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Is There a Limit To a Laser's Energy?

timothy posted about 6 months ago | from the of-course-there-isn't dept.

Shark 135

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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Photons aren't the problem (4, Funny)

Anonymous Coward | about 6 months ago | (#46911753)

It's the number of sharks you can fit into a given region of space.

Mmmm, spam (0)

Anonymous Coward | about 6 months ago | (#46911955)

Is it more or less than the times you can spam your own blog on Slashdot? Coz that's all I really care about.

Re:Photons aren't the problem (0)

Anonymous Coward | about 6 months ago | (#46912077)

what about 'angels on a pinhead'? isn't this the old problem?

Re:Photons aren't the problem (1)

davester666 (731373) | about 6 months ago | (#46914073)

No. photons actually exist. you can literally see them.

Re:Photons aren't the problem (1)

flyneye (84093) | about 6 months ago | (#46912171)

It may be the number of D cells you can fit into the units on the sharks.

No (0)

Anonymous Coward | about 6 months ago | (#46911757)

Except nex time. Later. Before then.

You keep using that word (0)

Anonymous Coward | about 6 months ago | (#46911767)

Infinity doesn't mean what you think it means.

Re:You keep using that word (3, Insightful)

Mitchell314 (1576581) | about 6 months ago | (#46912643)

We can outright throw it out, as there isn't an infinite amount of energy in the visible universe.

Re: You keep using that word (0)

Anonymous Coward | about 6 months ago | (#46913347)

Where is dark energy coming from? What is the limit of energy in the universe?

Re:You keep using that word (1)

superwiz (655733) | about 6 months ago | (#46912877)

Not having a finite limit means infinite. At least it does so in math. The summary got it right.

Re:You keep using that word (2, Insightful)

Anonymous Coward | about 6 months ago | (#46913109)

No, it doesn't. In maths, we use the phrase, "finite but unbounded". This describes the natural numbers, for example: each specific integer is strictly finite, but there's no finite upper limit.

Re:You keep using that word (-1)

superwiz (655733) | about 6 months ago | (#46913559)

Each integer very much does have a finite limit - itself. Integers, as a whole, do not have a bound. Which is why unbounded implies infinite. Oh, and there is no such thing as "maths". There is math, there is mathematics, and there is people who understand neither and are proud of it. But math is singular. There is only one math. Oh, and "we"? *I* do not. And I have a PhD in math. So do go on lecturing me on trivialities.

Re:You keep using that word (0)

Anonymous Coward | about 6 months ago | (#46913711)

You should probably send that back to whoever posted it to you.

Re:You keep using that word (0)

Anonymous Coward | about 6 months ago | (#46914589)

That we may have an unbounded number of photons in the same state is not the same that we can have infinitely many photons there. The theory may say that we could have n photons there for any natural number n, but that is still a finite number of them in each case.

So the way I read it, the summary might be wrong (unless the theory really says we can have infinitely many photons in the same state). I don't know if the universe can contain infinitely many photons at all.

Re:You keep using that word (1)

superwiz (655733) | about 6 months ago | (#46914929)

Infinity is not a number. It only exists as a formal notion to indicate lack of bound. One can talk about a "point at infinity", but to do so, one must adjoin to the set of numbers an object that is not numeric, but which only exists as a formal notion. Therefore, any set which contains a point at infinity is not a set of numbers. Yes, this is a "no true Scotsman" argument, but it has to be so because it is an argument about definitions.

Re:You keep using that word (0)

Anonymous Coward | about 6 months ago | (#46915005)

What I'm talking about here is cardinality. Maybe there can be a bounded number of photons at one place, perhaps unbounded but finite (what is what I was talking about in the post you replied to), could also be countably infinitely many (has the same cardinality as the set of natural numbers, which is already infinite), or even uncountably many.

Re:You keep using that word (0)

Anonymous Coward | about 6 months ago | (#46914779)

Oh, and there is no such thing as "maths". ... And I have a PhD in math.

How many years did you study math without ever encountering British English and "maths"?

Re:You keep using that word (1)

superwiz (655733) | about 6 months ago | (#46914843)

I resist as incorrect every time I have a chance. It only became the British short form fairly recently. And, to anyone who uses the singular short form "math", it makes a suggestion that mathematical conclusions are open to being overturned the way scientific conclusions are. Since mathematical conclusions are final, there is only one math.

