The Dipole Drive: A New Concept in Space Propulsion (aiaa.org) 84 points by cl42 66 days ago | hide | past | web | favorite | 67 comments

 Previously discussed: https://news.ycombinator.com/item?id=17428751I chewed on this idea for several days, eventually coming to the same conclusion others had: it won't work. This poster came up with a correct explanation much faster than I did: https://news.ycombinator.com/item?id=17432975Zubrin has taken a familiar approximation -- that the field near a flat charged plate is uniform -- and forgotten that it's only an approximation, and that the keyword is "near". As particles leave the vicinity of the drive, they will be attracted back toward it, reversing the momentum transfer.
 Agreee, although I think the arguments and calculations in that thread are a bit silly. If you take any finite distribution of fixed charges, and you fire a test particle from infinity at the charge distribution, one of two things happens: either it hits one of the charges, or it goes back out to infinity, possibly in a different direction, with the same velocity it started with. The idealized field distribution that accelerates everything going through does not exist in electrostatics.So either some time-varying fields are needed or interactions between space plasma particles is needed.Also, the efficiency calculation doesn’t account for particles hitting the grids and thus creating a short-circuit current. Which has its own issue: in the original paper, 500W is going somewhere, but there is no explanation of how, exactly, any current flows in the circuit, which is inconsistent with any power at all being used.
 > in the original paper, 500W is going somewhere, but there is no explanation of how, exactly, any current flows in the circuitI puzzled over that a bit too. I think the answer is that as protons are shot out the negative side and electrons out the positive side, they create a field that counteracts the field between the grids. Maintaining the field strength thus requires continuously adding charge to the grids. In the hypothetical case of infinite grids (being charged by an infinite number of generators), the areal charge density on the grids would increase without bound, so that the charge density difference, and therefore the field, across the infinite plane halfway between the grids would remain constant. Exactly how this would translate to the finite case I'm less clear on, but I suppose the behavior would be similar, at least shortly after the drive was first turned on. Sooner or later, though, I think the voltage across the gap would increase to the point that any generator you could actually build wouldn't be able to push any more charge onto the grids.
 > Maintaining the field strength thus requires continuously adding charge to the grids.I'm not sure what you mean. The whole system is running at steady state. For simplicity, pretend that the spaceship is very massive or is otherwise accelerating at negligible rate so that the spaceship's frame is inertial. [0] In the spaceship's frame, there is some electron current density at all points in space, some proton current density at all points in space, and some distribution of charge carriers on the grids. If we further assume an infinite work function on the grids, then there is no field emission and no plasma charges are captured, so there is no transfer of charge between the plasma and the grids. But then there is no current on the grid, so the driver circuit outputs no power, and the engine can't possibly do any work.So either my analysis is missing something or the paper is wrong. Or both.[0] This could be achieved exactly by firing an equal-but-opposite-thrust conventional rocket in the opposite direction. It doesn't really matter for my argument anyway, I think.[1] Or the grid could be made of metal filaments wrapped in perfect insulators, but then there's an issue of charges building up on the outside of the insulators. In general, the grid will capture some charge from the plasma, and this will be a parasitic loss on the system. I would certainly believe that a pair of solid parallel finite metal plates flying at high speed through a plasma generates a force, but that's not what's being described in the paper.
 Consider what happens when you've just turned the thing on and it has just barely expelled its first proton through the negative grid. As long as the proton is close to the grid, it may as well be on the grid: it cancels out the contribution of one electron to the field between the plates, which means the voltage across the gap is incrementally reduced. This reduction in voltage allows one electron to flow out of the generator into the grid.Of course, as the proton moves away, its contribution to the field between the plates diminishes at a rate depending on the ratio of its velocity to the radius of the grids. (This is why the infinite-grid case is easier to think about: the contribution doesn't diminish at all.) Eventually — and depending on the proton velocity and grid size, "eventually" might mean milliseconds — its contribution becomes negligible. At that point a steady state will have been reached, and (as you intuit) it won't be possible for the generator to push any more charge onto the grids without increasing its output voltage.
 That doesn't sound right to me. Imagine I have a proton and an anti proton, and I fire the proton at the anti proton along a perfect straight line. Classically, and ignoring nuclear forces, the proton will end up stationary and the anti proton will instead be launched away, just like a direct collision between pool balls leaving the cue ball stationary.
 GP was referring to a distribution of "fixed charges", so it's not possible to transfer momentum to them.
 Well a spacecraft is not fixed, you can push it around, which is the whole point of a reaction drive. An unmovable spacecraft would have to have infinite mass. So the assumption does not apply.
 Yes, as described with a constant electric field this shouldn't work, but the idea of using solar wind as a reaction mass seems reasonable. It seems like you'd want to concentrate it with strong magnetic fields first though.Here's one example of how you could use a dynamic electric field. A low voltage AC potential across a gap will make ions entering at slightly different times concentrate down to a bunch at some tunable focal length. From there you just have a normal LINAC to shoot them out the back.
