Lecture 2 – Blackbody radiation
Absorption and emission of radiation
What is the blackbody spectrum?
Properties of the blackbody spectrum
Classical approach to the problem
Plancks suggestion – energy quantisation
Bose-Einstein statistics
Objectives
Learn how classical theories cannot account for the spectral distribution of
light from blackbody objects such as stars etc.
Show that by making the assumption that light consists of small packets of
energy (photons) we can develop an expression which perfectly fits the
experimental data.
Blackbody radiation
Heated bodies radiate energy, but what is the mechanism? On an atomic
scale, heat causes the molecules and atoms of a solid to vibrate. As atoms
consist of electrical charges in the form of electrons and protons, it is the
vibration of these charges which is responsible for the emission of
electromagnetic radiation. A very hot object will emit visible light as the
electrons vibrate.
How do bodies absorb radiation?
In order to radiate energy, an object must first absorb it.
Suppose we shine a light on an object.
If we shine it on glass
Light passes through
If we shine it on a metal Light is reflected
If we shine it on carbon Light is absorbed
In glass the electrons are tightly bound to atoms and only oscillate at
certain frequencies outside the range of visible light. This makes glass
appear transparent as very little of the visible light is absorbed.
Blackbody radiation II
Metals conduct and have free electrons not bound to any particular atom.
These electrons oscillate in response to the light and then radiate light
themselves. This radiation is reflected light. Again, there is very little
absorption of light, most of it is reflected.
The electrons in carbon have a short mean free path, when they collide
their energy is transferred to the lattice. They are efficient absorbers of
the incident light, hence carbon appears black. Carbon and similar
materials are effective at converting incident light into heat.
In a reverse process, as the carbon atoms warm up and vibrate more
vigorously, more of the lattice energy is transferred to the free electrons,
thus carbon is also a good radiator of heat. It cools down much faster than
a metal as it is more efficient at converting the lattice energy into
radiation.
Measuring the distribution of emitted radiation
A simple spectroscope
If we pass white light through a prism we obtain a spectrum.
By measuring the intensity of each wavelength of light in the spectrum we
can plot the spectral distribution.
The blackbody spectrum
Consider a box with a small hole in the side. Radiation entering the hole is
scattered inside the box and only a small amount comes out again. If the
temperature of the box is in equilibrium the spectrum of light coming out
of the hole looks something like this.
-8
U(f) (arb. units)
2.0x10
T = 3000K
T = 6000K
-8
1.5x10
U f 
=
Intensity
of
radiation of frequency f .
-8
1.0x10
We find similar spectral
distributions if we measure
the light from stars.
-8
0.5x10
0
0
15
0.5x10
15
1.0x10
15
1.5x10
15
2.0x10
Frequency (Hz)
Typical blackbody spectra
From red hot to white hot
The colour of a hot object changes as the temperature increases
Properties of the blackbody spectrum
For small f, U is proportional to f 2 . But at some value of f there is a peak
before U falls to zero.
If we double the temperature the position of the peak doubles in frequency
( f peak ∝T ) (Wien’s displacement law), however the height of the peak is
multiplied eight times.
Why? U ∝ f 2 so if the temperature is constant U 2 f =4 U  f  , but we
doubled the temperature so in fact U  2T =8U T  .
The total energy radiated by the body, the area under the curve, increases
by a factor of 16, i.e 24 times more, when the temperature is doubled.
Stefans law of radiation: U = T 4 ,  = 5.67×10-8 W/m2/K4 - Stefans
constant. U is the energy radiated from 1m2 of black surface at
temperature T .
Wien’s displacement law
f
peak
=aT or  peak T =b , b =2.898×10-3 mK (Wien’s displacement law).
For frequency f
peak
=2.82 k B T / h .
k B = Boltzmann’s constant = 1.3807×10-23 J/K. Note that f
peak
× peak ≠c .
Relationship between emitted radiation and radiation
inside the cavity
The energy (electromagnetic radiation) comes out of the hole at a speed c.
Some of the light comes out at an angle and the hole appears smaller.
Emitted radiation = R, radiation inside the cavity = P
2  /2
R= ∫ ∫ cP cos θ
c
m
P
A = cos θ
0
0
 
d
, d  =sin θ d  d θ
4
θ
2  /2
2
R= ∫ ∫ cP cos θ sin θ d  d θ= ∫
0
0
0
cP
cP
d =
8
4
Energy incident at an angle θ on a
So R=cP / 4 .
hole of unit area.
The energy emitted from the hole is representative of the energy inside
the cavity.
Origins of the spectrum
The energy spectrum is formed by a continuous process of absorption and
re-emission of radiation by the atoms and molecules forming the walls of
the box. In this way the energy can shift from one mode to another. When
thermal equilibrium is reached the characteristic spectrum will be
established.
To form a theoretical description of the spectrum we need to determine
how many modes of oscillation have frequencies in a given energy range.
These oscillators are the electromagnetic waves inside the box, which can
be thought of as standing waves.
Modes in a box
The electric field at the cavity wall must be zero
Standing waves I
Pluck a string
n=1
n=2
n=3
n


