Perpetual motion?, Cyclic photon reflections

I had an interesting idea for a perpetual motion machine and wanted to see if anyone could shoot the idea down.

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

If reflected light exerts a pressure on an object, imagine two mirrors with light travelling between them (reflected light exerts double the pressure).

Imagine photons being reflected between them. Would each reflection continue to add more outward force? I have a feeling there's a quantum catch here, but I've never heard of light pressure varying depending on experiment set up.

Anyway, this acceleration could continue indefinitely, assuming the path of the photons was aligned correctly. The light will always travel faster than the mass, so no matter the distance, light will continue to cycle between them.

To make this a more realistic setup, just to show that this could provide a real gain (again assuming no quantum catches) imagine two mirrors with a rotating arm between them. If both sides of this arm had mirrors that were held parallel to the two surrounding and stationary mirrors, then you could reflect light on one side of the arm in one direction and one the other side of the mirror for the other side of the arm and generate a rotational force on the arm. Realistically you'd want to use large enough mirrors to get at least thousands of reflections, but assuming you had a source of photons that was more efficient than this (not very difficult), you should have a net gain of energy.

Ok, so does anyone know what the catch is, or is there one? I assume this effect, multiplying a reflective force for light could be verified rather easily.

Now something else to consider is that if there is a catch, it's interesting to consider what such a mechanism would imply about light.

Ok, let me help out a bit here:

1) Red shifting can't cause a significant effect because if you increase the masses of the mirrors, the acceleration and velocity is reduced, so even if you have a large number of reflections from mirrors moving away that are constantly red shifting the light, by increasing the mass, you reduce this effect, yet the force for each reflection should still be the same.

2) Losses in the mirrors can be minimized. I believe 99%+ reflective mirrors are possible, and even if we only got effectively 100 reflections of light, that energy should be absorbed somewhere as heat.

The issue seems to fundmentally be one of the two:

1) Either light loses some energy (and I assume becomes red-shifted over time) after each interaction.

or

2) A faster than light effect occurs and determines the path ahead of time and limits the number of real interactions that occur.

Hi StevenA,

I think the idea that photons are identical to a "bouncing" ball is wrong, photons "spread" as waves and ultimately they interact as a particle. Photons are not like billiard balls, they are actually entangled waves, we can see this from the Young's Double Slit Interference Experiment. I have given my "stringy" version of how to interpret the photon "event" and that is the way I think things occur.
http://forum.physorg.com/index.php?showtop...ndpost&p=112371
Consider two mirrors facing each other and the emission of a single photon event somewhere between the two "almost" perfect mirrors. You can see a "Hall of Mirrors" that indicate a history for this event. My view is that the ever expanding wavefront of that single event represents a single instant in time no matter where it has gone and what time our clocks show that it is there. This is because in the frame of the photon, there is no passage of time and the event "connects" a frozen "instant" as it spreads in "our" time. One instant, one wavefront, one photon. No matter how long the photon lasts and how far it spreads and how many images of it are created, it is still only a single instant in time. This is exemplified by the delayed choice quantum eraser experiment. It matters little in our time when an entangled photon is localized, the twin is "always" affected by apparent subsequent events in time of its mirror duplicate. This is because in the special rest frame of the photon everything is happening in the one instant of time and (through the most extreme form of "length contraction") everything is happening in almost the one place too (relativistic compression normal to the wavefront of the propagation "everywhere"). The "event" is compressed to a small action volume in time and space (almost infinitely small).

Naturally after a "long apparent time" (to the external observers) as a wave moving unseen in quantum space, the single photon will undergo an "interaction" and as far as we can tell it disappears from the Universe... or does it? Whatever... that will finalize the phenomenon from all observers points of view in our Universe can tell. Until then the total wavefront of the spreading photon (since it goes almost everywhere along almost every possible course) is a record of all influence the event will ever have.

Cheers

This post has been edited by Good Elf on Aug 15 2006, 03:51 PM

I admit this experiement would actually be fun to try out and not overly difficult. I've seen a radiometer before and they're rather simple, except you need a good vacuum and you'd have to hook it up a bit different for this case.

Well, I think you're correct that ultimately only one event at a time occurs but it would still be interesting to know if this even includes individual reflections ... it probably does, but it's just so wierd! (Wierd is good in this case)

Consider also that light travelling through space could be operating along a similar principle of being reflected or refracted along the path, so it could be that the same phenomenon occurs throughout the entire path of a photon ... so what would this pressure look like if it occured in space?

