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Richard Feynman in Lectures on Physics Vol. II Sec. 17-4, "A paradox," describes a problem in electromagnetic induction that did not originate with him, but which has nonetheless become known as "Feynman's Disc Paradox." It works like this: A disc (Feynman's spelling) that is free to rotate around its axis has set of bead-like static charges near its perimeter. The disc in it also has a strong magnetic field whose north-south axis is parallel to the rotation axis of the disc. The disc with its embedded static charges and magnetic field is initially at rest.

However, the magnetic field is was generated by a small superconducting current. The disc is permitted to warm up until the magnetic field collapses.

The paradox is this: Conservation of angular momentum says that after the field collapses, the disc must of course remain motionless. However, you could also argue that since the collapsing magnetic field will create a strong circular electric field that is tangential to the perimeter of the disc, the static charges will be pushed by that field and the disc will necessarily begin to rotate.

Needless to say, you can't have it both ways!

Feynman, bless his heart, seemed to have an extraordinarily optimistic view of the ability of others to decipher some of his more cryptic physics puzzles. As a result, I was one of many people who years ago discovered to my chagrin that he never bothered to answer his own question, at least not in any source I've ever seen.

In the decades since then, that lack of resolution has produced a surprisingly large number of published attempts to solve the Feynman Disc Paradox. Many of these are summarized in a paper that was written and updated a decade ago by John Belcher (MIT) and Kirk T. McDonald (Princeton) (Warning: I can see the paper, but it may have access restrictions for others.)

My problem is this: I more-or-less accidentally came up with what seems to be a pretty good resolution of the paradox, and it ain't the one described in any of the papers I've seen on it. But I can't easily back off, because the solution is a bit too straightforward once you look at it the right way. I think!

I also think that Feynman's solution was very likely to have been relatively simple, and not some kind of tremendously detailed exercise in relativistic corrections. He was after all trying to teach freshmen, and he honestly seemed to think they would all figure it out with a bit of thought!

So, help me out here folks: Does anyone know for sure what Feynman's solution to this little puppy was? Along those lines, is Laurie M Brown from Northwestern by any chance linked into this group? I can't imagine anyone who knows more about Feynman's published works!

I will of course explain why I think there's a simple solution, but only after seeing if there's something simple (but apparently hard to find) already out there.

I am always delighted when a question can be answered so specifically and exactly! @JohnMcVirgo uncovered the answer, right there in Volume II of the Feynman Lectures... only 10, count 'em 10, chapters later, in the very last paragraph of FLoP II 27, in Section 27-6 ("Field Momentum"), p 27-11:

Do you remember the paradox we described in Section 17-4 about a solenoid and some charges mounted on a disc? It seemed when the current turned off, the whole disk should start to turn. The puzzle was: Where did the angular momentum come from? The answer is that if you have a magnetic field and some charges, there will be some angular momentum in the field. It must have been put there when the field was built up. When the field is turned off, the angular momentum is given back. So the disc in the paradox would start rotating. This mystic circulating flow of energy, which at first seemed so ridiculous, is absolutely necessary. There is really a momentum flow. It is needed to maintain the conservation of angular momentum in the whole world.

Feynman hints at the above answer in earlier chapters, but never comes right out with a direct reference back to his original question.

John McVirgo, again, thanks. I'll review FLoP II 27 in detail before deciding whether to post that "other viewpoint" I mentioned. If Feynman already covers it, I'll add another addendum on why I think it's important. If the viewpoint is not clear, I'll need to do some simple graphics to explain how it may add some clarity to how the angular momentum conservation part works.

Well that was a short several weeks! @RonMaimon's additions to his initial answer, combined with his latest comment clarifying the difference between field momentum and "mechanical" momentum, demonstrate a deep understanding of the issues. Since @JohnMcVirgo already suggested the updated Ron Maimon text as the answer, I agree and have so designated it. I am still deeply grateful to John for pointing me to FLoP II 27, since without that clue I never would have found Feynman's answer in his own words.

I will at some point bring up my "other view" of Poynting problems as a new question. I now have two of those pending, since I am also still planning an updated Dual Cloud Chamber problem at some point.

