Conservation of Energy - Revisited:

Total Destructive Interference

With what certainty can we say that the laws of physics conserve energy? Certainly we have deduced a theorem to the effect that Existence has so structured Reality that it conserves energy everywhere and at all times, except, perhaps, in the region near the boundary of space. But that fact does not prevent us from harboring a suspicion that other exceptions may lurk among the various and many laws of physics and their combinations. After all, our deduction of conservation of energy did not proceed as cleanly as did our deductions of the conservation laws pertaining to linear momentum and angular momentum. So if some imaginary experiment tickles that suspicion, I say play it out and see whither it leads. In my experience that plan does not lead to any methods of violating conservation laws, but it does usually lead to new knowledge.

Even though in this section of the Map of Physics I feign not knowing a fundamental property of light, in this particular essay I want to examine that property for the possibility that it will enable us to violate the conservation law pertaining to energy. We do, if fact, know that light manifests itself as a sinusoidal wave of mutually supporting electric and magnetic fields and, thus, that it displays the phenomenon of interference. I will, therefore, postulate that, through the use of a simple interferometer, we can manipulate light in a way that exploits interference to destroy energy.

I first imaged this little experiment some time in the late 1960's when I was at UCLA and I have returned to it several times since then. In all cases I did not look into the subject intently enough or in sufficient detail to figure out the solution. At the end of August 1996 (according to my notes) I managed to figure out how to solve the problem, but I did not follow up with the appropriately detailed analysis. Recently though (as of February 2009) I have sat myself down and done the necessary work to solve the problem completely.

Imagine that a narrow beam of coherent, monochromatic light flies in the positive x-direction (left to right). The beam strikes a 50-50 beam splitter that crosses the beam’s path at a 45-degree angle, so half of the beam’s power bounces off the beam splitter and flies in the positive y-direction (up) and the remaining half passes through the beam splitter and continues in the positive x-direction. Each of those half-beams strikes a mirror oriented parallel to the beam splitter, so the upper beam, traveling in the positive y-direction, bounces off its mirror and flies in the positive x-direction and the lower beam, traveling in the positive x-direction, bounces off its mirror and flies in the positive y-direction. Where the two half-beams cross each other we put another 50-50 beam splitter, one made with a very special property.

We want to make that second beam splitter of such a thickness that the transmitted part of a beam striking the surface at a 45-degree angle follows an optical path that puts a whole number of wavelengths of the light between opposite faces of the beam splitter. With that beam splitter we can use the apparatus I described above to put to the proof the conservation of energy theorem as it applies to electromagnetic waves.

In theory (and imaginary experiments) we don’t need to do it, but in Reality we must adjust the length of the optical path that one of our half-beams follows between the two beam splitters. We can accomplish that task in a straightforward manner by putting two clear glass wedges together in such a way that their outward-facing surfaces are parallel to each other and then inserting that block into one or the other of the half-beams in our interferometer. If we slide one of the wedges over the other, we will make the block that they comprise thicker or thinner and thereby put more or fewer wavelengths into the optical path of the half-beam passing through the block. By that means we bring our apparatus into conformity with the following criterion.

Where the upper half-beam strikes the upper face of the second beam splitter the electric and magnetic fields of the part of the lower half-beam transmitted through the beam splitter exactly oppose the electric and magnetic fields of the reflected part of the upper half-beam, thereby canceling them. Because a whole number of wavelengths fit inside the beam splitter, the same condition prevails where the lower half-beam strikes the lower face of the beam splitter.

Thus, according to this analysis, two beams of light strike the beam splitter and nothing comes off it. If Reality conformed to that scenario, then that simple apparatus would violate the conservation of energy theorem by annihilating the energy coming into it. Worse, it would also violate the conservation law pertaining to linear momentum: the incoming beams of light carry a net linear momentum oriented parallel to the faces of the beam splitter. We know with absolute certainty that we cannot violate that latter law, so now we must ask what happens in this situation to uphold it and thereby answer the question about the conservation of energy in this case. Let’s start with a brief review of the electromagnetic theory describing the interaction between light and a flat surface of a dielectric material.

