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2.57 Fall 2004 – Lecture 11
1
2.57 NanotoMacro Transport Processes
Fall 2004
Lecture 11
4.1.1 Photons (continue)
First let us continue the discussion of photons. We have
/
1
(,)
1
B
kT
nfT
e
ω
==
−
=
.
The internal energy is
0
(,)()
xyz
kkk
uf
T
f
T
D
d
ωω
∞
∑∑∑
∫
,
where the density of states (two transverse electromagnetic waves) is
22
3
23
4
() 2
;
2
kd
k
Dc
k
c
d
L
πω
π
=×
=
=
⎛⎞
⎜⎟
⎝⎠
for light.
Thus
()
2
00
ω
1
ω
π
c
exp
ω
/
κ
T1
B
ud
u
d
∞∞
⎡⎤
−
⎣⎦
∫∫
=
=
,
in which we define
(
)
3
ω
ω
f
ω
,T
ω
D
ω
π
c
exp
ω
/
κ
B
u
−
=
=
=
.
Recall the Planck’s law as
2
1
,
/
5
(1
)
b
cT
c
e
e
λ
=
−
,
which has
5
but in
ω
u
we have
3
ω
. Noticing
uu
∆
=
∆
and
2/
ck
c
πλ
, we
obtain
2
2
dc
uuu
d
ωπ
λλ
, which accounts for the power difference in above two
expressions. Therefore, we have obtained the exact Planck’s blackbody radiation law.
The radiation intensity, defined as energy flux per unit solid angle and normal area, is
calculated as
4
cu
I
=
,
where 4
is the full solid angle (for sphere
2
/4
Area R
Ω=
=
), c is light speed.
With
4
0
4
d
T
c
σ
∞
∫
, the blackbody emissive power is expressed as
4
b
eI
T
, where
82
4
5.67 10
/
Wm K
−
⋅
.
4.1.2 Phonons
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2
In the Debye model, we assume
D
vk
ω
=
. The maximum frequency is thus
DD
v
a
π
=
,
which is different from original
ω
max.
For phonons, we have three polarizations (two
transverse, one longitudinal waves). Similar to photons, we have
()
3
D
2
2
v
2
•
3
Vd
dN
D
π
ω
=
ω
=
ω
.
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This note was uploaded on 01/12/2011 for the course ME 305 taught by Professor Wright,j during the Spring '10 term at Birla Institute of Technology & Science, Pilani  Hyderabad.
 Spring '10
 WRIGHT,J

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