Light Energy Puzzle

Eventually it will behave like an inflamed zit and just go kablooie. If you stand to the right side, you won't be splattered, so don't worry. See also Post 2.

Answer: the light will reduce to nothing and the prism will get warmer, releasing heat to whatever is outside it.

There is no need for "opinion"; the Laws of Thermodynamics say this will happen however many times the experiment is conducted.

Answer in similar post, http://cr4.globalspec.com/thread/80029#newcomments

Yes, bravo88. This is a very similar question. (The general concept is the same. Only the "device" is different.) I think that comments #24 and #25 give the answer on this issue.

No "drawings" visible! Maybe all the light escaped.

The material inside is not totally lossless. So you are right, it would eventually heat up until the rate of loss approached the energy input. It's the same idea as putting a resistor inside a perfectly insulated container and running a current through it. The energy doesn't go back out the same way it went in. Another example is a greenhouse, where light is converted to far infrared which cannot escape through the glass. I think in the real world, eventually something's got to give.

Photons are simply transfer vehicles for energy at a particular frequency or energy level.

In your hypothetical example you are simply pumping energy into the system on a continuous basis.

That energy goes somewhere, even trapped inside your "prison" of glass or fiber and does so in the form of heat.

So the system temperature will rise until it reaches homeostasis and the net outflow of heat energy equals the net input of the energy of photons.

That final steady state temperature will depend on the heat transfer efficiency of the system and the rate and energy of the photons put into the system.

Good question to follow that is, will there be light after which the source was cut and vessel was closed? Is the prism still illuminated or light is gone?

We normally describe an object as 'illuminated' when light shining on the object is reflected such that we can see (observe) the object, or when the object itself is emitting light.

In this case, the light is 100% contained within the vessel, so no light is observable from the outside of the vessel. Thus we would say it is not 'illuminated'.

It also would not be possible to directly view/observe the light from inside the vessel. If the faces of the prism/mirrors are perfectly oriented, all the photons will be circulating in a single 'ring' of straight line segments, and any sensor placed in that ring would be destroyed by the energy of the photons and also block the circulation.

If there is the slightest misalignment in the faces/mirrors, then the ends of the vessel must also be totally reflecting to avoid loss, and the photons will then be circulating in a 'sheet ring' as they migrate from one end to the other. This could reduce the intensity of the beam by orders of magnitude, but still, any sensor placed in that 'sheet/ring' would be destroyed by the energy of the photons and again block the circulation.

It should be possible to inject a sacrificial particle into the ring and observe the destruction of that particle.

this is an interesting post, I enjoy it greatly.

I like to add about the term you use of (observe). Which adds to a the original problem.

It's a quantum theory where observation effects reality hence the term Observer Effect.

Would be a catch 22 here. Without observation would it be illuminated or not.

Not to be mistaken for "Schrödinger's cat" but could be.

What if only a finite light energy will be captured in the prism which inwardly reflects photons and you have provided just a tiny bit of hole which has covered initially by a reflecting gate. Let us say the vessel or the prism is adiabatically insulated so that photons could not be converted to heat and will stay as it is. The reflecting gate would finally be open just to see if photons stay as they are in the vessel as it is gradually deteriorated through an infinitesimal hole (not reflectorized spot) on the prism . Would that make sense? I supposed photons are particles in duality.

it will eventually degrade.

Back in the 80's in some technical journal I long forgotten the name of, I had read that they were developing something similar to be used as a type of optic gyroscope. Using the apex of the sine waves as the positioning. Using this would be very accurate for long distance space travel.

Wish my memory serve me better, but that's about all I recall right now.

here there are, Ring laser gyroscopes.

http://www.airpower.maxwell.af.mil/airchronicles/aureview/1985/may-jun/siuru.html

http://www.airspacemag.com/flight-today/how-things-work-ring-laser-gyros-32371541/?all

This is the best info,

http://www.northropgrumman.com/Capabilities/LTN92/Pages/default.aspx

ps, I can see the pictures now.

What you've sketched won't work, and in general the concept won't work either.

The light beam you show hitting the prism refracts the wrong way. A light beam entering a denser medium refracts toward a line perpendicular to the surface struck by the beam - you show your light ray refracting away. So the sketch doesn't describe the physics properly.

In a more general sense, though, with your simple prism you're trying to violate a law of thermodynamics. If the light ray can get into the prism, then necessarily it can get out.

The fiber optic light pipe is a slightly more 'realistic' approach, but you still have the problem of injecting light into the pipe. If the light pipe is broken open to inject light, light will scape from the open end. Or if a 'y' connection is made like your sketch shows, the 'total internal reflection' rule is broken and light will escape through the wall of the light pipe in the vicinity of the 'y'.

