Section 2.4 Photons
The resolution of the ultraviolet catastrophe, reached in stages by Planck and Einstein, is that Maxwell's theory of electricity and magnetism is incomplete. In 1905 1 Einstein introduced the concept of a photon: a “particle” of light whose energy is related to the EM wave frequency via
where
is known as Planck's constant. Einstein's claim was that an EM wave of frequency \(f\) can be viewed as a gas of photons, each with this energy. Photons are not part of Maxwell's equations; they are in some sense a supplementary property of EM waves. 2
Example 2.2. Number of photons.
An EM wave of wavelength \(\lambda=420\Xunits{nm}\) has a total energy of \(1.80\Xunits{J}\text{.}\) How many photons are in this wave?
Each photon is contributing an energy \(E_\text{ ph } = hf\text{.}\) Using \(f = c/\lambda\) we can calculate the energy per photon,
The total energy \(E\) is given by \(N E_\text{ ph }\text{,}\) where \(N\) is the number of photons, so
Evidently, EM waves with macroscopic energies contain a lot of photons!
The idea of light being composed of particle-like photons is tremendously important with numerous implications throughout all of science and engineering. The photon picture also resolves the problem of infinite energy in a cavity. Imagine sorting all the EM wave modes in the cavity from lowest frequency to highest frequency. Since \(E_\text{ ph } =hf\text{,}\) the photons in the lowest frequency mode have low energy, while the photons in the high frequency modes have higher energy. But the total energy of each mode must be \(\textstyle{\frac{1}{2}}k_B T\text{,}\) which means each EM wave mode has exactly the same energy.
How can higher energy photons add up to the same energy? There must be fewer of them. So as we go to higher and higher frequency modes, we must have fewer and fewer photons in each mode. Eventually we will reach a high frequency mode with only one photon, and that's the end of the line. There can be no higher frequency modes. And so the ultraviolet catastrophe is averted.
In Problems 2.7.9 and Exercise 2.7.10 at the end of this chapter, you will do this quantitatively. You will find that the mode number \(j_\text{ max }\) where the number of photons drops to below \(1\) in a one-dimensional cavity is
For modes with larger \(j\text{,}\) there is not enough thermal energy to produce even a single photon. Given this result, we can find the total energy in the cavity. We have modes \(j=1\text{,}\) \(2\text{,}\) …, \(j_\text{ max }\text{,}\) each contributing an energy \(\textstyle{\frac{1}{2}}k_BT\text{.}\) That adds up to
This answer is a much nicer result for the thermal energy of the EM fields, because it's not infinite!
Let's reflect a moment on what we have just done. We argued that EM waves, which in classical physics can have any continuous value of energy, instead come in energy chunks. When the number of these photon chunks is large, some \(10^{18}\) or so, then we can ignore the chunkiness and treat the EM waves classically. Then the equipartition holds and we can use (2.2).
But for really high frequency (UV) waves, the photons have so much energy that there are only a few of them in a mode. At this point, the classical physics result (2.2) no longer holds. Essentially, there is simply not enough thermal energy available to create even a single photon for these modes. So we counted an energy of \(\textstyle{\frac{1}{2}}k_BT\) for each mode until we hit this wall, and then we stopped counting. The result is (2.7).
Note that this result only applies to a one-dimensional system. Real materials, of course, are three dimensional. We won't go through the full derivation for a three-dimensional cavity. That would require working with the thermodynamics of a photon gas, something that we do in our PHYS 317 Thermodynamics course.