Ask Ethan #35: Is there a limit to Lasers? (Synopsis)

“The atoms become like a moth, seeking out the region of higher laser intensity.” -Steven Chu

Sure, lasers are fascinating entities. By stimulating the right medium with the right conditions, you can induce the stimulated emission of radiation of the same exact wavelength in the same exact direction over and over again.

Image credit: Q-LINE Laser pointers, via Wikimedia Commons user Netweb01, under a c.c.-by-3.0 license. Image credit: Q-LINE Laser pointers, via Wikimedia Commons user Netweb01, under a c.c.-by-3.0 license.

But is there a limit to the amount of energy that can be produced by a laser? And if so, what is that limit, and might it be overcome by future technological advances, or are we simply running up against a limit inherent to the laws of physics?

Image credit: © 2014 Science and Technology Facilities Council, of the ALICE free electron laser, via http://www.stfc.ac.uk/ASTeC/17452.aspx. Image credit: © 2014 Science and Technology Facilities Council, of the ALICE free electron laser, via http://www.stfc.ac.uk/ASTeC/17452.aspx.

Hang on to your seats as a theoretical astrophysicist dives into the fantastic world of lasers!

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By James Edward Hughes (not verified) on 02 May 2014 #permalink

Interesting but slightly wierd article, I think maybe a little outside of Ethan's normal range of expertise.

For example, why the gripe about the acronym LASER? Nothing is really being amplified? What about the light! The spontaneous emission is being amplified inside the cavity. Seems like a very real amplification of light by stimulated emission to me.

We are told that lasers can be made of crystals, or even semiconductirs. Well, semiconductors are crystals!

By Waterbergs (not verified) on 02 May 2014 #permalink

@Waterbergs

A laser does NOT amplify light. The photons engage in a cascade of absorptive/emissive energy jumps. These become synchronized, polarized and emerge from the laser as a coherent beam with the same light energy that went into it. The "stimulation" is not amplification, it is synchronization and polarization.

Your "crystal" complaint is a misunderstanding of semantics. The first laser was a ruby crystal doped with a chromium corundum. That was one of a category of lasers made from crystalline or glass rods which are "doped" with ions to assist energy transitions. Those lasers are pumped with photons. That class of glass-rod lasers are called "solid-state" and are distinctly different from "solid-state electronic" semiconductor lasers

Semiconductor lasers (laser diodes) are electronic semiconductor crystals that are pumped with electrons (not photons) to produce the beam. The process is different.

As was mentioned, there are also other types lasers.

By MandoZink (not verified) on 03 May 2014 #permalink

Can a laser be twisted? and if it can, could a second laser be added to the twist without interfering with each other?

By Myxomatosis (not verified) on 03 May 2014 #permalink

In the Update at the end you say that photon energy was a limit factor. But a laser doesn't increase photo energy (hv) at all, only intensity.

What about the factor that Photons are matter. If you increase the intensity of light, you increase the amount of photons being emitted. A laser will then only be capable of emitting so many photons through the cavity, before there literally is no more space for a photon to fit through at that given moment in time.

Sorry, this just hit me as well. What about the source creating the photons? Wouldn't the same principle apply there as well, ie. Only a set amount of photons can occupy the space where a photon can be created, by the source. Photons take up space, so if the source were to create a photon at the position of another photon ( in the case scenario where all other locations around the photon as occupied by other photons), what would happen? A photon in a photon, or would there be the slightest unsimilarity between the photons such that the photons push each other away?

With reference to the question, what about the heat factor? The greater the intensity of the light, the more photons. The more photons being emitted, the more collisions there will be with atoms along the beam. A laser is constantly emitting photons, and the energy from the friction from the collisions will just keep adding up. Won't this reach a limit, for the surrounding material of the laser, before it starts to burn/melt it, and ultimately affect the laser to the state when it cannot function?

It looks like the post is incomplete. It is interesting but it poses questions that it does not provide answers to. Thanks to the other comments that make things a little bit more clear.

I think that the Eddington limit is more relevant.

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.

Now, you can make your laser from a bomb and evade such limits. This was basically Edward Teller's idea for gamma ray lasers for anti-missile defense, but I don't think such a one time use device quite fits the original question.

By Marshall Eubanks (not verified) on 04 May 2014 #permalink

Kaj,

Did you not read the article? Why do you think there is a limit to the number of photons that can be packed within a given region of space? Photons are not matter, at least not in the sense we normally think of. Photons don't have mass. They don't have volume. Ethan covered this issue well; photons are bosons, so they are not subject to the Pauli exclusion principle as are particles such as electrons, protons and neutrons that make up ordinary matter.

Also Kaj, why would there be friction when a photon collides with another particle? Sure, light energy carried by photons can be converted to heat, but that is typically a result of absorption of light. Reflective processes don't produce heat. Shine a very bright light on a mirror. Does it get hot from the "friction" of all the photons colliding with the mirror?