I can't see any picture or sketch....

Yep, it's the same old problem once more! If you use the dreaded Internet Explorer, then there are no diagrams. I have just logged on via Chrome and there they are. Explanation please!

I see perfectly good images for the Original Post in either Safari or FireFox on a Mac, although I nearly always use FireFox, since Safari does not show a CR4 Tool bar.

Hi Usbport,

"The light beam you show hitting the prism refracts the wrong way. A light beam entering a denser medium refracts toward a line perpendicular to the surface struck by the beam - you show your light ray refracting away. So the sketch doesn't describe the physics properly."

Yes, you are right (and that was my mistake) concerning a denser medium. But it is okay for a less dense medium. E.g. it could be a "prismoid air cavity" inside a piece of glass. I think that you could manipulate properly the angle of incidence and the exact shape of the cavity in order to get the result of total reflection.

"If the light pipe is broken open to inject light, light will scape from the open end. Or if a 'y' connection is made like your sketch shows, the 'total internal reflection' rule is broken and light will escape through the wall of the light pipe in the vicinity of the 'y'."

Yes, there is an "imperfection" on the exact 'Y' point which means that some of the light may escape through this point. However, I believe that this "light loss" will be relatively small compared with the vast amount of light that is injected inside the fiber (i.e. a lot more light "comes in" than "gets out").

There is no way to inject light into a prism so it can remain trapped, or into a bubble in the prism to keep the light trapped in the bubble. The requirement for total internal reflection is precise. Instead of just 'thinking' it will work, try making an accurate sketch where all the angles of incidence meet or exceed the TIR angle.

In this sketch a beam of light tries to come up through the glass below into air. The glass has an index of refraction of 1.52. If A equals or exceeds 41.13951, then A is 90 degrees or greater, and TIR occurs.

Get back to me when you've sketched a prism that will work like you want.

With your fiber optic light pipe, the situation is like pumping air into a tire without a one-way valve. You can pump a lot of air in, but once you stop pumping you'll have to close the opening very fast.

I think it is intended as a thought experiment... what happens if you keep injecting light into a system with total internal reflection?

In a real experiment the light would escape, yes, but put engineering to one side and view it like those exam questions that start off "A heater runs inside an adiabatic container...". Sure there's no such thing as an adiabatic container... but merely pointing that out is a quick way to fail your exam.

The OP is asking whether, in an ideal world, the material will heat up if there is 100% internal reflection (i.e. no absorption)? If so, how does it both reflect 100% and heat the material up? If not, what happens as you pump more and more light energy in?

(It's how many lasers work after all - trap the light between two parallel mirrors until all photons travelling exactly perpendicular to the mirrors become massively concentrated).

Photons are simply bosons and are massless. There is no limit to the number of bosons that can dance on the head of a pin.

So, hypothetically you could stuff an infinite amount of energy in a finite volume of your choice.

Maybe no limit, but plenty of things happen when there are enough.

One caveat... I should add that there is an upper limit to the energy that a photon carries where photon to photon interaction begins to take place. If your photons are gamma rays the interaction between photons in a high density space will produce a copious amount of elementary particles as they interact. Gamma-gamma colliders do this all the time.

Below the gamma-gamma energy threshold levels (I think that energy threshold is about 250 GeV) the interaction is only known as quantum scattering and no elementary particles are produced as a byproduct.

Keep in mind the energy threshold is for a pair of photons (each having ≥250 GeV) and not the sum total of energy in the "box" of photons.

So, in hypothetical reality the answer to the original question with perfect reflectors is really nothing more than adult diapers - it depends.

If the individual photon energy levels are below that interaction threshold of 250 GeV, then my original statement that there is no limit to the number of dancing bosons stands.

Okay, if particles could be created from the interactions of high-energy (gamma) photons then these particles probably could escape from the system, releasing the accumulated energy of the system. But for low-energy photons this is not the case.

"If the individual photon energy levels are below that interaction threshold of 250 GeV, then my original statement that there is no limit to the number of dancing bosons stands."

I strongly disagree. Photons have energy (=mass) and by increasing (infinitely) the energy inside a small, finite place the energy density -unavoidably- will reach a crucial limit creating a small black hole.

"I strongly disagree. Photons have energy (=mass) and by increasing (infinitely) the energy inside a small, finite place the energy density -unavoidably- will reach a crucial limit creating a small black hole."

So what is the mechanism by which energy is converted to mass at some density level?

Photons are massless, so you are postulating that at some density level they are converted to some other particle or particles, which would be what?

