|www.nortonkit.com||18 अक्तूबर 2013|
|Digital | Logic Families | Digital Experiments | Analog | Analog Experiments | DC Theory | AC Theory | Optics | Computers | Semiconductors | Test HTML|
|Direct Links to Other Optics Pages:|
|Basic Concepts:||[What Is Light?] [Light as a Wave] [Light as a Particle] [The Characteristics of a Photon] [The Photoelectric Effect] [The Transverse Electromagnetic Wave (TEM)]|
|Reflection and Refraction:||[Introduction] [Reflection, Part 1] [Reflection, Part 2] [Refraction, Part 1] [Refraction, Part 2]|
|Lenses:||[Introduction] [The Convex Lens]|
|Fiber Optics:||[Introduction] [Fiber Optics, Part 2] [Fiber Optics, Part 3] [Fiber Optics, Part 4] [Fiber Optics, Part 5] [Fiber Optics, Part 6]|
|The Transverse Electromagnetic Wave|
The basic transverse electromagnetic wave, as shown to the left, involves both a varying electric field and a varying magnetic field, appearing at right angles to each other and to the direction of travel of the wave. This figure represents a single photon traveling through space (and time).
Note especially that the electric and magnetic fields are not in phase with each other, but are rather 90° out of phase. Most books portray these two components of the total wave as being in phase with each other, but I find myself disagreeing with that interpretation, based on three fundamental laws of physics:
The total energy in the waveform must remain constant at all times. Any deviation from this condition constitutes a violation of this law.
The energy in a single photon is given by the expression:
|E = hν =||hc|
In this expression:
E = energy of the photon
h = Planck's Constant
ν = 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)
λ = The wavelength of the wave
If the photon is not moving through empty space, the index of refraction of the medium must be accounted for, since it reduces the effective value of c.
The above expression leaves no room for fluctuations or variations in energy over time; it implies that the energy of the photon is constant, and is determined specifically by the frequency (or wavelength) of the wave.
Furthermore, the phenomena of blackbody radiation and the photoelectric effect show that any electromagnetic wave is necessarily composed of independent photons, and is not really a continuously-variable phenomenon of its own.
Now, if the two component waves are assumed to be in phase with each other, then the total energy of the wave varies from some maximum value to zero, and then back up to the maximum value. This requires that each photon send all of its energy somewhere else twice per cycle, and then receive it back again. I have yet to see any satisfactory explanation of either where it goes or why it would come back to re-form the photon.
I've decided to end my argument page on the TEM dispute. The best assumption I ever got from anybody was that the energy was in "another part of the wave." But since the wave is necessarily composed of individual photons, that requires that photons trade energy back and forth with each other. This makes no sense anyway, and is quite impossible in a laser beam, where all photons are in phase with each other. Nevertheless, such photons must have the same properties as they do in random light or any other electromagnetic wave.
Beyond that, the electromagnetic wave is generated or emitted over time, so a different part of the wave is not only at a different physical location, it was also created at a different time. It can't very well be part of an energy transfer when it doesn't exist, or with another part of the wave that also doesn't exist. Rather, once energy has been emitted, the amount of emitted energy cannot simply change. We can emit more energy, but we cannot decrease the amount of energy already emitted.
This concept was originally stated by Ampere, and was later expanded by Maxwell. Thus it is generally known as the Ampere-Maxwell Law.
As an electric field moves through space, it gives up its energy to a companion magnetic field. The electric field loses energy as the magnetic field gains energy. Thus, we see a gradual transfer of energy from one form to another, but no loss or gain in the total energy of the wave.
This is Faraday's Law, and is exactly similar to the Ampere-Maxwell law listed above. A changing magnetic field will create and transfer its energy gradually to a companion electric field.
These last two laws define the fundamental principles behind all electric motors and generators, as well as transformers. In space or in air, however, they simply mean that the energy in each field transfers itself to the other field, draining itself as it does so. Thus, by setting the two fields as sine waves in quadrature (90° out of phase with each other), we can have a light wave that maintains a constant energy level at all times, and still has its energy constantly being shifted back and forth between electric and magnetic fields.
Incidentally, this phenomenon is not limited to light waves; any electromagnetic radiation behaves the same way. The relationship between the electric and magnetic fields in the wave is also in keeping with the phase relationship between voltage and current in a transmission line and in any antenna system the signal voltage (which produces the electric field) and the signal current (producing the magnetic field) are in quadrature.