=L
2
n=4
The natural modes of vibration of the string are standing waves with nodes
at the ends. In the same way, electromagnetic waves inside the box are
also standing waves.
Standing waves II
Amplitude
1.0
0.5
0
n=1
n=2
n=3
-0.5
-1.0
0
0.2
0.4
0.6
0.8
1.0
Position (x/a) (arb. units)
Standing waves inside a cavity of length a.
The standing waves have amplitude y= A sin (2 π x / λ)cos(2 π f t ) .
Let k =2 / , and =2  f then y= A sin (kx)cos(ω t ) .
If the box has sides of length a then a=n /2 for n =1,2,3…
The frequencies of these waves are f =c / =n  c/ 2 a .
So k =n /a and =ck .
K space
We can represent the standing waves in 3D space by a set of k vectors
k = k x , k y , k z =  l , m , n .
a
Each point can be associated with a cubic volume of space and represents
a frequency of =ck =c  k 2x k 2y k 2z , the volume of the associated space is
3
3
 /a .
For a fixed frequency,  , we obtain a set of values for k x , k y and k z
2

2
2
2
which lie on a spherical surface: 2 =k x k y k z with radius k =/c .
c
To get the total number of vibrational modes in the frequency range zero
to  we count the number of cubes contained in the sphere with radius
k =/c .
Ncubes=Volume between spheres / volume of cube.
K-space II
Each mode occupies a discrete volume of K-space.
Mode counting
The volume of the whole sphere is 4 k 3 /3 , but we are limited to positive
values of k , i.e. one octant of the sphere, so the volume becomes
3
3
1 4 k  k
.
V s= ×
=
8
3
6
3
8 3 f 3

3
Now k = 3 and =2  f so k =
.
c3
c
3
3
3
3
If V cube = / a , the total number of cubes, N , is V s /V cube
3
8 f a
, so N =
.
3
6c
Putting, V =a3 , differentiating with respect to f , and multiplying by two
because we can have two orthogonal transverse electromagnetic waves at
each frequency we get
N  f  8 f 2V
=
.
df
c3
Comparison with experiment
In the classical approach we assign each mode an energy k B T . The total
energy emitted at each frequency from a box of unit volume is given by
2
8 f
N  f ×k B T =U RJ  f = 3 k B T .
c
This is the Rayleigh-Jeans approximation (1900).
constant = 1.381×10-23 J/K.
k B is Boltzmann’s
We find good agreement between the Rayleigh-Jeans equation and the
observed results for low values of f , i.e. where U  f ∝ f 2 . The higher the
temperature the bigger the frequency range over which the agreement is
good. Doubling the temperature doubles the energy output at low
frequencies, as expected from the proportionality to T .
However, at higher frequencies the experimental and theoretical results
diverge. We expect more energy to be output at higher frequencies, but in
experiments the energy distribution falls to zero. This failure of the
Reyleigh-Jeans model is sometimes called the “UV catastrophe”.
Planck’s suggestion
Planck suggested that energy could only be emitted in chunks that are
multiples of hf , where h is Planck’s constant (6.626×10-34 Js).
Experimentally, we can see this reflected in the fact that f peak ∝T . By
doubling the temperature the number of modes that can radiate freely is
also doubled.
From this explanation it is clear that if the average energy per mode is
k B T and the value hf for a particular mode is, e.g., 5 k B T , that mode will
be unlikely to radiate. As the frequency increases, the probability of
radiation decreases.
Replacing U =k B T with U =hf in
the
Rayleigh-Jeans
equation
gives
2
8 f
U Planck  f = 3 ×hf × P BE  f  , where P BE  f  is the Bose-Einstein factor for
c
the average number of photons per mode at frequency f .
hf × P BE  f  is the average energy, 〈 E 〉 , of photons with frequency f .
The Bose-Einstein factor
We assume that the probability of occupying an energy level E is given by
−E / k T
. This was proved by Planck.
P  E = e
B
N3
E3 = 3hf
P3 = Aexp(-3hf/kBT)
E2 = 2hf
P2 = Aexp(-2hf/kBT)
E1 = hf
P1 = Aexp(-hf/kBT)
E0 = 0
P0 = A
N2
N1
N 1= N 0 e
−hf / k B T
N 2= N 0 e
N0
N n= N 0 e
−2hf / k B T
−nhf / k B T
Energy levels of a quantum oscillator
−x
−2x
−x
−2x
Let x=hf / k B T , then N 1= N 0 e , N 2 = N 0 e
, so N Total = N 0 1e e ... .
E n = N n ×nhf = N 0 e
−nx
×nhf = N 0 hf ne
−nx
−x
 and E Total = N 0 hf 0e 2 e
−2x
... .
The Bose-Einstein factor II
E Total
=
The average energy is given by 〈 E 〉=
N Total
N 0 hf
∑ ne−nx
n
N 0∑ e
−nx
=hfP BE  f 
n
∑ ne−nx d
d
1
1
1
−nx
P BE  f = n −nx =− log ∑ e =− log
=
=
−x
x
hf / k T
dx
dx
1−e
e −1 e
−1
n
∑e
B
n
2
8 f
hf
So our final result is U Planck  f = 3 × hf /k T
.
c
e
−1
B
Even though this is proportional to f 3 the Bose-Einstein term reduces the
energy to zero at high frequencies. At low frequencies the term on the
right approximates to 1, matching the Rayleigh-Jeans approximation.
As f  0 or T  ∞ the average energy approaches k B T , the same as the
classical result.
Various blackbody spectra
Conclusions
Classical physics, in which electromagnetic radiation is assumed to be a
continuous wave, cannot account for the blackbody radiation spectrum.
The assumption that electromagnetic radiation is emitted in quanta with
energy E =hf allows us to develop an expression which accurately
describes the spectral distribution of radiation emitted from a blackbody.