[Herra Tohtori]

ok, so I'd like to ask one of the more physics minded people a conservation question.

if I have a nulcear submarean that has suddenly been transported into outer space (and for some irrelivent to the larger point of the question reason the design of the sub allows it to continue functioning in a vacume). it is moveing at a constant velocity, yet inside it has a nuclear reactor wich is breaking apart uranium atoms for energy, as it does this the mass of the uranium and by extention the mass of the whole system (sub) it'self decreases (if I understand my nuclear physics properly). yet it's velocity remains the same, the propellers might spin in one direction or the other, but the velocity of the whole system remains constant yet the mass does not, by conservation of inertia, isn't this imposable? because I=M*V, the inertia of the system is going down, because the mass of the system is going down and the velocity isn't going up to counter this.

could it be posable that this situation is something similar?
I'll ask my physics teacher about this tomarrow and see if there is some odd thing in relitiveity that might allow this.

Well, easiest answer is that an object traveling at constant speed is in the middle of its own reference co-ordinate system, and if its mass is reduced by, say, slicing a piece of it away, the speed of the main body has no reason whatsoever to accelerate (change speed in any direction). Its momentum is reduced, but the sliced part of the body gets that momentum.

In a space ship, the "sliced" bit is fotons emitting at statistically every direction from the ship. The fotons have momentum, and they "steal" some momentum from the ship.


However, if we have a theoretical space ship that we all love, that does not emit anything and absorbs everything that the reactor produces, things become more interesting... but not that much.


In this scenario, particles and fotons radiated in the reactor core all absorb into the ship surrounding it, nothing escapes into space.

Thus, the energy of the ship-reactor system remains the same, and so does momentum. The thing which is important to notice here is that when nuclear reactions reduce mass of entire system, the momentum that these particles gained is absorbed into the surrounding hull. Again statistically, photons and mass particles are emitted into every direction from the reaction, so all parts of the ship get even dose of radiation (energy and momentum).

From this on, things become interesting.

If we consider the ship-reactor system on time t=0, the system has mass m, energy E = mc^2 and momentum p = mv... however, the momentum also has relation to ship's overall energy via the mass-energy relation, so momentum can equally be annouced as

p = v*E/c^2

At t=1, N amount of nuclear reactions occur in the reactor. They release particles and fotons at even directions. Now the reactor has less energy, less mass and less momentum than at t=0, but the velocity remains same - because it gave momentum to even directions.

At t=2, the radiation hits the spherical core around the reactor... mass particles come a bit later, but they have the same effect as fotons so it's no use to handle them separately.

What happens here is that the momentum of the fotons is transferred to ship's particles. The momentum of particles is then transferred to other particles that are close, and then to others... and others... And lo behold, the momentum originated in the reactor takes place as thermal energy in the core of the ship.

So, in a "closed ship" scenario the momentum of the system does conserve; it just changes form and becomes not-so easily noticeable. The ship heats, the heated matter has particles that have kinetic energy, which means that they have momentum... the very same momentum that came from the reactor.

So, all in all the energy of the ship remains the same, if nothing is removed from it, and thus the momentum and speed also remain same. Momentum is just dispersed along the ship so every bit of momentum pointing to direction is negated by other, random directed momentums. The sum of momentum is still the same that it was at t=0.

At least this is how I would handle this subject. Though my knowledge of physics is yet quite limited...

Of course, this kind of ship would be a death trap to all inside it, as it would rather quickly reach rather higher temperatures. In truth, most of thermal energy produced must be radiated out of a ship to retain the thermal equilibrium. To keep the crew quarters at static temperatures, the heating system must only work at same rate that the ship absorbs to space. Most reactors produce way more thermal energy... and when thermal energy is radiated away from the ship, it steals some of the ship's energy - most being energy from the reactor, energy which was converted from matter to radiation. And this does indeed reduce the momentum of the ship itself, even though the vvelocity remains the same.

[Bobboau]

I thought photons had no mass? if they have no mass how can they have (or 'steal') momentum? I can understand loseing energy, but not momentum.

[Col. Fishguts]

That's because mother nature is a *****.

Photons have zero mass but carry momentum anyway.

To be more exact, they have zero rest mass, but like any moving particle they have a relativistic mass (or just energy, depending which nomenclature you're using)

[Mars]

Dear God he's right

[Taristin]

I think dismissing it as a flight of fancy is a little off.

This is clearly a working technology, and while it won't become the core of a double decker bus tomorrow it does have potential to be developed.