How does it both create a pressure when reflected and yet still have a reflection visible if it's only one event?

(Dang, I wish I had my hands on some decent equipment to try this stuff out!)

A radiometer can measure the pressure exerted by light http://science.howstuffworks.com/question239.htm

If light was being uniformly exposed to this, the difference in pressure, I assume, is due to the doubled force on the mirrored side, so pressure is exerted on all faces but it's doubled for the mirrored sides and that creates the differential ... I wonder how much force a 20 mW laser would generate?

Ok, I've got to do some research here. This one is just bothering me too much

Ok here's some more info - the radiometer I provided a link to does not measure light pressure. This is misinformation that's been propogated. It doesn't work in a complete vacuum but relies on thermal energy instead and can even go backwards when cooled.

But there was a test using http://en.wikipedia.org/wiki/Nichols_radiometer to measure this pressure and here's an article also (salespitch I should say) http://66.249.93.104/search?q=cache:zt6gs0...+and+Hull&hl=en

But it appears there's more than a bit of skepticism around:

http://science.slashdot.org/article.pl?sid...tid=134&tid=160
http://www.hyperflight.com/lightintrinsic.pdf

Apparently others see the same problem with light exerting a force when reflected. My guess would be that light could exert a pressure when absorbed but not reflected.

Also, while doing this search I found a claim that, yes, light does appear to travel faster than light if viewed as following a path reflecting of the mirrors. It would be nice to actually see this stuff first hand but I think you nailed it, Good Elf. Congrats

I hope NASA isn't putting money into solar sails without experimental evidence that the idea actually works.

But on a deeper level, if reflections of light have no inertia, then the ideas of an equivalent momentum to light need to be rethought also. I tend to agree that light doesn't actually travel through space in any real sense, but that instead the way we view space is built upon correlating events as cause and effect. We "see" two things that interact as being located in the same location of space and group them according to what's most likely to interact with what and distances are comparisons of the order in which these interactions will occur but the "vacuum" or space that all these are envisioned as travelling through does not necessarily exist.

Hi StevenA,

I think you really have thought about this one and the analysis is "spot on". This seems to me as if we have the dual nature of photons coming to the fore. When the "event" is in motion according to our "external" concept of time, because of the Delayed Choice Quantum Eraser Experimental Result it is "painting" a single final view of the total event on the "hyper-surface" of our Universe once and for all time. This is just the "cosmic artist" using "his" timeless brush to paint this event into the framework of our Universe, the "artist" moves only at the speed of light but the individual event is "timeless" and "unchanging". The word "artist" is just a literary artifice... do not take this literally .. he he he!

It is true that the wavefronts could potentially criss cross themselves time and time again in the course of millions of years if those photons are traveling from far away to us. The "event" does have a beginning and an end when it is originally "emitted" and finally "absorbed", these are the only "particle events"... others are unseen wave phenomena that are executing in "hidden quantum space". To "see" this for an individual photon is to destroy it in a quantum demolition event shooting the "particle" down from its "hyperspace" into a local flash of scattered light from a hypothetical screen. These two "components" are not "exactly the same thing" since emission and absorption are not "instantaneous processes". This defines an exceedingly small interval of time in which a particle is in the process of emission and an exceedingly short period of time during the absorption... as well as an exceedingly small region of space ... the evanescent zone... in which this process is "confined". This means that in that "double ended" event the photon is experiencing a tiny quantity of time... not while traveling... but on "creation and destruction" as a particle. Without the progress of time and a localization of space... the particle processes of energy and momentum cannot occur... this is because all these "dynamic" processes require some infinitesimal time to elapse. All the multiple reflections are for free. In the real world if you "see" any of these wavelike behaviors then that is where those individual events (entering your eye) come to an end through quantum demolition.

If you want that exchange of momentum you would need to look carefully at these end processes to get the "deal" you need. Of course I have a "stringy interpretation" of this phenomena.

Yep!... I agree it is really interesting. But I think you can answer this now when you consider that only those "endings" involve any momentum or energy exchanges which necessarily involve some passage of infinitesimal time. I wonder if you have been following my arguments regarding "Kondo Phantoms"? This process supplies the alternative Universes required and insisted on by modern Quantum Theory and also the operation of Quantum Computers. Everything is "good".

