The Dipole Drive: A New Concept for Space Propulsion
centauri-dreams.org
centauri-dreams.org
It's just a parallel plate capacitor, with holes in the plates that allow the surrounding plasma to pass trough. The question is:
Why a plate capacitor with holes in the plates inside a diluted plasma is not a perpetual mobile?
Let's assume that initially the device is with speed 0 inside a huge bath of plasma that is also at speed 0. We charge the two plates of the capacitor instantly and magically. In this model and the calculations the ions never hit the plates because they have holes, so the plates never discharge. There is some energy in the electric field of the capacitor.
Now the device is operational and it starts to move, the ions also move. Now the devices and ions have some kinetic energy.
* If we believe the whole idea the ions just go away, and the plates never discharge, and new plasma enters the device, so it continues to accelerate forever. So the total energy of the system increase and increase, that is something impossible.
* If we don't believe the whole idea, the problem is that the ions accumulate outside the plates in some kind of ion cloud, even if they never hit the plates. Initially the effect is small, but eventually the charge of the ion clouds compensate the charge of the plates and the plasma inside the device feels no net force, so the device moves at a constant speed. Essentially, some part of the energy that was stored in the electric field was transformed into kinetic energy, but there is only a finite amount of it, so the device must reach a maximal velocity.
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There is some complications, because in the Solar system the plasma is not static, because there is solar wind. So, just imagine that you pick a reference frame (i.e. starship) that is moving at the same speed of the solar wind. In this reference frame (i.e. starship) initially the plasma looks static and the giant capacitor is moving. When it is turn on, there is some energy in the electric field and some kinetic energy. After a while, the charges are redistributed, and the speed of the giant capacitor changes, but only a finite amount, it can't continue accelerating forever. If you see the scene from the normal reference frame (i.e. the Sun) the device is initially static and after a while it is moving with a cloud of ions around it, but the total change in speed is the same.
I guess there is some drag, and the final speed in the starship reference frame is 0 and the final speed in the Sun reference frame is the same of the speed of the solar wind.
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About other comments:
* I agree that the momentum transferred to the protons and electrons is different. With some assumptions the ratio is 43, but in a moving device in a moving solar wind the calculation is more complicated. The exit speed of the electrons is 100 times faster than the solar wind, so it makes almost no difference. The exit speed of the proton is roughly the speed of the solar wind, so the calculation must be fixed, but there is a net effect that is no 0. (Until the ions cloud accumulate enough charge.)
* In an ion thruster the first step is to ionize the gas. Here they start with an ionized gas. The difference is that in a ion thruster they separate the positive and negative field.ions, so they waste some energy in the separation and then they accelerate only the positive ions. The separation of the ions is an active process, not just a static electric
* Hat tip for sandworm101 for the comment https://news.ycombinator.com/user?id=sandworm101
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About the article:
> The acceleration of the electrons is a form of drag, which is provided for by loss of spacecraft kinetic energy. It therefore could, in principle be used to generate electric power, partially compensating for the power consumed to accelerate the protons.
This makes no sense. The acceleration of the electrons create some thrust in the wrong direction, but it's not like a drag, it's like a thruster in the wrong direction.
> To see what the performance of a dipole drive might be, let us work an example, assuming a 500 W power source to drive the system. The electron current negates about 2% of the thrust (1/43rd) produced by the proton current. The maximum possible jet power is thus about 490 Wj. Assuming additional inefficiencies, we will round this down to 400 Wj, for a total system electrical to jet power efficiency of 0.8. The relationship of jet power (P) to mass flow (m) and exhaust velocity (c) is given by: P = mc^2/2
I'm not sure that mixing thrust and energy is a good idea, but anyway they assume that the efficiency is 80% and use the usual formula to get the speed. The problem is that assuming that only a 20% of the energy will be wasted is optimism, not a real calculation. How much energy is lost due to the collision of the ions and the plates? How much to recharge the plates? How much is lost in real drag? How much is lost as heat? ... There are a lots and lots of factors, and then you have to consider the nasty ions clouds.
You're refuting a claim that doesn't even exist.
And in doing so, you're also assuming the particles around it have "speed 0" when the explanation clearly describes individual particles as having non-zero velocity so that they can enter the space between the two plates on their own, as it is claimed that there would be no net electric field outside of them.
