# Does a Railgun's Current Violate Conservation of Momentum?

• I
• greypilgrim
greypilgrim
Hi.

I had a question about railguns, but I think I can formulate the underlying problem more clearly and concisely, hence I'm opening a different thread.
Consider the following rigid arrangement of three pieces of wire and two parallel capacitor plates:
There's an open switch somewhere in the wire and the capacitor is charged using an external voltage source, which is then removed. Then the switch is closed. This will start a current to flow in the direction indicated. The parallel pieces of wire both create a magnetic field that points downwards at every point of the transverse wire (maybe only approximately, since the wire pieces are finite). This causes a Lorentz force in the direction of the green arrow.

If I'm not missing anything, all other forces cancel due to the rigidity of the arrangement. This seems to violate conservation of momentum. I can see two ways out here:
1. The loop really start to move and there's opposite momentum carried away by the EM-field.
2. This arrangement reminds me suspiciously of the thought experiments that lead to displacement current. Since the electric field between the plates decreases, there is indeed displacement current there and it's directed opposite the current in the transverse wire.
Now, does Lorentz force also act on displacement currents? I have never heard of that, but that could potentially compensate the green force.
So, what's happening here?

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greypilgrim said:
At ##t=0##, the setup is at rest and the capacitor is charged. This will start a current to flow in the direction indicated.
If the capacitor is charged up, what causes a current to flow? Are you asking about a capacitor's discharge current? If so, what does this have to do with railguns?

Bystander
Oh, you mean the decreasing electric field inside the wire causes a displacement current there, that's exactly opposite the actual current?

So I guess you're saying the displacement current has exactly the same magnitude as the actual current, so there are no magnetic fields at all? I don't see why this should be the case. The discharge current is given by
$$I(t)=\frac{U_0}{R} e^{-\frac{t}{RC}}\ .$$
On the other hand, we have ##U=-\int \vec{E}\cdot d\vec{s}## and ##\vec{J}_D=\varepsilon_0 \frac{d\vec{E}}{dt}##.
Assume I make the wire a lot longer, but keep the resistance the same (different materials, cooling, etc.). Then the discharge current would be the same. But since ##U=-\int \vec{E}\cdot d\vec{s}## has to be the same as well but the integration path is much longer, ##\vec{E}## inside the wire must be smaller and its derivative too, no?

berkeman said:
If so, what does this have to do with railguns?
If that transverse segment of wire were instead a piece of conductive metal laid across the two parallel wires to make a connection between the two plates this would be a railgun - the crosspiece is the projectile. (we’d start it well before the ends of the parallel rails to give it a good long acceleration - and note that the faster it moves the greater the Lorentz force, which is what makes a railgun so much fun).

I think the question here is how a railgun recoils, although recoil it must.

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berkeman said:
If so, what does this have to do with railguns?
If you read my other, now closed thread, you'll see that I was asking why my suggested redesign of the projectile wouldn't significantly increase the Lorentz force. Then everybody was telling me that the magnetic fields created by those new parallel wires would cancel the fields of the rails.

This suspiciously sounded to me as if the projectile was causing a force on itself, violating conservation of momentum. Now I'm trying to recreate this as a thought experiment, leaving away everything external.

I thought you suggested in #2 that the magnetic fields created by "real" and displacement current exactly cancel, so there's no magnetic field at all, but I don't think that's true.

I'll rephrase the experiment:
1. Assume there's an open switch somewhere in the wire.
2. The capacitor is charged using an external voltage source, which is then removed.
3. The switch is closed. Current flows, the parallel wires create a magnetic field that is directed downwards at every point of the crosspiece (maybe only approximately, since the wire pieces are finite).
4. This causes the green Lorentz force which accelerates the projectile.
The question remains, how does this not violate conservation of momentum?

greypilgrim said:
If you read my other, now closed thread, you'll see that I was asking why my suggested redesign of the projectile wouldn't significantly increase the Lorentz force. Then everybody was telling me that the magnetic fields created by those new parallel wires would cancel the fields of the rails.

This suspiciously sounded to me as if the projectile was causing a force on itself, violating conservation of momentum. Now I'm trying to recreate this as a thought experiment, leaving away everything external.

