Relativity drive

[Qfwfq]

The real secret to where they are bamboozling us lies in the expression "Group Velocity". This is not a real velocity that should be used in the subsequent equations. It is a mathematical construct to help solve the equations and not anything real. To give you a clue as to why group velocity isn't anything like real velocity - it is perfectly possible to have a group velocity faster than that of the light photons or waves that make it up (which, obviously, travel at the speed of light themselves).

:( First, you're exchanging group and phase velocity, the former is that of the wave "packets" (and hence of the particles) and, second, it will be less than c; in a dispersive medium, due to the separation of the velocities, it will be even a bit less than c divided by the refractive index.

 

Of course, I'm not saying that the contraption could work. It's every bit as pointless as hoping to propel a spaceship with a jar of the same shape filled with hot gas.

 

I'd love to know what the real answer was.

The real answer is that you could never get the two beams to completely have exactly destructive interference from thence onward along the direction of propagation without getting any reflection. I'd say that physics lecturer hadn't thought it all the way through.

[Karnuvap]

: I'd say that physics lecturer hadn't thought it all the way through.

 

To give him his due he did mutter something about having to turn the thing on such that the beam's fourrier series wouldn't be pure and that this would result in there not being a perfect match for destruction so I think I may have besmirched him unduly.

 

I was only asking this because I was thinking about how it might be possible to make a real-life light sabre. My laser would be shon into a crystal and six or eight beams would emerge from the facets at slightly converging angles (such that they crossed about 1metre away). The clever bit was that half the crystal facets would be slightly thicker than the others such that the path length for the beams emerging from these ones would be a half wave longer than that of the beams that emerged from the other ones at the point where they crossed. Thus destructively interfering at precisely one metre away from the handle of my laser sword. Thus my sword would be 1 metre long and no more. << This doesn't work either>> Thanks for your comments though.

 

The Vap

[Pyrotex]

Could someone explain in plain english how this thing works? I have a ok idea, but I want to hear some other angles on it...

The engine apparantly works because the round side at one end of the truncated cone is larger than the round side at the other end. Therefore the radiation pressure on it will be larger.

 

The author totally ignores the "other" side: the conical cylinder that connects the top and bottom ends. If you calculate just that portion of the conical cylinder that is in the same plane as the two ends, I'm sure you will find that the area on top is equal to the area on bottom.

 

Perhaps he should call it, the "Null Drive" -- because it has no net force at all. :hihi:

[Qfwfq]

...he did mutter something about having to turn the thing on such that the beam's fourrier series wouldn't be pure...

actually that wouldn't be the problem in steady state. Even considering the perfect monochromaticity approximation (which is fine for the brunt of interferometric design, between a few cycles after reaching steady state and a few cycles before losing it) it still isn't possible without some weird non-linear propagation. As you're talking about photons in air and no material object at the sabre's tip, it's just nononono.

 

:confused:

[IDMclean]

Chinese are building it, apparently.

 

I figured I would update on this topic.

[DustinTheWind]

I didn't really buy the group wave velocity idea either. I thought about it some and this is what I think if any ideas how it might work. The tube works best off reflectivity. The Tube is shaped with a larger disk at one end and a smaller disk at the other with angled walls. The photons bouncing off the top plate impart all their momentum at once and in a shorter amount of time. It takes longer for the photons to impart their momentum through the side walls. There for you get a force in one direction for a short amount of time before it is canceled out. Now imagine this on a much larger scale and increase the scale by now allowing the reflections to continue more so as the Q of the cavity increases due to superconductivity. So basically the thrust may be due to a time delay in canceling out the momentum on the back reflection. Of course this is just an idea I had after having asked God how this thing might work.

 

I believe that group velocity can not transfer momentum as stated in the paper and this also was what bothered me.

[Pyrotex]

You guys are making this so complicated. :phones:

The presumed net force created my the Microwave Thruster is proportional to:

Area(of big circle) - Area(of small circle).

 

The MT assumes that the microwaves do not apply any force at all to the conical sides of the chamber.

And this is patently false.

 

Here is how to calculate the forces. However the microwaves are piped into the chamber, we assume that after some small time, t, the microwaves are in some sort of equilibrium in the chamber.