From the OP:

The wavelength that lasers can emit in range from extremely long radio waves to incredibly short X-rays, with gamma rays theoretically possible as well.

Point of clarification: your article makes it sound like these things are terms for different wavelengths of light. They are not. In modern parlance (not the way the Curies thought about it), 'gamma ray' refers to high energy photons produced by nuclear decays, while 'x-ray' refers to high energy photons produced by other means, but the wavelengths of the two can be the same.
So a gamma ray laser would be a laser that works by exciting nuclei and having them photodecay, while an x-ray laser might work by, say, using a big synchrotron as a source. But both things might produce exactly the same wavelength of light.

Odi @5: the limt Ethan is referring to is producing lasers of higher wavelength light, which is indeed an energy limit. As Einstein showed with the photoelectric effect, you can create as intense a beam as you want, and if it's too low energy, you won't blast apart what you want to blast apart. The specific limit he's referring to is that photons above 1.022 MeV can spontaneously convert into an electron-positron pair. This will work against the lasing, process, essentially "downconverting" any focused beam of higher photon energy into an unfocused set of 511 keV photons (and probably do a lot of other nasty things to your equipment, too). OTOH, I suppose the conversion might make it a lot easier ot produce a laser of 511 keV photon energy, if one wanted to do that. :)

@MandoZink
The actionof a laser can definitely NOT be described as simply "synchronization and polarization". The presence of photons stimulates the emission of extra photons that would not be otherwise emitted - this IS amplification of the photons that do the stimulating. In the absence of these photons the emission rate is just the spontaneous one - which is much less than the stimulated rate.

The process is actually really easy to see in "optical amplifiers" (no co-incidence - these "amplify" light!) If you take an optical amplifier (for example an erbium doped fiber amplifier) and you pass light of the correct wavelength through it it is massively amplified. You are probably reading this as a result of the action of just such an amplifier - as they are the basis for the operation of the internet over long distances.

This is exactly the same process that goes on in a laser - just via a single pass through the AMPLIFYING medium.

By Waterbergs (not verified) on 06 May 2014 #permalink

Thanks for pointing out that photons are bosons. I did extra reading on it and your right. The reason I why I thought photons have mass is because of how they are created : "A photon is produced whenever an electron in a higher-than-normal orbit falls back to its normal orbit. During the fall from high energy to normal energy, the electron emits a photon -- a packet of energy" and since energy is equivalent to mass, I made a link between the two. Also, on a side not, it hasn't been experimentally proven that a photon has no mass, it's only theorised.

Kaj:

Also, on a side not, it hasn’t been experimentally proven that a photon has no mass, it’s only theorised

I think every time we run a nuclear reaction we 'prove it' (or at least gather more evidence that it's true). These reactions obey laws of conservation of energy and momentum just like any other reaction, and the math works out when the photon mass is taken to be zero. If the photon mass were not zero, then we would also have to be making mistakes in the masses and energies of the other atomic particles in order to observe what we observe.

Let me use a very simple example to illustrate. An excited nuclei emits a photon. Before and after the emission, the center of mass of the total system must remain the same. So if the photon has 'hidden mass,' the nuclei will move differently on emission than it would if the photon has no mass. We watch how these things move, and they move the way they should if the photon has no mass. This is empirical evidence that the photon has no mass. Such nuclear reactions are probably tracked hundreds if not thousands a times a day at accelerators all over the world. I think it's pretty safe to say that if the photon had mass, we would've picked up some evidence of it in these reactions.

One can't really say it doesn't have mass. It does have energy, and it certainly obeys the mass/energy equivalence principle. What it's not is a fermion, nor is it affected by higgs field. So in that sense, it doesn't have rest mass.

Get hit by a high enough energy photon, and you will certainly feel it.

By Sinisa Lazarek (not verified) on 07 May 2014 #permalink

Thanks Eric and Sinisa for the insight. I'm sure every scientist out there can't wait for the day we unlock lights true secrets! I just have one more question. Light is essentially pockets of energy. Energy can be transferred, say, to the air molecules in the test environment. The area very close to the light source will have intense amounts of energy culminated at that spot, and could cause the nearby material on the laser to combust. I've read that a possible solution is to have the area close to the emission in a vacuum, but for practical applications, not every laser user is going to have access to a vacuum, and the container which keeps the vacuum would interfere with the light beam. So would the heat factor of the area very close to the emission, with a high intensity of energy from the photons, become a limiting factor, in practical applications?

Kaj,

To prevent combustion, you need not have a vacuum; purging the area near the emission with argon or nitrogen would work. There may be practical problems with this as well, though, so you still may have hit upon a practical limitation.