 This "debunking" oversimplifies the situation by only looking at individual charged particles.In reality, after this thing has been running for a bit, you'll have a cloud of free electrons ahead of the craft and a cloud of protons behind the craft. As protons ahead of the craft are decelerated by the leading positive plate, they will eventually slow to a low enough energy to combine with the free electrons, becoming neutral hydrogen. Only when they get inside the capacitor field will the voltage differential be great enough to re-ionize the hydrogen, at which point the electron will be shot forward to replenish the free electron cloud, while the proton will be accelerated out the back. Analogously, free electrons approaching from the rear will lose enough energy to recombine with the trailing protons, again becoming neutral hydrogen which no longer contributes to drag.It's really no different than if you had a purely mechanical funnel scooping up neutral hydrogen and using that as fuel for a conventional ion-engine.
 That conclusion is surprising to me. If the drive has a source of reaction mass then it works kind of like a jet-ski, it scoops reaction mass from the surroundings and shoots it out accelerated. In a way it almost has to work and it would take effort to screw it up.Sure the performance may be awful but it should be possible to get it above zero, unlike reactionless drives which violate established laws of physics.
 If the reasoning there is sound (and it makes sense to me), the closer analogy is trying to propel a canoe by rolling a cannonball along the bottom. It will work for a while, but eventually the cannonball hits the back of the canoe. No dice.
 I believe this canoe is picking up cannonballs along the way.
 It seems that way at first. The problem is that each of those cannonballs is tethered to the canoe by a long bungee cord.
 Ah, because the electrical field loops? Just thinking of the top of my head, couldn’t this pulse? Basically capture material then shoot it out?
 Maybe the best way to understand it intuitively is that the drive would actually work for a little while after you first turn it on, but as it shoots charged particles in opposite directions, it creates a field in its region of space that eventually gets strong enough to pull it backwards by the same amount that its thrust pushes it forwards. By continuing to add energy it could continue to move, but it can't continue to accelerate. The best it could do, perhaps, is to accelerate to a certain velocity, then turn off its generator and discharge its plates so it now has no net field, and then coast to its destination.Probably part of the confusion here is that a dipole does have a nonzero field even though the net charge on it is zero. Dipole fields are often negligible, particularly as they fall off faster than monopole fields -- 1/r^3 rather than 1/r^2 -- but ignoring the dipole field is another familiar approximation that doesn't hold in this case.
 > Zubrin has taken a familiar approximation -- that the field near a flat charged plate is uniform -- and forgotten that it's only an approximation, and that the keyword is "near".Near is all that matters in a plasma, since charged particles leaving the device will get screened and decouple from it within a few Debye lengths. Zubrin of course knows this. Read the paper.> As particles leave the vicinity of the drive, they will be attracted back toward it, reversing the momentum transfer.As particles leave the vicinity of the drive, they will get screened off and decouple from the drive.On top of that, there is a more general problem with the "correct explanation", even in a vacuum: it only considers the effect of the capacitor on the ion, but not of the ion on the capacitor.When the ion "climbs a big hill to get through the first plate", it loses kinetic energy. When the ion is accelerated between the plates, it gains kinetic energy. When the ion climbs back out of the potential on the other side of the capacitor, it again loses kinetic energy. Where is all that energy being transferred to?Right, to the capacitor. When the capacitor is absorbing energy, mutually repelling charges on each plate are being forced a little closer together. It's like loading a spring. When the capacitor is releasing energy, those charges are moving further away from each other again, like releasing the spring.Of course, this means that the capacitor's electric field is changing as the ion goes by.If there is no power source hooked up to the capacitor, the moving back and forth of charges on its plates will inevitably cause some losses; energy will be dissipated as heat, and in the end, you will actually have reduced the passing ion's speed.But of course, the capacitor in a drive is hooked up to a power source. It does work to keep the charges on the plates locked in place as the ion goes by. If it is perfectly successful (it isn't, of course, but it can get pretty close), the charges on the plates will not move at all, and the capacitor's field will be constant throughout the ion's passage.So the field is constant because you have a power source doing work, spending energy, to keep it constant while the ion is passing through.Where does all that energy end up in the end? Not in the field (hey, it's constant). Some of it is dissipated as heat, but the rest goes to the ion, as kinetic energy.
 That stance received some knowledgeable opposition in the quoted thread if I'm not mistaken. I'm in no position to assess the soundness of the arguments, but I noticed that the poster admitted to some flaws in his own reasoning at the tail end of it.
