h a l f b a k e r y
"More like a cross between an onion, a golf ball, and a roman multi-tiered arched aquaduct."

meta:

account: browse anonymously, or get an account and write.

 user: pass:
register,

# Propellant-less Propulsion System

Einstein's special theory of relativity mass velocity loophole for propulsion
 (+3, -10) [vote for, against]

Propellant-less Propulsion?

Ok, here is the basis of the problem I've run into:

Note all forces act on the particle for a time interval of 1ms and the system is not acted on by any external forces (floating in space).

1. Use a two axis linear accelerator to apply Force A to accelerate a proton up to 0.1Light in the X direction. 2. Then apply Force B1 to the proton in the Y direction, accelerating it to 0.01light in the Y direction. The proton will now be traveling in 0.1 Light X direction & 0.01Light in Y direction. 3. Then apply Negative Force A to decelerate the proton in the X direction. The proton will now only be traveling 0.01Light in Y direction. 4. Lastly apply Negative Force B2 in the Direction, decelerating the proton. The proton is now back to 0 velocity.

So basically the proton is starting from zero velocity, accelerating to the right, then up, then to the left and finally accelerating downward, back to zero velocity.

Now, Newton might say that in this example, Force A would be equal in magnitude to negative Force A and Force B1 would be equal in magnitude to negative Force B2. Hence the summation of forces acting on the proton to accelerate it from zero up to a velocity and then back down to zero, are equal to zero. Hence there would be zero net propulsion of the linear accelerator system. This result can be backed up using his motion equations.

However, consider Einstein's special theory of relativity equation (M=Mo/(1-(v^2/c^2))^.5)) which states that the mass of an object increases with the velocity of the object. Now you may find that the net acceleration of the above system does not cancel out to zero. Specifically, the accelerating force B1 would have to be greater than force B2. Very odd, but here is my reasoning:

The proton is more massive at 0.1Light then it is at rest (directly calculated with Einstein's equation). Hence in the above example where the proton is accelerated vertically when it is at an increased mass (when the proton is traveling at 0.1Light Direction) and then accelerated back down "after" the proton is at a lower mass (when the proton is at rest in the X direction). The result is a net accelerating force on the system of B1>B2. Therefore the entire system would have to be accelerating in the Y direction without expelling any propellant in the process.

How can this be??

Also note that this is not some sort of trick. Existing linear accelerators are now capable of accelerating a protons to velocities that increase their mass by several times. This easily proves Einstein's equation to be valid and accurate.

This may not be practical with existing technology. However I don't see any way to disprove the physics behind it yet.. Any ideas?

Kevin Ryan Project Engineer Engine & Control Systems, JSF Actuation Hamilton Sundstrand One Hamilton Road Windsor Locks, CT 06096 860.654.6383 KevinM.Ryan@hs.uts.com

 — Rhino1903, May 25 2007

Conservation of momentum http://en.wikipedia...rvation_of_momentum
[ldischler, May 26 2007]

More on momentum http://www.lightand.../2cl/ch04/ch04.html

EmDrive http://en.wikipedia.org/wiki/EmDrive
another 'reactionless drive' scheme under the guise of relativity. [xaviergisz, May 28 2007]

there are no paradoxes, energy and momentum are conserved even at relativistic speeds, even if you accelerate the partical so fast that it gets so massive that it becomes a black hole (I halfbakingly interpret this as the group velocity of the proton escaping light speed, but perhaps this is not possible either, well I maintain that perhaps it is possible)
 — quantum_flux, May 25 2007

 You cannot separate out your accellerations like that. A+B will always equal -A+B2 (why did you switch your variable names around like that?)

 The fact that you are applying forces A and B separately does not make any difference to your end equation. In your example force -A is going to be slightly greater in magnitude than force A (because your particle is moving faster at the time -A is applied) but force B2 will be slightly less than force B (because the particle is moving slower when the B2 is applied).