I thought you suggested in #2 that the magnetic fields created by "real" and displacement current exactly cancel, so there's no magnetic field at all, but I don't think that's true.

I'll rephrase the experiment:
1. Assume there's an open switch somewhere in the wire.
2. The capacitor is charged using an external voltage source, which is then removed.
3. The switch is closed. Current flows, the parallel wires create a magnetic field that is directed downwards at every point of the crosspiece (maybe only approximately, since the wire pieces are finite).
4. This causes the green Lorentz force which accelerates the projectile.
The question remains, how does this not violate conservation of momentum?
If I understand your question correctly:
1) Imagine we have an ordinary person firing a gun. Does the gun recoil? Of course.

2) Imagine we connect the gun to a really strong brace attached to a building. What happens? The brace will transmit the reaction force to the building, which is attached to the Earth. The Earth is not attached to anything, so the Earth moves! By how much does it move? Not much. The mass of the Earth is huge so, by conservation of momemtum, its velocity after firing the gun will be virtually the same as it was before firing.

Clearly this is a little bit of an exaggeration: the soil around the building will shift, some momentum will be transfered to air molecules as sound, the ground vibrates, etc. But you could systematically remove all of these effects in an "ideal" situation, and we are left with moving the Earth, and there is nothing we can do to stop that effect.

This sort of thing might seem weird, but it happens all the time. No one really bothers to think of it, but every time you take a step, technically, you are exerting a torque on the Earth's surface and you are changing the angular velocity of the Earth (by an immeasurably small amount.) In other words, when you run across a field you are actually changing the length of a day! You never think about it because you will never notice the change.

If this still bothers you, put the whole thing in intergalactic space. What happens when you fire the railgun? Of course it recoils.

-Dan

I don't see the link to the problem in this post. There just isn't anything around, no matter how heavy, and the orange structure seems to accelerate without throwing away parts.

greypilgrim said:
The question remains, how does this not violate conservation of momentum?
Or equivalently, what is the Third Law partner of the force on the transverse segment?

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greypilgrim said:
I don't see the link to the problem in this post. There just isn't anything around, no matter how heavy, and the orange structure seems to accelerate without throwing away parts.
As I said, in the end, if you make this problem "ideal" it is the Earth itself that experiences the recoil. That's the "part" that is "thrown away."

Or, as Nugatory is asking, is your question more along the lines of "What force could cause the railgun to recoil if there if the "bullet" doesn't physically touch the gun?" In that case, have you ever heard of Lenz's Law?

-Dan

This isn't about railguns anymore. There is nothing else in the universe than the orange structure. I don't see how this is unclear in the thread start.

greypilgrim said:
It’s still about railguns because that orange structure is an idealized railgun floating in empty space.

Search the Wikipedia article on railguns for the word “recoil”, there are some possibly helpful references.

But the whole thing is rigid. The transverse piece is fixed to the parallel wires!

greypilgrim said:
But the whole thing is rigid. The transverse piece is fixed to the parallel wires!
Yes, but it’s same force that needs a third law partner and that third law partner is the same force as the recoil force in a railgun. Your rigid structure is an instantaneous snapshot of a railgun at the moment of firing before the projectile has started to move.

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topsquark
So, in short, what happens when the switch is closed?
• Does it accelerate?
• Or does it not?

greypilgrim said:
But the whole thing is rigid. The transverse piece is fixed to the parallel wires!

The question does seem to be what could cause a force on the railgun. Again, look up Lenz's Law. The bullet contains charges (else the railgun wouldn't work at all.) The paddles are accelerating those charges, so the accelerating charges, in turn, will produce a "back EMF" in your paddle system. That back EMF is what produces the force on the railgun.

For the record, if something of the sort didn't happen, then it would technically be Newton's 3rd Law that would be violated, not Conservation of Momentum (which is derived from Newton's 2nd Law.)

-Dan

greypilgrim said:
So, in short, what happens when the switch is closed?
• Does it accelerate?
• Or does it not?
It does not.

topsquark said:
For the record, if something of the sort didn't happen, then it would technically be Newton's 3rd Law that would be violated, not Conservation of Momentum (which is derived from Newton's 2nd Law.)
I have only ever seen Newton's 3rd law linked to conservation of momentum (and they are equivalent, at least for "simple" mechanical systems).
Of course you can derive from the 2nd law that for a single mass ##\vec{p}=\text{const}## if ##\vec{F}=0##, but conservation of momentum is usually formulated for total momentum, I can't see how you would incorporate several particles into the 2nd law.

