Wolfram Alpha:

```Blackbody Radiation
-------------------

Blackbody radiation is a type of electromagnetic
radiation emitted by a blackbody (a hypothetical
body that completely absorbs all radiant energy
falling upon it, reaches some equilibrium temperature,
and then reemits that energy as quickly as it absorbs
it).  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.

The radiated energy can be considered to be produced
by standing waves or resonant modes of the cavity
inside a cavity with sides, L.  The standing waves
in the cavity have to satisfy the wave equation:

∇2E = (1/c2)∂2/∂t2.  With solution:

E = E0sin(2πx/λ)sin(2πy/λ)sin(2πz/λ)sin(2πct/λ)

This can be written as:

E = E0sin(kxx)sin(kyy)sin(kzz)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 condition
can be met by:

L = nλ/2 ∴ λ = 2L/n

Now k = 2π/λ

= nπ/L

In 3D we can write:

k2 = kx2 + ky2 + kz2 = (nx2 + ny2 + nz2)π2/L2

= n2π2/L2

The number of states, N, 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)4πn2dn

Now n2 = k2L2/π2 so n = kL/π and dn = Ldk/π

N(k)dk = (1/8)(4πk2L2/π2)(L/π)dk

= (1/2π2)L3k2dk

N(k) = (1/2π2)L3∫k2dk

= (1/6π2)L3k3

ω = ck gives:

N(ω) = (1/6π2)(L3/c3)ω3

ω = 2πν gives:

N(ν) = (4π/3)(L3/c3)ν3

k = 2π/λ gives:

N(λ) = 4πL3/3λ3

Now there are 2 polarizations allowed for the
photon.  Therefore, the constant needs to be
multiplied by 2:

N(ω) = (1/3π2)(L3/c3)ω3

N(ν) = (8π/3)(L3/c3)ν3

N(λ) = 8πL3/3λ3

This is the total number of modes.  To get the
distribution by ω, ν or λ it is necessary to
take the derivative of the number of modes with
respect to ω, ν or λ (i.e., the rates of change).
Therefore:

dN(ω)/dω = (1/π2)(L3/c3)ω2

dN(ν)/dν = (8πL3/c3)ν2

check:  ω = 2πν and dω = 2πdν

∴ (1/2π)dN(ν)/dν = (1/π2)(L3/c3)4π2ν2

∴ dN(ν)/dν = (8πL3/c3)ν2

dN(λ)/dλ = -8πL3/λ4

check:  ν = c/λ and dν = -c/λ2dλ

∴ (-λ2/c)dN(λ)/dλ = (8π)(L3/c3)c2/λ2

∴ dN(dλ)/dλ = -8πL3/λ4

In the latter case, the negative sign shows that
the number of modes decreases with increasing
wavelength.

Equipartion of Energy
---------------------

The theorem of equipartition of energy states
that each mode of radiation associated with a
classical harmonic oscillator has an energy equal
to KBT.  Therefore, the energy of the radiation
per unit ω, ν or λ per unit volume (≡ energy
density per unit ω, ν or λ) is:

Eω = (ω2/π2c3)(KBT)

or,

Eν = (8πν2/c3)(KBT)

or,

Eλ = (8π/λ4)(KBT))

Dimensional check:

LHS:  [E/ωV] = [mL2/t2/(1/t)L3] = [m/Lt]

RHS: [(mL2/t2)(1/t2)/L3/t3] = [m/Lt]

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.  Planck assumed
that the sources of radiation are atoms in a
state of oscillation and that the vibrational
energy of each oscillator may have any of a
series of discrete values but never any value
between. Planck further assumed that when an
oscillator changes from a higher state of energy
to a lower state of energy a discrete quantum of
radiation given by E = hν is emitted.

Planck was able to calculate the value of h from
experimental data.  His result, 6.55 x 10-34 J/s,
is within 1.2% of the currently accepted value. He
was also able to make the first determination of
the Boltzmann constant KB from the same data and
theory.

If we replace the classical oscillator with the
quantum oscillator we get:

KBT -> hν/(exp(hν/KBT) - 1)

Which results in:

Eω = (ω2/π2c3)[hω/((exp(hω/KBT) - 1)]

or,

Eν = (8πν2/c3)[hν/((exp(hν/KBT) - 1)]

or,

Eλ = (8π/λ4)[hc/λ((exp(hc/KBλT) - 1)]

Note:  For small ν, exp(x) - 1 = x so the formulae
reduces to the classical case.

A comparison between the Rayleigh-Jeans law and
Planck's law is shown below.

Stefan-Boltzmann Law
--------------------

The Stefan-Boltzmann law gives the total energy
radiated per unit surface area of a black body
across all wavelengths per unit time.  It is the

To find the radiated power per unit area from
a surface it is necessary to multiply the energy
density by c/4.  This comes about because in
thermal equilibrium the inward energy flow equals
the outward flow giving a factor of 1/2 for the
radiated power outward.  Then it is required to
average over all angles, which gives another
factor of 1/2.  Therefore, the radiated power
per unit area as a function of wavelength is:

Pλ = (c/4)(8π/λ4)[hc/λ((exp(hc/KBλT) - 1)]

= (2πhc2/λ5)[1/((exp(hc/KBλT) - 1)]

If we make the substitution x = hc/KBT and
dx = -(hc/λ2KBT)dλ, we can write:

Pλ = 2π(KBT)4/(h3c2)∫x3/(exp(x) - 1)dx

This is a standard integral equal to π4/15.  Thus,
we get:

Pλ = (2π5KB4/15h3c2)T4

= σT4

To get the total power in Watts it is
necessary to multiply the above equation by the
area, A.

For hot objects other than ideal radiators, the
law is expressed in the form:

Pλ = eAσT4

where e is the emissivity of the object (e = 1

If the object is radiating/absorbing energy to
its surroundings at temperature Tenvironment,
the net radiation loss/gain rate takes the form:

Pλ = eAσ(Tobject4 - Tenvironment4)

If Tobject > Tenvironment the object is radiating.

If Tobject < Tenvironment the object is absorbing.

Wien's Displacement Law
-----------------------

Eλ = (8πL3/3λ4)hν/((exp(hc/KBTλ) - 1)

The maximum value of this function can be found by
differentiating w.r.t. λ, setting the result = 0
and solving for λ.  This gives:

λT = (hc/KB)/5[1 - exp(-hc/KBλT)]

Which must be solved numerically to give:

λpeakT = 2.898 x 10-3 meter Kelvin

This is the WIEN DISPLACEMENT LAW.  It says that
as the temperature of a blackbody radiator increases,
the overall radiated energy increases and the peak
of the radiation curve moves to shorter wavelengths.
When the maximum is evaluated from the Planck
radiation formula, the product of the peak wavelength
and the temperature is found to be a constant.  This
is useful for the determining the temperatures of
objects whose temperature is far above that of their
surroundings, such as hot radiant stars, by looking
at their spectra. ```