 B2+ -A = -(B + A)

Now what you have done is moved your system while the accelerations are going on. But once the relative motion of the particle ceases, the motion of the entire system will cease.
 — Galbinus_Caeli, May 25 2007

I think your mistake is in thinking that the proton has a different relative X velocity to your B1 and B2 forces in the Y direction. How can you say this? B1 and B2 don't care, and cannot tell the X velocity. For those forces, the proton is the same mass.
 — Ling, May 25 2007

It's lucky, Kevin, that you brought this to us first, rather than to your boss. You can't avoid conservation of momentum by waving your hands. Any momentum gained by a particle imparts an equal an opposite momentum to the accelerator. Anytime you find otherwise, you can be sure your bookkeeping is off. In this case, you aren't accounting for the extra mass of the proton, which came from the accelerator, and which is now that much less massive. (Since, assuming this is closed system, the total mass hasn't changed.)
 — ldischler, May 25 2007

You don't get to cheat Ma Nature by unexpectedly switching between your Newtonian and relativistic jerseys. Those only serve to help you identify which method you're using to simplify the math. Nature doesn't do the math, it does the reality - and therefore never suffers a rounding error, a forgotten quantity, a dropped sign, or a failure to convert fully from one system to another. As [Idischler] said - you will find the flaw to be in your bookkeeping.
 — lurch, May 25 2007

Thank god someone deleted Treon's nonsense.
 — ldischler, May 25 2007

Here is my guess as to what would happen: You give the proton enough momentum in the Y direction so it goes at 0.1 light speed, then you give it some X momentum. Its relativistic mass increases, its momentum in the Y direction stays the same, so its speed in the Y direction decreases. (momentum is speed times mass). Similarly, at other times when you apply force on one axis, the momentum on the other axis doesn't change but the speed does. So after step 3, the proton is travelling more than 0.01 light speed in Y direction.
 — caspian, May 26 2007

 When a proton is a speed, it is more massive than is at rest. Right? (please see Einstein's equ)

 Hence, accelerating the proton (when at speed) will take a larger force than it would take to accelerate the same proton when it is at a lower velocity. Right?

 Think about it this way, it takes more energy to accelerate a 10lb ball than accelerating a 2lb ball. correct? yes

 Now substitute the ball with a proton. If the proton is traveling at .9999Light it will be 2Lbs (please see Einstein's equ). When the proton is traveling at .80Light it wil be 1lbs. (please see Einstein's equ)

 That's the loophole.

That's what I'm getting at..
 — Rhino1903, May 27 2007

If you accelerate the proton, you accelerate the accelerator in the opposite direction, thereby also increasing its mass due to increased velocity.
 — marklar, May 27 2007

what's up motherfucker!
 — Rhino1903, May 27 2007

you have too much free time bitches if got problem I ll kill you my name is casey
 — Rhino1903, May 27 2007

Rhino1903 is obviously drunk.

//If you accelerate the proton, you accelerate the accelerator in the opposite direction, thereby also increasing its mass due to increased velocity.//
You'd think, but no, the mass of the accelerator must decrease, as the mass of a closed system cannot change. The mass increase of the proton comes from some stored energy source associated with the accelerator, which energy has a mass equivalent.
 — ldischler, May 27 2007

 //you have too much free time bitches if got problem I ll kill you my name is casey//

 You know what the funny thing is... I shot this idea down less than 15 minutes after you posted it, it took me 1 minute to read it and 30 seconds to respond with the shoot down. You probably spent a hole day thinking this idea up in the first place! [-]

 //the mass of a closed system cannot change//

mass is not conserved at relativistic speeds or in nuclear reactions, closed system or not, it is energy and momentum that are always conserved.
 — quantum_flux, May 28 2007

 I would guess that the [Kevin] version of [Rhino1903] left his account logged in on a computer where it could be accessed by [casey], who hijacked the account and put his best foot, or other appendage, forward. If [Kevin] is really this ignorant of basic security considerations, we'd darn well better make sure he never gets his hands on anything that will accelerate anything to anything like relativistic velocities.

