1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
|
---
title: "Planck's law"
firstLetter: "P"
publishDate: 2021-09-09
categories:
- Physics
date: 2021-09-09T08:12:14+02:00
draft: false
markup: pandoc
---
# Planck's law
**Planck's law** describes the radiation spectrum of a **black body**:
a theoretical object in thermal equilibrium,
which absorbs photons,
re-radiates them, and then re-absorbs them.
Since the photon population varies with time,
this is a [grand canonical ensemble](/know/concept/grand-canonical-ensemble/),
and photons are bosons
(see [Pauli exclusion principle](/know/concept/pauli-exclusion-principle/)),
this system must obey the
[Bose-Einstein distribution](/know/concept/bose-einstein-distribution/),
with a chemical potential $\mu = 0$ (due to the freely varying population):
$$\begin{aligned}
f_B(E)
= \frac{1}{\exp\!(\beta E) - 1}
\end{aligned}$$
Each photon has an energy $E = \hbar \omega = \hbar c k$,
so the [density of states](/know/concept/density-of-states/)
is as follows in 3D:
$$\begin{aligned}
g(E)
= 2 \frac{g(k)}{E'(k)}
= \frac{V k^2}{\pi^2 \hbar c}
= \frac{V E^2}{\pi^2 \hbar^3 c^3}
= \frac{8 \pi V E^2}{h^3 c^3}
\end{aligned}$$
Where the factor of $2$ accounts for the photon's polarization degeneracy.
We thus expect that the number of photons $N(E)$
with an energy between $E$ and $E + \dd{E}$ is given by:
$$\begin{aligned}
N(E) \dd{E}
= f_B(E) \: g(E) \dd{E}
= \frac{8 \pi V}{h^3 c^3} \frac{E^2}{\exp\!(\beta E) - 1} \dd{E}
\end{aligned}$$
By substituting $E = h \nu$, we find that the number of photons $N(\nu)$
with a frequency between $\nu$ and $\nu + \dd{\nu}$ must be as follows:
$$\begin{aligned}
N(\nu) \dd{\nu}
= \frac{8 \pi V}{c^3} \frac{\nu^2}{\exp\!(\beta h \nu) - 1} \dd{\nu}
\end{aligned}$$
Multiplying by the energy $h \nu$ yields the distribution of the radiated energy,
which we divide by the volume $V$ to get Planck's law,
also called the **Plank distribution**,
describing a black body's radiated spectral energy density per unit volume:
$$\begin{aligned}
\boxed{
u(\nu)
= \frac{8 \pi h}{c^3} \frac{\nu^3}{\exp\!(\beta h \nu) - 1}
}
\end{aligned}$$
## Wien's displacement law
The Planck distribution peaks at a particular frequency $\nu_{\mathrm{max}}$,
which can be found by solving the following equation for $\nu$:
$$\begin{aligned}
0
= u'(\nu)
\quad \implies \quad
0
= 3 \nu^2 (\exp\!(\beta h \nu) - 1) - \nu^3 \beta h \exp\!(\beta h \nu)
\end{aligned}$$
By defining $x \equiv \beta h \nu_{\mathrm{max}}$,
this turns into the following transcendental equation:
$$\begin{aligned}
3
= (3 - x) \exp\!(x)
\end{aligned}$$
Whose numerical solution leads to **Wien's displacement law**, given by:
$$\begin{aligned}
\boxed{
\frac{h \nu_{\mathrm{max}}}{k_B T}
\approx 2.822
}
\end{aligned}$$
Which states that the peak frequency $\nu_{\mathrm{max}}$
is proportional to the temperature $T$.
## Stefan-Boltzmann law
Because $u(\nu)$ represents the radiated spectral energy density,
we can find the total radiated energy $U$ per unit volume by integrating over $\nu$:
$$\begin{aligned}
U
&= \int_0^\infty u(\nu) \dd{\nu}
= \frac{8 \pi h}{c^3} \int_0^\infty \frac{\nu^3}{\exp\!(\beta h \nu) - 1} \dd{\nu}
\\
&= \frac{8 \pi h}{\beta^3 h^3 c^3} \int_0^\infty \frac{(\beta h \nu)^3}{\exp\!(\beta h \nu) - 1} \dd{\nu}
= \frac{8 \pi}{\beta^4 h^3 c^3} \int_0^\infty \frac{x^3}{\exp\!(x) - 1} \dd{x}
\end{aligned}$$
This definite integral turns out to be $\pi^4/15$,
leading us to the **Stefan-Boltzmann law**,
which states that the radiated energy is proportional to $T^4$:
$$\begin{aligned}
\boxed{
U = \frac{4 \sigma}{c} T^4
}
\end{aligned}$$
Where $\sigma$ is the **Stefan-Boltzmann constant**, which is defined as follows:
$$\begin{aligned}
\sigma
\equiv \frac{2 \pi^5 k_B^4}{15 c^2 h^3}
\end{aligned}$$
## References
1. H. Gould, J. Tobochnik,
*Statistical and thermal physics*, 2nd edition,
Princeton.
|
