Home  www.playhookey.com  Sat, 04292017 

Direct Current

Alternating Current

Semiconductors

Digital

Logic Families

Digital Experiments

Computers

 Analog  Analog Experiments  Oscillators  Optics  HTML Test  

 Basic Concepts  Reflection and Refraction  Lenses  Fiber Optics   

What is Light?

Light as a Wave

Light as a Particle

Characteristics of a Photon

 The Photoelectric Effect  The Transverse Electromagnetic Wave (TEM)  
Characteristics of a Photon 

The original assumption made about electromagnetic waves was that their energy level was continually variable. That is, any change in the energy emitted by the source, from the smallest to the largest possible amount, would be precisely reflected in the energy in the EM wave.
There would seem to be evidence of this intuitive assumption. After all, a radio broadcast station can adjust its power output by very small amounts, and the effective range and coverage of the station changes accordingly, as the broadcast EM wave changes in energy.
But there was a problem with that assumption: it turned out that such an assumption was incompatible with the phenomenon called the photoelectric effect, and with the observed behavior of blackbody radiation. These phenomena could only be explained and mathematically modelled if the energy in an electromagnetic wave could only change in specific, discrete increments. The resulting investigation gave rise to the quantum theory.
Since most of the investigation involved visible and nearvisible light, the Greek word photo (meaning "light") was employed here, and the quantum unit of electromagnetic energy was called a "photon." However, the usage of the word photon has since been expanded to cover the entire electromagnetic spectrum.
Photons are often described as "wavelets" because a single photon covers only a very small amount of space. They are also designated as massless particles. But neither of these terms tells adequately just what a photon really is. Calling it "the quantum of electromagnetic energy" tells us nothing new, since the word itself was coined to identify this concept. Therefore, let's define it in a slightly different way:
The minimum "bundle" or "capsule" of energy needed to sustain the electromagnetic phenomenon at a particular frequency.
The above definition specifies that a photon is a selfsustaining "capsule" of energy, but doesn't tell us how much energy is involved. We also know that it is involved with both a magnetic field (commonly identified as the B field) and an electric field (normally called the E field). The figure to the right shows the B field only; the E field would be sticking straight up out of the screen at you, and alternately retreating back into the screen. Although we don't see it here, perhaps we can make some statements about what it must be.
The basic equation that specifies the energy of a photon is given generally as:
E = hf =  hc 
n 
In this expression:
E = energy of the photon
h = Planck's Constant
(= 6.62554±0.00015 × 10^{27}
erg·seconds)
f = The frequency of the wave
c = The velocity of light in open space
n = The index of refraction of the medium (n = 1 for open space)
= The wavelength of the wave
Planck's Constant can also be expressed as approximately 6.626 × 10^{34} joule·seconds or 4.135 × 10^{15} eV·seconds.
The above equation implies that the energy of a photon is constant over time, which seems intuitively logical. But we still need to determine what is required inside the photon, in order to accomplish this. How can we do this?
We've been thinking of a photon as it travels through space. However, that same photon also travels through time. That is, it is a phenomenon that has duration. It exists from moment to moment as it moves, not just from place to place. So let's look at the photon at different instants in time, rather than at different places in space.
The first thing we can say in this regard is that the photon exists entirely in the present. That is, it doesn't concurrently spread its energy over the past, present, and future; if we can isolate the photon at any instant of time, the entire photon will exist at that instant.
But how much time is an "instant?" The two fields are constantly changing over time. If we use even a short duration as our "instant," there will still be a change in those fields.
However, we can note that as we make the time interval of our "snapshot" sample shorter, the amount of change is measurably less. So we apply the concept of the limit here: If we keep reducing the time interval, it will approach (but never quite reach) zero time interval as a limit. In the limit, the energy in the electric and magnetic fields will also be unchanging, and we can examine the instantaneous energy content of both fields as a static phenomenon.
The figure above shows eight evenlyspaced "snapshots" of our photon over the duration of one cycle of its wave, at 45° intervals (that's π/4 radians between "snapshots.") We will now proceed to look at the magnetic field strength at each "snapshot" and determine, based on the equation for photon energy, how the electric field strength must vary.