Cheers

This post has been edited by Good Elf on Aug 16 2006, 02:38 AM

Your talking about trapping a perfectly uniform beam of light between two perfectly placed and 100% reflective mirrors. Good luck. Even then it wouldn't work. What it says at wackipedia is that electromagnetic radiation has force. When light is reflected it is absorbed then resent out. When the light is absorbed it imparts a force on what is absorbing it. When it is sent out (same photon with the same energy level) that force is once again imparted upon the object sending it out (for every action there is an equal and opposite reaction). The photon isn't traveling at twice the speed and so it will do the same with the opposing mirror applying an identical force as it did with the first mirror being absorbed and resent out back to the first mirror. You do have an endless cycle but the force doesn't increase (double) from mirror to mirror.



This post has been edited by Precursor562 on Aug 27 2006, 04:56 PM

No, if you read what I posted, I was even talking about typical optics and referring to even hundreds or less of reflections.

If light imparted a force being reflected, then it should lose energy and be red-shifted. Since when does light lose energy during a reflection?

It's not just wikipedia. Even NASA was researching solar sails. I'm not the one claiming reflected light creates a force. I'm the one pointing out the problem with such an idea.

Ok, so let's say you mounted a mirror on a large mass and reflected light off it. Do you believe you could get the object to move by shining light on the mirror?

If light is absorbed for some finite time, then a finite force is created and a finite amount of energy lost.

I'd assume reflections don't extert any force at all and NASA etc. are wrong. http://solarsails.jpl.nasa.gov/

Either that or we have free energy. They aren't simply assuming the non-reflected component provides a force but that the force is simply proportional to whatever equilavent momentum of light is reflected (double the absorbed value).

If I take a mirror and bounce light off at a 90 deg angle, would you expect to see a force at a 45 degree angle? If light is absorbed in one direction and reemitted in another direction, then these aren't equal and opposite.

Reflections must not exert a physical force at all otherwise we can use them for perpetual motion (not that I'd complain ).

I don't believe the particle component of a photon ever even travelled to the mirror at all.

Ok, if I take a beam splitter and run a photon through it then I recombine a reflection of this, doesn't the photon interact with itself at faster than light speeds?

How can a photon travel all these paths at the speed of light yet still take one path at the speed of light?

If you want to say it takes all these reflected paths simultaneously, then you're saying the same thing as I am - that light didn't actually follow each reflection, stop, exert a force and then bounce on to the next position but instead went effectively faster than light speed in determining where it was finally absorbed.

Personally, I don't think the photon was absorbed at all during the reflection or this seems to imply it had to physically interact with all these points at faster than light speed.

My personally view is that the path is determined at faster than light speeds (according to our interpretations of space) and that the particle itself simply hops from point to point instantly. This doesn't actually need to take any time at all. It only needs to take an average time relative to other such instantaneous events.

Ok, so you appear to agree that there's no reflective force from light also, or I'd assume you would have said the force increases.

Well in either event it's worth testing anyway. No matter what theories say, there's nothing like reality to give the final say. If reflected light does generate a force as a lot of mainstream views have portrayed then there's free energy sitting around and solar sails will work. If not, then the idea needs to be rethought and likely only absorbtion at the endpoint of a path for a photon yields a force.

This post has been edited by StevenA on Aug 27 2006, 06:42 PM

No it actually sounds like your talking about trapping light between two mirrors.

When light is "reflected" is is first absorbed then re-emitted by the material doing the reflecting. With 100% reflection for every photon absorbed one of identical energy level is emitted out. The photon impacting the atom imparts a force (the photon is absorbed upon impact) the same photon emitted will impart the same force provided it is the same photon (for every action there is an opposite and equal reaction). So the force is two fold on the mirror but nothing changes with the light. The light will hit the next mirror with the same force since it will be traveling at the same speed. That force too will be a double force with the same photon being emitted and the double force on the second mirror will be the same as that placed on the first mirror.

I'm not denying that electromagnetic radiation has force. Hell I believe light to be comprised of a photon particle with mass so of course it will impart a force if it collides with something.