In my description I change the reference frame because it's much easier to do the calculations. In the reference frame that moves at the same speed of the solar wind the plasma has speed 0, but it doesn't mean that all the particles have speed 0. It means that the number of particles that are moving in one direction is (almost) equal to the number of particles in the opposite direction. [2] It's like the air inside a quiet totally sealed room that is left alone for a few days. The air "doesn't move" but the oxygen and nitrogen molecules are moving very fast. The speed that they have is related to the temperature. But in bulk they don't move and if you put some small pieces of paper or strings they will not move (unless they are really tiny). If you have some container like a glass inside the room the molecules that are inside the container will change constantly, but the number of molecules inside the container will (almost) no change. The problem in my version of the experiment is not that the device will get empty.
[1] Assuming we don't need to change the current accepted laws of physics to explain this. It would be even more groundbreaking.
[2] There are more accurate technical description but this is good enough.
https://en.wikipedia.org/wiki/Electrostatic_deflection
Obviously CRTs don't contain perpetual motion machines. So there must be some way in which energy is lost in the plates.
(Note that when I say it's the same phenomenon, I'm talking about a force acting on charged particles. The initial direction of the particles is different, though. In this dipole drive, the charged particles are initially moving roughly perpendicular to the plates, while in a CRT the particles are moving parallel to the plates. But that shouldn't be relevant to the energy needed to accelerate them.)
Also, when the electrons are extracted from the cathode it gets a small positive charge, but this charge also circulate and is not accumulate. Moreover, it's the charge that eventually neutralize the charge of the electrons.
I think there are no special wires to make this happen, the current I very small and the charge is redistributed by the supporting parts.
A good comparison would be to use a CRT where the screen is split in two perfectly isolated parts. After a while the half you are sending the electrons two become too charged and it deflects the electrons.
I am NOT talking about the main electron gun. I am talking about the deflection plates.
Just to be perfectly clear, a CRT has three active components:
1. Electron gun to create a beam of electrons aimed in the direction of the screen.
2. Vertical deflection to move the beam up and down.
3. Horizontal deflection to move the beam side to side.
Here, you are talking about #1, and I'm talking about #2/#3.
If initially the position of the beam is in the center of the path , a charged #2 and #3 will change the direction and in a smaller amount the velocity of the electrons. The energy to change the velocity comes from the energy in the electric field, not from the plates.
You can say that this energy is energy of the electric field or that it is potential energy of the electrons in the electron beam. Both descriptions are accurate and give the same result. Just remember to not count it twice.
Now the question is where does the energy from the field or the potential energy of the electrons came from. The answer is that the repulsion has a small but not 0 effect outside the main part of the capacitor and this creates an electric field outside. To align the beam at the center you must overcome this small force, so you must add some energy to the electron (in this case it's easy to think in the potential energy of the electron). Perhaps the comment from ballenarosada is more clear https://news.ycombinator.com/item?id=17432975
In a normal CRT the electron bean is created from a high voltage source so it moves fast, and the electrons initially move inside wires, so you usually ignore the small potential field of the capacitor far away. If the charges move slowly and freely then you must consider the effect of the electric field outside the capacitor.
So you imagine that there is also an imaginary starships moving at a constant velocity in a constant direction. And you put an imaginary scientist doing imaginary measurements from the starship. A more serious name for the "imaginary starship" is "reference frame". All the reference frames that move at a constant speed are equivalent, they have the same laws of physics, so you can choose any of them to make the calculations.
In this case the calculations are more easy if you pick a reference frame that moves with the same speed and some direction that the solar wind (at least locally). It's only a mathematical trick, you imagine that there is an imaginary starship there just to do the calculations.
So if this device really works, it must work when you are measuring it from a imaginary starship that is moving at the same speed of the solar wind. You can't avoid the imagination. Also, all the other reference frames give the same result, but the calculation is much harder.
The advantage of using the reference frame that moves at the same speed of the solar wind is that in that reference frame the speed of the solar wind is 0. It doesn't mean that the particles doesn't move or don't interact, just that they don't move in any preferred direction. So the particles may enter the device from any direction.
If your initial setup is that the device is moving with some speed/direction that is not equal to the speed/direction of the solar wind, then in the reference frame we choose the device is initially moving in some direction with some speed.
Now, when you turn the device on, there is some kinetic energy because the device is moving and there is some energy in the electric field because the capacitor is charged. With the description in the site, the device must work even if it were a static capacitor with a fixed charge. So after a while, the ions around the device change their position and the device may start to move, but there is a maximal speed change. Initially the acceleration may be big, but then it starts to fade away. The initial energy is redistributed, part of the energy in the electric field is transformed in kinetic energy of the device, but there is only a finite amount.
When you see it from the usual reference frame, for example the one that is fixed to the Sun) the calculations are more complicated because now you have to consider the solar wind, and it's difficult to avoid some infinite amounts in the calculation of the energy. But the result must be the same that the result in the other reference frame. The device changes the speed a little and then the acceleration goes to 0.