For easy reference, I've repeated our figure to the right. The sine wave represents the changing magnetic field strength over time; the value of the sine wave at any single point represents the magnetic field strength at that single instant of time. The energy in the magnetic field is proportional to the square of the field strength. Similarly, the energy in the electric field is proportional to the square of that field's strength at any instant in time.
Now, at time t = 0, the B field has zero strength so it contains zero energy. If the photon is to hold any energy at all, it must all be in the electrical field. Therefore, we would seem to require the E field to be at its peak at time t = 0.
Now we move forward in time. The strength of both fields changes. At time t = a, we have traversed through 1/8th of a cycle. The B field has been increasing in strength, and therefore in energy content. To keep the energy of the photon constant, the E field must have been decreasing in strength over this same time interval.
Move forward again, to time t = b. Now the B field is at its peak, and contains its maximum possible energy. Therefore the E field must be at its minimum, which for a sine wave is zero as it crosses the axis and reverses polarity.
Going on to t = c and t = d, the B field is decreasing to zero, so the E field must be increasing to its peak over the same time interval. This action repeats itself, for the opposite direction of the B field, through the next halfcycle until we get to t = h. At this point, it starts all over again.
This describes the quadrature model of the photon. The peak energy in
either field is equal to the total energy of the photon, and energy is
constantly being transferred back and forth between the fields as each one
repeatedly gives up its entire energy to create the other as they move
together through time/space. Essentially, we have a sine wave and a cosine
wave forming the two fields. It doesn't matter which one you start with
for the picture to be complete and the math to work out. Since the two
fields are at right angles to each other (their planes of existence are
normal to each other), their individual energies are separate from each
other, and the total energy of the photon must be found as the sum of the
two at that instant in time. Since the energy in each field is
proportional to the square of the field strength, we must work with the
squares of these two waveforms. Squaring the instantaneous values of the
two waves requires us to make use of a trignonmetric identity:
Now, we must consider the inphase model of the photon, which is currently being taught as being in accord with Maxwell's equations. This model requires that the E and B fields be in phase with each other, so they will peak at the same moment and reach zero at the same moment.
If this model is correct, then at time t = 0 in our diagram, both fields contain zero energy. Then the energy in both fields rises until at time t = b both fields have reached their peaks and their energy is maximum. Then they both decrease again in strength until at time t = d their energy is again zero. And this continues with polarity reversing each halfcycle.
Now, looking at this model over the time span occupied by one cycle, we see that the energy of the photon is not constant, but is continually changing between zero and some maximum value. This would seem to defy the (E = hf) equation for the energy of the photon, which says the photon energy depends only on its frequency.
More importantly, it says that the photon is not a closed energy system in and of itself. Since the photon energy varies, any energy not in the photon must necessarily be outside of the photon temporarily, which means it must have been transferred to something else for this instant in time. This in turn gives rise to some obvious questions, which have never been answered to my satisfaction:
There are also some correlaries that would have to be covered, but this would answer the key elements required by the Law of Conservation of Energy.
One other question that has been asked of me is my description of the physical dimensions of a photon. On that topic, I must say that since the photon consists of changing energy fields, the photon encompasses that volume of space in which those energy fields have significant effect.
Thus, if a photon from a radio wave passes close enough to a receiving antenna, it will transfer its energy to the antenna, giving rise to an electrical signal even as the photon itself ceases to exist. The "size" of the photon is that volume of space within which this phenomenon can occur.
The higher the frequency of the photon, the higher its energy level. Therefore, such photons will have a correspondingly larger field of influence than their lowerenergy counterparts.


 
All pages on www.playhookey.com copyright © 1996, 20002015 by
Ken Bigelow Please address queries and suggestions to: webmaster@playhookey.com 