• IIRC, Feynman proved that magnetic fields have momentum of both kinds. Let me see if I can dig up more... – Manishearth Jul 6 '12 at 1:03
• From the title, I thought you meant the other Feynman disk paradox. I hadn't heard this one before. It's certainly interesting. – Mark Eichenlaub Jul 6 '12 at 1:17
• FWIW: Prof McDonald never did put any access restrictions on his examples. So if one is having trouble accessing the link, it is usually something else at fault. – Willie Wong Jul 6 '12 at 8:19
• Hmm, I think that understanding how exactly the fields are made of photons can help here (it had helped me, but this was a while back and ill need to refresh myself) Photons carry momentum, EM fields made of photons. If you play with a field in a weird way, it will lose the photons and thus momentum. – Manishearth Jul 6 '12 at 18:38
• This is not an answer but details around the genesis of this problem in FLP Vol II Section 17-4. TA Welton, in his Memories of Feynman published in Physics Today, Feb 2007 details how he posed this problem to Feynman in Los Alamos which then shows up in Feynman's lectures. See pg.49 of that issue here: bit.ly/11gO5zv. – Abulalia Jul 25 '13 at 18:28

Conservation of angular momentum does not predict that the disk stays motionless, because the field in this case has angular momentum. The charges produce an electric field, and the magnetic field is not parallel to it, so there is a Poynting vector going around in circles, and the field angular momentum is just converted to mechanical angular momentum when the magnetic field disappears. The motion of the disk is required to conserve angular momentum, since the angular momentum would otherwise be radiated out in the radiative field when the magnetic field collapses, and in this case, some of the angular momentum is absorbed by the disk.

Feynman includes a version of this puzzle in the Feynman Lectures Vol II, and resolves it. The one thing he doesn't emphasize is that field momentum and angular momentum are conserved during slow changes without regard to the radiative field, which is not excited.

### EDIT: On Parity with B fields

Reading the comments, it seems that you are concerned that the disk spins one way with one sign of charge, and spins the other way with the other sign of the charge. This looks strange, because the B field is completely rotationally invariant around the Z-axis and so is the E-field (assuming the point charges are small and dense), and it looks weird that the thing can spin in one direction--- how does it know to go one way and not the other?

The reason is that when you are thinking about parity (why one direction and not the other), the B vector is unnatural. The B-vector has a right hand rule in defining how it is made and how it acts. It looks like it violates parity, but when you use the right hand rule twice (once to make B and once to make a force), the result is invariant under parity. But the pictures look like they give weird unphysical forces. This is true for all B-fields--- even the B field making circles around a current carrying wire. How does it know to go one way and not the other?

The easiest way to resolve this is to draw the B-field not as a vector, but as a little swish, a swirl, in a plane perpendicular to the B-field direction. You should think of B (for parity purposes) as really living in the plane perpendicular to the B-field, and swirling in a certain direction. For the case of the disk, the B field coming up out of the disk is not really coming up at all, it's coming out in a swoosh twirling counterclockwise in the plane of the disk. This is reflecting the physical motion of the charges in the plane of the disk that give rise to the B-field in the first place.

The swooshing of the B-field removes any confusion regarding the sign of the rotation of the disk. When you add an E-field from surrounding static point charges, you generate field angular momentum because the pointing vector makes the B-field swooshing into an actual momentum flow with a definite angular momentum. This is where the angular momentum to spin the disk comes from.

You asked for a reference in the comments. The reference is Feynman lectures Vol II, where he discusses a circle of charges making a ring around a wire, and uses it to motivate field momentum and angular momentum. I forget the details, but it's the same sort of puzzle. These things were discussed in the late 19th century, when field momentum was discovered. Maxwell, Hertz, Pointing, Lorentz and other contributed, but I didn't read this original literature, since resolving Feynman's puzzles using modern formalism gives you the content of this literature most quickly.

Feynman often gave puzzles of this sort to summarize old and forgotten literature for a modern audience, to keep it alive. This was a wonderful service he did to previous generations, and it is one of the reasons he is so revered not only as a researcher but as a teacher. The puzzles are each really deep questions of a previous generation of physicists.