Ditto. The lot of you who are quick to dismiss something as bollocks are what's holding back a lot of potentially revolutionary discoveries. Just as radical thought is dangerous, such also is conservative.

... And we have seen how many of these threads, 5013, and how many have turned out to become actually working technology? We're lucky if even one.
The burden of proof is on the inventor to prove his concept right and working, and as the concept sounds wacky I will find skeptism very healthy.

So? By restriction spread of the idea and immediately writing it off as bollocks, you potentially crush the one in five thousand that does work.

[Bobboau]

ok, so as far as the subject of this propulsion system is concerned, is it posable for a photon to bounce off an particle without imparting momentum to it? since it seems to have momentum defined diferently for it. I'm guessing the answer is no, but if it's yes, then this might work.

[Bobboau]

[tangent]wait a sec, what about gravity? if we had some insainly huge black hole level gravity well (event horison about the size of a marble) and we fired a photon beam at it in just the right angle so that it would slingshot around and hit a reflecting surface, and back again forming a roughly u shaped path, would that do it? photons have no mass and thus nothing can be atracted to them but they can be diverted by gravity without effecting the masses involved (right?), so every interaction with the reflective surface would impart momentum and then it would get wiped around and hit the surface again. the momentum is haveing it's direction changed by gravity without induceing a opposite reaction.[/tangent]

[Herra Tohtori]

ok, so as far as the subject of this propulsion system is concerned, is it posable for a photon to bounce off an particle without imparting momentum to it?

No.

When an object reflects a photon, it actually first receives the momentum of the photon; then, when it sends it back on its journey, it gives the photon some momentum and gains itself equal opposite momentum, so in reflection situations, the reflector gains two times the momentum of the actual photon... that might seem weird at first, but if you want to, take a closer look at what happens in reflection situation.

When a photon "bounces" off from a particle, what happens is that it briefly absorbs to the particle, and then the particle emits a new photon to a direction defined by the particle's effect on electromagnetic wave frontier. Or, the photon may stay absorbed...

Reflective surfaces are interesting thing. They have uniform surface structure, which causes the impacting EM-waves to behave jsut like any other wave hitting straight surface. But the photons that are reflected are not the "same" photons that hit the reflector...

You see, a photon is a seemingly anomalous little bastard, though the strangeness fades a bit when you get to know the things behind phenomena (only to get new strangenesses to ponder).

A photon always travels at speed of light, related to any and all reference frames. That means that the photons experience no time, and thereby they also cannot change. From photon's point of view, the start of its life and journey and the end of them are the same.

That is also because a photon does not, and indeed cannot have rest mass - a photon cannot be at rest.

It a photon would be at rest compared to an object, it would not exist. Maxwell's electrodynamics contain this anomality, and that lead Einstein to state that every observer measures speed of light to be constant in all situations.


So, if a photon cannot be at rest and it always has constant speed, it also cannot change its direction during its life. Bouncing off a surface would involve changing the speed vector, which a photon cannot do as a particle. However, if the conditions are good enough to keep the wave front intact and the photon's energy is immediately emitted from an electron that absorbed it, the result is that the bouncing electromagnetic wave front gives some of the original photon's properties to the new one; most important being the direction, which follows Huygens' principles. That gives an impression of a reflecting photon; it's not really the same photon but it has the same wave length, same energy and logical reflection direction, so it can be handled as the "same" photon.

A good mirror reflects more than 90 % of all the photons that hit the electrons on its surface, and the reflection always has some scattering on it - the amount is inversely relative to quality of the mirror. The smoother it is, the less scattering. Matt surfaced objects reflect very scattered light; mirrors reflect very little scattered light.

Um... where was I? Yes, momentum.

A photon has momentum because it has energy; the momentum of photon is not, however, relative to the speed of it due to the fact that all photons have exact same speed regardless of their energy. As was already said, photons have a momentum mass because they have energy; the amount of mass used in momentum calculations is

m = E / c^2

The energy of a photon is

E = hf; thus the momentum of a photon is directly proportional to its frequency... and inversely proportional to its wave length:

p(f) = m*v = hf/c^2 * c = h f/c

...oh, and h is the Planck's constant, should it mean something to you. Whatever. Photon has momentum, always, period.

When the photon hits an object, it is absorbed into matter... usially the particle being an electron.

If the electron happens to be in stable state, it almost instantaneously emits a new photon back; this happens so fast that the quantum wave front of the photon is still there. If a new photon is emitted fast enough, it takes the reflection direction of the original photon's wave front.

