5. Thermal radiation
•These are humans seen by thermography
•Depending on human body temperature, the color is different (Red is hotter, blue is colder)
6. Temperature
•We first begins to consider “Temperature(温度)”
•We usually use degree(℃) as measure of temperature.
•In physics or astronomy, we often use Kelvin Temperature
instead of degrees
K(Kelvin) = degree Celsius (℃)＋273
For example, 0℃ is 273K, and water freezes.
•All thermal motion stopps at 0K (-273℃)
•Water boils at 373K(100℃)
8. Thermal radiation
•All macroscopic objects, such as fire, ice, people, and stars, emit radiation at all times.
•Thermal radiation（热辐射） is electromagnetic radiation(电磁波) generated by
the thermal motion of particles in matter.
Hot!!
9. Thermal radiation
Thermal energy is converted to electromagnetic
wave.
What is the relation between thermal
energy and electromagnetic wave?
Let’s consider it !
热🔥 电磁波
10. The blackbody spectrum
热🔥 电磁波 What is the relation between thermal
energy and electromagnetic wave?
Blackbody(⿊体谱): an object that absorbs all radiation and reemits the same amount of
energy it absorbs.
In order to consider this question, we consider blackbody (⿊体谱)
•The intensity(强度) of the blackbody depends
on frequency and temperature
热🔥
电磁波
14. The blackbody spectrum
•Once we determine temperature, the intensity of the blackbody is a function of frequency.
15. The blackbody spectrum
•Once we determine temperature, the intensity of the blackbody is a function of frequency.
16. The blackbody spectrum
•Once we determine temperature, the intensity of the blackbody is a function of frequency.
17. The blackbody spectrum
•Once we determine temperature, the intensity of the blackbody is a function of frequency.
18. The blackbody spectrum
•Once we determine temperature, the intensity of the blackbody is a function of frequency.
19. The blackbody spectrum
I(ν, T) =
2hν3
c2
1
ehν/kT − 1
The blackbody spectrum is expressed as
h : Planck constant
ν : frequency
T : temperature
•If frequency is low ( ),
hν/kT ≪ 1 I(ν, T) ∼
2ν2
kT
c2
k : Boltzmann constant
•If frequency is high ( ),
hν/kT ≫ 1 I(ν, T) ∼
2hν3
c2
exp
(
−
hν
kT )
20. •We have useful law to connect wavelength at the peak of intensity and temperature
The blackbody spectrum
λmax =
2.9 × 10−3
[m]
T
Wien’s law
21. •We have useful law to connect wavelength at the peak of intensity and temperature
The blackbody spectrum
λmax =
2.9 × 10−3
[m]
T
Wien’s law
22. •We have useful law to connect wavelength at the peak of intensity and temperature
The blackbody spectrum
λmax =
2.9 × 10−3
[m]
T
Wien’s law
λmax =
2.9 × 10−3
[m]
300
= 9.6 × 10−6
[m]
= 9600nm
23. •We have useful law to connect wavelength at the peak of intensity and temperature
The blackbody spectrum
λmax =
2.9 × 10−3
[m]
T
Wien’s law
λmax =
2.9 × 10−3
[m]
300
= 9.6 × 10−6
[m]
= 9600nm
24. Radiation law
•The total energy per unit area is given by Stefan-
Boltzmann law
•The total amount of energy depends on only its
temperature !
•In the case of blackbody radiation, we can calculate
total energy amount of radiation
If we measure the energy from stars or galaxies, we
can know their temperature of them!
25. Summary of the blackbody spectrum
Q:How do we measure the temperature of the stars and galaxies?
Remember our first question !
Answer
•The electromagnetic wave is related to temperature
via blackbody radiation.
热🔥 电磁波
•The energy of electromagnetic is determined by the
temperature in the case of a blackbody(Stefan-
Boltzmann law). Thus, we can determine the
temperature from energy.
F = σT4
[W/m2]
26. Temperature of the earth
F = σT4
[W/m2]
Let’s roughly calculate the temperature of the earth!
: The radius of sun,
R⊙ 7 × 108
m
1AU=
r = 1.5 × 1011
m
Tsun : The temperature of the sun,6000K
Rearth = 6 × 103
m
Hint 2:
(The energy which the earth absorbs)=(The
energy which the earth emits)
*We assume the earth reflects 30% of radiation
from the Sun.
Hint1: The energy which arrives at the earth
from the sun is L⊙/4πr2
Tearth = ?
27. Temperature of the earth
F = σT4
[W/m2]
Let’s roughly calculate the temperature of the earth!
: The radius of sun,
R⊙ 7 × 108
m
1AU=
r = 1.5 × 1011
m
Tsun : The temperature of the sun,6000K
Rearth = 6 × 103
m
Hint 2:
(The energy which the earth absorbs)=(The
energy which the earth emits)
*We assume the earth reflects 30% of radiation
from the Sun.
Hint1: The energy which arrives at the earth
from the sun is L⊙/4πr2
Tearth ∼ 265K ∼ − 8∘
C
Tearth = ?
28. Astronomical applications
(a) A cool dark galactic gas, the temperature is
around 60K.
(b) A young star with T=600K
(c)The sun with T=6000K.
(d)The Andromeda galaxy with T=60,000K
29. Summary
• In the astronomy, we use electromagnetic wave to
observe astrophysical objects and obtain information.
• The electromagnetic wave consists of radio waves,
infrared radiation, visible light, ultraviolet radiation, X-rays
and gamma-rays.
• Blackbody radiation is related to temperature. Given
temperature, blackbody curve is determined and we
measure intensity at given frequency.