E=MC^2 (2, Interesting)

Anonymous Coward | about 6 months ago | (#46911773)

Eventually the laser energy will create a black hole, provided some other exotic effect doesn't occur first. Realisitcally though it's not possible to attain those kinds of photon densities (nothing can reflect anywhere close to well enough for starters).

Re:E=MC^2 (0)

Anonymous Coward | about 6 months ago | (#46911883)

E = mc^2 doesn't mean that the photons will get mass at high energies.

Re: E=MC^2 (0)

Anonymous Coward | about 6 months ago | (#46911965)

They have zero rest mass, this does not mean zero mass when its not at rest. Good luck finding a photon at rest

Re:E=MC^2 (1)

K. S. Kyosuke (729550) | about 6 months ago | (#46912191)

Actually, it means precisely that. (Unless they lied to me in high school!)

Re:E=MC^2 (4, Informative)

Anonymous Coward | about 6 months ago | (#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 (1)

sillybilly (668960) | about 6 months ago | (#46914297)

It does not mean they have "rest mass", but they do actually have "mass", m=E/c2, that should create gravity around the photon ( if gravity is such as thing that gets created and is present around an object, and it's not just a straight "instantaneous action at a distance" between two objects. now my head is starting to hurt)

Re:E=MC^2 (0)

Anonymous Coward | about 6 months ago | (#46914681)

So Ronald Mallett was right, he just cant yet afford the power bill in his time machine? I Knew It!

Re:E=MC^2 (-1)

Anonymous Coward | about 6 months ago | (#46912533)

Hey bigmouth: You're being called out http://slashdot.org/comments.p... [slashdot.org]

Re:E=MC^2 (1)

Mitchell314 (1576581) | about 6 months ago | (#46912659)

It's been a long time since I last studied physics, but an extremely strong EM field can have mass.

Kugelblitz (4, Informative)

Guppy (12314) | about 6 months ago | (#46911953)

Eventually the laser energy will create a black
hole

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

http://en.m.wikipedia.org/wiki... [wikipedia.org]

That's a limit on energy DENSITY (3, Insightful)

Ungrounded Lightning (62228) | about 6 months ago | (#46913215)

Eventually the laser energy will create a black hole, provided some other exotic effect doesn't occur first.

That's a limit on energy density, not total energy in the laser. In principle you could use a very WIDE laser opterating below the black-hole thrshold and focus the beam externally (which, if it's powerful enough, it might do eventually, by self-gravitation, after leaving the cavity, even if the cavity geometry made it emit a colimated, rather than a converging, beam.) Thus, making a kugelblitz with a (very wide) laser might be theoretically possible (if "some other exotic effect" didn't make the required laser cavity to wide to be physically realizable).

I'd imagine "Some other exotic effects" might include the electric field component of the coherent light becoming strong enough to polarize the vacuum and create particle-antiparticle pairs from multiple photons, dissipating their energy, somewhere WAY below the threshold of gravitic-collapse effects. So you'd need a REALLY WIDE laser and REALLY GOOD optics to make your external-to-the-laser black hole.

Of course the question, being phrased in terms of Bose-Einstein vs. Fermi-Dirac statistics and "infinite" energy was really about energy density in the cavity - just poorly phrased. So you answered the question that was REALLY being asked.

Re:E=MC^2 (4, Informative)

MillionthMonkey (240664) | about 6 months ago | (#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.

Re:E=MC^2 (0)

Anonymous Coward | about 6 months ago | (#46914829)

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.

Except for the whole stress-energy tensor in GR depending on the energy density in terms of E/c^2, which means that even photons have a gravitational effect similar to as if you did apply E=mc^2 to get an effective mass. GR allows for a high enough photon density to form a black hole. The fact the light is traveling at c doesn't stop this, just like it is capable of forming sublight speed particles from pair production (with a little care when handling the momentum via a extra particle). The pedagogical handling of rest and relativistic mass is a mess, but you still have to be careful of over-correcting and resulting in a reasoning that doesn't actually work under GR.

ExactoMundo (2)

donnie Freyer (2881319) | about 6 months ago | (#46911779)

Napoleon, like anyone can even know that.

