Explaining Walter Lewin's "Complete Breakdown of Intuition"

Question

This Walter Lewin lecture hinges on the solenoid producing a non-conservative field.

I read in the Feynman Lectures that "all the fundamental forces in nature appear to be conservative" (Vol 1, 14.5) which comes decades after Faraday's Law of Induction so surely he would have known about it.

1) Who is right? Or, what did Feynman really mean?

2) In Lewin's final diagram there are two voltmeters which, apart from a section of wire, attach in the same place. What happens as you bring the two attachment points together to meet? I can't see, in a physical on-the-table experiment, how one voltmeter is "measuring in the other direction".

Answer 1

Feynman meant that conservation of energy always holds, so that if you have a static situation, the force field on a particle is conservative. For magnetic forces, you have moving (and changing) currents in the solenoid, so its not static, and if you extract energy from the field, you just weaken the current and extract energy from the system producing the field, doing work on it.

The fact that magnetically induced EMF is non-conservative is the basis of countless claims of perpetual motion machines, so it is good to say early that you can't do this.

Magnetic fields that are changing give rise to non-conservative forces, the integral around a loop is the change in flux inside, but the process of extracting energy from the EMF reduces the magnetic field, and the amount of energy stored in it.

Feynman discusses transformers and the EMF around a loop. He also discusses something else even more counterintuitive and not at all discussed by other people. He shows two moving charges, A and B, so that A is moving perpendicular to the line joining A and B and B is moving along the line joining A and B.

In this case, the force from A on B is not equal and opposite to the force from B on A! This shows you that the (nonradiative) field is carrying momentum, and is transferring momentum to the two charges as the E and B fields rearrange. The recognition that you need to include fields in the conservation laws was long in coming, and this example is just as useful as the transformer for explaining this. Feynman also discusses a case where the field is carrying angular momentum, a collection of charged balls with a current, and when you switch off the current, the balls start to rotated around.

Answer 2

Regarding part 2) of your question:

In elementary circuit theory, you learn that when the leads of a voltmeter are connected together, the (ideal) voltmeter will read zero volts. This is true as long as there isn't a changing magnetic flux through the surface bounded by the leads.

So, to simplify the example in the video, imagine connecting the leads of the voltmeter on the right together with the leads enclosing the solenoid. In this case, the voltmeter will read the emf of 1V.

Now, connect the leads of the voltmeter on the left together with the leads enclosing the solenoid. You'll find that this voltmeter reads -1V.

Why? Because the direction around the loop is opposite that of the voltmeter on the right. To find the emf, we integrate the E field around a closed loop. Assuming the direction is into the positive terminal of the voltmeter, we integrate clockwise for the voltmeter on the right and counterclockwise for the voltmeter on the left. This is the reason the two voltmeters read equal but opposite values.

Answer 3

The title of your question, it's a "“Complete Breakdown of Intuition”, refers to the reaction of EE and physics professors in Prof Lewin's department at MIT when he performed this experiment in front of them. Four minutes into part2 http://www.youtube.com/watch?v=1bUWcy8HwpM: "and some did not believe what they saw"... "some accused me of cheating on the demonstration".

The idea of a voltage between two points in a circuit only makes sense when magnetic fields are static because it's independent of the path taken. We can also say the voltage across the terminals of the voltmeter is the same as that across the ends of its leads from Kirchoffs law.

In non conservative circuits where there's changing magnetic fields from changing currents, the voltage between two points is path dependent and therefore meaningless. Neither can we say the voltage across the voltmeter terminals is the same as that across the end of its leads. We can only say the changing flux around the measuring loop induces an emf from Faraday's law, and therefore creates a current which creates a voltage across the resistance of the measuring terminals of the voltmeter, which is displayed by the voltmeter.

What's the voltage across the ends of an inductance L carrying a changing current i?

Most EEs and physicists would answer $Ldi/dt$. The true answer is close to zero volts because the inductance is constructed from a low resistance metal. The voltmeter across it displays the voltage $Ldi/dt$ dropped across its internal resistance from the induced current around the measuring loop.

All the above is explained more fully in:

What do ``voltmeters'' measure?: Faraday's law in a multiply connected region Romer, Robert H. American Journal of Physics, Volume 50, Issue 12, pp. 1089-1093 (1982).

A long solenoid carrying a varying current produces a time-dependent magnetic field and induces electric fields, even in the region exterior to the solenoid where ∂B/∂t and therefore curl E vanish. By paying attention to (a) what it is that a ``voltmeter'' measures and (b) the simplest properties of line integrals (e.g., under what circumstances the line integral of E is path independent), it is easy to use Faraday's law to predict the readings of voltmeters connected to various points in a circuit external to the solenoid. These predicted meter readings at first seem puzzling and paradoxical; in particular, two identical voltmeters, both connected to the same two points in the circuit, will not show identical readings. These theoretical predictions are confirmed by simple experiments.

