1. Lecture 14, Blackbody Radiation
Phys 371
Fall 2019
Prof. Steven Anlage
Lecture 14, Blackbody Radiation
Friday 27 September, 2019

5. Millikan also innovated by examining the dynamics of single
First Quantization
J J Thomson “discovered” the electron and determine it’s charge-to-mass ratio: e/m
Thomson tried to measure the charge of the electron using charged water droplets:
Charge up a large collection of water droplets and measure their total mass and charge
But the water evaporated during the measurement! - Big uncertainties
Robert Millikan switched to oil droplets to eliminate the evaporation problem
Millikan also innovated by examining the dynamics of single
individual charged oil droplets with varying charge and in a variable electric field

6. 𝑬 Millikan Oil Drop Experiment
The oil droplets are charged by the atomizer
They fall into under the influence of gravity into the capacitor
The droplets experience a drag force and fall at constant (slow) velocity
The charge on the droplets can change by means of ionizing radiation sent into the chamber
One can follow individual droplets through the microscope
The charge on the droplets can be measured from a balance of forces involved:
gravity
drag
electric 𝑬
Get the mass of the droplet from watching it fall under gravity and drag and measuring the radius. Get the charge by balancing the upward electric force with the weight.

7. Outcomes of the Thomson and Millikan Oil Droplet Experiments
After many observations of droplets in different charge states:
The charge of the droplets was found to be Quantized in units of 𝒆 =𝟏.𝟔𝟎𝟐× 𝟏𝟎 −𝟏𝟗 Coulombs
The electron is a “universal” particle that is found in all substances / atoms (follows from Thomson’s cathodes)
The mass and charge of the electron are quantized → First Quantization
Every electron is apparently identical to every other electron

8. Front row: Michelson, Einstein, Millikan at the Athenaeum at Caltech in 1931

9. Blackbody Radiation 𝑹 𝑻𝒐𝒕𝒂𝒍 =𝝈 𝑻 𝟒 𝑹 𝑻𝒐𝒕𝒂𝒍 𝑻 𝝈
All objects at finite temperature 𝑻 emit electromagnetic radiation
The total amount of radiation increases with the temperature of the object
𝑹 𝑻𝒐𝒕𝒂𝒍 =𝝈 𝑻 𝟒
𝑹 𝑻𝒐𝒕𝒂𝒍
Total radiated power per unit area of object (W/m2) 𝑻
Absolute temperature (K) 𝝈
Stefan constant
𝝈=𝟓.𝟔𝟕𝟎𝟑× 𝟏𝟎 −𝟖 𝑾/( 𝒎 𝟐 𝑲 𝟒 )

10. Blackbody Radiation is Widely Distributed in Wavelength / Frequency
Energy/
m^2/nm/sec
R(l) (W/(m2-nm))
Wavelength l (Å = m)
𝑹 𝑻𝒐𝒕𝒂𝒍 = 𝟎 ∞ 𝑹 𝝀 𝒅𝝀

11. This radiation spectrum is universal in the sense that all objects show the same emission spectrum as long as
they are the same temperature. Universality begs explanation …

12. Solar Blackbody Spectrum Measured Above and Below the Atmosphere
The discrete absorption “lines” suggest that atoms have quantized energy levels, but that will come up later …

13. Wien Displacement Law
𝝀 𝒎𝒂𝒙 ~𝟏/𝑻
𝝀 𝒎𝒂𝒙 𝑻=𝟐.𝟖𝟗𝟕𝟕𝟕𝟏𝟗𝟓𝟓× 𝟏𝟎 −𝟑 m-K

14. Blackbody Radiators 𝑹 𝝀 = 𝒄 𝟒 𝝆 𝝀 𝝆 𝝀 𝑹 𝝀
𝝆 𝝀 = Energy Volume∙Wavelength
𝝆 𝝀
𝑹 𝝀
= J 𝑚 3 ∙nm
𝑹 𝝀 = Power Area∙Wavelength
= W 𝑚 2 ∙nm

15. Commercial Blackbody Radiation Sources

16. Cosmic Microwave Background Radiation Spectrum
T = ± K
The radiation is isotropic to roughly one part in 100,000:
the root mean square variations are only 18 µK
𝟏 𝒄𝒎 −𝟏 ≈𝟑𝟎 𝑮𝑯𝒛

17. What is the Energy Density of Electromagnetic Modes in the Box, r(l)?
First do the classical calculation
Assume Equipartition
of Energy
Every mode acquires 𝒌 𝑩 𝑻
of energy, on average, for
a system in thermal equilibrium
𝝆 𝝀 = 𝒌 𝑩 𝑻 𝒏 𝝀
Number of modes per unit volume
at wavelength l
Energy per mode
What is the mode density of electromagnetic fields in the box of wavelength l, 𝒏 𝝀 ?