Well since light imparts a force onto the object it is being absorbed by and it also imparts a force in the same direction when the light is emitted back out. The problem is that light imparts a very small force and your sail will have to be huge but the larger the sail the more mass it has and the more inertia it will have so the force light applies will not be enough to make it move. Then there is the ship that will add to the mass and therefore the inertia. What would need to happen is to increase the current of high energy photons (number of photons passing a specified volume of space in a specified amount of time) while leaving the size of the sail fixed to give it (and the ship attached) a fixed mass. The question is how many high energy photons can we cram together and how crammed can we get them.

I swear if you don't spell things out.....

When light is absorbed by a material a force is imparted on it. Provided that the photon came perpendicular to (forming a 90deg angle with) the material doing the absorption than if the same photon were emitted back out in the direction in which it came (reflected) it would impart the same force once again in the same direction as before.

That makes absolutely no sense. First off a beam splitter splits the beam. Individual photons don't split. Lets say you shine a laser onto a surface, it takes so much time for the individual photons to make the journey from their source to the surface. Lets say mirrors are placed so that half the beam gets deflected in a different direction (beam splitter splitting the beam) while the other half continues to take the direct path. The reflected beam then reflects off three more mirrors so it ends up back next to the beam that took the direct route. If you were to track two separate photons that were emitted at the exact same time where one took the direct path and the other took the indirect path (following the mirrors) than the photon that took the direct path will reach the surface before the one taking the indirect route. If the indirect route is twice the distance as the direct route than the photon will take very little more than twice as long as the photon taking the direct route.

The distance alone would cause it to take twice as long but then you have to add the time it takes for a photon to be emitted after being absorbed and multiply this time by four (the number of mirrors doing the absorbing and emitting).

So for the one photon t=d/V
t=Time of travel
d=Distance
V=Velocity

The second photon t=d/V+4X
t=Time of travel
d=Distance
V=Velocity
X=Time it takes for the photon to be emitted after being absorbed


If you shine two lasers directly at each other then the speed of each photon is still only the speed of light but if two photons were to head directly toward each other the speed at which they are traveling toward each other is twice that of the speed of light (the sum of their individual speeds).

Car A is moving east along a road at 100mph and car B is headed west on that same road at 100mph. They may be moving at 100mph individually but are moving toward each other at 200mph.

Give car A a speed of 50mph and car B a speed of 150mph they are still moving toward each other at 200mph.

No I'm saying there is a force but that it just doesn't increase.

Here is what I'm not saying

Let's say that a photon applies a force of 1N on an object when absorbed and then applies this force again while being emitted for a total force of 2N.

So the photon contacts the mirror getting absorbed applying 1N of force. The mirror then emits the same photon applying an additional 1N of force in the same direction as before for a total of 2N of force on the mirror. The same photon then contacts a second mirror applying 2N of force while getting absorbed then applies an additional 2N while getting emitted for a total of 4N. The photon then goes back to the first mirror applying 4N to the mirror while getting absorbed then applies an addition 4N while getting emitted for a total of 8N.

I repeat "That is not what I am saying"

What I am saying is if the photon applies 1N of force to the mirror while getting absorbed it will apply an additional 1N when emitted for a total of two. When the photon contacts the second mirror it applies 1N of force once again and then applies an additional 1N when emitted for a total of 2N on the second mirror. Same as the first. It doesn't matter how many times the photon reflects back and forth the force applied to each mirror will be 2N. If you reflect the photon off 1 million mirrors you will still only have 2N of force applied to each mirror.

Hi Precursor562 and StevenA et al,

No quarrel here but just like to get your view on this one Precursor562... What if those 1 million mirrors were scattered around the galaxy? Does this one photon do work on all those 1 million mirrors?

Cheers

If some force is detectable from a reflection, then this force should be doing work and providing energy otherwise you couldn't measure the force if it provided no displacement.

So we'll assume this is a real acceleration and not simply a mathematical abstraction.

We also know photon energies are quantized into discrete units.

Now if I were to emit a single photon of the lowest possible energy, a single unit, and it reflected off something and was absorbed elsewhere, where do the energies do?

1) Did the emitter recoil? If so, how many units of energy were present in the recoil?
2) Did the reflector receive energy? If so how many units?
3) Did the receiver receive energy? If so how many units?

The only way any of this makes sense is simply if the receiver received a single unit of energy and the reflection (or even multiple reflections) received nothing.

This might sound unintuitive but consider the wavefunction for light is capable of being modulated and shifted at faster than light speed.

