Black Body Radiation
Electromagnetic Waves in a Cubical Cavity
Electromagnetic standing waves in a cavity at equilibrium with its surroundings
cannot take just any path. They must satisfy the wave equation in three dimensions:
The solution to the wave equation must give zero amplitude at the walls, since a non-
zero value would dissipate energy and violate our supposition of equilibrium. To form
a standing wave, the reflection path around the cavity must produce a closed path. The
boundary conditions can be met with a solution of the form:
Substituting this solution into the wave equation above gives
which simplifies to
How many modes in the cavity?
From the standing wave solution to the wave equation we get the condition
We need to evaluate the number of
modes which can meet this condition,
which amounts to counting all the
possible combinations of the integer n
values. An approximation can be made
by treating the number of combinations
as the volume of a three-dimensional grid
of the values of n, an "n-space". Using
the relationship for the volume of a
sphere, with the n values specifying the
coordinates along three "n" axes, gives
The Rayleigh scheme for counting modes.
After Richtmyer, et al.
This has a couple of problems, however. In using a sphere, we have used both positive
and negative values of n, whereas the wave equation solution uses only positive
definite values. Therefore we must take 1/8 th of the volume above. Another technical
problem is that you can have waves polarized in two perpendicular planes, so we must
multiply by two to account for that. Then the volume can be taken to be a measure of
the number of modes, becoming a very good approximation when the size of the
cavity is much greater than the wavelength as in the case of electromagnetic waves in
finite cavity. Using the relation obtained for the values of n, this becomes
How many modes per unit wavelength?
Having developed an expression for the number of standing wave modes in a cavity,
we would like to know the distribution with wavelength. This may be obtained by
taking the derivative of the number of modes with respect to wavelength.
The negative sign here reveals that the number of modes decreases with increasing
wavelength. Now to get the number of modes per unit volume per unit wavelength,
we can simply divide by the volume of the cubical cavity.
Note that this does not involve approximating a sphere with a cube! The sphere we
used in calculating the number of modes was a sphere in "n-space", allowing us to
count the number of possible modes. Also, the use of a cubical cavity for the
calculation just allows us to reduce the geometrical complexity of the development,
but the final result obtained is independent of cavity geometry.
How much energy per unit volume?
Assigning energy to the electromagnetic standing waves in a cavity draws on the
principle of equipartition of energy. Each standing wave mode will have average
energy kT where k is Boltzmann's constant and T the temperature in Kelvins. Letting
u represent the energy density:
This is an important relationship in classical electromagnetic cavity theory. It can also
be expressed in terms of the frequency ν by making use of the chain rule and the wave
relationship:
The minus sign here just reminds us that a decrease with wavelength implies an
increase with increasing frequency. The magnitude of the energy density dependence
on frequency is given by:
Note that this is the classical result which was used in the Rayleigh-Jeans Law, but led
to the ultraviolet catastrophe. It produces good agreement in the low frequency limit,
but for higher frequencies the Planck radiation formula must be used.
Radiated Energy as a Function of
Wavelength
If we consider energy radiated perpendicular to a small increment of area, then it must
be noted that half of the energy density in the waves is going toward the walls and
half is coming out if the system is in thermal equilibrium. Evaluating the power seen
at a given observation point requires a consideration of the geometry:
For perpendicular radiated energy
but at an angle θ, the effective area will be Acosθ and the effective speed will be c
cosθ, so the radiated energy will be reduced to
For a given observation point near a radiating surface, the power will be the average
power from all directions, and the average gives another factor of 1/2.
Having averaged over all angles, the calculated radiated power per unit wavelength is
finally
Note: More detailed development of the c/4 factor in below.
This is the Rayleigh-Jeans formula. The fact that it failed to predict the spectral
distribution from hot objects was one of the major unresolved issues in physics at the
beginning of the 20th century.
To express this in terms of frequency, an application of the chain rule as was done
above with the energy density yields a radiated power per unit frequency:
Comments on the Development of the
Rayleigh-Jeans Law
The Rayleigh-Jeans Law was an important step in our understanding of the
equilibrium radiation from a hot object, even though it turned out not to be an accurate
description of nature. The careful work in developing the Rayleigh-Jeans law laid the
foundation for the quantum understanding expressed in the Planck radiation formula.
In outline form, here are the steps which led to the Rayleigh-Jeans law.
Equilibrium standing wave electromagnetic radiation in a
Show
cubical cavity of dimension L must meet the condition:
The number of modes in the cavity is: Show
The number of modes per unit wavelength is: Show
The energy per unit volume per unit wavelength is: Show
The average radiated energy per unit wavelength is: Show
Which when expressed in terms of frequency is: Show
Rayleigh-Jeans vs Planck
Comparison of the classical Rayleigh-Jeans Law and the quantum Planck radiation
formula. Experiment confirms the Planck relationship.
More detailed development of the c/4 factor
Radiated Energy as a Function of Energy
Density
If we consider energy radiated perpendicular to a small increment of area, then it must
be noted that half of the energy density in the waves is going toward the walls and
half is coming out if the system is in thermal equilibrium. Evaluating the power seen
at a given observation point requires a consideration of the geometry:
Other elements of area on the surface send radiation through the specified surface
area, and that radiation has a component perpendicular to the surface. To evaluate that
contribution, consider a surface element at an arbitrary angle as shown below.
The factor c/4 developed here involves an integral over all angles. It is important
because it relates the radiated energy to the energy density given by the Planck
radiation formula.
