Watts. What is the luminosity of a common light bulb? How hot does
the filament get? AST 203 (Spring 2011) Luminosity and Flux
100W=10"ergs‘1 The filament in the lightbulb is typically a thin wire coiled up, with
an overall surface area of about 1 cm2 The temperature is then
T Z A 1/4 2 109 erg 8—1 1/4
GA 5.67 X 10—5 erg s—1 cm—2 K—4 1 cm2 T m 2000 K AST 203 (Spring 2011) Luminosity and Flux Sun's spectrum peaks at 500 nm, Using Wien's law, this corresponds to a temperature of 2.9 x 106 nm K
T : — = 5800 K
500 nm The radius of the Sun is R9 : 7 x 1010 cm , giving a luminosity of
L : 47r(7 x 1010 em)2 (5.67 x 105 erg s’1 errr2 K41) (5800 K)4
: 3.9 x 1033 erg 8’1 We call this the solar luminosity and denote it as LG We will use this as a reference luminosity when we compare to
other stars. AST 2031;8pring 2011) Luminosity and Flux Luminosity is invariant with distance, but flux falls off as 1/r2 Measure the same star at two different distances, the flux will be
different. The flux of the Sun at its surface is
f = (5.67 x 105 erg s—1 errr—2 K—4) (5800 K)4 = 6.4 x 1010 erg s—1 em—2 Earth is 1.5 x 1013 cm (1 AU) away —> flux of the Sun at earth is Lo
f 7 47rd2 —2 : 1.4 X 106 erg 8—1 cm This is called the solar constant. AST 203 (Spring 2011) Flux and 1/r2 Closer planets intercept more energy per second
mmswplanet's area) Think about the
brightness of a lightbulb
as you move closer or
farther from it. The
power is not changing. Luminosity distributed over a
sphere. Move further from the Sun,
the energy flux (W/m2)
decreases as 1/r2 The same total luminosity
(4 x 1033 erg/s) is
distributed over the surface
of both spheres. Flux is L/(4Trr2) A Note About Fluxes  Flux just means energy / unit time / unit area  We've now talked about flux in 2 different contexts  Flux at the surface of a star: — Blackbody, f = 0T4
 Flux received from some distant star: — f = L/(47rr2) ,where 7’ is the distance to the star.  This is the flux that enters into the magnitude equation. AST 203 (Spring 2011) Parallax Apparent magnitudes: measure of how bright the star appears to
be from earth. To compare different stars, we need to know the luminosity of
the star. This requires that we know the distance to the star. If a star has a luminosity Land is at a distance d, then its
observed flux is
L* ﬂ : 47rd2 (This assumes that there is nothing in between the star and us
that absorbs some of its radiation). AST 203 (Spring 2011) Parallax We can measure the distance to nearby star
using simple trigonometric parallax. Where do we observe this in everyday life? Over 6 months, the baseline for the
measurements will be 2 AU We define the parallax angle to be 1/2
of the total shift. 1AU
tanpzpzT 1AU AST 203 (Spring 2011) Parallax Parallaxes are measured in arcseconds (1/3600 of a degree).
We denote the parallax angle in arcseconds as p” Recalling 7r rad : 180°, ,, 180° 60’ 60”
6( ) _ 6(1raid)7T rad 10 T _ 206265 9(rad)
so the distance to the star is Note: this is written ,7 different than your
d 2 206265 < book to make the
1 AU p units clearer. Distance at which a star has a parallax angle of 1” is the parsec.
1 pC : 206265 AU : 3.09 x 1018 cm : 3.261y We use: 1” d
1190 p AST 203 (Spl ing 2011i Parallax Parallax is the only direct way to
measure the distances to the stars. There are other distance indicators
built upon the distances that we know
from parallax measurements. The Hipparcos satellite measured
parallaxes down to 0.001”, finding the
distances to over 120,000 stars. AST 203 (Spring 2011) Proper Motion Stars orbit around the galactic center —> causes shift in the position of a star Motion of “j”
overtime: proper motion. 3333;“ 
Must subtract the proper motion of the _ 33m“??? _
star from the parallax effect. ., ,
Barnard's star has a proper motion of 10.3” per year. l —
AST203(Spring2011) a H m: m Absolute Magnitudes Absolute magnitude: magnitude if object were 10 pc away. A star whose luminosity is Land is a distance d away will have
a measured flux of f* and apparent magnitude m L*
ﬂ : 47rd2
if instead it were 10 pc away, its flux would be
L* _
f0=47rdg d0—10pc and its magnitude would be its absolute magnitude, M AST 203 (Spring 2011) (uonmdmo 5U!lS!lqnd Ibis) Absolute Magnitudes Putting this together, we have: f0 d2 d
— M _ 2. 1 — _ 2. 1 — _— 1
m 5 0g (J: 5 0g d8 5 0g 10 pc We call m — M the distance modulus. Given any two of m , M, or d, we can compute the third.
We can estimate M for a star given its observed properties. homework: you will derive a relationship between L and M. AsT 203 (Spring 2011) Absolute Magnitudes Consider a star that has an apparent magnitude of 0.4 and a
parallax of 0.3”—what is its absolute magnitude? First, we compute the distance in parsecs,
d 177 177
: — —> d : 3.33 pc 1 pc p 0.3” Next we can find the absolute magnitude, d
m—M:5log<10pc> d
M = m — 5 log : —0.4 — 5 log(0.333) : 2.0
10 pc AST 203 (Spring 2011) Stellar Colors We've been talking about the flux and magnitude over the entire
EM spectrum—all wavelengths or frequencies. Usually called the bolometric flux or magntiude. Instruments usually measure the flux over a small wavelength
band. A standard set of filters is defined, that have a central
wavelength and an associated width. This gives us information about the color of a star. AST 203 (Spring 2011) Stellar Colors U B V 1.0 (emso one "01:20) 0.8 0.6 0.4 Sensitivity function S(/\) 0.2 0.0 l . l .
300 400 500 600 700 Wavelength (nm) Note that these filters have a rather complex shape. AST 203 (Spring 2011) Stellar Colors
fB =fux through the B filter
fv = fluX through the V filter magnitude difference: mB — my 2 2.5log B (usually written as B — V) This is a measure of the color of the star. Stars are good
blackbodies, so this determines of the temperature of the star. As T increases, fB/fv increases, so B — V decreases (because
the magnitude scale runs backwards). AST 203 (Spring 2011 :1 Stellar Colors Table 9.2. Spectral type, color, and
effective temperature.3 Main sequence Giants
Spectral type B — V T‘, (K) B — V Te (K)
05 70.45 35,000 i —
BO 7031 21,000 — 7
BS *0.17 13,500 — —
A0 0.00 9,700 — —
A5 0.16 8,100 — —
F0 0.30 7,200 — —
F5 0.45 6,500 — —
GO 0.57 6,000 0.65 5,400
G5 0.70 5,400 0.84 4,700
KO 084 4,700 1.06 4,100
K5 1.11 4,000 1.40 3.500
M0 1.24 3,300 1,65 2,900
M5 1.61 2,500 — i ‘ Adapted from C. W . Allen. Astrophysical Quantities.
h“...— (Shu) AST 203 (Spring 2011) ...
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