radiation phys390 (astrophysics) professor lee carkner lecture 3
Post on 22-Dec-2015
214 Views
Preview:
TRANSCRIPT
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 , 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:
maxT = 0.002897755 m K
Since short wavelengths look blue and long red: Blue stars = Red stars =
Stefan-Boltzmann
Stars are spheres, so A = 4R2
L = 4R2T4
is the Stefan-Boltzmann constant
=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
B(T) = (2ckT)/4
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 (h)
Result:
B(T) = (2hc2/5)/[e(hc/kT)-1]
Energy per unit time per unit wavelength interval per unit solid angle
Planck Function and Luminosity
Called the monochromatic luminosity, L dL d = (8R2hc2/5)/[e(hc/kT)-1] d
If we divide by the inverse square law we get the monochromatic flux, F d
F d = (L/4r2) d which is the flux for the small wavelength range d
top related