Some photons don't reflect; that's because the poor buggers are absorbed more permanently by the mirror surface and their energy is used to heat the mirror.

since it seems to have momentum defined diferently for it. I'm guessing the answer is no, but if it's yes, then this might work.

Huh?

The momentum of a single photon is always defined by

p = hf/c.

In reflection situation, the reflecting body gains two times the momentum of a photon; first time at arrival, second time at departure.

The weird thing in this particular claimed physics-breaker is that the photons are bouncing inside a confined space. It should be theoretically impossible for them to excert greater force on one end of a chamber than on other end, because statistically, they do not exit the chamber (except the occasional few), so the conservation of momentum is not valid if there indeed is a measurable force produced by this device.

It all boils down to whether or not the prototype works. If it works, it will become a subject of wild research, while it also can be of course used in applications even if it's not completely understood. People have done it before. Fire is a good example of that...


Oh, another message...

Well, gravity doesn't have effect on reflections, it's all wave motion principles and the fact that light have wave motion characteristics.

And, insanely huge black holes wouldn't be the size of a marble. They would be (and are) kilometres wide in event horizon apparent diameter.

I really can't say everything I know about black holes here and now, so sufice to say that if a photon hits the event horizon tangentially, it will forever and ever go round and round the event horizon's surface. If it goes in, it will forever spiral towards the nonexistent center of infinitely deep gravity well; if it swoops past the event horizon, it will not be emprisoned by the hole, but instead the straight path (geoide) in this area of space will take the photon onto journey around the black hole, much in a way you described.

Keep in mind though that if you apply General Relativity to matters, you have to apply it to all of them. The momentum of a photon diverted by gravity does not change, because its direction does not change and no forces are exerted upon it.

...As the annoying child in Matrix said "the spoon doesn't bend, you do" or something like that. The photon travels straight, period. What little disturbances there might be on the way, it just goes along them. Straight, no Chaser, as they say...

So the seemingly radical direction change around the black hole does not actually require change of momentum...

[Bobboau]

hitting some relitively minor things you seem to have misunderstood:

"huh?"
photon's momentum is defined as p = hf/c, matter's momentum formula I was always taught to be p = mv. so momentum for a photon is deined diferently... than it is for mater, is what I meant.

"And, insanely huge black holes wouldn't be the size of a marble."
I said insainly huge mass, I was thinking something the size that could fit on a table top, a black hole with a EH of a half inch across would have the mass somewere around that of the earth wich for something sitting on a table is insainly huge, though by blackhole standards is rather pathetic.

that hole post was a tangent thought about a relitivly diferent situation, trying to break conservation of momentum.

so if I were to fire a hypothetical laser at a hypothetical black hole in such a path that it were to make a pass around it and hit my laser (wich is made from some hypothetical substance that would reflect all light, thus not blow it'self up). when the beam comes back around and hit's the laser what would happen? both to the laser and the black hole.

sence the photon has no mass and thus no gravity, it's just folowing the bends in space, so it should have no effect on the black hole, at all, if I fired a quatrillion yattawatts worth of energy in the form of photons past an object you would not be able to tell unless the beam hit the object (in wich case it would probly explode), there would be no gravitational effects, and there would be no visable change to the object so long as none of the photons hit the object. right?

so a photon going around a black hole will not effect the black hole in even the slightest degree. but when it hits the laser wouldn't it interact with the atoms of the laser get absorbed and reemitted imparting momentum in what would seem the opposite diretion that wich it was fired from?
isn't this a loophole here?
the geometry of space is deformed such that it twists back over it'self  but in a diferent direction when the photon travles strightly through this deformed space it will end up going through the same reagon of space but in the opposite direction. the beam did not bend, truely, but it is hitting it's own emmiter, and imparting momentum in a direction that is, from the emmiter's point of view, in the opposite direction that it had fired the beam earlier. right?

now consiter a similar situation on a smaller scale were you have a small black hole in some sort of containment feild (held in place with electrostatic forces or something) inside a machine and a powerful laser is fired to just miss the event horison causeing the beam to double back (note: I understand the bending of space/photon travleing straight through non-straight space thing, but to simply describe it as it would be seen from an outside observer makes explaining the situation a lot clearer), like before, hitting the laser (without destroying it) this is just like the previous situation exept the black hole and the laser are held together so they move with each other.

from what you were saying it sounded like when the beam hit it would pull the laser in the direction opposite the beam was travleing, which makes me think I didn't follow you properly.

[Kosh]

It's like what I said about the Perpetual motion machine: I'll have to see it to believe it.