But just imagine that after it has been charged to the nominal value the energy source get's broken, and as an emergency measure you just blow up all the conductors and make the two capacitor plates fully isolated. Does the device continue to work? For how long? How do the capacitor discharge?
In their description the particles never hit the plates, so the plates don't discharge. In the real word it's more complicated, but it's an usual assumption for this kind of problems. The real word has these nasty details that reduce efficiency, but they may obscure the main ideas and problems.
If you still are not convinced, just make a hole in both plates and put a isolator tube between the holes. Also, put some isolator around the plates. Make all the plasma only go throw the tube.
Now the calculations are more difficult because it's no longer the electric field of a perfect plate capacitor, but any finite capacitor has problems at the border. If the hole is small enough the using the formula of a plate capacitor is good enough. Let's see a graphic, were the double lines are the insulator and the +++ and --- are the plates:
entry
╔══════════════╦ ╦═══════════════╗
║++++++++++++++║ ║+++++++++++++++║
║ ║ ║ ║
║--------------║ ║---------------║
╚══════════════╩ ╩═══════════════╝
exit
Moreover, in their calculation they assume that the 80% of the energy is transfer to the kinetic energy of the protons. Where does their calculation change in the model? The efficiency must be also 80%. (Perhaps as efficient, but with a smaller thrust.)
But I suspect that that's a situation like the way in which you can compress a gas in a cylinder without raising the temperature: "all you have to do" is to move the piston only when it will not contact a gas molecule. All the molecules will bounce off the piston while it's stationary, and the temperature won't go up. However, that requires a Maxwell's Demon to supervise the motion of the piston, and it's inevitable that the temperature goes up.
Likewise, in this situation, you simply cannot avoid having ions impact the grids, and that's where power is consumed.
However, in a normal gridded ion thruster, a lot of energy goes into ionizing the reaction mass, and you have to carry the reaction mass. This merely does away with the need for that.
And you can build a bigger device, where the distance of the plates is 200m/200yard and the holes have 10cm/5inches and use the same wires. Or multiply the design by 100x. A much bigger device is not technologically practical, but it shows that there is a theoretical problem.
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In their description they clearly explain that the electrons are reflected, and the protons pass though, so their design will not work. I think that if they absorb the electrons, use the protons as in a gridded ion thruster and then use the absorbed electrons to neutralize the ions, then it may work. You must use some energy to move the absorbed electrons from the front to the back of the ship because there is an electric field. This is not the device they are describing, but at least with this modification it may work.
This is all clear if you consider the ions falling through a potential field. 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.
I'm a fool. Don't speculate about E&M before you've had your coffee kids.
On the other hand, if the capacitor is finite, then the surface integrals over the plates are not equal.
No, but as you have shown it can be grossly overstated. The field outside a finite plate capacitor falls off as a power of distance >= 2 (details depend on the geometry), while the field inside it is constant. It can therefore safely be ignored for a first order estimate of the effect.
If you want to get fancy and claim that higher order corrections invalidate Zubrin's argument, you need to actually prove it. Also, don't forget to include other effects like plasma shielding.
Let A be the area of the capacitor, and dr the distance between the plates. Let c be the appropriate electrostatic constant for the coulomb force between a proton and the charge density on the plate. At a point a distance r from the capacitor, the field effect from the negative side is, ignoring curvature effects, about
cA/r^2
The repelling charge from the other plate will be about
cA/(r+dr)^2 = cA/(r^2 + 2rdr + dr^2) ~ cA(1/r)(1/(r+2dr))
So the difference between the coulomb forces, i.e. the net force, will be approximately
(cA/r)(1/r - 1/(r+2dr)) = (cA/r)((r+2dr - r)/(r^2 + 2rdr)) ~ cA(2dr/r^3)
So the net force drops off approximately as the third power of the distance, to a first order approximation. Integrating over the radius, we have that the potential goes as -1/r^2, with the approximation breaking down near r=0.
Actually inserting appropriate constants of integration would make this argument robust, but would also just reduce to the argument from potential at infinity. Either way it's clear that the effect can't just be ignored out of hand.
What really matters here is that with the force on the charge falling off as a power of distance, even if you integrate force * displacement from the screen out to infinity (which you shouldn't do in a plasma, because [2]), you get a finite contribution which can be made arbitrarily small relative to the work done inside the capacitor, where the force is constant, simply by increasing the size of the capacitor.
[1] http://student.ndhu.edu.tw/~d9914102/Teaching/EM/Paper/data/...