There are some... er, limits: (4, Informative)

theNAM666 (179776) | about 6 months ago | (#46911825)

Re:There are some... er, limits: (0)

Anonymous Coward | about 6 months ago | (#46912005)

I found this a better analysis
 
  Are Black Hole Starships Possible? [arxiv.org]

Re:There are some... er, limits: (4, Interesting)

drolli (522659) | about 6 months ago | (#46912089)

But long before that happens the question is if the laser can remain a laser.

A laser needs some kind of nonlinearity in the medium. Any nonlinearity introduces a scale. So the real question is: At which power does of-resonant driving cause transitions (e.g. Landau-Zener) or of-resonant shifts (Stark shift) and can you actually theoretically contruct a medium which fulfills the criteria to serve as a lasing medium for an arbitrary large scale of power?

As a starting point for an examination of such questions i recomment the Quantum Optics Toolbox for Matlab by Sze Tan.

Spoiler at the end. Answer is "No" (5, Informative)

Tablizer (95088) | about 6 months ago | (#46911831)

"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?

Re: Spoiler at the end. Answer is "No" (0)

Anonymous Coward | about 6 months ago | (#46911841)

Exactly. May the Fourth be with you.

Re:Spoiler at the end. Answer is "No" (1)

Zorpheus (857617) | about 6 months ago | (#46912009)

This is not only a problem for high photon energies on mirrors, but also for high energy densities in free space. If I remember correctly this starts at about 10^20 W/cm^2

Re:Spoiler at the end. Answer is "No" (1)

jovius (974690) | about 6 months ago | (#46912029)

So it wouldn't be possible to construct a reflective surface that's not solid in the traditional sense but a reactive field of energy, which would guide the process at the point of interaction - to bypass the limit?

Re:Spoiler at the end. Answer is "No" (1)

ixtapa (903468) | about 6 months ago | (#46912093)

Some sort of force field?

Re:Spoiler at the end. Answer is "No" (4, Informative)

Zorpheus (857617) | about 6 months ago | (#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:
http://journals.aps.org/prl/ab... [aps.org]
Generation in vacuum though seems to be shown only in models until now:
http://iopscience.iop.org/0295... [iop.org]
http://journals.aps.org/pra/ab... [aps.org]
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 http://en.wikipedia.org/wiki/N... [wikipedia.org] ).
Inside the laser cavity intensities are normally limited by the effects of nonlinear optics ( http://en.wikipedia.org/wiki/N... [wikipedia.org] ), which occur in all kinds of matter.

Re:Spoiler at the end. Answer is "No" (0)

Anonymous Coward | about 6 months ago | (#46912663)

That is energy per photon....you could just keep your photon energies below 1 MeV (which is gamma ray or at least hard X-ray, by the way), and just pile more low energy photons into the same space to increase the energy density...say visible light photons of ~1 eV.

Re:Spoiler at the end. Answer is "No" (0)

Anonymous Coward | about 6 months ago | (#46912693)

At least until they start recombining or scattering off each other....

Compton Scattering (2)

Roger W Moore (538166) | about 6 months ago | (#46913061)

This effect will not only kill at high energies but at high intensity too. With a high enough intensity you can have multi-photon interactions to achieve the same total energy. However pair production is not the only process you have to worry about compton scattering [wikipedia.org] will occur as well. This will impose an intensity limit well below pair production energies.

Essentially reflected photons will have less energy than incident photons and as the energy increases so too does this energy difference. It is caused by relativistic effects when a photon is reflected from an electron. At visible wavelengths the effect still occurs but is not noticeable but crank the energy up to several keV and your photon has an energy more comparable to the electron rest mass energy and the effect kicks in.

Re:Spoiler at the end. Answer is "No" (0)

Anonymous Coward | about 6 months ago | (#46913085)

So, an array of laser produces photons of more than 1 MeV. Use magnetic field for separating and focus into electon and positron beams. Focus and direct and no more Alderaan. Princess Leia cries and curses Grand Moff Tarkin. There some engineering problems here.

Re:Spoiler at the end. Answer is "No" (1)

tpjunkie (911544) | about 6 months ago | (#46913781)

Chad Orzel was my college physics professor. Cool dude.

Re:Spoiler at the end. Answer is "No" (0)

Anonymous Coward | about 6 months ago | (#46914719)

lol," Don't, I repeat, Don't use this laser to fill my house with popcorn! OK???? Funny guy!