In addition, since [Kevin] is working on the JSF project, he may well be halfbaking on the taxpayer's dime. I have noticed others doing this as well. After a good deal of careful consideration, I have been unable to come to a conclusion as to whether this an enviable acheivement, or felony theft. Or both.
 — lurch, May 28 2007

Re: Xavier's link to the EmDrive: I've been watching that one for a while. It's been featured in Eureka magazine recently, five years after its first appearance. It would appear to be making progress. I guess it's either rubbish or revolutionary; I'd love it to be the latter but we'll have to wait for a full independent test.
 — david_scothern, May 28 2007

[ldischler] I was not being all that clear, I was using 'its mass' in the same context as in the original idea, which is 'its mass relative to the other body in the system'. The overall mass remains the same, the masses of each of the 2 bodies relative to a fixed observer changes as velocity increases.
 — marklar, May 28 2007

 I've not checked, but this might well work - there are a variety of ways of achieving the same effect, but all of them have very low efficiency.

 The most obvious is a pair of electromagnets, each reversing direction at intervals at the same frequency, but out of phase in such a way that A is always attracted by B, but B is always repelled by A. This sounds ridiculous, and it is, but it could in principle be done if either the magnets were a long way apart, or they were switching at a very high frequency with a very accurately controlled phase difference, or a compromise between the two.

 It's inevitably extremely inefficient, because switching electromagnets fast consumes a lot of energy; and electromagnets that are a long way apart don't exert much force on each other.

 The reason it works is this: the system as a whole transmits a great deal of energy (in the form of electromagnetic radiation) in all directions, but it's not quite balanced: one magnet always reinforces the wave passing it from the other, while the other always weakens the wave passing it from the first. Thus more energy is transmitted in one direction, and less in the opposite one - photons (of low frequency but in enormous numbers) in all directions, but slightly more in one direction along the common axis. These photons carry some momentum.

 Shining a torch has the same effect...

[Rhino]'s machine may well also be sending out an unbalanced load of photons.
 — Cosh i Pi, May 29 2007

[Cosh i pi] I was just thinking of that electromagnet idea in response to "Nature [...] never suffers a rounding error." There's also the possibility of gravitational waves.
 — caspian, May 29 2007

Thanks [Lt_Frank]. Dear Master Rhino, when you accelerate the particle, of mass x to y velocity, and there is an increaese in mass, you are inputing the energy equivalent requirement for getting the higher mass to the same velocity. In your example the mass would increase throughout the acceleration period, not just in one great lump once it reaches velocity y. The reverse is true for the acceleration in the opposite direction, to reduce velocity to zero which conserves your 'spare' energy.
 — the dog's breakfast, May 29 2007

 [dog's breakfast]

 That doesn't affect his logic at all, because it's not accelerating on the y axis during the x axis acceleration.

He's not claiming any gain of energy for nothing - merely propulsion without propellant. He still needs an energy source.
 — Cosh i Pi, May 29 2007

 Rhino seems to have uncovered the inconvenient fact (not stressed in textbooks) that the Lorentz Transform doesn't constitute a closed group in anything more than one dimension. (It is inconvenient that the space transformation in a kinematic theory can't actually describe a transformation that forms a closed loop -- this doesn't seem much like an improvement over the Classical theories which have no such problems.)

 There was a time (back in the '30s) when the fact that a rotation must be added to a transform loop to complete it was used to "explain" some fine structure in atomic spectra -- but Schrodinger's theory came along and explained it without requiring any such "Thomas Rotations".

IMO, the whole thing is a logical category error (like concluding the road you're standing on doesn't exist because it isn't on the map) -- we all know things can move in circles. Special Relativity says otherwise -- it SR is right, then maybe you can build a space drive out of this, but I'm betting against it.
 — Optiker, May 29 2007

 [Optiker] The rotations associated with the 3-dimensional version of the Lorentz Transform are well understood and unsurprising. You might regard them as inconvenient, but they're fairly simple maths and precisely born out by experiment - any adjustments for General Relativity are truly minuscule on the scale of any experiment local to Earth.

 That mechanism's like [Rhino]'s can generate thrust without propellant is unsurprising. Your flashlight does it. The thrust is, however, extremely tiny - it's NOT what drives a Crooke's Radiometer, for example.