Do you have a link for the physics involved? I would be interested in understanding this.

Photons are massless only in their "rest state" (where, in a way, they don't exist any more... in fact, they can't be in any other state than travelling with v=c). Anyway, photons have energy. It doesn't matter if you accumulate energy or mass. Afterall, mass is "concentrated energy".

I think that assumption is wrong because you are using a classical or popular approach the E = mC^2 to assign mass to a photon.

That is really just an equivalence and not really assigning a mass to a photon. More like a model, which may be helpful is describing an object, but it can only do so incompletely.

The current thought on photons is that they have zero rest mass. If they did have mass there would be some problems with gauge theory. Particles that travel at C are called luxons and have no mass.

A modern approach is to assign every particle a rest mass independent of velocity.

Another thing about light. If you drop an object towards Earth from space its velocity would increase as it accelerated toward Earth. This is because both objects have mass.

What happens when light follows the same trajectory as the object on its path toward Earth? If light had mass it would also accelerate. Just as light projected from Earth would lose velocity...

However, what we have observed is that light always maintains the same velocity, C. What does change is the frequency (energy) of the photon. Photons approaching a mass gain energy (become bluer) and photons projected from Earth would lose energy (become redder).

So, I still am not convinced that at some critical density of photons they would collapse into a black hole as I can find no physical mechanism for that to happen.

Photons interact with gravitational fields (i.e. attracted by them). The proof of that is the "gravitational lens" phonomenon where the path of photons is curved by a stong gravitational field (e.g. a galaxy). This also means that energy (photons) behaves like a mass. The reason why the photons are not accelerated (or decelerated) by the Earth's (or any other) gravitational field is because the photons -by default- can travel only at speed c.

"The reason why the photons are not accelerated (or decelerated) by the Earth's (or any other) gravitational field is because the photons -by default- can travel only at speed c. "

Careful with this one - on a global scale photons get delayed by the presence of mass, which can be viewed in (say) Schwarzschild coordinates as a lowering of their average speed in a gravitational field (lowering while in-falling, increasing again when outgoing). It is only in inertial frames where their speed is always c. It is hence not generally true that photons "behave like a mass".

Because we do not have a quantum theory of gravity yet, we cannot even be sure that photons could have formed micro-BH's during/following inflation. We are reasonably sure that they had a strong gravitational influence in the early cosmic expansion, but they did not quite behave like matter. Although, some radiation-energy is likely to have been converted into matter.

A nice vague scientific answer...

-J

Thanks, Jorrie. So I think that we could say that energy (photons) behaves "not exactly" but "almost" as a mass.

The initial dissent between me and AH was the followng: I claim that photons produce gravity while he claims that they don't produce gravity. Which of these two claims is true?

Please, give us a clear answer.

GK, I think I have answered to some extent, but yes, concentrated photons produce spacetime curvature, which translates to a gravitational field. I base this on the accepted situation shortly after inflation, when radiation dominated the gravity component of the expansion equations.

All forms of energy gravitate, so if you could somehow create anywhere near that sort of energy density of photons inside a container of some sorts, I think it will create serious spacetime curvature around it. How do do it? I honestly don't know...

-J

Jorrie, thank you very much for your reply and your overall help. I just needed to be sure of that.

Practically you can't create such an accumulation of photons but we were discussing a "thought experiment" were photons are continously accumulated inside a (e.g.) fiber loop (see my initial presentation) with "ideal" (100%) total reflections.

Thanks again...

In Newton's world things fall.

However, Einstein showed us that objects do, in fact, travel in straight lines, but space becomes warped or curved in the presence of mass.

The gravitational lensing you are citing as proof is nothing more than the straight line vector through curved space-time.

The only interaction with gravity that photons have is a increase or decrease of frequency depending on whether they are vectored towards or away from the gravitational source.

"The only interaction with gravity that photons have is a[n] increase or decrease of frequency depending on whether they are vectored towards or away from the gravitational source."

Since E=hf, that sure sounds to me like increasing or decreasing energy of the photon.

... and it seems to me a matter of semantics whether the change in frequency is due to the curvature of space or due to gravitational attraction, since the curvature of space is due to the presence of the (attracting) mass.

Please elaborate!

The point is if a mass is attracted to another mass you will get a change in both velocity and direction of the two masses. It's a mutual attraction.

Not so with a massless boson that happens near a gravitational field. The boson's velocity does not change, but as you pointed out the energy will.

That's not a semantic difference.

Thanks. I think that helps...

Now this is really getting interesting! I'm an old (relative term), mostly classical physicist, trying to get a bit more up-to-date.

