7 ms·

Previously discussed: https://news.ycombinator.com/item?id=17428751 https://news.ycombinator.com/item?id=17428751 I chewed on this idea for several days, event

Previously discussed: https://news.ycombinator.com/item?id=17428751 https://news.ycombinator.com/item?id=17428751 I 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=17432975 https://news.ycombinator.com/item?id=17432975 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". As particles leave the vicinity of the drive, they will be attracted back toward it, reversing the momentum transfer.

starpilot 7y ago
So the particles follow the path of a doublet?
stareatgoats 7y ago
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.
- Steuard 7y ago
  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-a-new-concept-for-space-propulsion/ 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.]
  - heavenlyblue 7y ago
    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.
    - lowdose 7y ago
      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.
  - T-A 7y ago
    > 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 https://news.ycombinator.com/item?id=17432843 > I'm a physics professor Sadly, 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.
    - Steuard 7y ago
      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.)
amluto 7y ago
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.
- doubleunplussed 7y ago
  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.
  - jbay808 7y ago
    GP was referring to a distribution of "fixed charges", so it's not possible to transfer momentum to them.
    - doubleunplussed 7y ago
      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.
- ScottBurson 7y ago
  > in the original paper, 500W is going somewhere, but there is no explanation of how, exactly, any current flows in the circuit I 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.
grogers 7y ago
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.
foobarian 7y ago
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.
- rrss 7y ago
  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.
  - aey 7y ago
    I believe this canoe is picking up cannonballs along the way.
    - ScottBurson 7y ago
      It seems that way at first. The problem is that each of those cannonballs is tethered to the canoe by a long bungee cord.
      - aey 7y ago
        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?
- ScottBurson 7y ago
  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.
cryptonector 7y ago
Can there be a differential like in gravity assists?
jjk166 7y ago
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.
T-A 7y ago
> 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.

starpilot 7y ago

So the particles follow the path of a doublet?

stareatgoats 7y ago

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.

Steuard 7y ago
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-a-new-concept-for-space-propulsion/ 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.]
- heavenlyblue 7y ago
  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.
  - lowdose 7y ago
    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.
- T-A 7y ago
  > 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 https://news.ycombinator.com/item?id=17432843 > I'm a physics professor Sadly, 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.
  - Steuard 7y ago
    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.)

amluto 7y ago

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.

doubleunplussed 7y ago
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.
- jbay808 7y ago
  GP was referring to a distribution of "fixed charges", so it's not possible to transfer momentum to them.
  - doubleunplussed 7y ago
    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.
ScottBurson 7y ago
> in the original paper, 500W is going somewhere, but there is no explanation of how, exactly, any current flows in the circuit I 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.

grogers 7y ago

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.

foobarian 7y ago

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.

rrss 7y ago
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.
- aey 7y ago
  I believe this canoe is picking up cannonballs along the way.
  - ScottBurson 7y ago
    It seems that way at first. The problem is that each of those cannonballs is tethered to the canoe by a long bungee cord.
    - aey 7y ago
      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?
ScottBurson 7y ago
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.

cryptonector 7y ago

Can there be a differential like in gravity assists?

jjk166 7y ago

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.

T-A 7y ago

> 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.