Finally! (0)

Anonymous Coward | about 6 months ago | (#46911847)

Something worth reading.

Re:Finally! (0)

Anonymous Coward | about 6 months ago | (#46911857)

Gotta use the cache for now.

Not even for "normal" matter (1)

DMiax (915735) | about 6 months ago | (#46911861)

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.

Wrong, unless you assume space is discretized, which might happen around Planck's length, but has never been proven theoretically nor experimentally.

Bosons vs Fermions. (0)

Anonymous Coward | about 6 months ago | (#46911865)

If you did physics like me then it's not that mysterious. There are some problems. The particles would burn through the cavity long before 'it' got 'infinite,' not that that sentence makes any sense whatsoever. Also where the !@#k are you going to get infinite power to pump the stupid thing?(!)

Re:Bosons vs Fermions. (1)

Anonymous Coward | about 6 months ago | (#46911901)

  1. 1.Put up a site
  2. 2.Link to slashdot
  3. 3.Use energy of the burning servers.
  4. 4.Profit!

Re:Bosons vs Fermions. (2)

ledow (319597) | about 6 months ago | (#46911913)

The power generation isn't one billionth as hard as - how the hell do you get that energy (presumably electrical) to the device in the first place in a usable format? You can alway just build a nuclear fusion plant, then another, then another, then another, then another, in close proximity.

But somewhere, somehow, you have to transport or convert that amount of energy in a non-light way, which is going to involve some humungously gigantic amount of heat on a physical component, or some monstrously huge device to attempt to dissipate the heat.

The problems of generation are solveable - we just need a way to harness something like a Sun (e.g. Dyson spheres). The problem you really have is how do you concentrate that energy onto a point such that it generates a laser?

Re:Bosons vs Fermions. (0)

Anonymous Coward | about 6 months ago | (#46911943)

The problem you really have is how do you concentrate that energy onto a point such that it generates a laser?

In theory you can put a beam in orbit around a black hole. That way you can accumulate energy over a long time.

I guess it should be possible to "slingshot" the beam between two black holes too if you want the energy to be accessible.

Re:Bosons vs Fermions. (0)

Anonymous Coward | about 6 months ago | (#46914853)

Photons orbiting a black hole would be in unstable orbits and would not stay there long, assuming you could even get them into anything close to an orbit in the first place.

Re:Bosons vs Fermions. (1)

ultranova (717540) | about 6 months ago | (#46912031)

But somewhere, somehow, you have to transport or convert that amount of energy in a non-light way, which is going to involve some humungously gigantic amount of heat on a physical component, or some monstrously huge device to attempt to dissipate the heat.

Why? If raw power is all you're concerned with, just use gamma rays and/or neutrons from an exploding nuclear bomb to pump your lasing media. Sure, you'll only get a single pulse, but that's all you'll need.

Heck, you could surround the bomb with primary converters to capture as much energy as possible, and focus them on a secondary converter to get a single beam, Death Star style.

Re:Bosons vs Fermions. (1)

Kjella (173770) | about 6 months ago | (#46912051)

The problems of generation are solveable - we just need a way to harness something like a Sun (e.g. Dyson spheres). The problem you really have is how do you concentrate that energy onto a point such that it generates a laser?

Well, if you have a Dyson sphere you can shoot a laser from every point on the surface facing the target, they won't be perfectly aligned but they will pass through the same volume of space, like a magnifying glass effect with lasers. That should make a rather nice bug zapper.

Re:Bosons vs Fermions. (0)

Anonymous Coward | about 6 months ago | (#46912777)

The problem you really have is how do you concentrate that energy onto a point such that it generates a laser?

You don, you build a very physically large laser that would then need a low energy and power density. Or use one that has a broadband amplifier to you can split a short pulse into its constituent frequencies, amplify them separately, then recombine to get a short pulse, as is currently done for ultra-fast lasers.

Re:Bosons vs Fermions. (3, Funny)

Hognoxious (631665) | about 6 months ago | (#46911917)

Also where the !@#k are you going to get infinite power to pump the stupid thing?(!)

Two or three of Elon Musk's batteries of sheer awesomeness should do it.

Make it four, in case one catches fire.

