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Radiation PHYS390 (Astrophysics) Professor Lee Carkner Lecture 3
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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
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Questions 3) Which looks brighter, a star with m bol = 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 m bol =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)
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Light Properties Light is both a particle and a wave Where: c = 3X10 8 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, or E
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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
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Wien’s Law Given by Wien’s Law: max T = 0.002897755 m K Since short wavelengths look blue and long red: Blue stars = Red stars =
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Stefan-Boltzmann Stars are spheres, so A = 4 R 2 L = 4 R 2 T 4 is the Stefan-Boltzmann constant =5.67X10 -8 W m -2 K -4
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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 T e The temperature of a perfect blackbody that emits the same amount of energy as the star
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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 B (T) = (2ckT)/ 4 Where k = 1.38X10 -23 J/K Leads to ultraviolet catastrophe Energy goes to infinity as wavelengths get shorter
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Planck Function but only if he assumed that energy was quantized (h ) Result: B (T) = (2hc 2 / 5 )/[e (hc/ kT) -1] Energy per unit time per unit wavelength interval per unit solid angle
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Planck Function and Luminosity Called the monochromatic luminosity, L d L d = (8 R 2 hc 2 / 5 )/[e (hc/ kT) -1] d If we divide by the inverse square law we get the monochromatic flux, F d F d = (L /4 r 2 ) d which is the flux for the small wavelength range d
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Next Time Read: 5.1-5.3 Homework: 3.9e-3.9g, 3.17, 5.1, 5.4
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