 None of that second commenter's objections even began to address the core of the original poster's point: whatever the exact details of the field configuration might be, it's inescapably true that the electric potential is zero at infinity, rises to a positive maximum at the + plate, falls quickly to a negative minimum at the - plate, and then returns gradually to zero at infinity again.That means that if an ion's initial speed is "v0" well before it encounters the device, it may speed up and slow down while interacting, but its final speed will be "v0" again once the device has passed by. If its direction is unchanged (which is the intended case for the thrust-producing protons, as illustrated in Fig. 1 of https://www.centauri-dreams.org/2018/06/29/the-dipole-drive-...), that means that the ion will experience zero net change in momentum due to the device. By conservation of momentum (and/or Newton's 3rd law), the device won't have any net change in momentum either. (And if an ion reflects back, as the electrons are intended to in that illustration, the device is equivalent to a single charged plate.)The debate in that earlier comment thread was all about how to correctly model the non-zero fields outside of the parallel plates: how quickly do they fall off to zero? But in the end, that doesn't matter. However small the fields outside may be, it is mathematically guaranteed that they will also extend over a large enough distance so their integrated effect from infinity to the nearer plate is exactly opposite to their integrated effect from that plate to the center of the gap between the plates.[Background: I'm a physics professor. I teach basic E&M every year and upper-level undergraduate E&M occasionally.]
 I haven't read the paper, so I'm purely arguing against your argument here: You said "whatever the exact details of the field configuration might be, it's inescapably true that the electric potential is zero at infinity, rises to a positive maximum at the + plate, falls quickly to a negative minimum at the - plate, and then returns gradually to zero at infinity again."This is true, for a charge at rest. It is equally true for the gravitational potential. Yet it is possible to extract momentum from other massive bodies through a gravitational slingshot. So I don't think one can conclude purely from that argument that a static field distribution can't extract momentum from other charged particles.The fact that the charges are moving means you have to worry about which frame you do the calculations in. The slingshot works because it preserves momentum in the center-of-mass frame, not in the solar system frame. The same applies in E&M, When you have charges in motion you have to worry about electrodynamic effects.My E&M is far too rusty for me to say whether those effects change the conclusion of your argument, but I believe you have to include them before making a valid conclusion.(Background: I don't teach E&M. I just argue about it with my wife, who sometimes does... ;-)
 Hmm, this is a very interesting line of argument: I am very much focused on electrostatic reasoning here, and I haven't given significant thought to frames of reference at all. My instinct is still that it won't change my overall conclusions here, but I'd love to be proven wrong: that would be pretty cool.For what it's worth, gravitational slingshots (like everything else) have to conserve total momentum in any frame: (e.g.) when Voyager speeds up, Jupiter must correspondingly slow down (though Jupiter's change would be nearly zero, of course). The CM frame is certainly very handy for "elastic collision" calculations like this, but the laws of physics (like momentum conservation) can't care what frame you're in.But that by itself doesn't say anything about whether the changes in motion of these charged objects might change the potential reasoning. Hmm... honestly, it's late at night and I don't think I have time to really sort through this in detail. (I tried to sketch an argument for the simple case of circular parallel plates with all motion limited to be along their central axis. The electric field parallel to the relative motion of the primed frame should be unchanged, but I just don't have the time to sort out how to reconcile that with the inevitable Lorentz contraction of positions along that axis. Glancing at tables of field transformations, it looks like charge density and potential should both be increased by a factor of gamma while distances along the axis are reduced by the same factor: that should mean that the integrals to find potential come out the same, at least for the cases that I've used in my argument. But there are some nagging details bugging me about it: I think they'll sort themselves out, but if you or your wife or someone else can use this reasoning to change the result, that would be neat to see.) (Also, I'm completely ignoring any effects of radiation due to accelerated charges in this story. I haven't seen anyone say that's relevant and my impression is that the calculations used in the original paper are based on much more basic physics, and I'd be shocked if somehow radiating away energy wound up being the secret to making this work. But again, if someone does a more detailed calculation and discovers that radiation really does make it work, that would be fascinating.)
 I sometimes wonder if the true strength of a physics professor isn't about the pure intellect but about the ability to answer questions seriously lacking in fundamentals concisely and accurately.
 At least in physics it is known what we don't know. So the physics Professor will not go all out on a theory based on his pet peeve. Especially in the social sciences this margin of error is so large nobody actually knowns what they know.
 It's far too late to edit my original post, but I want to add a disclaimer here. I still don't see how this system can work, but (as suggested by T-A in a brief comment in that earlier thread) there's at least some chance that the details of plasma physics make it function. My current sense is still that one way or another it doesn't work out (I'll spell that out in a bit more detail in a response to one of my later comments here where I was flat out wrong about how plasma screening might work), but I am not remotely an expert in plasma physics: maybe there's some subtle dynamical effect that really does save the day. (If that's the key, though, then I'm pretty sure that the plasma details are essential for making it work: none of the arguments I've seen in that earlier thread or in this one make the case.)