History.... (4, Informative)

beheaderaswp (549877) | about 6 months ago | (#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.

Putting the cart before the horse? (3, Interesting)

Chas (5144) | about 6 months ago | (#46911893)

Okay. Interesting on a theoretical level.

The main problem with testing this is "how does one generate infinite or near-infinite energy" to power something like this?

Of course, if we've answered that, we're ALREADY in a place where we've either wiped ourselves out (accidentally or otherwise), or we've basically solved the greatest real-world problem in the history of humanity.

Re:Putting the cart before the horse? (2)

wisnoskij (1206448) | about 6 months ago | (#46912245)

I think the idea if that you could create a stable laser, that exists in some reflective box. And you would keep adding power to it forever.
Or at least until it exploded.

Re:Putting the cart before the horse? (0)

Anonymous Coward | about 6 months ago | (#46912565)

Black holes are the answer to everything. The light..wait a minute, or a year. Put that a millennium.

Oblig. XKCD (0)

Anonymous Coward | about 6 months ago | (#46911935)

I got nothing.

But what a great what-if question you could forge from it.

Re:Oblig. XKCD (4, Interesting)

FatLittleMonkey (1341387) | about 6 months ago | (#46911951)

Yes [xkcd.com] .

Wrong interpretation of energy (4, Informative)

AchilleTalon (540925) | about 6 months ago | (#46911949)

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.

Re:Wrong interpretation of energy (0)

Anonymous Coward | about 6 months ago | (#46911999)

I thought he was talking about the energy density at the focal point, not the energy of a single photon.

Re:Wrong interpretation of energy (0)

Anonymous Coward | about 6 months ago | (#46912045)

> I thought he was talking about the energy density [...]

Then he chose the wrong units. Energy density would be in MeV per cubic furlong or whatever.

Overall, a very sloppy article (including the point where he cites Pauli's exclusion principle ("no two fermions ... in the same *state*") and moves nonchalantly to "no two fermions may occupy the same *volume of space*". Is state the same as space? How are they related?

Of course I wouldn't expect a rigorous explanation (most of us wouldn't understand that anyway), but at least a mention in-passing.

Re:Wrong interpretation of energy (5, Informative)

poodlediagram (1944244) | about 6 months ago | (#46912037)

Exactly.

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).

Re:Wrong interpretation of energy (1)

grimJester (890090) | about 6 months ago | (#46912357)

(IAAP working on this topic).

I demand proof in the form of a youtube video of your laser!

Re:Wrong interpretation of energy (1)

Attila the Bun (952109) | about 6 months ago | (#46912055)

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

He is indeed talking about 1 MeV per photon. He's discussing the theoretical limits of photon power density in a hypothetical gamma-ray laser with an adjustable wavelength. An ordinary laser pointer stores more than 1 MeV of energy in its lasing cavity, although a physicist would not typically use eV to describe the combined energy of a light beam.

Re:Wrong interpretation of energy (2)

Too Much Noise (755847) | about 6 months ago | (#46913031)

He is indeed talking about 1 MeV per photon.

he is jumbling together a lot of nonsense, imnsho.

He starts with the idea of an ordinary laser. Those are not even in the X-ray range, nevermind the MeV gamma-ray range. Then he wants to 'compress' the lasing cavity to *ahem* reach black-hole level of energy densities. While you can transfer energy to the radiation field (thus shifting up photon energies from the visible/UV range) you'll need a HECK of a fast compression to reach the electron-positron generation threshold. So that's nonsense.

Second, lasing does not happen in effing vacuum. Your first problem if you increase photon density, assuming your mirrors do not start to degrade before that, is nonlinear effects. Both in the lasing medium and in the mirrors. You start losing photons via multiple photon absorption that will give you back a higher energy photon that most likely escapes your cavity (goes in the wrong direction most of the time, and when it goes in the right direction the decreased mirror reflectivity and absorption/reemission x-section will not keep that energy contained for long). He never even sees that one coming.

Third, his armchair laser building scenario conveniently ignores all the losses that a real laser system has to contend with. The most obvious part being heat dissipation. Your pretty 99.999% reflective mirror will start to degrade rather quickly if you increase too much the incident radiation density without keeping it adequately cooled (this goes back to several things - normal absorption coefficient in that 0.001% that does not get reflected, having a lasing medium inside the cavity that loses energy to walls, nonlinear absorption effects in the mirror, etc.). Once that happens, you start to say goodbye to the containment properties of your lasing cavity, and thus to your 'bajillion increase in laser field energy density' plans for taking over the world. Try again tomorrow night, Brain.