The Fresnel Equations

A beam of plane-polarized, monochromatic light consists of a sinusoid-shaped electric field combined with a sinusoid-shaped magnetic field such that at any given point the electric field vector and the magnetic field vector meet at a right angle and the electric sinusoid and the magnetic sinusoid are in phase with each other. If we look at the center of an old-fashioned analogue clock and we see the electric-field vector pointing toward the twelve and the magnetic-field vector pointing toward the three, then we know that the wave is propagating away from us and into the clock.

If we treat light propagating through clear air as though it were propagating through vacuum, we won’t go discernably wrong if we restrict our consideration to short distances. That assumption aids us in devising a relatively simple description of what happens when a plane electromagnetic wave strikes a flat interface between air and a dielectric material, such as glass. We assume further that both of the media in this experiment (air and glass) have the same magnetic permeability, equal to that of vacuum. And we assume that both media are perfect dielectrics, so only their different electric permittivities contribute to making their indices of refraction differ from each other.

Imagine a block of some transparent dielectric material
with its upper surface perfectly horizontal, perfectly flat, and perfectly
smooth. Imagine further a ghostly grid of Cartesian coordinates so established
within the block and its surroundings that we can, at least in concept, use it
to make measurements. At a point O on the top surface of the block construct a
vector **N** of unit length and oriented perpendicular to the surface and
call it the normal unit vector.

A ray of light strikes the point O from somewhere above the block. At that point the electric and magnetic fields that constitute the light have the form

(Eq’ns 1)

On the surface of the block the incident ray splits into two components, one
that is reflected by the surface and one that is transmitted through the
surface. The equations describing the electric and magnetic fields in the
reflected and transmitted rays have the same form as Equations 1, but with the
variables associated with the fields subscripted with r and t respectively and
with the magnetic permeability associated with the transmitted ray represented
by μ_{1}
(of course, by assumption in this case,
μ_{1}=μ_{0}).
The components of the electric and magnetic fields that lie tangent to the
surface of the block must remain continuous across that surface, because they
represent conservative forces: the fields can only satisfy that boundary
condition if the angular frequency of all three rays has the same value (ω_{i}=ω_{r}=ω_{t},
the subscript i referring to the incident ray) and if the propagation vectors
satisfy the relation

(Eq’n 2)

To parlay that boundary condition into a fuller description of the relation
among the three rays we adjust the coordinate grid to put its origin on the
surface of the block in such a way that the coordinate vector **x** and the
normal unit vector combine in the dot product as **N∙x**=0. Thus, by the **B**(**A∙C**)-**C**(**A∙B**)
rule of vector multiplication, we have **N**×(**N**×**x**)=-**x**.
We can combine that result with Equation 2 to get

(Eq’ns 3)

Because **N**×**x** does not equal zero and because the dot product in
the second part of each of Equations 3 does not equal zero, we must have as true
to our mathematical description of light

(Eq’ns 4)

In terms of magnitudes we can express the propagation vector of each ray as k=nω/c, in which n represents the index of refraction of the medium through which the ray propagates, and we can thus express Equations 4 as

(Eq’ns 5)

The first of those equations tells us simply that

(Eq’n 6)

which is the law of reflection. Theta represents the angle between the propagation vector and the normal unit vector, so that equation tells us that if we project a ray at the surface of our block at a given angle, the reflected ray will come off the surface at the same angle, like a ball bouncing off a flat surface.

The second of Equations 5 tells us that

(Eq’n 7)

which is Snell’s law. It means that if we project a ray down onto the surface
of the block at a certain angle, the part of the ray penetrating the block will
descend at a steeper angle, θ_{t}<θ_{i}.

As an aside I’ll note that we could have obtained Equations 6 and 7 by using the optical analogue of the principle of least action. In that case we determine between two points the path of least time.

