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 , 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

Next Time

Read: 5.1-5.3 Homework: 3.9e-3.9g, 3.17, 5.1, 5.4