Fourth ... bah, why bother. This is pretty much a jumbled collection of ideas that you'd expect from someone taking a first course in a given field and imagining things without an effective reality check. Perfect /. front page material.

Re:Wrong interpretation of energy (1)

epine (68316) | about 6 months ago | (#46913809)

Then he wants to 'compress' the lasing cavity to *ahem* reach black-hole level of energy densities.

It seems pretty clear to me—I took that same first course—that a neutrino is just a white hole (moving at the speed of light) made up of photons which such a strong self-interaction they can't escape from themselves and thus refuse to interact with much of anything else.

This all seemed to fit with the gravitational contribution of the EM Stress Energy Tensor until I saw a post from Lubos on Stackexchange about the non-zero photon pressure and their Tii spatial components in GR, so I'm now back to looking for a different way to pretend I have a clue.

Re:Wrong interpretation of energy (0)

Anonymous Coward | about 6 months ago | (#46912175)

At one level that the EM field is quantised into discrete excitations we call photons. (frequency more characteristic of energy than wavelength, E = hbar*omega, wavelength varies under various conditions) Interactions/resonances with other systems occur in these quantised amounts. However there is uncertainty in the energy contained within small volumes of the field, particularly over short time/length scales in a dE*dt > hbar/2 Heisenberg sense and at some point you have enough energy density such that 2 * 0.51 MeV/c^2 positron/electron pairs are likely to be created & destroyed (pair production). Dirac predicted these troubling particle creations & destructions when merging special relativity into the QM wave equations. Once the energy in a volume c*dt across is high enough the system behaves relativistically & particle number is not conserved. Energy can even be "borrowed" from uncertainty momentarily so this occurs with some probability at any energy. Photons aren't little billiard balls they're just a representation of the quantisation of modes of the field and not the only picture of what's going on. Quantum field theory / Quantum Electrodynamics are better descriptions of what goes on at his level.

A more troubling question that brings up for me is why other energies of photon aren't produced. Perhaps that's where power broadening comes in. I think the resonant cavity would be one factor that would strongly suppress other modes.

Re:Wrong interpretation of energy (0)

Anonymous Coward | about 6 months ago | (#46912513)

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.

While I agree with the first statements, the last one is not necessarily true. Due to Heisenberg's Uncertainty Principle a particle's position (x) and momentum (p) cannot be known beyond a certain limit: delta_x*delta*p >= h/(4*pi)

In the case of light this uncertainty, also called the Fourier limit, expresses itself in terms of frequency (f) and time (t): delta_f*delta_t >=1/(4*pi)

So for a continuous wave laser delta_t tends to infinity and the frequency variation goes to zero but in the case of pulsed lasers, an ultrashort pulse with a duration of 10E-14 seconds (10 fs) corresponds to a frequency variation of about 8 THz, so if your pulse is centered around 700 nm in wavelength that means that your pulse width is about 190 nm or ~80 nm if you take the full-width at half-maximum and assume a Gaussian emission spectrum.

Multi-photon interaction (0)

Anonymous Coward | about 6 months ago | (#46912661)

all photons collectively cannot create an electron-positron pair.

This is false. Multiple photons can participate in an interaction and work with their combined energy. The probability of this happening is rather low compared to a single photon (and for pair production, it is rather in-efficient until you get above a couple MeV anyway), but typically the effect is also quite non-linear in terms of intensity. This is seen with laser induced breakdown of material. No individual photon has the ionization energy, and the particular material does not have a transition level that matches the photon, so it is not a matter of exciting the atoms first. Instead, there is a small chance that two photons (or more) will interact at the same time to combine energy and ionize the atom.

This effect is important and useful because of the non-linear response to intensity, so that if you focus down a beam you can get a very small region of activity, instead of a gradual increase in activity as you approach the focus point.