To determine the amplitudes of the fields in the reflected
and transmitted rays we begin by reviewing the boundary condition: the
components of the electric and magnetic fields that lie tangent to the surface
of the block must remain continuous across that surface. We know that when we
calculate a vector cross product the product vector points in a direction
perpendicular to both of the multiplier vectors, so if we calculate **N**×**E**,
we will obtain a product vector that lies tangent to the surface of the block as
a projection of the electric field onto that surface. Thus we can write our
boundary condition as

(Eq’ns 8)

For convenience we use the second of Equations 1 to express the second of Equations 8 in terms of the electric field,

(Eq’n 9)

Because we can represent any vector as a sum of two mutually perpendicular component vectors, we have two ways in which we can solve Equations 8 for the amplitudes of the fields in the reflected and transmitted rays as functions of the amplitudes of the fields in the incident ray.

Start with the case in which the electric vector lies
entirely parallel to the surface of the block (**N∙E**=0). We then have the
magnitudes of the cross products in the first of Equations 8 as |**N**||**E**|sin(π/2),
so that we have from that first equation

(Eq’n 10)

Remembering the **B**(**A∙C**)-**C**(**A∙B**) rule and our
assumption that μ_{1}=μ_{0}
in this case, we have Equation 9 in the form

(Eq’n 11)

Again using the fact that k=nω/c, we can write out the dot products explicitly, divide out the factors common to all three terms, and get Equation 11 as

(Eq’n 12)

In that equation I have substituted n_{0}=n_{i}=n_{r}
and n_{1}=n_{t} for the indices of refraction in this case. We
can now solve Equations 10 and 12 for E_{0r} and E_{0t} and get
the Fresnel equations for this case as

(Eq’ns 13)

We obtain the equivalent equations for the associated magnetic fields by applying the second of Equations 1 and get

(Eq’ns 14)

The second case is the one in which the magnetic vector lies entirely
parallel to the surface of the dielectric block (**N∙H**=0). Using the
relation

(Eq’n 15)

we have Equations 8 as

(Eq’ns 16)

Solving those equations gives us the Fresnel equations for this case,

(Eq’ns 17)

We obtain the description of the electric fields associated with these magnetic fields by applying the relation of Equation 15 to Equations 17.

For the imaginary experiment yet to come we want to calculate the rate at which energy flows through the rays. To that end we want to calculate the Poynting flux in the rays, so we use the Poynting vector,

(Eq’n 18)

which gives us watts per square meter when we use MKS units. Because the
fields in each ray keep a right angle between them, the sine in the vector cross
product equals one and the magnitude of the Poynting flux in the ray equals the
product of the magnitudes of the electric and magnetic field vectors. If we
substitute the second of Equations 1 into Equation 18 and apply the **B**(**A∙C**)-**C**(**A∙B**)
rule, we get

(Eq’n 19)

which tells us that the energy in the ray flows in the direction of the propagation vector, as we expect. Substituting Equation 15 into Equation 18 gives us

(Eq’n 20)

Equations 19 and 20 give us the instantaneous values of the Poynting flux. In order to carry out those calculations we must substitute into those equations the field descriptions given in Equations 1, complete with the exponentials. But we need only the time average of the Poynting flux, so we exploit our knowledge that the average value of a squared sinusoidal function equals half of the maximum value of that squared function. Thus we have Equations 19 and 20 as

(Eq’ns 21)

But those equations describe the energy flux in the incident ray. To obtain a
description of the energy fluxes in the reflected and transmitted rays in the
two cases given above we simply substitute from Equations 13 or 17 into the
appropriate Equation 21. Recalling that we have
μ_{1}=μ_{0}
and taking the case in which **N∙E**=0, we substitute Equations 13 into the
first of Equations 21 and obtain

(Eq’ns 22)

Given those equations, can we say that the system
described above conserves energy? To answer that question we must multiply the
power fluxes by the cross sectional area of the rays to calculated the total
power in each ray and then add up the results for the reflected and transmitted
rays to see whether the sum equals the total power flowing in the incident ray.
If we have the cross section of the incident ray as A_{i}, then for the
reflected ray we know that A_{r}=A_{i}. The calculation for the
transmitted ray is a bit more complicated.