I fully understand (I think...) the Newtonian interaction between two masses. Now what I think I'm hearing is that the presence of the mass adds energy to the boson as it approaches the mass, yet the boson has no effect on the mass, unless they actually collide.

Am I once again way off base?

You raise a good question. The conservation of energy must still be obeyed, so there may be some energy transfer in the process, but I am awfully sketchy as to the mechanism. I believe this is another one of those quantum world things.

Since photons are simply a carrier for energy, you are right that the energy of the photon will be transferred to the mass if it strikes the mass.

Thanks! I don't feel quite as old now...

That was about the best condensed explanation I have heard. Thanks!

Does also explain the refraction of light? After all, matter is considerably more dense where the η>1. Maybe charge wells should be looked at as well, by distorting the space where the e-m wave travels? Everything is a wave, and everything is a particle. We know that photons "colliding" with a target (in a vacuum) will turn the paddles on a spindle (especially if the paddles are light and silvery on one side, but dark on the other). This indicates that photons have impulse to impart on the paddles, and by conservation of momentum, photons must have momentum, and that by being absorbed (detected) they lose their momentum. That takes care of the question of mass, since at rest, indeed they are massless and without velocity.

Gravitational lensing is then related to refraction in that something is distorting the space-time fabric traversed by the photons-waves. If energy is decreased, as you stated, it has to be by loss of frequency (increase of wavelength), but how sure are you that locally, the speed of light (in the space near a massive object) is not decreased similarly as in the case of refraction within a dense medium? Yes, when the photon-wave exits the distortion out the other side, the frequency goes back to normal. I am totally blown away by all this.

"Does also explain the refraction of light?..."

No, James. The apparent slower speed of light -as it travels inside a (denser than air) material- is just a delay because of the absorbtion-retransmission of the photons by the material's atoms. Between the atoms, the photons still travel at v=c. On the other hand, "gravitational lensing" is due to the interaction of the photons with a strong gravitational field.

Then why does the change in speed so accurately predict the angles of refraction? Can you make the same predictions based on the absorbtion-retransmission delay?

sorry, but "retransmission", really? If the light is absorbed how does "it" know where to retransmit? hmmmm?

I am sorry, but at this point, I regret I must unsubscribe to this.

James, threre are plenty of physics forums out there and you could ask them how the 'reflection' and 'refraction' works.

In short, photons are not small balls which bounce on the walls. A photon is absorbed by an atom and the atom is stimulated on an upper energy level. Right after that, the atom returns to its previous lower energy level by retrasmitting a photon. In general, that's the way the 'reflection' and 'refraction' works. In the case of refraction, this process introduces a delay leading to a decrease of photon's speed inside a tranparent material.

If you are not yet convinced then see:

http://en.wikipedia.org/wiki/Speed_of_light

Read in "In the medium" paragraph:

... However, this represents absorption and re-radiation delay between atoms, as do all slower-than-c speeds in material substances...

Concerning the angle that the 'new photon' is transmitted by the atom (during 'reflection' or 'refraction' process) I don't know the answer why it as it is (I'm not a physisist afterall to know every detail), but I'm sure that if you do a research and a serious study on the specific issue you'll also find the answer on this question.

Go back and read that Wiki Article (your link) a couple more times.

You left out an important detail: In the sentence preceding the text that you quoted was "In exotic materials like Bose-Einstein condensates near absolute zero, ..."

This does not apply in ordinary materials at ordinary temperatures.

Hi dkwarner. If you don't believe me (and as I said to James Stewart), there are plenty of Physics forums out there where physicists could give you an answer. Maybe they'll be able to convince you about that. Photons cannot travel at a speed lower than c. In any inertial frame and inside the vacuum, the speed of light is always (and only) c. Now, think about it again. Between the atoms of a material there is nothing else than vacuum. So, between the atoms, the photons are traveling at speed c. Again, only the average speed of light inside a material is lower than c (as I have already explained).

Balderdash! I will still with the physics I learned over 40 years ago, I know that still works. I don't work with Bose-Einstein condensates at or near 0 K. I simply refuse to hire them, as they are exceptionally needy, and work on union schedule. I don't have time for that.

Another couple of points: It is totally annoying when certain posters assume their own erudite knowledge supersedes that of all others "in the room", and further they seem to imply that he/she is conversing with uneducated country hicks (some of which are probably way more intelligent than the poster), rather than well-educated scientists and engineers. It is not diplomatic to make a general practice of insulting the audience.