Re:Wrong interpretation of energy (0)

Anonymous Coward | about 6 months ago | (#46914851)

Thermodynamics states that energy can neither be created nor destroyed so If you do figure out a solution to the power problem, you might want to christen the "Laser" with this song, because it is what you would be doing...

http://www.youtube.com/watch?v=xNnAvTTaJjM&feature=kp

Even if there is no limit (1)

Vinegar Joe (998110) | about 6 months ago | (#46911961)

Would it be effective against a Dikironium cloud creature?

How much energy would you need... (0)

Anonymous Coward | about 6 months ago | (#46912111)

How much energy would one need in order to fire a laser so powerful the atmosphere would just instantly ignite.
I just need to know for, hmm, educational purposes. I am not an evil-doer, honest. That guy over there is, get him.

Really though, the amount of power to come out of these lasers to be SUSTAINED beams would be immense.
We are talking fusion-power generations here. And we are trying to ignite a fusion reactor WITH a powerful laser! The madness!
Hopefully one day. Fusion has the ability to change everything about our society once we get around the initial hurdle of making the damn thing work and getting it smaller and more optimized.

bomb-pumped X-Ray lasers (1)

dltaylor (7510) | about 6 months ago | (#46912129)

If you can channel the energy of a fusion explosion into many lasing-while-ionizing rods (think "Real Genius"s death ray laser, but MUCH larger) you could pack so many X-Ray photons into a burst that the impact (momentum transfer) alone destroys the target's armor, at least according to David Weber.

Yes (0)

Anonymous Coward | about 6 months ago | (#46912147)

42

Old news? (4, Interesting)

KClaisse (1038258) | about 6 months ago | (#46912169)

There was an article from 2010 that talked about the theoretical limit to laser beam energy. From the article:

"At high laser intensities interaction of the created electron and positron with the laser field can lead to production of multiple new particles and thus to formation of an avalanche-like electromagnetic cascade"

Here's the link to the article in question: http://physicsbuzz.physicscent... [physicscentral.com]

That article was ultimately using this [nytimes.com] article as a source.

Please (0)

Anonymous Coward | about 6 months ago | (#46912183)

Stop creating these stupid web layouts. With the full screen opening page and the rest below.

Photons Under Pressure (0)

Anonymous Coward | about 6 months ago | (#46912227)

Theoretically there is a definable limit to the number of photons of a given energy that can be constrained within a specific region. It comes from the fact that photon interactions with normal matter transfer energy see https://en.wikipedia.org/wiki/Radiation_pressure. So that containing photons by reflection in present laser equipment does exert an infinitesimal force outwards on the mirrors.

So as I see it the limit is one of how well made the laser containment equipment is manufactured, but there is a limit where no material could resist the tendency for the device to tare itself apart, I leave it to the reader to do the actual engineering math to work out what that is.

Silly article (0)

Anonymous Coward | about 6 months ago | (#46912233)

The article outlines how lasers work but it's basically this: If we could build powerful lasers, which we do not know how to do due to technical issues, we had powerful lasers! Awesome!

Car analogy: There is no theoretical limit how far you can drive with your car! If we could build cars that drive further, we could drive further with our cars! Even an infinite mileage seems possible! Awesome!

Overcoming the "black hole" problem (1)

MobyDisk (75490) | about 6 months ago | (#46912341)

It is well-known that once you exceed a certain energy density you create a black hole. This is why the Death Star superlaser consists of multiple small lasers that combine.

Of course there are limits. (0)

Anonymous Coward | about 6 months ago | (#46912467)

That's why ZPMs eventually go flat.

File this under "smart people are stupid" (1)

NemoinSpace (1118137) | about 6 months ago | (#46912551)

IANAP. Which means, i probably have a better understanding of the subject than they do. Theoretically.

Eddington Limit (2)

mbone (558574) | about 6 months ago | (#46912597)

As a certain energy density, the radiation pressure from the photons will be stronger than the tensile strength of the optical cavity, and the laser will blow apart. In astronomy, a similar limit is called the Eddington limit, so this is really the Eddington limit for a laser.

The radiation pressure is (ignoring all factors of 2 or cos(incidence)) E / c. A tensile limit, T, of 500 mega pascals (reasonable for steel) thus would imply an energy intensity of c T, or 1.5 x 10^17 Watts/m^2. If the total cavity had an area of 1 m^2, then that's ~ 10^17 Watts.