 > there's at least some chance that the details of plasma physics make it function. My current sense is still that one way or another it doesn't work outYou really should learn to read papers before trying to debunk them. Zubrin's paper has Debye length all over it; it's a crucial consideration.All right, weekend fun: let's look up the closed form expression for the electric field of a capacitor immersed in a plasma. This is old stuff, it was worked out many decades ago, and can be found e.g. inhttps://link.springer.com/article/10.1134/S1064226910050116The internal field is given by Eq. 12 and depicted in Fig. 2 (looks roughly like a catenary, sagging between the plates), the external field is given by Eqs. 6 and 8, Fig. 1.We only need to concern ourselves with the simple case of zero frequency, in which case the external electric field E at distance r from a plate with surface field E_0 isE = E_0 * exp( -r / l_D )with l_D denoting Debye length. The internal field at distance z from the left plate (at the origin) for plate spacing d isE = E_0 * ( sinh( (d - z) / l_D ) + sinh( z / l_D ) ) / sinh( d / l_D )It sags more in the middle for larger d, but it has the nice property that it never reverses sign (as can happen for non-zero frequencies) so there's no risk of the ion getting trapped uselessly inside the capacitor.To get the energy lost by a positive unit charge entering through the anode and then exiting through the cathode, integrate the external field from r = 0 to infinity and multiply by two. Total loss: 2 * E_0 * l_DTo get the energy gained by the same unit charge between the plates, integrate the internal field from z = 0 to d. Total gain: 2 * E_0 * l_D * ( exp( d / l_D ) - 1 ) / ( exp( d / l_D ) + 1 )Subtract loss from gain and you have the net kinetic energy gain,2 * E_0 * l_D * ( exp( d / l_D ) - 2 ) / ( exp( d / l_D ) + 1 )which is positive for d > l_D * ln(2) ~ 0.693 * l_D and grows asymptotically toward 2 * E_0 * l_D as d -> infinity (in practice, there is little difference beyond d = 5 * l_D).So, it works, and you can space the two "plates" (actually meshes of conducting wires, each 0.1 mm wide, in Zubrin's paper) as far apart as you want, but there is little to be gained (other than the flexibility of working across a larger altitude range) from spacing them more than 5 Debye lengths apart.
 ... except I just realized that the final expression for gain minus loss is wrong: redoing the calculation I get-4 * E_0 * l_D / ( exp( d / l_D ) + 1 )So charges traversing the whole thing lose energy. You still get a net momentum due to the mass difference between electrons and protons, but in the opposite direction. :/
 >> Background: I'm a physics professor. I teach basic E&M every year and upper-level undergraduate E&M occasionallyCool. I've been wanting to talk to someone like you about some physics stuff. If you're up for it, drop me a gmail - same id.
 Alas, I don't think I should make promises of free time that I'm unlikely to be able to keep: my fall term starts in just a couple weeks, and my available time is going to rapidly fall to zero.
 Evil thought: you should use this thing as an exam question :-)
 ELI5: Given this, how does an ion drive work?
 The exhaust of an ion drive is electrically neutral [0]. The drive doesn't send nuclei one way and electrons the other, nor is there any reason it should have a significant large-scale electric field of its own.
 Right, the exhaust is electrically neutral, but only the positive ions are accelerated and generate thrust. So if the ions return eventually to their original potential (as 'Steuard described), there cannot be any net change in momentum. I don't think the electrons injected into the exhaust make any difference in this - as far as I know, that is only necessary to avoid the craft accumulating negative charge.Perhaps my question could be simplified if we assume the propellants are already ionised - how can a craft generate thrust electrostatically with two grids if a dipole can't generate thrust?
 I don't know much at all about ion drives. But in (e.g.) an x-ray tube, electrons are accelerated to high speeds by arranging for them to begin in a location with non-zero potential (a hot negative filament) so that they will gain speed as they go away. So at a very vague guess, I'd expect that ion drives arrange to ionize the atoms in a region at a high potential, after which the positive ions quickly accelerate toward some negatively charged but porous "target" and then (mostly) pass through and continue at constant speed.My broad take on the distinction (assuming I have the right general idea about ion drives!) would be that if you're the one providing the material to be ionized, you get a lot of control over its initial conditions. But if you're planning to use matter that's externally available, your control is a lot more limited. In the schematic that I've seen for this dipole drive, the ions begin far away (so, at zero potential) and end far away (again at V=0), so I don't see a way to avoid that.
 > None of that second commenter's objections even began to address the core of the original poster's point […] By conservation of momentum (and/or Newton's 3rd law), the device won't have any net change in momentum either.There was no need to address that, since the full derivation was given in a previous comment, also by me:https://news.ycombinator.com/item?id=17432843> I'm a physics professorSadly, I have no problem believing that. You may or may not teach basic E&M, but you failed to apply conservation of momentum correctly in the general case of two different masses.