 

Now we add up all the upward vertical components of the forces on the chamber walls.

 

Then we will add up all the downward vertical components of the forces.

 

As drawn, we can assume the microwaves are piped in from anywhere. (The answer will be unchanged) So, for symmetry, we assume they are piped in from the center of the chamber.

 

Waves going toward the large circular side hit it, and have a vertical component that depends upon:

energy flux of emitted waves;

1 over-distance-squared from the center of the chamber;

cosine of the angle between wave and flat upper circular side.

 

If you integrate the tiny elements of force on every small area over the entire upper circular side, and over all angles between wave and flat upper circular side, You will get a total force that is proportional to:

The energy flux of emitted waves;

the cross-sectional area of the chamber as seen from above. (a large circle)

 

Waves going toward the lower circular side AND the conical sides, hit them, and have a vertical component that depends upon:

energy flux of emitted waves;

1 over-distance-squared from the center of the chamber;

cosine of the angle between wave and flat lower circular side & between wave and conical sides;

cosine of the angle between conical sides and flat lower circular side.

 

If you integrate the tiny elements of force on every small area over the entire lower circular side AND the conical sides, and over all angles between wave and flat lower circular side, You will get a total force that is proportional to:

The energy flux of emitted waves;

the cross-sectional area of the chamber as seen from below. (a large circle)

 

Net force = force up minus force down.

 

BUT, the chamber has the SAME CROSS-SECTIONAL AREA seen from above AND below!!!!

 

The lower circular side may be smaller than the upper circle, but it's irrelevant. What is relevant is the total cross-sectional area of the entire chamber as seen from top and bottom -- and they are identical.

 

The force up is exactly equal to force down. changing the shape of the chamber CANNOT violate this simple fact (unless you put holes in the chamber)

 

Once you "GET IT" that the partial force upward or downward (actually, in ANY arbitrary pair of directions) is determined by the total cross-sectional area, and not the area of any select piece of the chamber, then it becomes dirt-obvious that no net force can arise.

[DustinTheWind]

You guys are making this so complicated. :lol:

The presumed net force created my the Microwave Thruster is proportional to:

Area(of big circle) - Area(of small circle).

 

The MT assumes that the microwaves do not apply any force at all to the conical sides of the chamber.

And this is patently false.

 

Here is how to calculate the forces. However the microwaves are piped into the chamber, we assume that after some small time, t, the microwaves are in some sort of equilibrium in the chamber.

 

Now we add up all the upward vertical components of the forces on the chamber walls.

 

Then we will add up all the downward vertical components of the forces.

 

As drawn, we can assume the microwaves are piped in from anywhere. (The answer will be unchanged) So, for symmetry, we assume they are piped in from the center of the chamber.

 

Waves going toward the large circular side hit it, and have a vertical component that depends upon:

energy flux of emitted waves;

1 over-distance-squared from the center of the chamber;

cosine of the angle between wave and flat upper circular side.

 

If you integrate the tiny elements of force on every small area over the entire upper circular side, and over all angles between wave and flat upper circular side, You will get a total force that is proportional to:

The energy flux of emitted waves;

the cross-sectional area of the chamber as seen from above. (a large circle)

 

Waves going toward the lower circular side AND the conical sides, hit them, and have a vertical component that depends upon:

energy flux of emitted waves;

1 over-distance-squared from the center of the chamber;

cosine of the angle between wave and flat lower circular side & between wave and conical sides;

cosine of the angle between conical sides and flat lower circular side.

 

If you integrate the tiny elements of force on every small area over the entire lower circular side AND the conical sides, and over all angles between wave and flat lower circular side, You will get a total force that is proportional to:

The energy flux of emitted waves;

the cross-sectional area of the chamber as seen from below. (a large circle)

 

Net force = force up minus force down.

 

BUT, the chamber has the SAME CROSS-SECTIONAL AREA seen from above AND below!!!!

 

The lower circular side may be smaller than the upper circle, but it's irrelevant. What is relevant is the total cross-sectional area of the entire chamber as seen from top and bottom -- and they are identical.

