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Black Body Radiation
--------------------
Black body radiation is the type of electromagnetic radiation.
It can either be within or surround a body in thermodynamic
equilibrium with its environment. It can also be emitted by
an opaque and non-reflective 'black' body held at constant,
uniform temperature. The radiation has a specific spectrum
and intensity that depends only on the temperature of the
body.
Consider electromagnetic radiation inside a cavity with sides
L. The standing waves in the cavity have to satisfy the wave
equation ∇^{2}E = (1/c^{2})∂^{2}/∂t^{2}. With solution:
E = E_{0}sin(2πx/λ)sin(2πy/λ)sin(2πz/λ)sin(2πct/λ)
This can be written as:
E = E_{0}sin(k_{x}x)sin(k_{y}y)sin(k_{z}z)sin(2πct/λ)
where k = 2π/λ
The solution must give zero amplitude at the walls, since a
non-zero value would dissipate energy and violate the condition
of equilibrium. This requirement can be met by:
L = nλ/2 ∴ λ = 2L/n
Now k = 2π/λ
= nπ/L
In 3D we can write:
k^{2} = k_{x}^{2} + k_{y}^{2} + k_{z}^{2} = (n_{x}^{2} + n_{y}^{2} + n_{z}^{2})π^{2}/L^{2}
_{ } = n^{2}π^{2}/L^{2}
The number of states in n space occupies 1/8 of a sphere.
The volume of a shell of radius n and thickness dn is:
N(n)dn = (1/8)4πn^{2}dn
Now n^{2} = k^{2}L^{2}/π^{2} so n = kL/π and dn = Ldk/π
N(k)dk = (1/8)(4πk^{2}L^{2}/π^{2})(L/π)dk
= (1/2π^{2})L^{3}k^{2}dk
= (1/2π^{2})(Vω^{2}/c^{3})dω since ω = ck and dk = dω/c
Now there are 2 polarizations allowed for the photon therefore,
N(ω)dω = (Vω^{2}/π^{2}c^{3})dω
Therefore, the energy density of radiation per unit frequency
using the classical harmonic oscillator is:
E_{ω} = (ω^{2}/π^{2}c^{3})(K_{B}T)
This is the classical RAYLEIGH-JEANS law that says the energy is
proportional to the square of the frequency. Since there was no
reason to believe that their existed a maximum frequency inside
the cavity, the energy would increase towards infinity as the
frequency is increased. However, experimentally this did not
happen. This was referred to as the ULTRAVIOLET CATASTROPHE.
It wasn't until the advent of Quantum Mechanics that this paradox
was solved. If we replace the classical oscillator with the
quantum oscillator we get:
E_{ω} = (hω^{3}/π^{2}c^{3})(1/(exp(βhω) - 1))
This is PLANCK'S RADIATION LAW.
A comparison between the Rayleigh-Jeans law and Planck's
law is shown below.
We can find the total energy density by integrating the Planck
formula over all frequencies:
E_{Total} = (h/π^{2}c^{3})∫ω^{3}/(exp(βhω) - 1)dω
If we make the substitution x = hβω and dx = hβdω we can write:
E_{Total} = (h/π^{2}c^{3})(1/hβ)^{4}∫x^{3}/(exp(x) - 1)dx
This is a standard integral equal to π^{4}/15. Therefore, we get:
E_{Total} = (h/π^{2}c^{3})(1/hβ)^{4}(π^{4}/15)
_{ } = {π^{2}K_{B}^{4}/15c^{3}h^{3}}T^{4}
_{ } = σT^{4}
This is the STEFAN-BOLTZMANN LAW. It represents the total
energy radiated per unit surface area of a black body across
all wavelengths per unit time. To get the total power it is
necessary to multiply the above equation by the area, A.
Return to the Planck radiation law:
E_{ω} = (hω^{3}/π^{2}c^{3})(1/(exp(βhω) - 1))
Now ω = 2πc/λ and dω = -2πc/λ^{2} so we can write
E_{λ} = (16π^{2}hc/λ^{5})(1/(exp(βh2πc/λ) - 1))
_{ } = (8πhc/λ^{5})(1/(exp(hc/K_{B}Tλ) - 1))
If we differentiate this w.r.t. λ and set the result = 0 we get
the following relationship:
λT = (hc/K_{B})/(5[1 - exp(-hc/K_{B}/λT)]
Which must be solved numerically to give:
λ_{peak}T = 2.898 x 10^{-3} meter Kelvin
This is WIEN'S DISPLACEMENT LAW. This law is useful for
determining the temperature of stars by looking at their
spectra. The observed temperature will always be slightly
less than the actual temperature because the photons have
to overcome the gravitational field of the emitting object.