Having said that, I will now refer you to my disproof of your statements about "normal" light propagation in a medium. (1) temporal coherence of the light transmitted would be lost. A short temporal pulse of non-coherent (i.e. non-laser) light being transmitted would be broadened due in randomness in the supposed atomic absorption/emission (spontaneous only) of the light transmitted. (2) spatial coherence of the light is also lost due to spontaneous emission of the light in 4π. This is completely contradictory to simple observations. To have a process as you have mentioned for "normal" light transmission, one would not be traversing "normal" matter.

Have another glass of Ouzo.

Agreed, with both paragraphs!

Also, to the second paragraph, add: (3) this supposed absorption/emission would destroy the directional properties of the waves, making solid lenses and prisms useless.

True. Agree fully with your remarks. In fact, I am beginning to wonder how human or other animal vision would be supposed to work under these "new" rules.

"I am beginning to wonder how human or other animal vision would be supposed to work under these "new" rules."

Ask Richard Feynman http://www.feynmanlectures.caltech.edu/I_31.html

I would defer to Dr. Feynman the credit he is due. Certainly, the explanation (he wrote) that the oscillating E-M field results in acceleration of the charged particles in the medium, and this "retards" the phase of the light as it propagates appears to be "reasonable" explanation enough. Certainly, we all know that as an absorption band is approached from one wavelength limit or the other, the refractive index of the material (and dispersion of said index) changes much more drastically than in optically transparent regions of the spectrum. This relates to critical behavior in that at an absorption line (if the spectrum is narrow and intense), the index of refraction approaches infinity.

Yes, it is perfectly valid to consider space and time to be 'more dense' near a massive body and that refraction and lowering of the speed of light happen, but only if measured from 'outside' of the gravitational field.

When we talk "local speed of light", or "speed of light as measured locally" we must remember that the observer and her apparatus are in the same gravitational field as the photons (i.e. "inside the medium"). Jill gets the standard value c, but if Jack, in a different part of the gravitational field, tries to interpret her test using his own clock and distance measures, his idea of effective speed will be different. He will get a higher or lower value than c, depending on if he is in a stronger or weaker gravitational field.

If Earth and Pluto are on the same side of the Sun, with Jack sitting on Pluto and Jill on Earth, his radar pulse to Earth and back will take longer to arrive than Jill's radar pulse to Pluto and back. Jack will get a lower overall speed of light than c, and Jill will get a higher average speed than c. However, each measuring the speed of light in their own backyards will get exactly the normal c.

Photons have momentum, quantified by p = h/λ, with energy E = hc/λ, but remember this momentum and energy are strictly an observer-dependent quantities, i.e. measured locally.

Hope it eases the "blown-away" issue...

-J

Hi AH. I recently found out that, indeed, the continuous accumulation of photons (or energy in general) in a closed, ideal system could lead (theoretically) to the formation of a Black Hole. This "special Black Hole" (which is formed by the accumulation of energy instead of matter) is called Kugelblitz.

https://en.wikipedia.org/wiki/Kugelblitz_(astrophysics)

So, my assumption (about the creation of a Black Hole in such an ideal system) was correct.

I think Kugelblitz (ball lightning) is not yet a confirmed phenomena and if existing, they almost certainly are not small black holes. There are many speculations as to what they could perhaps be.

That said, it is obvious that if one could (hypothetically) concentrate enough energy in a small enough volume, a black hole must form, even if just for an instant before Hawking radiation destroys it again.

-J

I say do not feed the bears! Take all the pictures you want, but do not feed the local fauna.

Ball lightning has been observed by many people, and I have heard it discussed by at least one scientist of some renown (Henry Eyring as an example), and it keeps popping up in stories about unexplained lights associated with (or not associated with) lightning storms.

Instead of requiring a black hole every time something comes up we do not understand, or simply brushing it aside, and stating "it is not yet a confirmed phenomena", try observing nature, and also consider the aether (zero point energy) of the universe, which some have estimated to be upwards of 10¹³¹ J/cm³. And consider how some disturbance might cause the fluctuations of instantaneous particle formation/annihilation to be biased in favor of stable trapping of this energy within a "system" to persist for variable amounts of time.

The truth is stranger than fiction.

There are two definitions of Kugelblitz. One is ball lightning:

https://en.wikipedia.org/wiki/Ball_lightning

One is a black hole formed from a concentration of photons:

https://en.wikipedia.org/wiki/Kugelblitz_%28astrophysics%29

I think I implied this more or less in #77. Although enough concentrated energy can surely form a BH, I'm pretty sure there is no evidence of one formed by pure radiation. So in that respect, the two types of Kugelblitz are both speculative.

"Photons have energy (=mass) and by increasing (infinitely) the energy inside a small, finite place the energy density -unavoidably- will reach a crucial limit creating a small black hole."

