derive expression for stefan boltzaman law of radiation
Answers
Answer:
B(T)=∫∞02hc2(kTxh)3(1ex-1)(kTh)x=2k4T4c2h3∫∞0x3dxex-1. 11-z. ∫∞0x3dxex-1=∫∞0x3(∞∑m=1e-mx)x.
Essential Radio Astronomy
A Fourier TransformsC Special Relativity
Appendix B
Mathematical Derivations
B.1 Evaluation of Planck’s Sum
Planck’s sum (Equation 2.83) for the average energy per mode of blackbody radiation is
⟨E⟩=
∞
∑
n=0 nhνexp(-
nhν
kT
)
∞
∑
n=0 exp(-
nhν
kT
)
.
It is convenient to introduce the variable α≡1/(kT), so
⟨E⟩=
∞
∑
n=0 nhνexp(-αnhν)
∞
∑
n=0 exp(-αnhν)
.
Next consider the quantity
-
d
dα
[ln
∞
∑
n=0 exp(-αnhν)].
Using the chain rule to take the derivative yields
-
d
dα
[ln
∞
∑
n=0 exp(-αnhν)] =-[
∞
∑
n=0 exp(-αnhν)]-1
d
dα
[
∞
∑
n=0 exp(-αnhν)]
=
∞
∑
n=0 nhνexp(-αnhν)
∞
∑
n=0 exp(-αnhν)
.
Thus
⟨E⟩=-
d
dα
[ln
∞
∑
n=0 exp(-αnhν)].
Then,
∞
∑
n=0 exp(-αnhν)=1+[exp(-αhν)]1+[exp(-αhν)]2+⋯
has the form 1+x+x2+⋯=(1-x)-1, so
∞
∑
n=0 exp(-αnhν)=[1-exp(-αhν)]-1
and
⟨E⟩ =-
d
ln[1-exp(αhν)]-1
dα
=-[1-exp(-αhν)](-1)[1-exp(-αhν)]-2hνexp(-αhν)
=
hνexp(-αhν)
1-exp(-αhν)
=
hν
exp(αhν)-1
=
hν
exp(
hν
kT
)-1
.
B.2 Derivation of the Stefan–Boltzmann Law
The Stefan–Boltzmann law for the integrated brightness of blackbody radiation at temperature T (Equation 2.89) is
B(T)=∫
∞
0
Bν(T)ν=
σT4
π
,
where
Bν(T)=
2hν3
c2
1
exp(
hν
kT
)-1
is Planck’s law and σ is the Stefan–Boltzmann constant. Although the Stefan–Boltzmann law and constant were first determined experimentally, both can be derived mathematically from Planck’s law. For simplicity, define
x≡
hν
kT
,
so
B(T)=∫
∞
0
2h
c2
(
kTx
h
)3(
1
ex-1
)(
kT
h
)x=
2k4T4
c2h3
∫
∞
0
x3dx
ex-1
.
The quantity
1
ex-1
=
e-x
1-e-x
=e-x(
1
1-e-x
)
can be expanded in terms of the infinite series
∞
∑
m=0 zm= 1+z+z2+z3+⋯
= 1+z(1+z+z2+z3+⋯)
= 1+z
∞
∑
m=0 zm,
∞
∑
m=0 zm=
1
1-z
.
Thus
1
ex-1
=e-x
∞
∑
m=0 e-mx=e-x+e-2x+e-3x+⋯
and the integral becomes
∫
∞
0
x3dx
ex-1
=∫
∞
0
x3(
∞
∑
m=1 e-mx)x.
Each integral in this series can be integrated by parts three times:
∫
∞
0
x3e-mxx =
x3e-mx
-m
|
∞
0
-∫
∞
0
3x2e-mx
-m
x=
3
m
∫
∞
0
x2e-mxx,
∫
∞
0
x2e-mxx =
x2e-mx
-m
|
∞
0
-∫
∞
0
2xe-mx
-m
x=
2
m
∫
∞
0
xe-mxx,
∫
∞
0
xe-mxx =
xe-mx
-m
|
∞
0
-∫
∞
0
xe-mx
-m
x=
1
m
∫
∞
0
e-mxx=
1
m2
,
to give
∫
∞
0
x3e-mxx=
6
m4
and
∫
∞
0
x3dx
ex-1
=∫
∞
0
x3(
∞
∑
m=1 e-mx)x=6
∞
∑
m=1
1
m4
.
The sum
∞
∑
m=1
1
m4
=
1
14
+
1
24
+
1
34
+
1
44
+⋯=1+
1
16
+
1
81
+
1
256
+⋯≈1.082
converges quickly and is the value of the Riemann zeta function ζ(4)=π4/90≈1.082. Thus
∫
∞
0
x3dx
ex-1
=
π4
15
.
(B.1)
Finally, the integrated brightness of blackbody radiation is
B(T)=
2k4T4
c2h3
∫
∞
0
x3dx
ex-1
=
2k4T4
c2h3
(
π4
15
)=
2π4k4
15c2h3
T4=
σT4
π
,
so
σ=
2π5k4
15c2h3
≈5.67×
10-5
erg
cm
2
s
K4
(sr)
is the value of the Stefan–Boltzmann constant.
Similarly, the integral
∫
∞
0
x2dx
ex-1
is needed to evaluate the number density nγ of blackbody photons:
nγ=
8π
c3
∫
∞
0
ν2dν
exp(
hν
kT
)-1
=
8π
c3
(
kT
h
)3∫
∞
0
x2dx
ex-1
.
Following the derivation above,
∫
∞
0
x2dx
ex-1
=∫
∞
0
x2(
∞
∑
m=1 e-mx)x
and
∫
∞
0
x2e-mxx=
2
m
∫
∞
0
xe-mxx=
2
m
(
1
m2
)=
2
m3
,
so
∫
∞
0
x2dx
ex-1
=2
∞
∑
m=1
1
m3
=2(
1
13
+
1
23
+
1
33
+⋯)≈2.404.