What's the dimension that you're proposing to increase of the capacitor? The total work done across the capacitor will be fixed regardless of distance across.
Since you insist: your derivation goes wrong right at the start by, as you say, "ignoring curvature effects", i.e. by considering radial distance only. By doing that, you are effectively imposing spherical symmetry; you are not doing parallel plates, you are doing concentric spherical shells. That makes the whole exercise pointless, since it's obvious from symmetry alone that such a device could never produce a net thrust: there is no preferred direction for the thrust to act along.
To answer your final question, just look at Eq. (19): make the plate radius (R) larger.
That answer should also be perfectly obvious from the limit case of infinite plates. A correct derivation for the general case must reproduce that result in that limit. Yours does the opposite; it gets worse the larger you make the capacitor. In the infinite limit, it is infinitely wrong.
You're also right that all that is moot from sheathing. But an ion still begins it's journey out at the bottom of a large potential well; one which is particularly steep because of the debye length, but still just as deep.
Cathode Ray tube comparison below isn't entirely valid, as the goal of a CRT isn't to generate net thrust.
Does this follow? I thought the force of an E-field on a particle is proportional to charge and field strength-- so the electrons are lighter, but will be accelerated with the same force to a much higher velocity. So wouldn't the reactive force be the same for both?
An electron will shoot through that distance in no time flat and accrue very little momentum. A proton will plod through and transfer far more.
For a 250V potential difference with 20m between screens, you can check it yourself here:
https://www.wolframalpha.com/input/?i=(1+electron+charge+%2F...
https://www.wolframalpha.com/input/?i=(1+electron+charge+%2F...
The ratio of those two values is 42.8.
This also makes it easy to see why heavier ions full of neutrons are _awesome_. They give a ratio of 5343.
W = F * d / q
where
d = distance from the electron's start position to the positive screen
q = the electron's charge
All this work is now kinetic energy carried by the electron, so we also have
W = (m_e/2) * v_e^2
where
m_e = mass of the electron
v_e = final speed of the electron
Therefore,
v_e = sqrt( 2 * F * d / (m_e * q) )
If we repeat the experiment with a proton, it accelerates in the opposite direction, toward the negative screen, and its mass m_p is larger, but other than that it all goes as with the electron; the absolute values of q and F (which only depends on electric field strength and q) are the same. So,
v_p = sqrt( 2 * F * d / (m_p * q) )
Divide the two equations to get
v_e / v_p = sqrt( m_p / m_e )
Momentum is conserved, so we also have
M * v + m_e * v_e - m_p * v_p = 0
where
M = mass of the screens (and rest of the spacecraft)
v = speed gained by the screens (and rest of the spacecraft)
and the minus sign accounts for the opposite direction of travel of electron and proton. Use the expression for v_e / v_p to eliminate v_p, solve for v, and find
v = (sqrt(m_p / m_e) - 1) * (m_e / M) * v_e
In numbers, m_p ~ 1837 * m_e, so
v ~ 41.86 * (m_e / M) * v_e
and Zubrin's argument is essentially correct.
I am struggling to come up with an explanation as to why I haven't heard about this before other than sneaky DOD stuff.
Who would like to bet me a large sum of money that there are currently operation military sats employing this......
EDIT: as epicureanideal pointed out, commenters in the article raised valid points discrediting most of the author's analysis. +/- charged ions get same net kick in the field, comparing proton mass to electron mass is a red herring.
Overall concept still passes sniff test, just needs further analysis. It's going to be all about the angle - you need one ion species to spend more time in the field than the other to get a net thrust
A magnetic scoop with a normal ion thruster might be more practical.
EDIT 2: After further consideration, my concern here is what happens outside the plates. This works for infinite parallel-plate capacitors - you just need to have the field strong enough so one ion species undergoes more net acceleration. In practice the field outside the plates will not be zero and that could spoil everything
EDIT 3: I know nothing. E&M is hard. There will be a current around the outside as you're pumping positive ions to one side and negative to the other. This will induce a magnetic field affecting the ions being pumped through the field. I have no clue what the net effect will be.
Note: After reading post by nickparker here, note that most of the commenters on the linked website who were claiming mistakes with the math seem to be wrong. Nickparker here posted links to what seem to be valid calculations showing the math works out.
Seems like this would be a fairly straightforward proof of concept experiment to fly outboard on the ISS. Basically a couple of charged screens with force transducers on the screen mounts. It would seem this is within the bounds of Figure 1, The Dipole Drive Accelerating Within a Magnetosphere.
Wish this paper had been published a year or two before SpaceX launched Elon's Tesla. They could have tested a new interplanetary drive.