We know that all three rays have the same projected area
onto the dielectric interface, the surface of the block. We also know that the
area of the projection A’=A_{i}/cosθ_{i}
(that’s simple geometry) and that A’=A_{t}/cosθ_{t},
so we have

(Eq’n 23)

Multiplying Equations 22 and the first of Equations 21 by the appropriate areas then gives us

(Eq’n 24)

That inequality fails to achieve the equality required by the conservation of
energy theorem because the second term on the left of the inequality sign is too
large by a factor of n_{1}/n_{0}.

That problem goes away when we take into account one final factor. We know that the electric fields in the rays remain continuous across the dielectric interface, so the part of the incident ray’s electric field that goes into the transmitted ray does not change. Using magnitudes only, we have Faraday’s law manifested in our rays as

(Eq’n 25)

which gives us

(Eq’n 26)

For the electric field of the transmitted ray we have E_{0i}'=E_{0t},
so we have the ratio of the corresponding magnetic fields as

(Eq’n 27)

The corresponding Poynting vectors differ in the same ratio, so we correct the second of Equations 22 by multiplying it by that ratio. Thus we have what we might call the Fresnel equations of the Poynting vector as

(Eq’ns 28)

These are the equations that we will use below.

The Beam Splitter

A beam splitter consists simply of a thin, flat plate of some transparent substance whose index of refraction differs from that of the medium in which the plate is immersed. (Actual beam splitters are cubes made by gluing triangular prisms together, but I want to use the basic flat-plate version for a reason.) An incident ray of light striking one face of the plate, typically at a 45-degree angle, gets split into a reflected ray and a ray that passes into the plate and thence out the opposite face to continue along a trajectory parallel to that of the incident ray. A beam splitter is rated according to the proportions in which it shunts the power in the incident ray into the reflected ray and the transmitted ray.

We calculate the ratio of the reflected power to the transmitted power, R(r/t), by multiplying Equations 28 by the cross sections of the appropriate rays and then dividing the first equation by the second. Thus we get

(Eq’n 29)

We can use Snell’s law,

(Eq’n 30)

to eliminate the transmitted angle from the equation. Square Equation 30, replace the squared sines by the equivalent expressions using the cosines, and carry out a little algebraic manipulation to get

(Eq’n 31)

Substituting that result into Equation 29 would get very messy, but we
already know that we intend to make
θ_{i}=45
and n_{0}=1, so Equation 31 then becomes

(Eq’n 32)

Substituting that into Equation 29, along with n_{0}=1 and cosθ_{i}=(1/2)^{½},
gives us

(Eq’n 33)

But instead of starting with a given index of refraction and calculating the
corresponding ratio, we want to start with a given ratio and calculate the index
of refraction needed to produce that ratio, so we must solve that equation for n_{1}.
Equation 33 looks like it will give us a depressed cubic, which we can solve,
but the square root of f(n_{1}) dividing the n_{1}-cubed term
makes it impossible to solve with known techniques. However, if instead of the
ratio of reflected to transmitted power flux, we use the ratio of reflected
power flux to incident power flux, we can simply use the first of Equations 28
and, using the criteria stated above, we obtain

(Eq’n 34)

In that equation we have the denominator and numerator as perfect squares, so with a little algebraic rearrangement we can rewrite that equation as

(Eq’n 35)

in which we define for convenience

(Eq’n 36)

Exploiting the fact that A^{2}-B^{2}=(A-B)(A+B), we split
Equation 35 into two factors and note that one of them must equal zero. We thus
have

(Eq’n 37)

which gives us

(Eq’n 38)

and thence

(Eq’n 39)

And we also have

(Eq’n 40)

which gives us

(Eq’n 41)

and thence

(Eq’n 42)

Only one of the solutions of Equations 39 and 42 can be correct, but we have
a simple criterion for choosing the correct one: for all values of r between 0
and 1, n_{1} 1. For a 50-50 beam splitter we have r=0.5, so Equation 39
gives us n_{1}=4.18 and Equation 42 gives us n_{1}=0.71744. So
Equation 39 gives us the index of refraction of the material we must use in our
beam splitter.