 I'm happy to discuss this stuff, but I think the conversation will be more pleasant if you avoid the snark, and maybe if you're a bit more willing to entertain the possibility that someone who's spent their career studying a subject might have spotted things about it that you haven't recognized yet.In this particular case, my impression is that (like the paper's author) you haven't understood the core of the objection: that the non-zero field outside of the gap between the plates is important. In your derivation, you only address the exterior field long enough to say that it "quickly drops to zero". But we know that the electric potential far from the (finite!) plates approaches zero, even though it is high at the positive plate. So it's clear that even if the field outside is relatively small, its integral is not negligible: it adds up to exactly enough to cancel out all of the speed gained by an ion that started out exactly between the plates.If you're really determined on this, I can type up a sketch of the full calculation for a pair of circular plates. But the energy considerations ought to be enough to show what the ultimate result will be. (And, to be clear, I'm not even looking at pairs of charged particles with different masses: it doesn't even require that level of analysis. This is true for any one charged particle considered on its own.)
 A very similar exchange with T-A happened back in the comment thread from last year: https://news.ycombinator.com/item?id=17432975As you and several others pointed out last year, the energy considerations here are good enough to invalidate the scheme.
 > As you and several others pointed out last year, the energy considerations here are good enough to invalidate the scheme.The only "energy consideration" in that thread is this, from "ballenarosada":"The potential is 0 at infinity, positive at the first plate and negative at the second. An incoming ion starts at 0 potential, climbs a big hill to get through the first plate, then falls down below 0. Then on the way out it has to climb back to 0 potential at infinity. So the ions gain energy inside the plates but lose it all back on either side."In order to believe this to be correct, you must fail to understand that the device is electrically neutral. As you move away from it, the distance between the plates becomes negligible compared to your distance from it, and all you see is a neutral object exerting a negligible attraction on the ion.This is, of course, the principle behind multipole expansions [1]."balenarosada" thought he or she could just use the dipole term of the expansion to model the capacitor, maybe misled by the name "dipole drive". In hindsight, maybe Zubrin should have gone with "capacitor drive" to prevent this misunderstanding.
 > the conversation will be more pleasant if you avoid the snarkAfter you.> maybe if you're a bit more willing to entertain the possibility that someone who's spent their career studying a subjectYou mean someone like Zubrin [1], who's spent decades designing space propulsion systems? Does it really seem plausible to you that he wouldn't understand conservation of momentum, as used e.g. to derive the rocket equation [2]?> In this particular case, my impression is that (like the paper's author) you haven't understood the core of the objection: that the non-zero field outside of the gap between the plates is important.That was the subject of the entire subthread which you just dismissed by instead incorrectly invoking conservation of momentum.> If you're really determined on this, I can type up a sketch of the full calculation for a pair of circular plates.Not really, because once you graduate from basic E&M, or just read the work which you decided to criticize without first bothering to read it, you learn about things like plasma screening [3], which (as also pointed out in the subthread which you dismissed) implies that you should not integrate out to infinity.
 I need to make a correction to my comments on plasma screening here, after a 4am realization that my mental model of what would happen to the potential was in fact pretty solidly wrong. That said, I feel like fixing my mistake just leads this proposed dipole drive to fail in a different way.So: It is entirely possible for a negative screening charge to accumulate right next to the positive plate/grid that will completely eliminate the exterior field (and similarly on the other side), and that will not generate the extremely strong field at the edge of the plate that I was imagining in my comment above. Specifically, if the negative screening charge is more or less in the shape of an adjacent parallel plate of plasma (and a positive parallel plate of plasma right next to the negative plate/grid), that will indeed perfectly cancel the field outside. At that point, the most you'd find for the field in between the "plasma plate" and the real one would be a simple parallel plate capacitor field: the same magnitude that was originally expected between the real plates. Because it would only act over a very small distance, this would only lead to a negligibly small change in potential. BUT! In this scenario, you have essentially neutralized both of the original plates: both sides are at roughly zero potential, and the net effect of these "four parallel plates" in the interior region is to have zero field there, so there is no longer any field in the middle to accelerate plasma particles and the drive doesn't work.On the other hand, if you constantly keep pumping charge to maintain the desired potential difference between your original plates/grids, that precisely means that you are not allowing your plates to be fully screened: you get your drive field back in between the plates, but now you're basically back to the unscreened case and it seems that my original objections hold. (I assume that as charge piled up, eventually the narrow fields between the grids and the surrounding plasma would become strong enough that electrons would leak between them, creating an awkward circuit.) And I suspect that any intermediate situation between these two extremes would just interpolate between their two different modes of not working.That said, I am not even remotely an expert on plasma physics. I can't rule out the possibility that there's some fascinating effect in plasma dynamics that would turn the nitty-gritty details of all this into a functioning drive system. But I'm pretty convinced that if it does work, the justification will depend strongly on detailed plasma dynamics: the simple arguments we've been making about how parallel plate capacitors accelerate individual ions (or even individual electron/proton pairs) won't capture the heart of it.
