Download Lesson 7 - Blackbody Radiation and Luminosity


Recall that light is EM
radiation and is therefore
characterized by its
wavelength.



Imagine we had the “perfect”
emitter of EM radiation. We
would call that a “Blackbody”.
The Sun (and all stars) give
off nearly every wavelength
of EM radiation.
But what wavelength
dominates?????
During this Solar Eclipse (1991)
you can see the Corona of the
Sun. It is extends hundreds of
thousands of Kilometers into
space

The simplest and most
common way to produce
EM radiation is to heat up
an object.

Lets consider the
welder here.

As the temperature
increases what can we
say about ...
 The
color?
 The amount of Heat?
BLACKBODY CURVES

The Blackbody curve for
the Sun

When we compare the
curve for a star to the
curve of a blackbody at
a given temp, we can in
effect...
Take the temperature of a
star from millions/billions
of miles away!!!

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The equations that
describe these curves are
rather complicated.
However, we can deduce
2 useful expressions. The
first one relates
temperature to wavelength
of maximum emission.
aka Wien’s Displacement
Law
(PSRT)
2.9x10-3
max (meters) 
T ( kelvin)

Example: Determine
the Surface
Temperature of the
Sun.

Assume λmax is 500 nm
(PSRT)
2.9x10-3
max (meters) 
T ( kelvin)
Thought Experiment. Consider a hot
metal rod and a burning match at the
same temp.
 Which gives off more energy???
 We must consider the amount of energy
emitted per surface area. ENERGY FLUX


The energy flux is just a fancy way of saying
“The amount of energy emitted from 1m2 of an
object’s surface per second.”
[J/m2/sec] or [W/m2]


So for the match and metal bar, which one has a
higher energy flux? Neither. But clearly the
match will not do as much damage to your skin.
So what’s the difference?

In 1879, Josef Stefan
showed that

Five years later Ludwig
Boltzmann derived the
coefficient that related the
2 quantities.

Recall Luminosity is the
power output of a star.
Therefore....
Energy Flux  T
4
4
4
Energy
Flux


T
L  AT
-8
2 4
where

=
5.67
x
10
W
/
m
K
where
 = 5.67 x 10-8 W / m2 K4
L = luminosity
A = cross sectional area
T = temp in Kekvin
Sample problem: F1 (c) M02 exam
Antares A is part of a binary star system. The companion
star Antares B has a surface temperature of 15 000 K and a
luminosity that is 1/40 of that of Antares A. Calculate the
ratio of the radius of Antares A to Antares B.
Antares A has a surface temperature of 3000 K. This info
was provided in an earlier part of the problem (not stated).
In addition, you must know the formula for the surface area
of a sphere to solve these types of problems.
A  4 r
2
LA
LB 
use Stefan-Boltzmann Law
40
2
4

(4

r
)
T
A
A
 (4 rB2 )TB4 
40
2
4
2
4
40rB (15000)  rA (3000)
2.025 10 r  8.110 r
18
2
B
rA
 160 (2 SF)
rB
13
2
A

Worksheet on
 Apparent
Brightness
 Wien’s Law
 SB Law