Now, I said Newton's 3rd law and conservation of momentum are equivalent for "simple" mechanical systems. And apparently this is just not true for magnetism in general, as the following simple situation from StackExchange illustrates:

Clearly, only particle 1 feels a magnetic force, particle 2 does not. On StackExchange, this is explained by the magnetic field carrying momentum as well.

And this situation does not look totally dissimilar to my orange structure, where there are charges moving at right angles to each other as well.

Nugatory said:
It does not.
Thank you. Now, please help me understand this. Where in the sketch would you draw force vectors compensating the green vector? On what parts of the structure do they act? And by which fields are they produced?

topsquark said:
The paddles are accelerating those charges, so the accelerating charges, in turn, will produce a "back EMF" in your paddle system.
Even if there is a "back EMF", Lorentz forces can only act onto moving charges or currents. So where in this system would this back EMF create force, compensating the green arrow?

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greypilgrim said:
I have only ever seen Newton's 3rd law linked to conservation of momentum (and they are equivalent, at least for "simple" mechanical systems).
Of course you can derive from the 2nd law that for a single mass ##\vec{p}=\text{const}## if ##\vec{F}=0##, but conservation of momentum is usually formulated for total momentum, I can't see how you would incorporate several particles into the 2nd law.
Here's a brief derivation. You can look up the picky details on the net or a textbook.

Newton's 2nd, in the form that (I believe) he originally published it:
##F_{net} = \dfrac{dP}{dt}##

If we have several particles to the system, then ##F_{net}## is the total sum of the forces on each particle, and P is the sum of the momentum of all particles in the system.

In short, if ##F_{net} = 0##, then ##dP/dt = 0## and the net momentum of the system does not change.
greypilgrim said:
Thank you. Now, please help me understand this. Where in the sketch would you draw force vectors compensating the green vector? On what parts of the structure do they act? And by which fields are they produced?Even if there is a "back EMF", Lorentz forces can only act onto moving charges or currents. So where in this system would this back EMF create force, compensating the green arrow?
Did you look up Lenz's Law? See here. (Again, briefly) What happens is that the rail gun uses an electromagnetic field between the paddles to accelerate the charges in the bullet. But the charges in the bullet, in turn, create a new electromagnetic field by being accelerated, which by Lenz's Law (essentially a result of conservation of energy), acts in opposition to the electromagnetic field between the paddles. This, in turn, will act to force the paddles in the opposite direction of the motion of the bullet.

Unfortunately, Lenz's Law is a shortcut result. It's been measured, and all, but to really understand where it comes from you'd have to solve the Maxwell equations for the situation and, even if I could sit down and recall how to do it, I probably wouldn't be the best source to talk you through it. (It's due to the changing magnetic flux through a surface... Faraday's Law, if I remember correctly.) The best I can say (again, briefly!) is that if the bullet caused an EMF in the paddle system that pushed it in the direction the bullet was moving in, the system could theoretically be self-sustaining (and in fact push the bullet faster and faster) and that would violate conservation of energy. So, the Lenz's Law reaction force has to be in the opposite direction to the acceleration of the bullet.

I'm not sure why Nugatory is saying that nothing would happen when the switch is flipped but, then, I have not actually looked at the original thread, so I'll take his/her word for it. But if the bullet does actually move, the system will have a reaction force in the direction opposite to the direction of the acceleration of the bullet. To put it on your diagram, the force would be acting on the whole circuit, more or less, but if you had to put it in one place, I'd put it on the paddles.

-Dan

I do know Lenz's law, but I have used it so far only for external fields, not fields that are produced by the system itself.
topsquark said:
This, in turn, will act to force the paddles in the opposite direction of the motion of the bullet.
I assume by "paddles" you mean the capacitor plates? How would an electric field between them cause a force (there's nothing in between!), and why would it be directed opposite the green force? Electric forces are always directed parallel or antiparallel the field lines.