Answer 4

The "trick" in Lewin's lecture, almost a "fraud" I'd say is that he uses a common assumption from experience, namely that a conductor in a conservative electric field has the same potential at every point on it. We expect, for example the conductive wires of a voltmeter to have the same voltage on them at the meter as they do at the probes. We do not expect the voltage to change as we move along the wires. But in a non-conservative fields highly conductive wires actually become "batteries" producing a voltage difference between different locations.

Consider the wire in a transformer. It is highly conductive copper. Yet when turned on there is voltage that appears between the ends of the secondary that is NOT shorted out by the low resistance of the copper wire.

So how is a non-conservative electric field made? It's made from a changing current. About a current there is created a magnetic vector potential. That potential (called A) is in the same direction as the current that created it. This field penetrates conductors and is found is space about the current. And an electric field, a non-conservative electric field, is found when that magnetic vector potential is changing with time. Or dA/dt. And that electric field integrated over some path (Integral E dot dl) give the potential found between any two end points of that path. In effect the conductive wire becomes a battery with a potential difference between its terminals.

And that is the trick of Lewin's demo. He points to the wire connecting the resistors and says, Look, these two voltmeters are connected to the same point! And the casual observer falls for it assuming the potential at one end of the piece of wire is the same as at the other end. But it is not.

If Lewin's demo had been wired with transformer secondaries replacing the pieces of wires nobody would have been fooled. He'd say look both these voltmeters are connected to the same point, and you'd say, No they aren't! There is a transformer secondary which is a voltage source between the resistors and the junction so it's no wonder I get two different readings!

Answer 5

Instead of using voltage and voltmeters let's use just the fundamentals like Electric field, charges, and force. Get rid of the voltmeters and explain what the electric fields are inside resistor R1, R2 and the ideal wires. Before we inserted the current carrying wires, the electric field lines formed circular loops uniform across the loop. Once current carrying loop including R1 and R2 is introduced the electric fields change completely although the integral around the same circular loop is still equal to the rate of change of the magnetics flux through it. Explain why did the electric fields around the same circular loop changed. Is there a charge build up across the resistors that provide the field across the resistor? Is there any electric field across the ideal wire which will cause infinite current?

If you can explain the above, then it is simple to explain the voltage readings without reference to voltmeters. You can connect a capacitor across R1 and another across R2 and if they build up opposite and unequal charges I can see how the scope will show different voltages.

I know this is not really an answer and I should perhaps word it as a question. But it might help to answer this question as well.

Answer 6

Just two comments regarding Lewin's experiment:

1 -If he uses two voltmeters and puts their leads very close to each path (in order to measure the voltage across each path as he claims), then he will end up measuring the voltages on the resistors only, not the path between A and B. The path between A and B has induced voltages that will be cancelled by those induced on the voltmeter leads.

That is why when many other people put the voltmeter leads normal to the loop plane, they measure the "same" voltage difference between A and B.

2- the correct way to do the demonstration as he intended, is to remove the resistors and to use the voltmeter leads only. Just fix two points in air, and change the positioning of the voltmeter leads between them to show path dependency and the non-conservative nature in AC fields. Unfortunately, connecting the two points by any wire will build up a conservative voltage difference due to charge accumulation on the wire ends, which may deteriorate the results. The conservative counterpart can be demonstrated in the vicinity of some static charge to show the nature of the conservative field. Doing it using a DC circuit, then translating into AC case involving a voltmeter used out of its natural scope of usage, is very confusing even to the demonstrator himself.

In short, "Lewin does not measure what he claims to measure in the demonstration".

Answer 7

And... six years later I saw the video, but there is something about the interpretation that can't be right. Let's think that the two voltmeter are exactly in the same point, "red lead one" almost touching "red lead" of meter 2, and the same for black leads, even cross them if you want. How each meter is going to know what path it has to measure? They are not intelligent not talk to each other, so each should pick up a path just by chance. So, it would be expected to see 4 possible results: (path 1, path 1), (path 1, path 2), (path 2, path 1), (path 2, path 2), or even worst, they could be changing from one path to the other, why not?. Reality is not like that, so there is something wrong with the explanation. Each piece wire provides electrons to be accelerated by the induced electric field, so, assuming that the resistors are physically small (and are not good conductors), all the accelerated electrons are coming only from the wire, each inch count, so the "misisng volt" has to be on the wire between the lead from the meters. I bet that if a third meter were installed between the "plus or red" leads of the meters one and two (assuming negative been in touch), we were able to read it.