[Janos]

I think dismissing it as a flight of fancy is a little off.

This is clearly a working technology, and while it won't become the core of a double decker bus tomorrow it does have potential to be developed.

Ditto. The lot of you who are quick to dismiss something as bollocks are what's holding back a lot of potentially revolutionary discoveries. Just as radical thought is dangerous, such also is conservative.

... And we have seen how many of these threads, 5013, and how many have turned out to become actually working technology? We're lucky if even one.
The burden of proof is on the inventor to prove his concept right and working, and as the concept sounds wacky I will find skeptism very healthy.

So? By restriction spread of the idea and immediately writing it off as bollocks, you potentially crush the one in five thousand that does work.

No I do not. My skeptism is just skeptism - it does not define reality, it does not restrict the spread of idea. Only thing it - combined with others' skeptism - is to place a burden of proof on this guy. If he can prove it, fine - then this guy is a genious and I am wrong. If he can't, well - no one has lost anything.

However, I don't take Superb Science Breakthrough Science Magic Threads any more seriously than Homeopathy, Body Toxins and stuff like that, because 99,9% of time it's just complete crap and the rest just suits the margin of error. If someone told you that this very night you can observe UFOs and he had told you that 1000 times earlier and every single time he had been wrong, would it really be dumb to write him off this time? Bad analogy but still.

Edit: I see
VVVV

[Descenterace]

Well, the business with existing particle accelerators proves that the theory behind this engine needs investigating. Whether it actually works or not in a practical sense has yet to be seen.

What's happening in this engine is that kinetic energy is imparted to the cavity via microwaves. Energy is converted from electrical to kinetic with minimal intermediary stages. Conservation of energy is satisfied.

[Colonol Dekker]

Electric to Kinetic transfer is one of the most innefiecient transfers of energy though.

[Herra Tohtori]

hitting some relitively minor things you seem to have misunderstood:

photon's momentum is defined as p = hf/c, matter's momentum formula I was always taught to be p = mv. so momentum for a photon is deined diferently... than it is for mater, is what I meant.

Actually... If you want to get technical, the definition of momentum is the excact same for both photons and particles with mass.

For a particle, momentum is always

p = mv.

Photon's mass comes from equation E = mc^2 -> m = E/c^2.

Photon's energy is E = hf.

Photon's speed is c.

So, photon's momentum is

p = mv = E/c^2 * c = E/c = hf/c = h*lambda

lambda there is photon's wave length...


Alternately, because all particles have also characteristic wave length (deBroglie wave length), EVERY particle's momentum can also be defined as

p = h * lambda

where lambda is particle's deBroglie wave length.

"And, insanely huge black holes wouldn't be the size of a marble."
I said insainly huge mass, I was thinking something the size that could fit on a table top, a black hole with a EH of a half inch across would have the mass somewere around that of the earth wich for something sitting on a table is insainly huge, though by blackhole standards is rather pathetic.

Right. 

that hole post was a tangent thought about a relitivly diferent situation, trying to break conservation of momentum.

so if I were to fire a hypothetical laser at a hypothetical black hole in such a path that it were to make a pass around it and hit my laser (wich is made from some hypothetical substance that would reflect all light, thus not blow it'self up). when the beam comes back around and hit's the laser what would happen? both to the laser and the black hole.

The laser would gain momentum backwards when it sent the photon on its way; then it would get the same momentum again when the photon, after going round the black hole, hits it and is absorbed.

However, as to whether or not the black hole is affected by photon through gravitation, well that's more difficult question. Theoretically, all energy bends space-time continuum - and since photon has energy, it should also do that... but I must admit that in this phase things are starting to get a little past my current knowledge. I have perceptions but not actual knowledge at this level.

sence the photon has no mass and thus no gravity, it's just folowing the bends in space, so it should have no effect on the black hole, at all, if I fired a quatrillion yattawatts worth of energy in the form of photons past an object you would not be able to tell unless the beam hit the object (in wich case it would probly explode), there would be no gravitational effects, and there would be no visable change to the object so long as none of the photons hit the object. right?

You must then remember that in fact every object in curved space time just travels along straight line.

An object seemingly under gravitational influence does not in fact feel any force excerted upon it in any situation. Gravitation force in general relativity is much similar to centrifugal force; it's an apparent force created by differences in co-ordinates... in centrifugal force, rotating reference frame causes apparent force; in gravitation, twisted reference frames (that appear to be straight!) cause apparent gravitational effects such as falling.

