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Recently I've had several non-physicist friends ask me, independently of each other, about an experiment where two collinear laser beams destructively interfere along a certain length. Everybody wants to know "where does the energy go?"

Answering that question is not the problem, but I would be more convincing if I knew what experiment they were talking about! None of them can recall where they read it, but it seems to have made the rounds of popular science websites sometime in the last year. I am especially surprised that this has been published now, since similar experiments have been done for decades, so I'm guessing there must be some other twist to the experiment that didn't register in my friends' memories.

Can anybody point me towards a published paper or a popular science article?

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[see at PSE anti-laser-how-sure-we-are-that-energy-is-transported] (…) and post your answer there. – Helder Velez May 9 '11 at 11:01
up vote 8 down vote accepted

If we have the same popular press in mind, the official name of this device - informally known as anti-laser - is a "coherent perfect absorber". See

It was proposed in early 2010 by A. Douglas Stone and collaborators:

It was already observed in late 2010, see e.g.

The anti-laser is just a time-reversal of a laser, so whatever problem with energy conservation you could see exists for both devices. Obviously, there is no problem - the energy of light is being taken from, or deposited to, the atomic excitation energies in the cavity.

There are roughly 10 papers about this subject only

so one shouldn't buy some popular media's hype suggesting that it's one of the hottest things in physics.

If your question were not about anti-laser but about ordinary interference between two laser beams, well, I would leave it as no comment. Of course that the energy is conserved at the end. One may get destructive interference in a big region but this is always compensated by the existence of constructive interference in other regions - e.g. on the boundary of the beams. Maxwell's equations may be showed to imply energy conservation so the total energy can never get lost.

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Thank you very much! This sounds like something that might have registered in my friends' minds if they read it in the popular press, but the way they explained it I thought it was about ordinary destructive interference. Now that I know what people are talking about, I'll be able to answer questions better. – ptomato May 9 '11 at 11:39

The electromagnetic wave energy is an integral notion. The energy space density may depend on particular place but it is not the total energy. As Lubosh says, according to the Maxwell's equations the total energy is conserved in absence of charges. If it is not conserved, then the filed is not governed with the Maxwell's equations.

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EDIT add
The papers are correctly posted in Lubos answer and, if this is all you need, than you can skip this answer
EDIT add end

If two coherent light beams are counter propagating in perfect phase opposition the Poynting vectors are of equal magnitude and directly opposed. The total field is null everywhere those conditions apply and energy is null. This comply with Maxwell equations and the Superposition Principle. The issue that Energy+Energy=0 derives from Superposition Principle.

I think that the Nonradiation_condition uses the same superposition principle to cancel the energy in a region outside a finite one. Paul Ehrenfest, Goedeck, Haus, and Douglas Pinnow. And goes like this (from the same WP-link):
"a distribution of accelerated charges will radiate if and only if it has Fourier components synchronous with waves traveling at the speed of light."

I suspect that the charge do not have a sensor to see if the conditions of radiation are met or not met. The charge always radiate electrostatic field and due to the relativity of motion the radiated pattern can be constructive or not. When it is desctructive (as in the sense of the nonradiation condition) we measure and obtain 0 energy outside the region but the particles always radiate energy at the same rate.

The motion of the particle induce the waves in the radiated electrostatic field.
And when it cancels: energy is not there. 'Cancel', 'Vanish', 'Destroyed'.

This Nonradiation condition, and also the anti-laser experiment work the same way in the vacuum where there are no absorbers, no heat, no particles, ...

I'm clearly stating that the only radiation of any charge is originated only on its continuos electrostatic field.

There is a magical thinking when we say that the particle radiates if it is in motion or other particles in the vicinity are in motion. To be physical It can not be like this, simply.

Explore visually this Radiation2D.exe and its formulation "NEW MATHEMATICAL METHOD FOR RADIATION FIELD OF MOVING CHARGE" by T. Shintake.

To see how we like to complicate the radiation issue see : "The simplest, and the full derivation of Magnetism as a Relativistic side efect of ElectroStatics" by Hans de Vries;

(I've explored this issue PSE-here, but here I've added the nonradiation condition argument)

EDIT add: time-reversal and information loss
The experiment is not time-simetric because after the cancelation any information that could be equally encoded in both of the beams are definitelly lost.
EDIT ad end

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Thank you for the effort, but you're not really answering the question. I asked which article was being referred to. Luboš pointed me to it. – ptomato May 9 '11 at 14:20
Exactlly because I knew you already have read Lubos answer I ommited it from the answer. If you folow the last sentence link you will find it linked there since 23/Feb. But I will edit to add it. – Helder Velez May 9 '11 at 15:34
I'will extend in the body of answer the time-reversal argument and information lost. – Helder Velez May 9 '11 at 15:56

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