The thrust is real. It's not a practical spacecraft drive, however.
 — Cosh i Pi, May 29 2007

[Cosh i Pi] Yes it is. It Must be noted that the change in direction of force applied is irrelevant. if you have one push that accelerates it to a certain velocity (and increases the mass) then the force applied during the next push will be more than the last. In this bit ""Now, Newton might say that in this example, Force A would be equal in magnitude to negative Force A and Force B1 would be equal in magnitude to negative Force B2. Hence the summation of forces acting on the proton to accelerate it from zero up to a velocity and then back down to zero, are equal to zero. Hence there would be zero net propulsion of the linear accelerator system. This result can be backed up using his motion equations."" is where the error lies.
 — the dog's breakfast, May 29 2007

 I don't think there is an error though. In Newtonian mechanics, there would be no nett thrust, each thrust being matched by a corresponding equal and opposite one at a different part of the cycle. Conservation of momentum requires that the nett thrust is zero in Newtonian mechanics.

What I'm not sure about is whether there's a nett thrust under Special Relativity, but there's no paradox caused if there is, because photons carry momentum, and accelerating charges emit photons, so momentum is conserved. The reason I'm not sure is that the situation isn't a simple rectangular circuit - it's a rectangle in the velocity-time diagram, but a nasty shape in location-time, and even nastier under Special Relativity. In particular, the angle between the X and Y axes is dependent on one's velocity relative to them. I don't think that matters, though, because we're doing everything relative to an inertial frame (the machine, not the accelerated mass).
 — Cosh i Pi, May 29 2007

 [Cosh] Rhino's argument works equally well for neutral masses (he only postulated charged particles for practical reasons) -- however, your explanation (photon thrust) requires charged masses. His argument also depends on the Lorentz transform not being separable in two dimensions (equivalent to my comment).

The maths may be simple (which is irrelevant), but I challenge your claim that Thomas Rotations have been "precisely born out by experiment" -- I would be willing to bet that you know of no experiments that have accomplished this.
 — Optiker, May 30 2007

 [Optiker] On neutral masses and photon thrust: agreed, I was realizing this flaw in my argument last night. Photon thrust (as in my electromagnet device and as in flashlights) does achieve the (pointless) purpose of [Rhino]'s device, but you're right, [Rhino]'s doesn't.

 I'd not heard of Thomas rotations, but the rotations (whether these are the same rotations as Thomas rotations, I don't know) implicit in 3-dimensional analogues of the Lorentz transformation certainly do work out consistently, and consistently with experiment. They're precisely mathematically equivalent to Special Relativity.

However, now you mention the question of experiment - what I'm confident of is that they've not been contradicted by any experiment; but it's dashed difficult to see how you could actually test them by experiment. We can't accelerate anything big enough to detect its orientation to a high enough velocity. So "consistently with experiment" is (probably) true, but misleading.
 — Cosh i Pi, May 30 2007

1: He said "proton" you all said "photon"

2: A flashlight exhibits backwards thrust because it is not a closed system. Shine it on a wall and you have a net 0.

3: Momentum must be conserved. Picture three people tossin a ball around in a triangle, all standing on a big piece of flat plastic, which is on a frictionless surface. A throws to B. That gives thrust purpendicular to C. B throws to C. That cancels out the sideways component of the A->B throw and some of the forward component. C throws to A. That cancels out all momentum, as the ball is back where it started. In the mean time the whole system has rotated a bit, but it is now at rest. Total momentum is 0. You can accomplish the same tiny bit of movement by throwing a ball back and forth inside the space station. If the ball started on one end and finished at the other, the movement of the ball moved the space station the opposite way a little.

Now lets tackle the high-speed-higher-mass part of it. When A threw to B, he accelerated the ball. The ball was heavier when it left his hand. Lets call X the initial weight and Y the moving weight. A was pushed backwards by the amount of momentum given to the ball: Y*speed. B must completely stop the ball, and the total energy pushed B backwards by Y+speed. B throws to C, and again the net energy is 0.

