8 ms·

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

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.)
  - T-A 7y ago
    > the conversation will be more pleasant if you avoid the snark After you. > maybe if you're a bit more willing to entertain the possibility that someone who's spent their career studying a subject You 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. [1] https://en.wikipedia.org/wiki/Robert_Zubrin https://en.wikipedia.org/wiki/Robert_Zubrin [2] https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation#Derivation https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation#De... [3] https://en.wikipedia.org/wiki/Electric-field_screening https://en.wikipedia.org/wiki/Electric-field_screening
phkahler 7y ago
>> Background: I'm a physics professor. I teach basic E&M every year and upper-level undergraduate E&M occasionally Cool. 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.
- Steuard 7y ago
  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.
lutorm 7y ago
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... ;-)
- Steuard 7y ago
  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.)
ScottBurson 7y ago
Evil thought: you should use this thing as an exam question :-)
rrss 7y ago
ELI5: Given this, how does an ion drive work?
- ScottBurson 7y ago
  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. [0] https://en.wikipedia.org/wiki/Ion_thruster https://en.wikipedia.org/wiki/Ion_thruster
  - rrss 7y ago
    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?
    - Steuard 7y ago
      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.
Steuard 7y ago
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.)
- T-A 7y ago
  > 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 You 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. in https://link.springer.com/article/10.1134/S1064226910050116 https://link.springer.com/article/10.1134/S1064226910050116 The 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 is E = 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 is E = 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_D To 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.

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.)
- T-A 7y ago
  > the conversation will be more pleasant if you avoid the snark After you. > maybe if you're a bit more willing to entertain the possibility that someone who's spent their career studying a subject You 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. [1] https://en.wikipedia.org/wiki/Robert_Zubrin https://en.wikipedia.org/wiki/Robert_Zubrin [2] https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation#Derivation https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation#De... [3] https://en.wikipedia.org/wiki/Electric-field_screening https://en.wikipedia.org/wiki/Electric-field_screening

phkahler 7y ago

>> Background: I'm a physics professor. I teach basic E&M every year and upper-level undergraduate E&M occasionally Cool. 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.

Steuard 7y ago
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.

lutorm 7y ago

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... ;-)

Steuard 7y ago
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.)

ScottBurson 7y ago

Evil thought: you should use this thing as an exam question :-)

rrss 7y ago

ELI5: Given this, how does an ion drive work?

ScottBurson 7y ago
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. [0] https://en.wikipedia.org/wiki/Ion_thruster https://en.wikipedia.org/wiki/Ion_thruster
- rrss 7y ago
  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?
  - Steuard 7y ago
    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.

Steuard 7y ago

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.)

T-A 7y ago
> 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 You 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. in https://link.springer.com/article/10.1134/S1064226910050116 https://link.springer.com/article/10.1134/S1064226910050116 The 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 is E = 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 is E = 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_D To 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.