 > On the other hand, if you constantly keep pumping charge to maintain the desired potential difference between your original plates/grids, that precisely means that you are not allowing your plates to be fully screened: you get your drive field back in between the plates, but now you're basically back to the unscreened case and it seems that my original objections hold.What happened to your previous realization that> Oh! I hadn't understood that the ion is what's being screened, but of course it would be.?As I already explained: pumping electrons between the plates will (partially) restore unscreened conditions around them, but as the ion leaves, it goes into the surrounding plasma and gets screened, i.e. decoupled from the device, thus invalidating your original objections.As an aside, I find it odd that you decided to post this comment above the earlier subthread where I explained screening to you. Maybe you just don't know any better, but it creates an impression that you are trying to rewrite the history of this discussion.
 I suspect that we're only talking to ourselves at this point, so perhaps it's not worth a lot more time on this conversation. But a few notes:Regarding the infinite plate limit: Locally, it is absolutely true that as you increase the size of the plates (relative to the distance between them) the fields must approach the result for an infinite plate. But as soon as you want to consider global questions like the behavior of the potential at infinity, there is a true qualitative difference between a large-but-finite capacitor and an infinite one. Any finite charge distribution will result in a potential of zero at infinity (or rather, will allow you to consistently make that choice), but an infinite charge distribution will not in general allow that. For the case of an infinite parallel plate capacitor, if you set the potential to zero out to infinity on one side, it will be non-zero (and constant) out to infinity on the other. One upshot of this is that it's misleading at best to use the infinite plate approximation for any sort of scattering analysis (like what we're discussing here).Regarding multipole expansions: I love multipole analysis, it's a beautiful way to study charge distributions. That said, I think it's massive overkill to actually use it in this case, since at least for circular plates along the axis that process is far more complicated than just using the exact fields themselves.Regarding momentum transfer: You've claimed that "It is the functional form of the potential out to infinity which determines how much momentum the ions transfer back to the device". I agree that if you want to know the rate of momentum transfer at any given point, you need to know the exact functional form of the potential (and/or the field). But if we're only interested in the total momentum transferred in the interaction, we don't need to know the details of where and when the momentum transfer happened: all we need to know is the net result. So in particular, if your ion starts far in front of the device with (relative) velocity v0 in the -x direction, and we can show from energy conservation that it ends far behind the device with the same velocity v0 in the -x direction, that's sufficient to prove that the ion has experienced no net change in velocity and thus that no net momentum transfer has occurred (even though there may have been substantial transfers along the way).With that in mind, if we specialize to the simple case of an ion moving along the axis of symmetry of two circular plates, we can find exactly this result explicitly from an energy calculation. So to heck with it, let's just do this. (And apologies if this is repetition, but I don't think I was able to get your link in that other thread to work, or perhaps I tried to follow the wrong one.) Along the axis of a uniformly charged circular disk of radius R,`````` E_x(disk) = K * (1-|x|/sqrt(x^2+R^2)) * sign(x), `````` where K is a constant (K=sigma/(2 epsilon_0) for surface charge density sigma). Assembling two of these with opposite charges a distance D apart (with the + plate at x=0 and the - plate at x=-D) gives for x>0:`````` E_x(x>0) = K * [(x+D)/sqrt((x+D)^2 + R^2) - x/sqrt(x^2 + R^2)] `````` This has the properties you'd expect: for example, it approaches D/R->0 as R approaches infinity, matching the infinite plate limit. (The expression's clearly different between the plates.) You can integrate this (or rather, its negative) to get the electric potential:`````` phi(x>=0) = K * [sqrt(x^2 + R^2) - sqrt((x+D)^2 + R^2) + D] `````` The constant +D was chosen to maintain the convention phi=0 at infinity. But then we find that the potential at x=0 is`````` phi(x=0) = K * [R - sqrt(D^2 + R^2) + D], `````` or approximately K * D in the limit of large R (and this actually decreases as R gets smaller). [This is consistent: by symmetry, phi=0 at the midpoint between the plates (x=-D/2), and in the large R limit the electric field inside is a constant sigma/epsilon_0 = 2K, so 2K * D/2 = K * D.] No matter how large you make the plates, the total change in the incoming ion's energy and thus the total effect of the exterior field on the ion's speed will not approach zero, even though the magnitude of the field at any given point does approach zero as R grows (as it must, to match the infinite plate limit). And no matter how large you make the plates, the total effect of the field on the speed of an ion that moves from infinity to the midpoint of the plates will be zero (and similarly on the negative/outgoing side). You're welcome to solve equations of motion to find the exact functional form of x(t) for this particle, but that's a lot of work, and energy conservation guarantees that there will be no net change in speed.Finally, regarding plasma screening: My comment about positive ions being strongly repelled outside of the positive plate was perhaps unclear, and certainly I'd welcome corrections if I'm misunderstanding screening somehow. I was speaking specifically of ions that came closer to the positive plate than the length scale over which the screening brings the outside field to zero. The screening charges can't change the electric potential at the position of the plate itself (unless you allow them to discharge the whole thing), so rather than falling off to zero very slowly over a long distance as in the unscreened case (giving rise to a small but wide electric field as the gradient), the potential must fall to zero very rapidly over the short screening distance (giving rise to a large but short-ranged electric field). Either way, the height of the potential energy barrier at the plate (relative to infinity) can't change. So again, in the screened case I would think that incoming positive ions would feel essentially zero field for most of their approach but would then feel a very strong repulsion in the narrow region immediately before the plate.