• Ron, do you have a reference on where the resolution is that you mention? I'm familiar with his discussion of inertial and fields, but I never recall him saying point-blank that "it spins" or "it does not spin," and why. In fact, I recall being left with the distinct impression that Feynman was not entirely happy with his tentative resolution, which I agree seems to have been "it will spin." I played with that for some time and was left decidedly unhappy. If nothing else, it produces a really odd relationship between charge sign and angular momentum, one that seemed awfully arbitrary. – Terry Bollinger Jul 7 '12 at 15:20
• @TerryBollinger: There is no recent reference except Feynman Vol II, and Feynman knew the resolution for sure. The rotation isn't arbitrary--- the Pointing vector tells you which way it goes, and it doesn't break rotational invariance, because B is a pseudovector. You can think of what you are doing as transferring the angular momentum from the currents to the disk. Feynman is repeating famous puzzles of the late 19th century, which were resolved with the introduction of field momentum, but which everyone else, who doesn't know the history like Feynman, forgot about. – Ron Maimon Jul 7 '12 at 19:04
• It's interesting that Feynman himself is clear that it's not his puzzle, although in later literature it ends up getting referenced as his. I'll post something later this weekend about a somewhat different way to frame the question, one that I like because it seems (if I have it right) to give a cleaner visualization of how the conventional concept of angular momentum remains conserved. (No Poynting vectors will be harmed!) – Terry Bollinger Jul 7 '12 at 19:18

Chapter 17 precedes chapter 27 which covers field momentum and so he's looking for a simple explanation involving mechanical angular momentum. The initial angular momentum of the system is carried by the initial current in the coil, and so there's no paradox.

Note also that the magnetic field can't collapse immediately, but has to disspiate the stored magnetic energy into the resistance of the coil which will transfer the angular momentum of the current into the coil into the disk, causing it to rotate.

• John, thanks, I'll take a look. I don't recall what's in Vol II Ch 27 just from the number (but then I never do...). Update: Good ref!, reviewing it now... – Terry Bollinger Jul 7 '12 at 23:35
• Delightful!! I tip my hat to you sir, you found exactly what I was looking for and had never found: Feynman's own answer, point blank and without any ambiguity, to his version of the disc-with-charges paradox! Answer to you, and I'll add an addendum with Feynman's quote. All this time it was right there in my book, way off in another part of Volume II! – Terry Bollinger Jul 7 '12 at 23:49
• @TerryBollinger I didn't know Feynman stated the momentum of the field made up for the missing angular momentum in the disk problem in section 27-6. Since this comes later, I thought he was looking for a more mundane explanation as the angular momentum coming from the angular momentum of the moving charges in the current. I think you should remove my answer as being correct, and add your own answer based on that section, or make Ron's answer correct which is similar. However, the angular momentum of the current certainly contributes and I'm surprised Feynman doesn't mention this! – John McVirgo Jul 8 '12 at 13:35
• @JohnMcVirgo: The angular momentum of the current does contribute, but a negligible amount, as the mechanical momentum in current flows is always negligible compared to field momentum. The mechanical momentum energy is is never taken into account when you consider these things. – Ron Maimon Jul 8 '12 at 18:05
• @TerryBollinger: Yes, FLoP II is the meatiest stuff. It has a description of electromagnetic Green's functions, certain fluid flows, method of images for spheres (this is enough to induct how conformal maps work in general), and a lot of other things described in the optimal way. I think the index of refraction calculation is not great, and the derivation of the classical optics lens-law is not the best one (there are simpler demonstrations which are easier to carry in the head), but the intuition in these two is still important. – Ron Maimon Jul 8 '12 at 18:29

the current in the coil can hardly account for the momentum as the same current could be provided by opposite charges going in the opposite direction. These charges would have an opposite angular momentum to the first one, whereas they produce the same current. If you want to check out where go the whole momentum you should take a look at a paper called "hidden momentum", which involves relativity. However I think not the whole paper is correct, he still doesn't think about the whole system and miss some parts.

Returning to the Feynman paradox, you can see the situation described is half complete. One never ask how the charges came into the magnetic field, or how the magnetic field came onto the charges. Imagine the charged disc was already here, surrounding a passive coil with no current, then when one turn on the current, one creates a variable magnetic field, and thus an electric field, that's just induction, therefore the charges must move and the angular momentum of both field + charges equal 0 (radiations appart, but it's linked). Then if charges are at rest in a magnetic field, that just means they already dissipated the angular momentum they got from the induction.