GK, while this may be true in principle, I do not think there is any practical means of achieving it. The universe could perhaps have done this right after inflation (or post-bounce), because most versions do expect micro BHs to have formed and then evaporated at that stage. The dominant energy then is thought to have been photons, so yea, maybe photons can in principle create BHs.

-J

Thanks, Jorrie...

Thanks for weighing in, Jorrie.

It (the prism) is like the A.C.A. you have to pass it (or in case of prism, smash it) to find out what is in it.

Sounds reasonable, but the original poster spoke of the system melting down due to the ever increasing energy, so that hints at something different, as you pointed out.

If the material reflects 100% of the energy, then the end state is open as it will continue to acquire more and more energy as the photon soup in the prism continues to increase toward infinity (E=h•ν).

With a true 100% reflectivity you could hypothetically capture an infinite number of photons , which are simply bosons. Essentially, infinite energy density is hypothetically possible.

"(It's how many lasers work after all - trap the light between two parallel mirrors until all photons travelling exactly perpendicular to the mirrors become massively concentrated)"

That's not how lasers work. The photons are not trapped 'until ... massively concentrated'. For those lasers that use mirrors (some do not) one of the laser mirrors is only a partial reflector, and a portion of the photons leave the laser through this mirror right away. Some people have the misconception (which you seem to) that a threshold number of photons is needed to 'overcome' the mirror, as though the mirror is a barrier. That common misconception is simply ill-informed and wrong.

I disagree that this is a good answer.

Hi Usbport - Thanks for your input and demotion of my answer. You should consult to the Department of Energy, University of Colorado Physics Dept or Lawrence Livermore National Laboratory (and thousands of other labs and universities) - you could set them straight too as many of them seem to be labouring under the same misapprehension as me.

Would you believe that many physicists are silly enough to think that photons in a laser are bounced back and forth between two mirrors until the low-energy beam has been amplified more than a quadrillion times? And that some of them publish this nonsense on the web? How do they keep their jobs?

Here's an example of the LNL charlatans spreading their mischievous lies across the interwebs:

• (https://lasers.llnl.gov/education/how_lasers_work) Mirrors at both ends of the glass amplifier cause the photons to travel back and forth through the glass, stimulating more electrons to drop to their lower energy states and emit photons. This process produces huge numbers of photons of the same wavelength and direction-an extremely bright and straight beam of light. In this way the initial low-energy pulse is amplified by more than a quadrillion times to create 192 highly energetic, tightly focused laser beams that converge in the center of the Target Chamber.

Actually I did not demote your answer. I thought the first part of your response (#20) was fine, I disagreed with the 2nd part. So I neither added nor deleted a GA from it.

Your response above is playing with semantics. I never said that photons don't bounce back and forth between the mirrors, nor did I say that lasers don't build energy as the photons traverse the lasing chamber. What I said is that your answer (#20) implied that the laser pulse builds until suddenly it acquires enough energy to overcome the mirror 'barrier' - and that is not what happens. The link you cited does not address this common misunderstanding.

Since you feel slighted about losing a GA, I'll go back and vote you a GA for your answer.

"Get back to me when you've sketched a prism that will work like you want."

Hmm... ... Yes, it seems that there is no way to insert a beam into a cavity (from the outside) and keep it inside the cavity. Also, it seems that the light beam can be "prisoned" inside a prism only if the light source (laser) is incorporated inside the prism too. [Of course, this should be a problem in the 'thought experiment' case (ideal 100% reflections / no heat transmission from the system) as the continuous accumulation (increase) of the energy inside the system will destroy the light source (leading the experiment to an end).]

"With your fiber optic light pipe, the situation is like pumping air into a tire without a one-way valve..."

No, this is not the case. The photons that are inserted into the fiber have no way to go back and escape through the 'entrance'. Neither the photons which already circulate inside the fiber can do that. This could happen only if there is a kind of obstacle inside the fiber -right at the 'Y' point- which can reflect the photons back to the 'entrance'.

I think a lot of the questions you have were discussed in an earlier thread on this subject. If you haven't read through the comments there, it might be worth your while to do so.

http://cr4.globalspec.com/thread/80029#newcomments

Yes, Usbport. Bravo88 on post #3 had already guided me to this thread. I read the GAs and I already replied to him that comments #24 & #25 give the right answers. However, this thread of mine leaded us to more complex thoughts as we discussed this concept in an "ideal form" (as an experiment thought).

I am not an optics expert, but if memory serves, the schematic symbol shows partial reflection as in a beam splitter.

The idea of a photon being trapped forever is a false premise.

no surface is a perfect reflector and that photonic energy will "leak" out in the form of heat. Photons are only carriers of energy, not a ping pong ball particle.

