# Destructive interference

Take the most simple academic example for interference. Since it is not any real experiment, one can have shocking contradictions.

For example: 2 monochromatic plane waves with (parallel) amplitudes propagating in the same direction. The Poynting vectors of the 2 waves without superposition are always constant. Once superposed, the resulting Poynting vector is constant, but is dependent on the phase difference.

Thus, how can we explain the energy balance? If there is an energy redistribution it may be easy perhaps, but when the three values are constant in space... What is the right explanation? –

I do not think "energy flow" is a well defined variable for electromagnetic waves.

The single photon at a time double slit experiment, shows that the energy from the dark lines has gone to the bright ones

1. Single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1’000, and 500’000 frames.

So there is no problem of conservation of energy.

This video from MIT open courses helps in understanding the complexity of interference patterns from two beams : Optics: Destructive interference - Where does the light go? . It experimentally shows that the light from the dark fringes goes back to the source of the collimated beams. So overall energy is conserved.

One has to have a specific experiment, including the source of the beams, in order to answer in detail about conservation of energy, and see where the energy flows during interference.

• Hi @anna v can you explain what you mean by “I do not think "energy flow" is a well defined variable for electromagnetic waves”. I know you are aware of the Poynting vector so you must be making some more subtle point here
– Dale
Nov 9 '20 at 12:58
• Energy flow is well defined for EM waves. It is the $T^{0i}$ component of the energy-momentum tensor.For free waves this tensor is symmetric, in which case the energy flow equals the momentum density divided by $c$. In gauge invariant electromagnetism it is $1/c$ times the Poynting vector. Nov 9 '20 at 23:00
• @Dale while I cannot speak for AnnaV I also believe that while the Poynting vector itself $\mathbf{S}=\mathbf E \times \mathbf{H}$ is well-defined the real energy flux is not in the sense that one may add to $\mathbf{S}$ any solenoidal vector field without changing anything measurable. This goes along with the observation that only the work rate $\dot W_e = \int E\frac{\partial D}{\partial t} dV$ is a directly measurable quantity by instruments and not its "volumetric density" $\dot w_e = E\frac{\partial D}{\partial t}$, same observation for the magnetic energy/work. Nov 10 '20 at 0:38
• @dale , I am commenting on the " energy flow". flux and flow are two different words. Nov 10 '20 at 5:17
• @my2cts can you give a link for this definition? I could not find it by google. Nov 10 '20 at 5:19

The energy balance is indeed an interesting problem. For a monochromatic plane wave the source is an infinite sheet of sinusoidal current.

It is not trivial but is straightforward to calculate the Poynting vector for this arrangement. When you do so you find that energy propagates away from the current sheet with equal power density on both sides of the sheet. When you further calculate $$\vec E \cdot \vec J$$ at the current sheet itself you find that the work done by the current is equal to the radiated power. So the conservation of energy holds.

Now, Maxwell’s equations are linear and translation invariant, so you can simply shift the current sheet some distance to get two current sheets. The total field from the sum of the two current sheets is simply the sum of the fields from each sheet.

However, although the fields add linearly, the energy is not linear. So you could take a current sheet which by itself produces waves with some given power density $$P_1$$ and a second sheet which by itself produces a power density $$P_2$$ and when you add them together you get waves with a power density $$P\ne P_1+P_2$$.

The key is to recognize that the two sources affect each other. If you calculate the work done by the first sheet you will find that $$\vec E \cdot \vec J \ne P_1$$. In other words, the presence of the second source changed the work needed by the first source to produce the same current.

Such sources are called coupled, and this coupling can be damaging to RF power amplifiers driving coupled antennas. The power density of the two waves is different from the sum of the original waves, but it matches the power produced by the coupled sources.