18. What is the Mode Density 𝒏(𝝀)?
Imagine all the resonant oscillation modes of a string of length L
The wavelengths of standing waves on the string are given by
𝑳=𝑵 𝝀 𝟐 , where 𝑵 is an integer, 𝑵=𝟏,𝟐,𝟑,𝟒,…
We can ask the question: For a given wavelength 𝝀, what mode number
does it correspond to? The answer is 𝑵=𝟐𝑳/𝝀.
Doing this in 3 dimensions for a cube of size 𝑳×𝑳×𝑳 gives:
𝑵=𝟒𝝅 𝑳 𝟑 / 𝟑𝝀 𝟑 .
The number of modes with wavelengths between 𝝀 and 𝝀+𝒅𝝀 is given by:
𝒅𝑵= 𝟒𝝅 𝑳 𝟑 𝝀 𝟒 𝒅𝝀.
The number of modes per unit volume per unit wavelength is:
𝒏 𝝀 = 𝟖𝝅 𝝀 𝟒 (assuming 2 polarizations for each wavelength)
+ many more modes!

19. What is the Energy Density of Electromagnetic Modes in the Box, r(l)?
Still doing the classical calculation
𝝆 𝝀 = 𝒌 𝑩 𝑻 𝒏 𝝀
𝝆 𝝀 = 𝟖𝝅𝒌 𝑩 𝑻 𝝀 𝟒
Rayleigh-Jeans Law
The ultra-violet catastrophe ensues…
Somehow we have to limit the occupation
of the short-wavelength modes …

20. What is the Energy Density of Electromagnetic Modes in the Box, r(l)?
The birth of Quantum Mechanics
The atoms in the walls of the box have electrons, and when these electrons vibrate at frequency 𝑓 they emit electromagnetic waves at frequency 𝑓.
Likewise light at frequency 𝑓 can be absorbed by the atoms in the walls and start oscillating at frequency 𝑓.
In equilibrium there is an equal energy flux from the walls to radiation and from radiation back into the walls.

21. What is the Energy Density of Electromagnetic Modes in the Box, r(l)?
The birth of Quantum Mechanics
Max Planck (1900) made two (largely unjustified and revolutionary) assumptions:
The atoms in the walls of the box have discrete energy levels given by 𝑬 𝒏 =𝒏𝜺, where 𝒏=𝟎, 𝟏, 𝟐, … Hence the atoms can only interact with light of energy 𝜺, 𝟐𝜺, 𝟑𝜺, etc.
2) The energy of light is directly related to the frequency of oscillation of the EM fields as 𝜺=𝒉𝒇, where 𝒉 is a fudge factor with units of 𝑱/𝑯𝒛 or 𝑱−𝒔. Surprisingly, this energy is independent of the intensity of the light.
Why did Planck take 𝜺=𝒉𝒇  and not some other dependence?

22. How Does This Get Rid of the Ultraviolet Catastrophe?
The likelihood of finding an atom in the walls of the box being excited to energy state 𝑬 is given by:
𝒈 𝑬 =𝑨 𝒆 −𝑬/ 𝒌 𝑩 𝑻 (the Boltzmann factor from classical statistical mechanics)
𝒈 𝑬 is the fraction of atoms in the wall of the box that are excited to a state of energy 𝐸
𝑻 is the temperature of the walls of the box (K)
𝑨 is a normalization constant
This shows that the highly energetic atom energy states (𝑬 >> 𝒌 𝑩 𝑻) will be occupied with near-zero probability
Hence in equilibrium there will be very few highly energetic electromagnetic modes excited in the cavity
In other words, equipartition of energy breaks down for the energetic modes of the box
𝒌 𝑩 𝑻~𝟎.𝟏 𝒆𝑽
For example:
𝑬~𝟐 𝒆𝑽
𝒆 −𝑬/ 𝒌 𝑩 𝑻 ~ 𝟏𝟎 −𝟗

23. What is the Energy Density of Electromagnetic Modes in the Box, r(l)?
First step: Find 𝑨
We can find 𝑨 by using the discrete nature of the energy states of the atoms: 𝑬 𝒏 =𝒏𝜺
In this case we have the discrete fractions: 𝒈 𝒏 =𝑨 𝒆 − 𝑬 𝒏 / 𝒌 𝑩 𝑻 =𝑨 𝒆 −𝒏𝜺/ 𝒌 𝑩 𝑻
Find 𝑨 by demanding that the total probability of finding the atom in any state is unity:
𝟏= 𝒏=𝟎 ∞ 𝒈 𝒏 = 𝒏=𝟎 ∞ 𝑨 𝒆 −𝒏𝜺/ 𝒌 𝑩 𝑻 =𝑨 𝒏=𝟎 ∞ 𝒆 −𝜺/ 𝒌 𝑩 𝑻 𝒏
We recognize the sum as that for 𝟏 𝟏−𝒙 , 𝒇𝒐𝒓 𝒙<𝟏 (note that 𝒙=𝒆 −𝜺/ 𝒌 𝑩 𝑻 <𝟏):
𝟏 𝟏−𝒙 = 𝒏=𝟎 ∞ 𝒙 𝒏
S𝐨𝐥𝐯𝐢𝐧𝐠 𝐟𝐨𝐫 𝑨, 𝐰𝐞 𝐠𝐞𝐭 𝑨=𝟏− 𝒆 −𝜺/ 𝒌 𝑩 𝑻