 

The force up is exactly equal to force down. changing the shape of the chamber CANNOT violate this simple fact (unless you put holes in the chamber)

 

Once you "GET IT" that the partial force upward or downward (actually, in ANY arbitrary pair of directions) is determined by the total cross-sectional area, and not the area of any select piece of the chamber, then it becomes dirt-obvious that no net force can arise.

 

What I just pointed out in the post just before you which you might have missed is that there may be a time delay in the time it takes to cancel out that momentum. It takes 2 or more reflections to cancel out the momentum on the back reflection than it does for the top plate (1 reflection for top plate). There for you might get a force for a short duration towards the large plate before the force cancels out. Now imagine if you increase the amount of reflections before the photons are converted to heat. This is what the author is doing by using superconductivity to increase the Q or reflectivity of the metal. Not only that it would also be proportional to the amount of photons bouncing around in the sealed chamber.

 

The only problem with my hypothesis is that I don't know that this explains what is going on in the wave nature of light but from a particle perspective this seems like it might be a plausible explanation of what is going on.

 

So yes if you don't consider time it would seem the forces would cancel out. However if you consider the passage of time then there is a chance this might work. If so the force might be at least 1/2 that of the force of light reflecting. Reflecting light provides 2 times its force since the particle is being caught and thrown back. But if you increase the % of photons in that state and increase the reflections you would take that force generated by the power output times the number of reflections. ... I think.

[Qfwfq]

So yes if you don't consider time it would seem the forces would cancel out. However if you consider the passage of time then there is a chance this might work.

uh, brownian motion?

[DustinTheWind]

I don't believe it would be Brownian motion.

[Pyrotex]

.... However if you consider the passage of time then there is a chance this might work. If so the force might be at least 1/2 that of the force of light reflecting. Reflecting light provides 2 times its force since the particle is being caught and thrown back. But if you increase the % of photons in that state and increase the reflections you would take that force generated by the power output times the number of reflections. ... I think.

Nope. Still won't work.

 

Consider the Relativity Drive as a closed system. Say, it's built inside a spherical hull of your spaceship. The hull has no holes. Now turn your Drive on. Microwaves are generated and get reflected and absorbed and slapped around and bounced and jiggled and any thing else you care to do to them, inside the Drive chamber which can have any geometry that you can think of.

 

Now, let's go back outside the ship. Before Drive on, the ship is in free fall in deep space. No accelleration in any direction. Now turn Drive on. Nothing leaves the ship. Nothing gets through the hull. Therefore the total momentum of all radiation in the ship, stays in the ship. For every dF (tiny force) in one direction there is a -dF. Every photon is generated inside the Drive and is absorbed in the Drive. Net total momentum is zero.

 

The ship experiences no accelleration.

 

If it DID, say it experienced an accelleration up to a velocity V(d) in the d direction, and the ship has mass M(s) then it gains momentum equal to

V(d) * M(s) in the d direction.

 

Therefore, something ELSE, a particle let's say with mass M(p) must have been emitted in the -d direction so that

V(d) * M(s) + V(-d) * M(p) = 0

 

This is conservation of momentum, which is a Law in most civilized universes.

At least, those universes worth talking about.

 

But since nothing gets out of the Relativity Drive or the ship's hull, then there IS NO "particle".

Which is to say, M(p) = 0 and therefore the resulting velocity of the ship, V(d) must be zero.

[DustinTheWind]

Nope. Still won't work.

 

Consider the Relativity Drive as a closed system. Say, it's built inside a spherical hull of your spaceship. The hull has no holes. Now turn your Drive on. Microwaves are generated and get reflected and absorbed and slapped around and bounced and jiggled and any thing else you care to do to them, inside the Drive chamber which can have any geometry that you can think of.

 

Now, let's go back outside the ship. Before Drive on, the ship is in free fall in deep space. No accelleration in any direction. Now turn Drive on. Nothing leaves the ship. Nothing gets through the hull. Therefore the total momentum of all radiation in the ship, stays in the ship. For every dF (tiny force) in one direction there is a -dF. Every photon is generated inside the Drive and is absorbed in the Drive. Net total momentum is zero.

 

The ship experiences no accelleration.

 

If it DID, say it experienced an accelleration up to a velocity V(d) in the d direction, and the ship has mass M(s) then it gains momentum equal to

V(d) * M(s) in the d direction.