In #17, there was an example of a very simple system (two particles only) where conservation of momentum is violated if one ignores that the field can carry momentum. What makes you so sure that this doesn't happen here, that there's no exchange of momentum with the field?

topsquark said:
But if the bullet does actually move, the system will have a reaction force in the direction opposite to the direction of the acceleration of the bullet.
There's no bullet here - the transverse conductor is rigidly attached to the two parallel conductors so nothing moves and we have two equal and opposite internal forces.

topsquark
Nugatory said:
we have two equal and opposite internal forces
And where is the second one?

greypilgrim said:
And where is the second one?
Check out the links in the wikipedia article that I mentioned above. It is exactly the same force that would cause a recoil if the transverse conductor were free to move (and then we would have the apparatus moving in one direction and the transverse conductor moving in the other with equal and opposite momenta), exactly the same force that causes recoil in a railgun, and exactly the Lenz's Law force that @topsquark pointed out in posts 9, 15, and 18 (at least).

topsquark
greypilgrim said:
I do know Lenz's law, but I have used it so far only for external fields, not fields that are produced by the system itself.
Lenz's Law usually deals with a circuit with a loop of changing area. But the concept is easily extendable.
greypilgrim said:
I assume by "paddles" you mean the capacitor plates? How would an electric field between them cause a force (there's nothing in between!), and why would it be directed opposite the green force? Electric forces are always directed parallel or antiparallel the field lines.
A charged particle in an electric field experiences a force. Coulomb's Law.

I haven't looked at your original thread but, in general, there's an EM field between the plates of the capacitor (produced by a changing electric field) that acts on the charged particles in the bullet. Why shouldn't the accelerating charges in the bullet (which produce EM waves) produce a force on the panels of the capacitor?
greypilgrim said:
In #17, there was an example of a very simple system (two particles only) where conservation of momentum is violated if one ignores that the field can carry momentum. What makes you so sure that this doesn't happen here, that there's no exchange of momentum with the field?
First, why should we ignore that the field can carry momentum? If we ignore some feature of the problem it does not surprise me that we can draw incorrect conclusions. I have not heard of any (non-Quantum) system that can violate conservation of momentum. (And even in the Quantum case, it balances back out almost instantly.)

Second, we have far more than two particles here, we have the bullet and a circuit. The bullet moving past the paddles (capacitor plates) creates a change in the magnetic flux in the area the circuit encompasses and, by Lenz's Law, this affects the entire circuit, not just one charge in the circuit. You say you are familiar with Lenz's Law, so I encourage you to look at the derivation again, if you haven't already.

Like I said, a better analysis would be a full out derivation using the Maxwell equations. Lenz's Law provides a convenient shortcut, but if this is not convincing you then that's probably your next step.

-Dan

Addendum: You keep implicitly insisting that it should be possible for conservation of momentum to be violated. I (and some others) are saying "no." If you like, do a Journal search to find references where such forces were measured. (Stock Exchange is nice, but not a Journal.) But you could probably find references in just about any Junior year level Physics text that teaches Lenz's Law. I have no problem with trying to explain the reasoning, but you seem to be more interested in repeating your claim that momentum conservation can be violated, rather than asking questions about Lenz's Law. If you are going to keep insisting on that, please just tell me, and I'll go away quietly.

topsquark said:
You keep implicitly insisting that it should be possible for conservation of momentum to be violated. I (and some others) are saying "no."
Then you're arguing against Feynman, that is if you're saying Newton's 3rd law has to hold for the involved particles. There are magnetic situations – even very simple ones, check the link – where this is simply not true. But nobody questions conservation of momentum – I certainly do not! –, this all gets perfectly resolved if we also consider that the EM field itself carries energy and momentum as well (Poynting vector).

Something I already asked before: How do I know that my experiment is not a situation where some momentum runs off with the field, when such phenomena already happen with only two particles?

topsquark said:
I haven't looked at your original thread but, in general, there's an EM field between the plates of the capacitor (produced by a changing electric field) that acts on the charged particles in the bullet.
This isn't about my original thread and was never meant to be. If you just scroll up to the thread start you'll see that there's nothing inside the capacitor.
I opened this new thread to focus on that and I'd rather you didn't bring up railguns anymore. By the way, somebody included "rail gun" in the thread title, I didn't write that.