This reminds me of this old classic.

You have a sealed, perfectly insulated room, the only thing penetrating the wall is an electrical cord leading to the refrigerator inside the room with its doors open. When you turn the power on, what will happen to the temperature in the room?

Temperature will increase.

Everyone knows when you open the refrigerator door the giraffe will step out.

The increase of temperature however is dependent of molecular property of the prism or fiber optics or any.

It will heat up even more if the light source is a laser beam.

The cat will die. Don't open them.

"a "black body" which absorbs all the incident electromagnetic radiation and gradually is heated up (till it "glows" in an equilibrium state)" It won't be black any more, will it?

How do you know it will ever reaches an equilibrium state if it absorbs everything?

Enlighten me please.

It is well known that a black body -which absorbs external radiation continuously- reaches an equilibrium state, where the absorbtion of the e/m energy equals the energy emitted as heat (infrared radiation) by the black body itself. The spectrum and intensity of the radiation depends on the temperature of the black body (which depends on the intensity of the incoming radiation).

The only known body that really absorbs everything without ever reaching such an equilibrium state is a 'black hole'. (However, it emits 'Hawking radiation' which depends on the mass of the black hole and not on the occasional incoming energy.)

This read like a episode of the Big Bang Theory from TV. Way to go all!

You keep putting things into black boxes, and the rabbit will die!

A black body is nothing more than "the perfect radiator". This means that it does not distort the emitted light at a given temperature by absorbing any part of it. There are no "perfect" black body radiators in nature, although some are amazingly near to perfection. Basically a perfect black body is one with unit absorbtivity and unit emissivity at the same time. Whether absorption or emission wins out is simply a matter of temperature.

As relates to the question at hand: (1) even if light could be "trapped" in a loop, eventually the light would be absorbed by the medium, since refractive index is mathematically related to the absorption of radiation, in that it has been put forth that dη/dλ→∞ when k (absorbtivity) is at its maximum. Therefore, k>0 at any time when η>1. (2) any dissipation of energy will heat up the medium, and either it will lead to optical changes that permit light to escape at a sufficient rate to allow a steady state, or that if light be being loaded in faster than can dissipate, the medium will suddenly (and violently) self-destruct.

In your (thought) experiment, one thing that will happen is that the combined electric fields from the trapped photons will approach the Schwinger limit, producing particle-antiparticle pairs as predicted by QED. Photon-photon collisions (fermion-type behaviour) have already been observed, one of the effects seen when electric fields become non-linear around 1.3e18 V/m. Axion production from photons is also an area of active research. Several current experiments are approaching the Schwinger limit. Appleton-Rutherford Lab's 10 petawatt laser, focussed to a region spanning a few tens of microns produces electric fields in this range. Other experiments also.

petawatt! That is a watt I would not want to pet.

Those are insanely high electric field values! WOW. Now if I can just calculate how a carbon (charged) particle moves in water in an electric field, and transfers its momentum to a molecule or atom, then I will get an insanely high temperature! Not me personally, just my experiment..I can't tell you what I am working on, but it is not rocket science, yet.

If your drawing showed the proper angle, then as the light enters the prism it would not go back to where it started. Once the light enters it is slightly diffused and instead of going back to where it started it would ricochet around in the prism and get gradually weaker. the idea that it would be a perfect union when the beam comes back around to itself, both being of equal strength and building energy would classify it as a perpetual motion device, as such a buildup of energy would only occur if there was zero loss when the two beams came together, meaning the original beam could be shut off and the circular motion of the light would continue without it.

Seems to me that if you put enough light into a prism like that, aside from all the other things that everyone has mentioned, you would also get a lot of pressure on the inside surfaces of the prism just from the light bouncing off it. I think it would physically explode if it didn't vaporize from the absorbed heat first.

One key issue in the super laser fusion initiation facility in California involves considerations of how much light can be transmitted through a given material before it explodes, but it is not based on light pressure inside the medium, it is based upon heat in versus heat out. Basic stuff. No one beam traverses the media with the final energy, but rather a lot of beams are manipulated to strike the fusion target at all angles at the final incident power to achieve ignition of fusion (for a very brief instant). None of the laser light beam are continuous beam radiation, but all are pulsed in synchrony to arrive a the target at the correct point in time.

This reminds me of a question that I have had for a long time, and this may be the place to ask it.

There are two scenarios. One is that you have two black holes of equal mass, one made of matter and the other made of antimatter. They collide, and form one black hole with twice as much mass (minus some loss due to the process of falling into each other's gravitational field.) Is it any different from if they were both composed of conventional matter? What happens when the two singularities meet at the center of the combined black hole? (Do they actually meet?) I would think they turn to pure energy in the form of photons but that nobody outside the black hole would see any difference. Is there any difference at all?