Note that it is common in pulsed lasers to have a lot of energy in a very short pulse (so the actual power during the pulse is very high). If your pulses were a microsecond in length, then the Eddington limit per pulse would be about 10^11 Joules, equivalent to 24 tons of TNT.

Re:Eddington Limit (0)

Anonymous Coward | about 6 months ago | (#46912813)

You can get around that by just making the laser bigger, keeping density down, then focusing the beam down at the end. Or like ultrafast lasers, take the frequency spread and use that to spread a pulse out in space or time with a diffraction grating, then recombine after amplifying. High power lasers using amplifiers, or even oscillator stages that are single pass, don't require any reflections to work too.

Re:Eddington Limit (1)

mbone (558574) | about 6 months ago | (#46913337)

The first part is irrelevant, as he asked the limit for "a cavity" (i.e., inside a specific sized object), and not on the target, but the second is not. You can, in fact, make your laser from a bomb and evade such limits. This was basically Edward Tellers idea for gamma ray lasers for anti-missile defense.

Already covered (0)

Anonymous Coward | about 6 months ago | (#46912767)

They've already ran into this issue with fusion power generation via lasers. When they increase the laser output too much, the laser light decreases. Turns out the photons are spontaneously turning into matter at high energy densities. This issue has been known about for years.

Re:Already covered (0)

Anonymous Coward | about 6 months ago | (#46912929)

The lasers used in inertial confinement research are many orders of magnitude lower in peak power than ultrafast lasers, and don't have a problem with pair-production, especially within the laser. There is not much to be gained (in fact, it would be worse) by having shorter pulses in such a set up, but even though the total energy is large in something like NIF, the pulse times of ultra-fast lasers is millions of times faster. That gives much more intense power available for non-linear and exotic effects, and even then they don't have pair production in the laser itself, only when hitting special targets after the beam has been focused.

E=mc^2 (3, Interesting)

ILongForDarkness (1134931) | about 6 months ago | (#46912933)

At high enough energies particles are spontaneously created. They in turn will obey Pauli Exclusion (at least if they have spin I think). So enough photons and you make matter that will prevent you from making more particles ie pumping more energy into the space.

Re: E=mc^2 (0)

Anonymous Coward | about 6 months ago | (#46913509)

And you'll get a Big Bang?

"physics" is multidisciplinary (2)

Goldsmith (561202) | about 6 months ago | (#46913359)

"Physics" is not just one thing anymore. The guy writing TFA, Ethan Siegel, is a bonified professional physicist. Reading the comments, you can see he just didn't know this one thing as well as he thought. How does that happen?

I don't know that there's any physicist going through training today or in the last 20 years who really understands "all" of physics.

Physics PhDs learn most of physics up to about 1910 (even that is a stretch, but at least the complete fields up to that point are introduced and sketched out), and the next 100 years are based on your specialty. The limits of energy density for photons are usually in this realm of "introduced only if directly important to your specialty."

It's up to the individual to fill in the gaps after formal classes, and it can be very hard to figure out what you don't know. It's particularly hard because of the oversimplified way physics is generally taught in undergrad, even to physics majors. Your old reference books may not actually be correct. I'm sure I've got a physics textbook around which claims almost exactly what Ethan said in his blog; the "why" of pair generation is just too distracting.

On a tangent note... (1)

Twinbee (767046) | about 6 months ago | (#46913373)

On a related note, I asked this quessie at a laser pointer forum a while back. Would still be interested in hearing a real answer: http://laserpointerforums.com/... [laserpointerforums.com]

Please stop posting medium.com urls. (0)

Anonymous Coward | about 6 months ago | (#46913437)

They're unreadable without a tablet.

3 words: Bose-Einstein Condensate. (2)

Paul Lester (3318297) | about 6 months ago | (#46914291)

Bose-Einstein Condensate! In more detail, fermions cannot be crammed together but in certain conditions, Bosons can. Photons are a type of Boson but not the only one. The Pauli exclusion principle does not apply to Bosons! Looks like a non-specialist needs to read some books on this concept. I won't even go into deeper details without this point being crystal clear!

There are a few things that might get in the way (0)

Anonymous Coward | about 6 months ago | (#46915103)

The inability to focus light to a small enough region of space. At a high enough intensity it will start pair production of electron and positron and that will dissipate the energy. If you can get pass the pair production problem then there is the mini black hole problem.

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