About Lenz's law: I have applied it lots of times before, but always in the form that a change in magnetic flux created a current that would both
1. create a magnetic field that opposes the change in flux
2. exert a mechanical force which opposes the motion
So, where is that current here?

About my general intention: I'm really not questioning mainstream physics in any way. But I'd just like to know how the mechanics exactly works – i.e., which fields and currents produce those forces – rather than referring to general principles like Lenz's law.

greypilgrim said:
Something I already asked before: How do I know that my experiment is not a situation where some momentum runs off with the field, when such phenomena already happen with only two particles?
greypilgrim said:
About my general intention: I'm really not questioning mainstream physics in any way. But I'd just like to know how the mechanics exactly works – i.e., which fields and currents produce those forces – rather than referring to general principles like Lenz's law.
I've already told you that as well. If Lenz's Law will not convince you, then you need the full mechanism of the Maxwell equations.

Clearly, you don't want to listen to me. Fair enough. I'm gone.

-Dan

greypilgrim said:
Something I already asked before: How do I know that my experiment is not a situation where some momentum runs off with the field, when such phenomena already happen with only two particles?
Yes, I think your notion that "momentum runs off with the field" is correct. I propose that momentum is conserved in your structure simply because it's essentially an antenna for emitting EM radiation that precisely balances any mechanical momentum imparted to the device. Your concept is similar to at least two conventional antennas (especially the MED):
@greypilgrim railgun-inspired antenna:

Magnetic loop antenna (MHz):

(https://www.angelfire.com/planet/sft_accommodation/magnetic_loop_antenna.htm)

Magneto-electric dipole antenna (GHz):

(https://www.researchgate.net/figure/Concept-of-magneto-electric-dipole-MED-antenna_fig5_284019198)

By inserting an open switch into your loop, the plates may be charged to high voltage, thereby establishing a large electric-field ##E## in the gap between the plates (that's the "electric dipole" portion of your antenna). Closing the switch rapidly discharges the plates by establishing a large current in the loop (the "magnetic dipole" portion), along with a concomitant magnetic field ##B## surrounding the loop. The whole configuration amounts to a paralleled capacitor (##C##) and inductor (##L##), oscillating at frequency ##\left(LC\right)^{-\frac{1}{2}}##, and damped by conductor- and radiation-resistance losses. These oscillating ##E## and ##B## near-fields develop into EM radiation in the far-field, and by Poynting's theorem said radiation carries exactly the energy and momentum necessary to compensate for any changes in the energy and momentum of your structure.

renormalize said:
I propose that momentum is conserved in your structure simply because it's essentially an antenna for emitting EM radiation that precisely balances any mechanical momentum imparted to the device.
So you're saying there is EM radiation emitted carrying momentum, but there's still no net momentum change of the device? So the EM radiation is emitted symmetrically such that the total momentum of the EM field is still zero?

Does that mean the Poynting vector in the top half of the picture is upwards and downwards in the bottom half? Because that's the only symmetry I could think of given the structure of the device.

But again, what makes you think that momentum of the device itself is conserved, while it is not in Feynman's example if just the particles are considered?

greypilgrim said:
So you're saying there is EM radiation emitted carrying momentum, but there's still no net momentum change of the device? So the EM radiation is emitted symmetrically such that the total momentum of the EM field is still zero?
Not at all. What I said was that the radiation compensates for any change in the momentum of the antenna. Most all antennas, including yours presumably, are non-isotropic and emit more radiation in some directions than others. So think of an antenna as a "photon rocket". If the radiation-pattern is unbalanced (like from a parabolic dish antenna) and the antenna-directivity points east, there will be a reaction force on the antenna pushing it west. But if the radiation pattern is balanced in every direction, like from a dipole antenna, there is no net force tending to push the antenna left or right, up or down, fore or aft:

Okay, then I misunderstood.