Another scenario is when two masses collide, one matter and one antimatter, that are each almost massive enough to be black holes, like two neutron stars. Do they convert to photons and radiate away before they have a chance to form a black hole? Seems like at the point of contact, at the exact center of mass of the system, there is no net gravity, so nothing to keep the photons in one place. Would the energy of the conversion be so great as to push the two masses apart before they can form a black hole? Or is the combined mass of the system so great that they form a black hole just by being close to each other?

There may have been micro black holes in the early universe made of anti-matter only, together with some, perhaps an eqaual amount, made of normal matter. These micro-BHs would have evaporated into bursts of gamma rays, but some 'opposites' may have collided before evaporation, forming a larger BH and an internal annihilation, which would have produced strong radiation inside the event horizon, possibly with density near the Planck scale.

I think Hawking radiation would have evaporated these 'Planck density' BHs rather quickly and have produced strong gamma ray bursts. Maybe this could be part of mechanism causing matter to dominate over anti-matter in our portion of the universe, but this is just my speculation...

-J

Okay, I will bite on that one. If one postulates the Big Bang, followed by inflation, then as matter begins to form, is the universe still in a state of near symmetry, as in a spherical shock wave, or is there already some anisotropic gravitational field causing density waves in energy-matter? Suppose the micro black holes formed with equal density of antimatter ones to matter ones. Further suppose that all types of collisions are allowed (i.e. -etc. 1:2, 1:1, 2:1, etc.), would we not have to find the most probable mode of collision? Would there not already have to be anisotropy for there to be any collisions between MBH- and MBH+ (for anti, and normal matter)? In fact, for matter to form, would there not have to be energy anisotropy that arises from energy density in some critical state where the critical parameter(s) result in chaos from order (2nd law of thermodynamics)?

So assuming something about symmetry (lack thereof), would this not already bias the expanding universe in favor of normal matter, else most of the matter would have cancelled out to some later point in the expansion?

It seems plausible, just barely. The part I do not get is that is as long as any 'opposites' are colliding, the net mass of the system is steadily decreasing. As soon as the two M and AM micro black holes approach (but still not touching), one could assume two point masses, in a tight spiraling orbit (about the center of mass), since there is no known mass, anti-mass construction. As this orbit decays further and further, it necessitates emission of radiation of higher and higher frequencies up to some unknown limit (gamma ?). This is one possible evaporation mechanism without actual annihilation? Certainly, if space-time is an engineer, then the two masses will be near enough to undergo one super-energetic final annihilation (for all practical purposes), but if space-time is a pure physicist, will they ever annihilate?

Is this what you (and Hawking) are referring to? The next question is: "Is there an astrophysical diagnostic to distinguish between a single singularity, and a dual one, in actual practice?"

If all the matter in black holes is no longer "normal" matter, with protons, electrons, and neutrons, what is it? How can one be anti-matter (if charge is irrelevant), and the other be matter?

These are actually nice and blow-minded questions (although out-of-topic). I'll dare to express my humble thoughts.

In the case of a BH+ and a BH- collision, I think that a larger BH will be formed (with an overall mass somewhat less than the sum of the masses of the original BHs). The larger event horizon of the new BH will be formed before the two singularities meet each other. The actual masses of these two BHs will be merged when the two singularities will meet each other (as, essentially, all the mass of a BH is accumulated right on its singularity). Then the annihilation of the "matter" and "antimatter" will take place producing a vast amount of photons which will not be able to escape. The problem is, of course, that the singularity is not consisted of ordinary matter. So, I don't know if it's right to even talk about "matter and antimatter annihilation" (i.e. we don't have -e.g.- electrons and positrons any more that could be annihilated). In fact, we don't know even the laws of physics that govern the singularity itself.

In the case of a NS+ and a NS- collision, there will be annihilation at the point of contact as they both consist of ordinary matter. Probably (as you said) the photons will be able to escape and the annihilation of the whole bodies will take place (before a BH has the chance to be formed). I don't know. That's a good question for the Physics forum.

I asked the same question in the following thread in the Physics Forum (PF):

https://www.physicsforums.com/threads/two-black-holes-collision-thought-experiment.872599/

Concerning your 1st question, it seems that my reply in my post #91 is valid.

Concerning your 2nd question, it seems that this issue is more complicated and it needs simulation in order to get a definite answer.

An interesting video (concerning your 1st question) is the following:

https://www.youtube.com/watch?v=6zw5DuWAyco