 > Regarding multipole expansions: I love multipole analysis, it's a beautiful way to study charge distributions. That said, I think it's massive overkill to actually use it in this case, since at least for circular plates along the axis that process is far more complicated than just using the exact fields themselves.Yes, yes, in this case the multiple expansion is a conceptual tool which can be used to understand why things work as they do (and it originally entered the discussion with ballenarosada because he or she was trying to use just the dipole term to model a plate capacitor).The exact solution can be looked up e.g. in the pedagogical presentation which I linked to in that thread. I see now that the original link no longer seems to work, but this one does:https://www.researchgate.net/publication/228807623_Electric_...Cutting to the chase: you are still not accounting for plasma screening. As the ion leaves the plate, it's surrounded by a cloud of oppositely charged ions which reduce its effective charge, i.e. its coupling to the plate's electric field. The further away it moves, the smaller its charge seems to be.To account for that, include a Debye factor: the coupling between electric field E and ion with naked charge q goes from E * q toE * q * exp(-r/l_D)where r is distance from the plate and l_D is the Debye length [1].As you say, integrating the equations of motion for the correct E is a lot of work, but we don't need to: it's quite enough to agree that the field is falling by some power of distance from the plate. The exponential then wins in very short order. At a separation of just ten Debye lengths, the ion's effective charge has been reduced by a factor 0.000045; at a separation of one hundred Debye lengths, the factor is 3.72e-44, for all practical purposes zero.When the Debye length is short compared to the field gradient (i.e. to the size of your capacitor) you can just let E be constant. It's then trivial to integrate the work done by the electric field as the ion moves out; even if you let the ion run all the way to infinity, the work done is justE_0 * q * l_Dwhere E_0 is the electric field at the plate.Compare with the work done inside the capacitor (where E is actually constant) and you see that even using this maximally pessimistic approximation of constant external field, you've only shaved off one Debye length's worth of the capacitor's plate separation from the total work done on the ion.In low Earth orbit that's one to a few millimeters [1].Regarding this,> rather than falling off to zero very slowly over a long distance as in the unscreened case (giving rise to a small but wide electric field as the gradient), the potential must fall to zero very rapidly over the short screening distance (giving rise to a large but short-ranged electric field)I think the easy way to think about it is to view the ion as being screened. The plates will of course also attract oppositely charged ions, but they are connected to a generator (solar panels, nuclear reactor...) which is constantly pumping electrons accruing around the anode to the cathode, where they recombine with protons accruing around the cathode. As long as that's running, you have a steady state situation where the net charge of each plate is somewhat reduced relative to the vacuum case, but the field can still be approximated by the vacuum solution (using the reduced net charge). For the ion, this is clearly not true; it is not connected to a generator and will just behave like any other ion in the plasma.
 Oh! I hadn't understood that the ion is what's being screened, but of course it would be. That begins to sound a lot more promising. I still have big questions, but maybe there are answers. Like......does the screening not have the same effect between the plates? I expect they'd be more than a few millimeters apart....won't this screening simply act to essentially diffuse the ion's charge into the surrounding plasma? And if so, won't that mean that (e.g.) in the case of a proton, the device's external field will still be exerting an attractive force on the net-positive plasma surrounding the it? I kinda suspect that would ultimately lead to the same cancellation of thrust that I was worried about in the absence of screening.
 In inverse order,> ..won't this screening simply act to essentially diffuse the ion's charge into the surrounding plasma?I guess that's a reasonable description, but remember you are diffusing ions (well, protons and electrons) of opposite sign, in equal quantities, so diffusion is good; they will eventually mix back together.> ...does the screening not have the same effect between the plates? I expect they'd be more than a few millimeters apart.This is a good question. For this thing to work, plasma must be allowed into the capacitor and yes, there will be screening there too. The obvious dumb way I can think of to counter that is to have not just one pair of plates, but a sequence of more closely spaced ones, linear accelerator style.
 Can there be a differential like in gravity assists?
 So the particles follow the path of a doublet?
 General note to naysayers: before you present yet another "obvious" argument why Zubrin's proposal can't possibly work, you may want to check how it applies to this:http://news.mit.edu/2018/first-ionic-wind-plane-no-moving-pa...https://www.nature.com/articles/s41586-018-0707-9It's essentially the same thing, except it has to ionize the medium (air) on its own rather than just pick up the ions from its surroundings.(Yes, it does so in front of the vehicle, so on average stripped-off electrons get less work done on them by the electric field than positive ions, but the electrons would carry less momentum than the more massive ions even at kinetic energy parity, with the difference going to the vehicle, as Zubrin correctly says - please let's not beat that dead horse again.)