But then you're saying there IS a net force on the device, leading to an acceleration (contrary to what everyone else in this thread was trying to convince me about)? Is it in the direction of the green arrow?

greypilgrim said:
But then you're saying there IS a net force on the device, leading to an acceleration (contrary to what everyone else in this thread was trying to convince me about)? Is it in the direction of the green arrow?
I haven't said that either. You'd have to calculate the radiation-patterns and predict the radiated-power of the device over the entire frequency spectrum of current flow expected as the capacitor discharges and the current begins to oscillate and decay. At first glance that appears to be quite a formidable task. But you could check the literature for the similar MED antennas to get an idea of radiation patterns at the frequencies where they operate to see how directive they are.

But as a guess, more radiation may be emitted backward in the direction of the capacitor (since that portion is a break in the magnetic-dipole loop), than in the in the forward-direction of the green arrow. If that's true, your antenna would accelerate forward like a "photon rocket" until the energy originally stored in the charged capacitor (the "fuel") is expended and the radiation ceases, after which it would coast at constant speed. In other words, everything would happen in accord with Newton's laws.

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renormalize said:
You'd have to calculate the radiation-patterns and predict the radiated-power of the device over the entire frequency spectrum of current flow expected as the capacitor discharges and the current begins to oscillate and decay.
For the moment I'm focusing on the forces acting at very first moment when the left plate is positive and current starts to flow clockwise.

renormalize said:
In other words, everything would happen in accord with Newton's laws.
I'm sure it's in accord with conservation of momentum, but is there really a Newtonian 3rd law formulation (i.e. with forces) when it comes to photons? Anyway, that's just semantics.

renormalize said:
If that's true, your antenna would accelerate forward like a "photon rocket" until the energy originally stored in the charged capacitor (the "fuel") is expended and the radiation ceases, after which it would coast at constant speed.
Some energy will also dissipated as heat due to the resistance of the wire. Can there be made a a very, VERY rough estimate about the percentages of energy radiated and dissipated as heat in the limit ##t\rightarrow\infty##? Or are there calculations/measurements for real antennae?

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greypilgrim said:
If I'm not missing anything, all other forces cancel due to the rigidity of the arrangement. This seems to violate conservation of momentum
My point is that the device will not accelerate in any direction because the law of conservation of momentum cannot be violated. Even from Lenz's law and the Lorentz force, there is no sufficient reason to infer that the entire device is accelerated in any direction due to the forces exerted on the transverse wire.

Note that the wires in this structure are not infinitely long. The direction of magnetic field created by the current flowing through these wires slowly changes with distance and angle.

The direction of the magnetic field generated by the two wires connecting the two plates of the capacitor inside this approximately ring-shaped structure is downward, but this does not prove that this downward magnetic field will extend to evenly wrap the entire transverse wire.

If you think that's roughly the case, how do you prove that the magnetic field generated by the current in the transverse wire does not produce the same reaction force on the two wires connecting the capacitor?

alan123hk said:
The direction of the magnetic field generated by the two wires connecting the two plates of the capacitor inside this approximately ring-shaped structure is downward, but this does not prove that this downward magnetic field will extend to evenly wrap the entire transverse wire.
Not evenly, but enough to produce a force in the direction given. That's how railguns work.

alan123hk said:
If you think that's roughly the case, how do you prove that the magnetic field generated by the current in the transverse wire does not produce the same reaction force on the two wires connecting the capacitor?
Most forces cancel due to the (anti-)symmetry of the structure, since the currents in the parallel wires go opposite ways.

But yes, maybe the finiteness of the wires leads to non-cancelling contributions that add up go an exact counterforce. I might try and calculate that at some point, but I assume that could become very ugly.

May I suggest Feynman Lectures II 27 about field momentum. I will not try to recapitulate the master's explanation. As the slider accelerates it creates nonzero field momentum in its wake which then interacts with the rails and capacitor making them "recoil". It is left as an exercise for the interested student to verify Newton. I will rely on physics faith here.

vanhees71
Also one should note that, when taking into account field momentum to get the conservation laws right, you need to calculate everything relativistically. This gives rise to the somewhat unfortunate notion of "hidden momentum", though it's just momentum done relativistically right. There are nice examples in Griffiths's E&M textbooks and (imho even better) on McDonald's website:

http://kirkmcd.princeton.edu/examples/

The most simple "magnetostatic" example is

http://kirkmcd.princeton.edu/examples/penfield.pdf

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