 

Therefore, something ELSE, a particle let's say with mass M(p) must have been emitted in the -d direction so that

V(d) * M(s) + V(-d) * M(p) = 0

 

This is conservation of momentum, which is a Law in most civilized universes.

At least, those universes worth talking about.

 

But since nothing gets out of the Relativity Drive or the ship's hull, then there IS NO "particle".

Which is to say, M(p) = 0 and therefore the resulting velocity of the ship, V(d) must be zero.

 

You ignored my suggestion earlier about considering the passage of time. There is a shorter time for momentum on one side and a longer time for momentum to cancel on the other side. Your ignoring time is convenient for your argument. Imagine balls bouncing around inside in which one bounce carries momentum forward. While it takes twice the distance to cancel the forward momentum then repeat the process repeatedly. However now consider them photons. There would be a delta T in which there would be forward momentum for a short period of time for each cycle of bounces.

[Pyrotex]

You ignored my suggestion earlier about considering the passage of time. There is a shorter time for momentum on one side and a longer time for momentum to cancel on the other side. Your ignoring time is convenient for your argument. Imagine balls bouncing around inside in which one bounce carries momentum forward. While it takes twice the distance to cancel the forward momentum then repeat the process repeatedly. However now consider them photons. There would be a delta T in which there would be forward momentum for a short period of time for each cycle of bounces.

Sorry. I ignored your suggestion, because it holds no water at all.

I was hoping you would see that the Law of Conservation of Momentum trumps your argument.

Well, perhaps "trumps" is not the right word.

How about, "stomps it into the dirt"? :)

 

We can simplify your model thusly:

 

We have a spherical ship that emits no matte or radiation. We have a R-Drive inside and we turn it on. Over a short period of time (measured in nanoseconds) there is an excess of momentum in one direction. After that, equilibrium is established, as it must, because nothing leaves the ship.

 

The net effect of this excess of momentum is that some "mass" (M) was shifted a distance D within the ship. To conserve momentum, the ship itself (mass m) moved a distance d in the opposite direction. d = - M*D/m

 

So, you turn the R-Drive on. The ship moves a few nanometers and stops.

 

You turn the R-Drive off. The moves back to its original position and stops.

[Qfwfq]

Pyro, if one is convinced that some wierd mechanism could perhaps violate momentum conservation then it isn't conclusive to say: "That's impossible, due to momentum conservation." To him it's only begging the question, like saying there can't be one kind of fish that breaths air cuz fish don't breath air.

 

There is a shorter time for momentum on one side and a longer time for momentum to cancel on the other side. Your ignoring time is convenient for your argument. Imagine balls bouncing around inside in which one bounce carries momentum forward. While it takes twice the distance to cancel the forward momentum then repeat the process repeatedly.

Sorry but I can't discuss your argument because I don't get the semantics of it, I would need to understand exactly where you think the effect might arise. Lacking this, I presumptuously adhere to the holy dogma that momentum is conserved, no matter how wierd a contraption might be devised. As long as each of the elements in play complies, there appears to be scarce chance of the overall thing not being compliant.

 

If instead you are able to clearly explicate exactly what you believe could cause the effect, then I might participate in a critical discussion of it.

[Pyrotex]

...If instead you are able to clearly explicate exactly what you believe could cause the effect, then I might participate in a critical discussion of it.

Me, too!! :)

[DustinTheWind]

I thought I had already quite clearly but from what I can see it was not absorbed. A hint would be to consider what I said and look at the shape of the cone then draw lines of impact. Then measure the lines length from equal angle reflections and consider the differences in length off the walls. Then consider the difference in time between momentum exchange.

[Qfwfq]

A hint would be to consider what I said and look at the shape of the cone then draw lines of impact. Then measure the lines length from equal angle reflections and consider the differences in length off the walls. Then consider the difference in time between momentum exchange.

I'm not sure exactly what you said that we should consider, this doesn't really make it clear.

 

Do you agree that, at each reflection, overall momentum is conserved? I still don't see how these differences in length and time will give a [imath]\Delta p[/imath] in the long run. The position and velocity of the box will jitter around a bit but, if you count the total of balls and box, and if you agree momentum is conserved at each impact, it doesn't really get anywhere.