 I found a non-paywalled discussion here:
 6 milli-Newtons per kW is not a lot of force relative to the electricity consumed... but this is sort of the early alpha version of this technology. I expect future versions of this to be greatly improved. So, all in all, a brilliant set of ideas with a lot of future potential!
 The dipole drive uses the protons and electrons of the solar wind for propulsion. They are reflected in an electric field and the paper claims that velocities higher than the solar wind speed can be achieved.But take the proposal with a grain of salt, because the whole mechanism highly depends on the plasma density of the solar wind which is rather small in the interplanetary space. It really works well only close to planets or stars where the plasma density is high.
 Should be an interesting read. Can't reach the site. Is it up or down? Tried loading it several times but it didn't load for me.
 I feel as though this is one of those points where a picture is worth a thousand words.
 Isn’t this basically electrohydrodynamics?
 > the dipole drive can be used to accelerate a spacecraft at velocities greater than that of the solar wind [0]and> [the Dipole drive] therefore offers potential as a means of achieving ultra-high velocities necessary for interstellar flight.Sorry, but I couldn't come up with anything else than: warp-drive!! (if it can deliver on that promise that is). I have long hoped/expected we would crack the challenge of propellant-less acceleration through space, not sure if this is it though.
 Nothing that accelerates at a tiny fraction of a percent of 1g is within miles of a "warp drive", no matter how high the max speed is.
 The point of a warp drive is to bypass the speed of light limit by warping space. Acceleration is kind of irrelevant.
 Theoretically, in addition to the warp drive, you also need some kind of secondary propulsion, no matter how slow, to move the ship through the warp bubble/vortex/portal/curved space, etc.Later versions of warp will be able to move the bubble/vortex/portal of curved space itself -- around the ship and from front to back, thus obviating the need for motion relative to the warp field.Technology-wise think of the steps like this:1) Figure out how to create a warp bubble2) Figure out how to move it, relative to what's projecting itWithout #2, secondary propulsion is required. #1 is the first technological achievement, #2 would follow upon that. Until #2 is achieved, secondary propulsion is required...#2 is a little bit wierd to do, because once you've changed your location in space, you've changed the location of what's projecting the warp field -- you need to include that in your warp field projection calculations for #2... and that's IF it could be done...In fact, the initial version of warp might look something like this: Ship A, stationary in space, projects the warp bubble, Ship B goes through it. Why? Because otherwise you've got all kinds of relative motion that would need to be compensated for...Of course, all of this is theoretical...
 Then again, if you can actually generate a field that warps spacetime, you can probably arrange for the spacetime inside the warp bubble to be curved such that it causes objects inside to accelerate.
 > 1) Figure out how to create a warp bubbleHow hard can it be? My guess is "very" :-P . My bet is more along the lines of constant acceleration, which is not "warp drive" per se, but holds a better promise of interstellar and possibly even intergalactic travels.
 The fact that the abstract does not mention laser driven sails, the currently most viable approach to relativistic propulsion (ie, the current breakthrough starshot plan), does not give me much hope in this researchers credibility in the rest of the paper.Furthermore, the performance of this engine is abismal. To put it into perspective,a modest ion engine would produce about a million times more thrust per power. While the ion engine of course requires carrying fuel, and is thus governed by the tyranny of the rocket equation and not directly comparable in performance, this engine still requires a power source and thus effectively requires fuel anyway.Also worth mentioning is that the least efficient possible massless thruster, a simple blackbody radiator with reflector to aim all of the photons behind you, is only about 1000x worse than this unit in terms of propulsion force per power, and essentially infinitely better than the unit in terms of propulsion force per mass, since this structure is incredibly massive where as you can usually generate blackbody radiation directly from you power source.
 The author is Robert M. Zubrin. To write it off because the abstract doesn't mention a specific technique seems a bit silly, especially given that Zubrin personally knows a lot of the people involved in Breakthrough Starshot and surely is aware of it, including its massive caveats.It's worth noting that the paper specifically discusses the need for methods that are not limited in travel direction, and in that respect Breakthrough Starshot has the same caveats as a solar sail and so treating laser driven propulsion separately is pointless.Another major difference that is so obvious as to not require a mention of every specific alternative is that it is talking about a propulsion method that avoids the need for us to provide a source of thrust.Laser driven systems doesn't require the ship to carry propellant, but we're still paying the cost to provide thrust, while this design is basically 'just' relying on there being a source of protons and electrons with a speed differential relative to the current speed and direction of travel of the ship that it can 'scoop up'.That makes for fundamentally different tradeoffs relative to classical sail based propulsion.
 > Furthermore, the performance of this engine is abismal. To put it into perspective,a modest ion engine would produce about a million times more thrust per power.Would it, though? The proposal mentions 14mN/kW. What ion engine produces 14N/W ? I think that such an engine would be a massive improvement of current designs. VASIMIR, for instance, is designed with 1N/14kW.

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