24. What is the Energy Density of Electromagnetic Modes in the Box, r(l)?
Second step: Find the average energy of the atoms in equilibrium
𝑬 = 𝒏=𝟎 ∞ 𝑬 𝒏 𝒈 𝒏 =𝑨 𝒏=𝟎 ∞ 𝒏𝜺 𝒆 −𝜺/ 𝒌 𝑩 𝑻 𝒏
Trick to simplify the sum: The pre-factor in the sum, 𝒏𝜺, appears in the exponent. Hence take a derivative
with respect to −𝟏/ 𝒌 𝑩 𝑻 to bring it down from there. Then move the derivative outside the sum!
𝑬 =𝑨 𝒅 𝒅(−𝟏/ 𝒌 𝑩 𝑻) 𝒏=𝟎 ∞ 𝒆 −𝜺/ 𝒌 𝑩 𝑻 𝒏
Now the sum can be easily performed!
𝑬 =𝑨 𝒅 𝒅(−𝟏/ 𝒌 𝑩 𝑻) 𝟏 𝟏− 𝒆 −𝜺/ 𝒌 𝑩 𝑻
Finally we get
𝑬 = 𝜺 𝒆 𝜺/ 𝒌 𝑩 𝑻 −𝟏

25. What is the Energy Density of Electromagnetic Modes in the Box, r(l)?
𝑬 = 𝜺 𝒆 𝜺/ 𝒌 𝑩 𝑻 −𝟏
This is the average energy of the atoms making up the walls of the box
They are in equilibrium with the electromagnetic modes.
Therefore we use Planck’s assumption that 𝜺=𝒉𝒇 to find the energy of the light
𝑬 = 𝒉𝒇 𝒆 𝒉𝒇/ 𝒌 𝑩 𝑻 −𝟏 = 𝒉𝒄/𝝀 𝒆 𝒉𝒄/𝝀 𝒌 𝑩 𝑻 −𝟏
using the fact that 𝝀𝒇=𝒄 for light
This is the average energy in a mode of wavelength 𝝀.
Find the Total energy density of the light by multiplying by the mode density calculated above:
𝝆 𝝀 = 𝑬 𝒏 𝝀 = 𝑬 𝟖𝝅 𝝀 𝟒
𝝆 𝝀 = 𝟖𝝅𝒉𝒄/ 𝝀 𝟓 𝒆 𝒉𝒄/𝝀 𝒌 𝑩 𝑻 −𝟏
Planck Blackbody Radiation Formula

26. 𝝆 𝝀 = 𝟖𝝅𝒉𝒄/ 𝝀 𝟓 𝒆 𝒉𝒄/𝝀 𝒌 𝑩 𝑻 −𝟏 Planck Blackbody Radiation Formula
𝝆 𝝀 = 𝟖𝝅𝒉𝒄/ 𝝀 𝟓 𝒆 𝒉𝒄/𝝀 𝒌 𝑩 𝑻 −𝟏
Planck Blackbody Radiation Formula
This equation fit the data very well!
The prediction remains finite as 𝝀→𝟎
𝝆 𝝀 ≈ 𝟖𝝅𝒉𝒄 𝝀 𝟓 𝒆 −𝒉𝒄/𝝀 𝒌 𝑩 𝑻
This predicted exponential suppression
at short wavelength is a characteristic
property of blackbody radiation data

27. Planck has overcome the Ultraviolet Catastrophe, but at what price?
The value of Planck’s constant
𝒉=𝟔.𝟔𝟐𝟔 × 𝟏𝟎 −𝟑𝟒 𝑱−𝒔
=𝟒.𝟏𝟑𝟔 × 𝟏𝟎 −𝟏𝟓 𝒆𝑽−𝒔
The reduced Planck constant: ℏ=𝒉/𝟐𝝅
Planck was obsessed with justifying these two assumptions in the 10 years or so that followed
He was a “reluctant revolutionary”
He was relieved when a young Albert Einstein picked up these ideas about light and showed that
they could be used to understand the Photo-electric Effect (a totally different phenomenon), and
determine the value of Planck’s constant with high precision!
Thanks for Listening!