Erasing the Poynting Flux

Now we come to the payoff. We want to use a beam splitter of appropriate thickness to make two laser beams cancel each other perfectly. Once we have achieved that goal we ask what happens to the energy in the beams: does it get annihilated or does some other phenomenon emerge to save the conservation law? Well, let’s see.

We have so set up our beam splitter that it slants from
southwest to northeast on our imaginary optical bench. We have two plane
monochromatic waves of coherent light approaching the beam splitter, one from
due south and the other from due west. We have also imposed a plane polarization
upon the waves in such a way that their electric fields point up or down (again
**N∙E**=0). The Poynting vector allows us to keep track of the electric and
magnetic field vectors through their relation with the direction of each wave’s
propagation: the description of the Poynting vector as the cross product of the
electric and magnetic fields tells us that if I curl the fingers of my right
hand in through the smaller angle from the electric field vector to the magnetic
field vector, my thumb points in the direction of propagation. Now let’s look at
a single instant of time, with the waves frozen in place.

The west wave strikes the beam splitter at a designated point on the northwest face and we see that its electric field at that site points up. That fact necessitates that the incoming wave’s magnetic field at the site points due south. At that place the incoming wave gets split into the reflected and transmitted waves, each of which takes one half of the incoming wave’s power. In the reflected wave the magnetic field points due east, because the direction of the electric field has not changed. In the transmitted wave the magnetic field points 35.26 degrees west of south.

On the southeast face of the beam splitter, opposite the first designated point, we have a second designated point where we see the south wave splitting into reflected and transmitted parts. At that point the electric field of the wave points down, so the magnetic field points due west. In the reflected wave the magnetic field points due north and in the transmitted wave the magnetic field vector points 35.26 degrees south of due west.

At the first designated point the transmitted part of the south wave emerges from the beam splitter with its electric and magnetic fields oriented 180 degrees away from the corresponding fields in the reflected part of the west wave. We expect the two waves to cancel each other through the mutual destructive interference of their fields, so we expect to see nothing coming northward off the beam splitter. Likewise at the second designated point the transmitted part of the west wave emerges from the beam splitter with its field vectors oriented 180 degrees away from the corresponding fields in the reflected part of the south wave. Again, we expect the two waves to cancel each other out, so we expect to see nothing coming eastward off the beam splitter. Thus we expect to see two beams of light going into the beam splitter and nothing coming out.

What happens to the energy carried by the incident waves? Does it, as implied above, simply vanish, in blatant violation of the conservation law? Or must we look for some other phenomenon?

To answer that last question we must look inside the beam splitter. Relative to the straight line connecting the two designated points (a line running northwest to southeast), the propagation vector of each of the transmitted waves makes a 9.74-degree angle. Crossing each other at a 19.48-degree angle, the two waves interfere with each other and manifest a standing-wave pattern inside the beam splitter. The extra half-wavelength in the thickness of the beam splitter ensures that on the beam splitter’s faces the two transmitted waves differ in phase by 180 degrees.