I just stumbled upon this old post, and thought I'd share another answer I'm familiar with -- from Feynman himself. I really love how Feynman introduces big concepts early on, and keeps hinting at them so that when you finally get there, you're excited and prepared! In chapter 15 -- just before the paradox -- Feynman introduces the vector potential $$\mathbf{A}$$. This is basically what Ron Maimon was explaining above with the momentum of field and Poynting vector (I like to remember the magnetic field as a swoosh too, thank you!). But, now that I read it, Feynman seems to really drag himself through the approach of chapter 27 -- you can tell its not his preferred physical idea.

With the vector potential fresh in our minds, along with his hint to generalize to the idealized situation in the original problem statement, perhaps he was leading us here...

Chapter 21 The Schrödinger Equation in a Classical Context: A Seminar on Superconductivity

This is the last chapter of the whole dang thing, and a crown jewel at that. From what I've learned, superconductors were kind of frustrating for him not long before the Lectures time, and he was a little sore when the BCS theory came to light (it was such a nice solution perhaps he wish'd he'd thought of it himself).

Anyway, it seems like a crown jewel chapter not only because it's last, and a little more advanced explanation of things, but also because he gets to tie a lot of things together.

My understanding is that we can treat the system in an ideal sense, like a superconductor, and apply our results to single particles (or charged spots on a disc) affected by the turning on (or turning off) of $$\mathbf{A}$$ (or flux of $$\mathbf{B}$$ thru a surface). The whole chapter is a real treat so I wont go through it all here, but the gist is that the amplitude for a charged particle's location is changed in the presence of $$\mathbf{A}$$ exponentially, and this results in a change to the momentum operator in the Hamiltonian from $$\hat{p}=\frac{\hbar}{i}\nabla$$ to $$\hat{P}=\frac{\hbar}{i}\nabla-q\mathbf{A}$$. So a sudden change in $$\mathbf{A}$$ doesn't change the wave function immediately. With whatever momentum the particle started out with (say, $$mv=0$$), it must now move with an impulse (change in $$mv$$) equal to the vector potential times the charge in order for local conservation of momentum to hold during the brief time of collapse. That is, it's total p-momentum associated with $$\hat{p}=\frac{\hbar}{i}\nabla$$ becomes $$\mathbf{p}=m\mathbf{v}+q\mathbf{A}$$. This maintains the conservation of momentum locally, since $$\mathbf{P}=\mathbf{p}-q\mathbf{A}$$.

With the local conservation of probability density $$\psi\psi^\ast$$, if the probability density decreases in one place, it must increase in another. That is, there must be a probability density current! This is where the superconducting bit comes in. For a wavefunction describing a bunch of charged particles in a single state (like cooper pairs of a SC), $$q\psi\psi^\ast$$ describes a real electrical charge density. So a current of probabilty density is just a real electrical current density for a superconductor.

Bringing it all back, a solenoid made of superconducting wire (a superconducting ring) with a current flowing through it will have a flux threaded through opposing the current (really its the other way around -- the current is opposing the applied flux!). When the temp increases and the current in solenoid goes to zero due to restored resistance in wire, an electric field is generated around the solenoid by accelerating charges (spots on a disc) that were initially (if artificially) at rest, responding with a change in $$m\mathbf{v}$$ equal to $$q\mathbf{A}$$. So the disc rotates. Phew!

As you know that a changing magnetic field produces induced electric field which is tangential to the perimeter of the disc.So when the magnetic field collapses there should be an induced electric field.

Now since the initial angular momentum of the system is zero,and no external force is acting on the system,one should think that the final angular momentum of the system should remain zero but instead the disc rotates.This is because induced electric field is a part of electromagnetic wave which has both momentum and energy(A good example for this is that laser which is produced by stimulated emission of electromagnetic radiation is used to cut diamonds.This proves that em radiations have momentum).So as soon as the magnetic field collapses,the induced electric field transfers it's angular momentum onto the bead and the system starts rotating.

What if there were no beads on the perimeter,then the induced electric field would have transfered it's momentum to the surrounding medium particles and the energy would have dissipated.