For example, quantum tunnelling, and Cherekov radiation are suppose to be due to these effects.

In small clusters of cesium atoms, charged particles can travel at up to 300 times light speed for short distances.

http://www.space.com/scienceastronomy/gene...n_c_000719.html

Here's a more in depth description:
http://www.theness.com/articles.asp?id=34

Of course on average for long distances (in uncontrolled, non-accelerating, average vacuum of space, for round trip measurements) you can't exceed the speed of light for long but the group velocity for the wave could potentially be infinite.

I believe the resolution is simply that the wavefunction is affected by reflections but this doesn't mean the particle has to physically bounce off the mirrors to do so, nor need to deliver a force. It can simply hop to whereever it gets absorbed and delivery the energy there.

This post has been edited by StevenA on Aug 28 2006, 05:35 AM

If it reaches all of them without colliding with anything first.

I think there is a confusion here. One is the photon's energy level (eV) while the other is the photons momentum.

Simply put, a photon has mass so when it decelerates (upon being absorbed by the atom it collides with) it exerts a force upon the atom. The atom being part of a greater body (sail) will feel this force. The emitting of a photon will exert a force upon the atom once again in the opposite direction of that in which the photon leaves. The the direction of both forces are the same.

The photon carries a charge (eV) and when absorbed by the atom causes electron(s) to increase in energy level. The atom then drops in energy level and a photon is emitted.

Two different scenarios that are closely related. I believe it would be best to think of the atom as the perfect spring. Throw a base ball (photon) at a spring (atom) mounted on a wall (greater body). The ball contacts the spring causing it compress (atom getting absorbed) but as it does the the ball slows down to a stop prior to changing direction. As the spring gets compressed by the ball a force is place upon both the ball and the wall. Both the ball and the wall receive the same force in opposing directions so the wall receives the force in the same direction the ball would be headed had the wall and spring not be there where the ball is receiving the force in opposition to the direction in which it is headed and is the force causing it to slow down. At this point the kinetic energy (the eV of the photon) the ball had went into compressing the spring (increasing the energy level of the atom causing one or more electrons to jump to a higher energy level). Now the spring will 'spring' back to it's original state (the atom will return to its original energy level) by exerting force upon the ball. However the force the ball receives by the spring re-asserting itself is also applied to the wall opposite to that of the ball's direction.


With the ball and the spring the spring will continue to apply pressure to both the ball and the wall for as long as it is compressed. This is unlike the atom and the photon where the pressure ceases. So it would be like the ball compressing the spring which will apply pressure but as soon as the spring reaches maximum compression (the point where the photon is absorbed) it no longer applies any force to neither the wall nor the ball. However it once again applies pressure to both when the ball is ready to leave (when the photon is released).

So the photon impacts the atom. The photon rapidly slows to near stop due to a force applied to it. The same force is applied to the atom in the opposite direction. The photon is then absorbed adding its eV to that of the atom causing one or more electrons to jump energy levels. The one or more electrons then drop back down and the photon is sent back out in the direction from which it came. It gets up to speed at the same rate at which it slowed before getting absorbed and so the same force is applied to the photon once again in the same direction as before. An equal and opposite force is applied to the atom once again as before. The perfect spring.

This post has been edited by Precursor562 on Aug 28 2006, 11:06 PM

Well it wouldn't be lossless for the photon if it delivered some momentum to the mirror. The photon would then have less momentum as it transferred it to another mass.

So how would this loss of momentum be witnessed in a reflected photon?

I know light becomes polarized reflecting at an angle off of metal and so it could lose some component of energy in one direction from this, but that's actually due to an imperfect reflection and some energy being absorbed on an axis. I've never heard of polarization from a reflection off a mirror or red shifting before. So what component of light provides this energy for a reflection? How can it deliver a force to a surface and not obviously have lost any energy in the process? Is there some unobvious mechanism that transfers energy?

99.99% reflective mirrors aren't difficult to make and some mirrors are claimed to be 99.9999% reflective and could potentially amplify such a force over 1 million times with possibly even better mirrors in the future. This seems like free energy.

This post has been edited by StevenA on Aug 28 2006, 11:43 PM

The short answer...eV.

Photon hits atom with momentum.

Collision places a directional force on the atom.