On the center plane of the beam splitter the transmitted waves interfere with each other in such a way that the vector sum of the waves’ magnetic fields points in a direction parallel to a line running northwest to southeast. Combining that resultant magnetic field with the vector sum of the transmitted waves’ electric fields in a cross product yields a Poynting vector oriented in the northeast direction. That Poynting vector represents a flow of electromagnetic energy vibrating at the same frequency as that of the transmitted waves, so we expect to see light of that frequency emerging from the northeast edge of the beam splitter. That light draws energy out of the transmitted waves, thereby diminishing the amplitudes of their fields and making them less able to cancel out the reflected waves. Thus we expect that the energies in the reflected waves and in the northeast wave add up to an amount equal to the sum of the energies in the west wave and the south wave, in accordance with the conservation law.

But now we have one more challenge to the conservation law. Given the apparatus and operation described above, we introduce a third incident wave through the southwest edge of the beam splitter. We so adjust that wave that its field vectors point in the directions 180 degrees out of phase with the corresponding field vectors in the northeastward wave that we just created. Thus we expect that third wave and the northeastward wave will cancel each other out, which means that half of the energy from the west wave and the south wave and an equal amount of energy from the third wave simply vanish.

To understand how to interpret that proposition and come up with a full, correct description of what happens, we must recall to mind a fundamental fact about the nature of light. What we have been calling an electromagnetic wave is simply an aleatric field that tells us the probability of finding a photon of a given energy in a certain volume of space at a given time. Our description of the electrotonic field, the source of both the electric and magnetic components of the wave, corresponds to the state function of the quantum theory, in this case the one that appears in the Proca equation. Inside the beam splitter the photons get absorbed and re-emitted by the atoms in their glassy array in accordance with the propagation of the electrotonic field.

In the first cancellation experiment described above the electromagnetic waves coming into the beam splitter from the northwest and southeast interfere with each other and produce an electromagnetic field whose Poynting vector points in the northeast direction. The photons in the streams guided by the waves get absorbed and re-emitted in the northeast direction under the guidance of the probabilities associated with that Poynting vector.

In the second cancellation experiment the photons in the northeastward wave appear to vanish, wiped out of existence by the third input wave. We can deduce how that happens if we recall to mind the fact that the photon is its own antiparticle. Inside the beam splitter, then, we have a weird form of matter-antimatter annihilation occurring: photons and antiphotons cancel each others’ existence. That cancellation requires that each photon/antiphoton pair interact with the atomic structure of the beam splitter: the atoms absorb the energy, linear momentum, and spin carried by the photons, thereby upholding the three conservation laws pertaining to those quantities. We see that a material medium is necessary to align the beams of light in a way that enables perfect destructive interference, so atoms will always be available to uphold the conservation laws.

Appendix: Optical Logic Gates

We have established the proposition that if two laser beams strike a beam splitter as described above, a third beam emerges from one of the edges of the beam splitter. That device performs the same function as does the AND gate in a digital computer.

In essence, the AND gate multiplies two inputs, ones and zeroes, together. When one of our laser beams is turned on it corresponds to a one and when it is turned off it corresponds to a zero. When one laser beam is turned off and the other turned on, we have 0x1=0 and there is no output. When both beams are turned on, we have 1x1=1 and we get a beam from the northeast edge of our beam splitter. This phenomenon is useful in the Boolean algebra used as the foundation for digital computer operations.

If we add the third input beam, as described above, we convert the AND gate into a NAND (Not AND) gate, the reverse of the AND gate. The device then produces a nonzero output if one or neither of the lasers is turned on and a zero output if both lasers are turned on. It simply reverses the action of the AND gate.

Now imagine that we have combined the beams from our two lasers into one beam and then split that combined beam into the two input beams that strike our beam splitter. If one or both lasers are turned on, there will be a nonzero output from the beam splitter’s edge, and if both lasers are turned off, there will be no output. In this case the device works as an OR gate. And by adding the second beam splitter we can turn the device into a NOR gate.

This device can provide all four of the functions that a digital computer requires, so we can imagine optical computers in which specks of glass and filaments of light weave the tapestry of calculation and knowledge. We can thus imagine computers smaller, faster, and cheaper than anything we have today. And we got all of that from questioning a conservation law.

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