The traditional equations for field strength of the electric (E) and magnetic (B) fields in the electromagnetic wave are:
In these expressions,
E = instantaneous electric field strength
Em = maximum or peak electric field strength
B = instantaneous magnetic field strength
Bm = maximum or peak magnetic field strength
kx = an arbitrary offset to the same point in any cycle of the wave
ω = 2πf = wave frequency in radians/second
t = time in seconds
c = the speed of light
The two sine functions are quite standard when dealing in sinusoidal waveforms as in this case. We could just as easily use cosine functions; the result would be effectively the same. It has simply become standard practice to use the sine function. If we were dealing with only a single field (E or B, but not both), it would make no difference which function we used. The end result would be the same. The only difference is whether we start at a peak of the wave (cosine) or a zero crossing (sine).
But in this case we have two related energy fields. By arbitrarily assigning the sine function to both of them, traditional physicists have equally arbitrarily assumed the two waves are in phase. I contend that it is just as valid to assign a cosine function to one while the sine function is still assigned to the other. This gives us the following equations:
The only difference this makes is in the assumed phase relationship between the two waves, putting them in quadrature with each other. The magnetic field does not change its properties in any way; only its phase relationship with the electric field.
What this change does accomplish, however, is to make the total energy in a single photon a constant value throughout its entire cycle. This now complies with the Law of Conservation of Energy and with the equation for the energy of a photon.
To see this, we look at the energy in each field. Theoretically the energy of any electric or magnetic field occupies all of space, although as we move away from the center of the field it rapidly becomes unmeasurable. Therefore, we consider instead the energy density, or energy per unit volume, of the field. This is much easier to deal with. The relevant equations are:
The "| |" symbols are used to indicate that although E and B are vector quantities, we are only interested in their magnitudes for these equations. For a single photon in open space, the equations become (assuming the fields are in quadrature -- we'll look at the problem posed with the fields in phase shortly):
|ηE||=||1||ε0 |Em|² sin²(kx - ωt)|
|ηB||=||1|||Bm|² cos²(kx - ωt)|
At the same time, we note that the total energy in a photon is constant. Thus, a photon of red light has an energy of about 1.7 eV. This is not an "average" energy or "peak" energy; it is the total energy of that photon at all times, no matter where either field might be in its cycle. For the same reason, the energy density of this photon is also constant. But the energy density of the photon is necessarily the sum of the energy densities of the two fields that comprise the photon. Therefore, the sum of these two energy densities is equal to some constant K, in some units indicating energy/volume. We can express this mathematically as:
|1||ε0 (|Em|sin(kx - ωt))²||+||1||(|Bm|cos(kx - ωt))²||=||K|
Algebraic manipulation of this expression is straight-forward:
|ε0|||Em|²||sin²(kx - ωt)||+||1|||Bm|²||cos²(kx - ωt)||=||2K|
||Em|²||sin²(kx - ωt)||+||1||cos²(kx - ωt)||=||2K|
|c²||sin²(kx - ωt)||+||c²||cos²(kx - ωt)||=||2K|
|c²(sin²(kx - ωt) + cos²(kx - ωt))||=||2K|
Note that because of the trigonometric identity sin²ωt + cos²ωt = 1 we find the trigonometric functions disappear from the equation, leaving us with only constants to define the constant energy of the photon. This is possible only if the two fields are in quadrature, as I contend. If the two fields were in phase, the cos function above would be replaced by another sin function, and we would have to contend with a term of 2sin²(kx - ωt) on the left side of the equation. There is no way this term can be made constant, so there is no way such a photon could possibly have constant energy. Since it is experimentally provable (see The Photoelectric Effect) that each photon is self contained and does have constant energy, it cannot be true that photon energy varies through the cycle.
With the two fields in quadrature, however, we can express the constant energy density K in terms of either field:
The maximum energy in each field occurs when that field is at its peak value. This can be expressed as:
Furthermore, the maximum energy will be the same for both fields, since the same energy is being transferred back and forth from one field to the other. Thus, ηE(max) = ηB(max). Therefore, we can equate the other halves of these equations:
Remembering that ε0µ0 = 1/c², this expression can easily be rearranged to:
All pages on www.nortonkit.com copyright © 1996, 2000-2009 by
Er. Rajendra Raj
Please address queries and suggestions to: firstname.lastname@example.org