The ball started at x weight and ended at x weight. You did cause more movement by throwing it very fast, but you did not add momentum to the system. Now if you threw the ball very, very fast such that the moved weight is very high its true that you can move the platform more than one would expect. But as long as the ball returns to A, you have added no momentum.
 — Voice, Jul 17 2008

 //The overall mass remains the same, the masses of each of the 2 bodies relative to a fixed observer changes as velocity increases.//

Is this the reason this won't work? If this is true, then shouldn't the idea that it is impossible to exceed the speed of light due to increased mass be flawed? (or am I misunderstanding?)
 — Zimmy, Jul 22 2008

 Do I understand this correctly? ........North West..object..East ........South.

 Say "object" has a mass of "x" and it is accelerated Northward by accelerator to the South at a significant fraction of the speed of light, thus gaining mass of say (x+1) and imparting a force of "A" to the South accelerator. "object" is then also accelerated by the West accelerator w/ a force of "A+1" towards the East, thus gaining a mass of (x+2). "object" is then decelerated by the North accelerator imparting a force of "A+2" on the North accelerator, with the object losing mass in the deceleration and having a new mass of (x+1) "object" is then decelerated by East accelerator w/ a force of "A+1" to the East accelerator, thus giving the object a mass of "x" due to the deceleration.

 (A+1) - (A+1) cancel each other out for the East & West reactions, but A - (A+2) = -2 for the North & South reactions resulting in a "thrust" of "2" in the North direction on the entire system.

 It seems like it shouldn't be able to work, but I can't figure out why. Perhaps because I don't entirely understand some of the other annotations.

It seems to me that even if you consider the increases / decreases in force required during the accelerations (due to the increasing mass due to speed) you still result in a net "thrust" as these cancel each other out.
 — Zimmy, Jul 22 2008

 Well, take into the other special relativistic forces into consideration. The distance also required to decelerate is also less. It all balances itself out somehow. You can't really do that old math trick in the physics world (which doesn't even work in math anyway):

 x=0.9999.... (define x)

 10x=9.99999.....

 9.99999....- 0.99999....=9.00000....

 10x-x=9x

 9x=9 x=1! (<- not what we started with!)

The properties of infinite series tells you that this solution to the problem is just wishful thinking, kinda like trying to defy relativistic mechanics by speeding up and then slowing down the same object.
 — WhereYouAt, Jul 22 2008

 I'm wondering if the relativistic equation for Momentum has been getting ignored in this discussion? Perhaps the cause of the problem presented here is the use of the Newtonian version instead of the relativistic verson.

New topic:
Some years ago I read about a different and much simpler way to accelerate things "reactionlessly". Take an electron gun and mount it on a kind of "plate" and shoot electrons away from the plate in a vacuum. The gun moves, with the plate. in the opposite direction. However, now mount appropriate magnets on the plate; use their magnetic fields to cause the beam to curve in a nice 180 degree arc. The moving electrons then impact the plate, and impart their momentum to the plate, ALSO in the direction that the reacting gun is making the plate move. Note the electrons can be collected and recycled to keep the beam going. I suppose you could do this with protons, instead, for 1836 times the effect (protons are that much more massive than electrons).
 — Vernon, Jul 23 2008

 My head really hurts. I read now that Mass is relativistic & that it is probably not the reason FTL is impossible, but a violation of casuality (is that right?) is.

But the proof of the casuality violations do not seem to include time dialations.
 — Zimmy, Jul 23 2008

 // I read now that Mass is relativistic & that it is probably not the reason FTL is impossible, but a violation of casuality (is that right?) is.//

I just always figured if mass increases as the object approaches the speed of light, the closer you get, the object needs more and more energy to accelerate it to a higher velocity. In theory, the mass would be practically infinite at c, so it would take about an infinite amount of energy to accelerate it to c. There isn't that much energy anywhere in the Universe, so FTL travel is impossible.
 — WhereYouAt, Jul 23 2008

 but I am reading that rest mass (invariable mass) is constant & that relative mass might be a way to describe momentum added to particles when not at rest.

I am uncertain that I grasp this correctly. If I do, then the invariant mass remains constant, while the relative mass would change, relative to an observer, but not in the objects own frame. If that IS the case, then it closes the loophole mentioned in the idea, but, to me, opens up the possiblility of FTL - probably only until I can better understand how they say it would require time travel, but I don't yet.
 — Zimmy, Jul 23 2008

 [annotate]

back: main index