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Shark Science

Is There a Limit To a Laser's Energy? 135

Posted by timothy
from the of-course-there-isn't dept.
StartsWithABang (3485481) writes "For normal matter — things like protons, neutrons and electrons — there's a fundamental limit to the number of particles you can fit into a given region of space thanks to the Pauli exclusion principle. But photons aren't subject to that limit; in theory, you could cram an infinite number of them into the same exact state. In principle, then, couldn't you create a laser (or lasing cavity) with an infinite amount of energy inside? Perhaps, but there are some big challenges to be overcome!"
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Is There a Limit To a Laser's Energy?

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  • by Tablizer (95088) on Sunday May 04, 2014 @04:36AM (#46911831) Homepage Journal

    "Update: After a conversation with Chad Orzel, it looks like although there's no limit to the photon energy you can produce, you will at some point--above about 1 MeV in photon energy--start spontaneously producing matter-antimatter pairs of particles whenever your photon interacts with a reflective surface. So at extremely high photon energies, your laser light begins to resemble a matter-antimatter thermal bath rather than merely coherent light."

    So it would act like more Star Wars weapons?

  • History.... (Score:5, Informative)

    by beheaderaswp (549877) * on Sunday May 04, 2014 @05:07AM (#46911885)

    Billions and billions of years ago, even before lord Xenu, there was a scientist who pulled this off.

    Blext Telfrawd, an A type Hixoid, did get an infinite number of protons into a finite space. Then the containment field faltered, obliterating the iteration of his universe..

    Most historians agree this was tragic for it ended his universe, and created one with Justin Bieber. Sentients who were able to achieve trans-dimensional universital access, send a message to you from the past: It's just too risky to repeat the so called "Bieber Event",

    You've been warned.

  • by AchilleTalon (540925) on Sunday May 04, 2014 @05:56AM (#46911949) Homepage

    The energy of a photon is characterized by its wavelength. In a laser, the wavelength is constant. You have a large amount of photons which are coherent but at an almost single wavelength. When the article is talking about 1 MeV, it falsely interprets this as if the laser is emitting a single photon at 1 MeV. That is not what happen. It emits many photons in coherence which the sum of energy of all the individual photons will reach 1 MeV or more. Each photon cannot create an electron-positron pair and all photons collectively cannot create an electron-positron pair.

    A 1 MeV photon would be a gamma ray photon and it is not true at all, your laser doesn't change its wavelenght as more more "energy" is emitted. In fact, we should instead talk about the power of the laser rather than its energy. The power being the amount of energy emitted by unit of time.

  • Kugelblitz (Score:4, Informative)

    by Guppy (12314) on Sunday May 04, 2014 @05:57AM (#46911953)

    Eventually the laser energy will create a black

    There is a specific term in astrophysics for such a theoretical object: []

  • by poodlediagram (1944244) on Sunday May 04, 2014 @06:47AM (#46912037)

    This is the usual muddle up between energy and intensity.

    There is no apparent upper limit to the energy of a photon. The galaxy Markarian 501 emits photons in the teraelectronvolt (TeV) range.

    The question here is about intensity. The relativistic energy-moment dispersion, E^2=(mc^2)^2+(pc)^2, which applies to all on-shell particles, has a gap when m>0. This gap, which is about 1 MeV for electrons and positrons, can be overcome when the electric field (generated by a sufficient number of photons, irrespective of their energy) approaches the Schwinger limit of about 1.3 x 10^18 V/m. At this point, virtual electron-positron pairs can be created in abundance because the mass gap has been overcome, and electromagnetism then becomes non-linear. Pumping in more photons after this simply creates more virtual e-p pairs.

    Hope that helps.

    (IAAP working on this topic).
  • by Zorpheus (857617) on Sunday May 04, 2014 @07:25AM (#46912115)
    I just searched for an answer to this question. Seems that pair generation by irradiation of matter (e.g. a mirror) is shown experimentally and can reach quite high intensities: []
    Generation in vacuum though seems to be shown only in models until now: [] []
    Seems that the reaction rate is much lower, so maybe this is not a limiting factor for building a laser.
    Normally high intensities are achieved by building a pulsed laser. This produces a beam of laser pulses, which is then focussed into a tiny spot. Intensities in this spot can be alot higher than inside the laser cavity. You could achieve higher laser intensities just by building a larger laser (like [] ).
    Inside the laser cavity intensities are normally limited by the effects of nonlinear optics ( [] ), which occur in all kinds of matter.
  • Re:E=MC^2 (Score:4, Informative)

    by Anonymous Coward on Sunday May 04, 2014 @08:12AM (#46912247)

    They did. Photons will gain an increasing energy density, and a corresponding increasing pressure, which both have an impact on the gravitational field and, if high enough in a concentrated area, could indeed cause the appearance of a black hole. That does not mean they have "mass". The photon energy is basically E=hbar*omega or E=pc, depending how you want to write it. Here omega is the angular frequency, p the momentum, while hbar and c are Planck's constant and the speed of light respectively.

  • Re:E=MC^2 (Score:5, Informative)

    by MillionthMonkey (240664) on Sunday May 04, 2014 @03:55PM (#46914679)

    E = mc^2 specifically applies only to objects that have nonzero mass and are at rest with respect to the observer. Photons are massless and move at the speed of light.

    The general equation is E = sqrt((mc^2)^2 + (pc)^2) for rest mass m and momentum p. If a particle has mass and is at rest, then p=0 so E=mc^2. If a particle is massless, then m=0 so E=pc.

    (The "m" here refers to rest mass m0, not the "relativistic mass" m* which is defined as m* = m0 / sqrt(1-(vc)^2)). Relativistic mass is best thought of as a fake concept to hide the ugly sqrt denominator. People can imagine things getting heavier when they're moving, and can keep saying "Einstein discovered E=mc^2". But it still has division-by-zero problems with massless particles, and things don't really "get heavier" when they move, so if you try to avoid thinking in terms of m* you won't get as confused. Neither m nor m* makes E=mc^2 work with photons.

    Imagine if a bundle of photons could gather and form a "black hole". The hole and its event horizon would be constrained to move at the speed of light, which you can't, since you have mass. so you might easily escape its event horizon- you wouldn't have time to fall in before the thing was gone. Real black holes have mass and don't move at the speed of light relative to anybody.

Stupidity, like virtue, is its own reward.