Photon gets absorbed by the atom causing the atom to jump energy level(s)

Atom returns to previous energy level releasing a photon.

The photon is released in the opposite direction from which it came (reflection)

The photon is released with the speed of light so a force must be applied to the photon to allow for such a velocity.

Same force gets exerted on the atom in the same direction as the previous force.


No energy is lost. The photon gives up its energy to the atom which then returns it to the photon. In a perfect reflection 100% of the energy is returned. When you look at something that is red, purple, green or whatever color, photons of all energy levels (white light) collides with the atoms making up the colored object. They get absorbed but then released in a controlled manner. Where photons of all energy levels get absorbed by the atom only photons of a specific energy level gets released giving you the color of the object. So the atom takes the energy of the photons whose energy level is higher than those of the color and gives it to the photons whose energy level is lower than the color. You still end up with no energy lost. However in the case of something that is purple, Violet photons have the highest of the energy levels in the visible light spectrum. So lets say that only photons of the visible light spectrum shines on the purple object. The violet photons will be re-released but those photons that are lower in energy level will be combined within the atom then release as violet photons. When an atom absorbs a photon the photon is essentially gone (although it still exists as an increase in the atom's energy level). So when the atom drops in energy level it releases a photon whose energy level is equal to the atom's energy drop. So in the case of colored objects the atoms that make up those objects will only drop a set amount of energy level. That set amount is the color.

Well the energy and momentum are supposedly related by e=p*c, which would make sense if you assume the photon is travelling at light speed.

So at least if this is true, then the entire momentum is delivered to the atom when it's absorbed and not just a fraction of it.

Now if the photon is then reemitted, then this momentum absorbed by the atom should be driving force behind it as the atom wouldn't spontaneously emit that energy photon under other circumstances (it only emits the photon from a reflection, so it couldn't have that momentum available to pass onto anything else, because if it did, then it would no longer be in an excited state and reemit the photon).

So it must consume this momentum instead of creating a duplicate and reinforcing copy of the momentum.

I'll draw a little text picture here. The photon heads from the left toward atom1. Atom 1 is bound to atom2 by EMF forces.

photon -----> atom1 - atom2

Now if atom1 absorbs the photon, then it would have a momentum to the right. I'm certain we both agree that if the photon wasn't reflected, then at least part of this momentum could be passed on to atom2 and so the momentum is detectable as a force.

But the question is over the reflection.

If atom1 passes momentum on to atom2, then atom1 no longer has the same momentum and it can't reemit a photon of the same energy.

If atom1 instead consumes the momentum received by the photon in order to reemit it, then the energies remain the same and no momentum is left in either atom.

Now you might say that atom1 absorbs the momentum from the photon, then reemits another photon and again absorbs reinforcing momentum from the reemission of the photon so that now it has twice the momentum of the single photon.

What would stop atom1 from passing some of this overabundance of energy into both atom2 and creating a second photon. Or if atom2 received some of the energy, then it could also emit another photon and gain even more energy etc. The two atoms could pass momentum back and forth and emit photons continually from the extra energy as far as I can tell.

I believe what happens is that for a reflection, the light wave reflects before the particle component ever gets there.

For example, if the wave function drops to 0, a particle will never be detected there. If you look at the mirror as reversing the wave function 180 degs, then you could have a perfect phase cancellation of the wave on the immediate surface of the mirror and the photon would never physically reach that point.

When a photon is instead absorbed, there's no reverse component to the wave and hence no cancellation so absorbtion doesn't have a problem.

I don't believe this 180 deg phase shift of the wave is seen using mirrors, but consider that we only see the absolute magnitude of the wave and not the polarity of it so -amplitude^2 = amplitude^2 so it wouldn't be obvious ... though you'd still assume this would create a problem with cancellations after using a mirror as distances of wavelengths wouldn't be preserved accurately. (does anyone know off hand if this is a real effect that's observed? The conflict I'm thinking of is that a laser should have a standing wave passing through 0 at the mirrors, but it takes self interference to notice this, but that could be because the mirrors aren't perfectly spaced so some phase shifting occurs .. but that should still servo to a steady state I'd think)

I've seen something similar to this before. If you place a mirror to directly reflect a laser back into the emitter, you can kill the beam. I don't know if it's due to an effect exactly like I described, but it could be.

This post has been edited by StevenA on Aug 29 2006, 01:10 PM