As Mr. E noticed: You don't feel your weight when you are falling...

so a photon going around a black hole will not effect the black hole in even the slightest degree. but when it hits the laser wouldn't it interact with the atoms of the laser get absorbed and reemitted imparting momentum in what would seem the opposite diretion that wich it was fired from?
isn't this a loophole here?
the geometry of space is deformed such that it twists back over it'self  but in a diferent direction when the photon travles strightly through this deformed space it will end up going through the same reagon of space but in the opposite direction. the beam did not bend, truely, but it is hitting it's own emmiter, and imparting momentum in a direction that is, from the emmiter's point of view, in the opposite direction that it had fired the beam earlier. right?

now consiter a similar situation on a smaller scale were you have a small black hole in some sort of containment feild (held in place with electrostatic forces or something) inside a machine and a powerful laser is fired to just miss the event horison causeing the beam to double back (note: I understand the bending of space/photon travleing straight through non-straight space thing, but to simply describe it as it would be seen from an outside observer makes explaining the situation a lot clearer), like before, hitting the laser (without destroying it) this is just like the previous situation exept the black hole and the laser are held together so they move with each other.

from what you were saying it sounded like when the beam hit it would pull the laser in the direction opposite the beam was travleing, which makes me think I didn't follow you properly.

I suppose that the beam of light does in fact have effect to momentum of black hole, at least if the conservation of momentum holds... but the effect only is measured in laser's reference frame.

What happens is that when laser emits a photon, it gives photon momentum and gains itself same amount of momentum to opposite direction.

When the photon rotates the black hole, it does a loop in laser's reference frame, so from laser's point of view the black hole alters the photon's path; thus in laser's reference frame, the photon must also affect momentum of black hole. In black hole's point of view, the photon doesn't affect it; in photon's POV the black hole doesn't affect it. Only in laser's reference frame (and others, but we are on laser frame measuring things) there is an apparent interaction between two objects.

But, in photon's point of view it goes along straight path, the starting point and ending point just happen to be the same ones in Laser's reference frame.

In GR, there are much of these seemingly paradoxal "physical loopholes", but eventually when matters are viewed from all co-ordinates, things usually sort out mathematically, even if they wouldn't do it logically... I've not yet had the pleasure to actually familiarize myself with actual mathematics of general relativity, and I suspect it won't be just dancing on the roses...

AS I said, it gets muy complicado at these levels of physics, even if the basic principles are quite simple...

[Nuke]

it looks like he found a way to exploit one of theese physics loopholes. this device might be the first of many usefull devices that will exploit them. we must find them all and use them to our advantage, and hope we dont break the laws of physics in the process.

[Herra Tohtori]

it looks like he found a way to exploit one of theese physics loopholes. this device might be the first of many usefull devices that will exploit them. we must find them all and use them to our advantage, and hope we dont break the laws of physics in the process.

Laws of physics can easily be broken.

Laws of nature can not be broken by definition. Never ever.

Physics is a composition of man-made concepts that try to describe actions and reactions occurring in nature... if it cannot describe something accurately, or something which obviously happens seemingly breaks the "laws of physics", than the obvious solution is that the particular "law" of physics was inaccurate... or downright wrong.

In a way, quite a few applications use "loopholes" in old theories.

[Nuke]

but of course, contrary to popular belief, im not exactly stupid

[Janos]

but of course, contrary to popular belief, im not exactly stupid

how would we know

BAMM ZINGG BURRRRN

j/k bro

[DeepSpace9er]

At first glance, it sounds like another cold fusion project. But, as im just a layman with pretty much 0 knowledge in physics besides basic stuff (if i trip there is this strange force that pulls me to the ground) i cant say one way or the other. It would be cool though... kind of like the fly-by-wire of propulsion.

[Herra Tohtori]

On the other hand, I wouldn't be too quick to judge this project... creating thrust from electricity isn't actually that new a discovery.

Biefeld-Brown effect in action (Jean-Louis Naudin's lifters)





That's something really intriguing.... The concept consists of an asymmetric capacitor that apparently generates lift when power is turned on to create a electric potential (charge, if you will) between upper and lower part of the capacitor... They demand about 30-40 kV voltage, but they really seem to be flying... 

I don't know how the hell it does it, really. It's apparently not even propelling air or anything downwards, it's just... lifting.

It feels like baron von Munchausen lifting himself by tugging his hair and thus saving himself from drowning in a swamp. I wonder if they have tested the effect in vacuum? At least they have proposed the use of asymmetric capacitors as maneuvering thrusters for satellites... They are in fact the same thing as those lifters. The same thing can be found here.