l max T = 0.002897755 m K
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Transcript l max T = 0.002897755 m K
Radiation
PHYS390 (Astrophysics)
Professor Lee Carkner
Lecture 3
Questions
1) Which would look brighter, a star with
magnitude 20 or a star 100 times brighter than
magnitude 26?
Answer: m=20
Explain: 100 times brighter than m=26 is m=21.
Smaller magnitude brighter
2) Which would look brighter, a star with m=10
or a star that has M=10 and is at 20 pc?
Answer: m=10
Explain: A M=10 star would have m=10 at 10pc and
be fainter than m=10 at 20pc
Questions
3) Which looks brighter, a star with mbol = 10 or a star
with V = 10?
Answer: V=10
Explain: the V=10 star has the same luminosity in just one
band as the mbol=10 star has over all wavelengths, so if you
include the bands other than V it looks brighter
4) Which looks brighter, a star with B = 10 or a star
with V = 10?
Answer: It depends
Explain: It depends on the shape of the blackbody curve. The
B=10 star might be brighter than 10 in the V band (if it is a red
star) or fainter (if it is a blue star)
Light Properties
Light is both a particle and a wave
Where:
c = 3X108 m/s
h = 6.626X10-34 J s
Long wavelength (low energy) –
Short wavelength (high energy) –
We can often think of light as a stream of
photons, each with an l, n or E
Blackbody Curve
Blackbodies have a
very specific emission
spectrum
A rapid fall off to short
wavelengths
Gradual Rayleigh-Jeans
tail to long wavelengths
Higher temperature
means more total
emission and peak at
shorter wavelengths
Wien’s Law
Given by Wien’s Law:
lmaxT = 0.002897755 m K
Since short wavelengths
look blue and long red:
Blue stars =
Red stars =
Stefan-Boltzmann
Stars are spheres, so A = 4pR2
L = 4pR2sT4
s is the Stefan-Boltzmann constant
s=5.67X10-8 W m-2 K-4
Stefan-Boltzmann and Stars
T is more important than R for determining L
If we know L and T, we can find R
Stars are not perfect blackbodies so we often
write T in the equation as Te
The temperature of a perfect blackbody that emits the
same amount of energy as the star
The Blackbody Curve
We need an equation for the shape of the
blackbody curve
Blackbody curve as a function of wavelength due to
temperature T
Bl(T) = (2ckT)/l4
Where k = 1.38X10-23 J/K
Leads to ultraviolet catastrophe
Energy goes to infinity as wavelengths get shorter
Planck Function
but only if he assumed that energy was
quantized (hn)
Result:
Bl(T) = (2hc2/l5)/[e(hc/lkT)-1]
Energy per unit time per unit wavelength
interval per unit solid angle
Planck Function and Luminosity
Called the monochromatic luminosity, Ll dl
Ll dl = (8pR2hc2/l5)/[e(hc/lkT)-1] dl
If we divide by the inverse square law we get the
monochromatic flux, Fl dl
Fl dl = (Ll/4pr2) dl
which is the flux for the small wavelength range dl
Next Time
Read: 5.1-5.3
Homework: 3.9e-3.9g, 3.17, 5.1, 5.4