Planck Radiation Formula
From the assumption that the electromagnetic modes in a cavity were quantized in
energy with the quantum energy equal to Planck's constant times the frequency,
Planck derived a radiation formula. The average energy per "mode" or "quantum" is
the energy of the quantum times the probability that it will be occupied (the Einstein-
Bose distribution function):
This average energy times the density of such states, expressed in terms of either
frequency or wavelength
gives the energy density, the Planck radiation formula.
Example
Example
The Planck radiation formula is an example of the distribution of energy according
to Bose-Einstein statistics. The above expressions are obtained by multiplying
the density of states in terms of frequency or wavelength times the photon
energy times the Bose-Einstein distribution function with normalization constant A=1.
To find the radiated power per unit area from a surface at this temperature, multiply
the energy density by c/4. The density above is for thermal equilibrium, so setting
inward=outward gives a factor of 1/2 for the radiated power outward. Then one must
average over all angles, which gives another factor of 1/2 for the angular dependence
which is the square of the cosine.
Blackbody Intensity as a Function of
Frequency
The Rayleigh-Jeans
curve agrees with
the Planck radiation
formula for long
wavelengths, low
frequencies.
Applications of the Planck Radiation
Formula
Blackbody Radiation
"Blackbody radiation" or "cavity radiation" refers to an object or system which
absorbs all radiation incident upon it and re-radiates energy which is characteristic of
this radiating system only, not dependent upon the type of radiation which is incident
upon it. The radiated energy can be considered to be produced by standing wave or
resonant modes of the cavity which is radiating.
The amount of radiation emitted in a given frequency range should be
proportional to the number of modes in that range. The best of classical
physics suggested that all modes had an equal chance of being produced, and
that the number of modes went up proportional to the square of the
frequency.
But the predicted continual increase in radiated energy with frequency
(dubbed the "ultraviolet catastrophe") did not happen. Nature knew better.
A Nickel's Worth of Radiation
As an example of the kind of things you can model with the Planck radiation
formula, consider the following questions:
1. How much radiant energy comes from a nickel at room temperature
per second?
2. How many photons per second leave the nickel?
3. What volume of air would it take to have energy equal to 1 second
of radiation from the nickel?
4. What volume of the room would have radiation energy equal to 1
second of radiation from the nickel?
Measured properties of the nickel are diameter = 2.14 cm, thickness 0.2 cm, mass 5.1
grams. This gives a volume of 0.719 cm3 and a surface area of 8.54 cm 2. This gives a
calculated density of 7.09 gm/cm3 compared to the density of 8.9 gm/cm3 for the pure
element from the periodic table. So either the material is not pure nickel, being
alloyed with some lighter element, or the measurement of thickness is off, which is a
real possibility.
1. How much radiant energy comes from a nickel at room temperature per
second?
The radiation from the nickel's surface can be calculated from the Stefan Boltzmann
law. The room temperature will be taken to be 22°C = 295 K. Assuming an ideal
radiator for this estimate, the radiated power is
P = σAT4 = (5.67 x 10-8 W/m2K4)x(8.54 x 10-4 m2)x(295 K)4 = 0.367 watts.
So the radiated power from a nickel at room temperature is about 0.37 watts.
2. How many photons per second leave the nickel?
Since we know the energy, we can divide it by the average photon energy. We don't
know a true average, but the wavelength of the peak of the blackbody radiation
curve is a representative value which can be used as an estimate. This may be
obtained from the Wien displacement law.
λpeak = 0.0029 m K/295 K = 9.83 x 10-6 m = 9830 nm, in the infrared.
The energy per photon at this peak can be obtained from the Planck relationship.
Ephoton = hυ = hc/λ = 1240 eV nm/ 9830 nm = 0.126 eV
Then the number of photons per second is
N = (0.367 J)/(0.126 eV x 1.6 x 10-19 J/eV) = 1.82 x 1019 photons
3. What volume of air would it take to have energy equal to 1 second of radiation
from the nickel?
The translational kinetic energy of an air molecule can be obtained from equipartition
of energy and the definition of the kinetic temperature.
Kinetic energyaverage = 3/2 kT = 3/2 (8.617 x 10-5 eV/K)(295 K) = 0.038 eV
The number of molecules to equal one second's worth of radiation is then
Nmolecules = (0.367 J)/(0.038 eV x 1.6 x 10-19 J/eV) = 6.0 x 1019 molecules
The volume per air molecule can be found by using the fact that a mole of an ideal gas
occupies 22.4 liters at STP. Using the ideal gas law
Vair molecule = (22.4 x 10-3 m3)(295K/273K)/(6.02 x 1023 molecules/mole) = 4.02 x 10-
m
26 3
Volume of air to have 0.367 J =(4.02 x 10-26m3)(6.0 x 1019 molecules) = 2.41x 10-6m3
So the small volume of about 2.4 cm3 of air will have kinetic energy equal to the
amount of energy radiated from the nickel per second.
4. What volume of the room would have radiation energy equal to 1 second of
radiation from the nickel?
The Stefan-Boltzmann relationship is also related to the energy density in the
radiation in a given volume of space. The energy density is
4σAT4/c = 4(5.67 x 10-8 W/m2K4)x(295 K)4/(3 x 108m/s)
Energy density in radiation = 5.73 x 10-6J/m3
So the volume required is (0.367 J)/(5.73 x 10 -6J/m3) = 64000 m3. This is the volume
of a cube with dimension 40 meters per side.
The air molecules in a volume of air about three times the volume of the nickel will
have kinetic energy equal to the radiation from the nickel, but it takes the radiation
content of a large auditorium to equal that amount. The ratio of the energy per unit
volume in the molecular kinetic energy is more than 10 10 times that in radiation.
