#562 – Electroboom!

[Sredni]

"A toroidal vacuum chamber encircles the core of a large magnet. The magnetic field is produced by pulsed coils; the magnetic flux inside the radius of the vacuum chamber changes with time. Increasing flux generates an azimuthal electric field which accelerates electrons in the chamber.
In the absence of an air gap, there is little magnetic flux outside the core."

Emphasis mine, this is what I was referring to. It's the electric field the electrons interact with.

But it's not independent from the changing magnetic field. The way I see it is they are two manifestations of the same phenomenon. (Like in the propagation of an EM wave, it's not that the changing B field causes a changing E field that causes... ). But anyway, we can consider this just nitpicking.
It appears to me that you now understand Faraday's law in full.

I cannot help imagining Lewin stroking his hands with satisfaction while, in a black hood and cloak, reads your exchange with jesuscf and mutters "Well done, my apprentice... Hihihihi.... Now, strike him down with your path integral!"

May the nonconservative force be with you.

[bdunham7]

IT DEPENDS ON THE PATH

Of course it does.  And if I were measuring an AAA cell and I looped one test lead through a core with dB/dt, I might not get 1.5V.  But the problem I posed specified two points out of any external fields.  The voltage across those should be unambiguous if your path is also out of those fields.  This was all in response to bsfeechannel's assertions regarding partial turns and the electroscope example was to show that there is an observable potential even without a closed path. 

Stop. This is wrong.
It is NOT the EMF. We always end up here: forgetting the role of the displaced charge. The EMF -  or better, the induced electric field Eind, is compensated in the conductor by the coloumbian field Ecoul. If there is a gap and no current is flowing, this is always exactly true.

Well, reading the first line of Wikipedia on 'EMF' I'm not sure they agree, but since you've been using E_ind and E_coul and those are very nice, clear terms, I'll use them instead and I'll avoid the term 'EMF'. 

So, it is not Eind that is acting on the carriers. It is the combined effect of the EMF (represented by Eind) and the displaced charge (represented by Ecoul).

To put this in context, I was comparing three phenomena- a conductor in a static, irrotational conservative field, a conductor in a rotational non-conservative field derived from dB/dt and a conductor moving across magnetic lines as in a generator, where the charges experience the Lorentz force.  So in the first case, I think we all agree with your assertion.  Apparently, and don't attribute this assertion to me, the third case does result in a potential gradient over the length of the conductor.  Is that right or wrong?  And if it is right, then how do you differentiate the effect of the Lorentz force from the local E_ind force in the second example?
[/quote]

As for the rest of what you responded to, you may have misunderstood--I was asking bsfeechannel if what I wrote was his understanding, and since you appear to disagree with the statement, if he disavows it as well then we're all on the same page.

We may be straying from the Lewin experiment a bit in these discussions but I think we need to in order to understand where the dispute lies--which, if you remember, was my original intent.  So if you have an open ended conductor and over some portion of that conductor you apply an electric field along it(E_ind, the charges instantly rearrange to oppose and cancel the field.  How is that different from inserting a battery in the conductor?  Won't the charges also rearrange to oppose the battery voltage?  Is the conductor equipotential?  And before you accuse me of being a KVLer or a battery moron, I'm not claiming battery equivalence and I do have a further point that I'm working on.

[rfclown]

...
The circuit had just two resistors and nothing else. And his model predicted exactly what happened in practice. KVL failed on the board. And then failed on the bench.
...

Wow. What a strange thread. It isn't "just two resistors and nothing else", it has a closed loop of wire in a varying magnetic field. It has a transformer (as has been pointed out over and over here). If the model doesn't include the transformer, it is an incomplete model.

[bsfeechannel]

The core is split, it's two toroids which both carry about half of the total flux each. His "Lewin loop" is around the center column where the two toroids join, same as the yellow reference loop, they both experience the total magnetic flux. However, when he measures the voltage across the "red" parts, his probe leads and the red wire form a loop around one of the two toroids, encompassing half of the magnetic flux. So what he actually measures is not the voltage across that red wire, it is the EMF in his measurement loop, which is, since it experiences about half of the total flux, half of the voltage induced in his reference loop.

Exactly. What he had was NOT Lewin's setup. The meters that were measuring the voltages across the wires were forming additional loops around half of the magnetic field in the opposite direction. No wonder he got negative voltages.

PS: this would be a brilliant experiment: Use a split core with different permeability in each of the toroids, so that the flux is no longer evenly split. I wonder if the voltages would still add up

What is funny is that he was told by user Copernico Felinis to use a torus, but the poor bugger refused saying that the leads of the meters would develop a voltage that would mask the real voltages he thinks are in the wires.

This unfortunate experimenter doesn't know that if he measures the voltages the same way he did for the EI core transformer, they will actually add up to zero. Only that the bottom meter will exhibit zero volts, while the top meter will show 200mV, proving that the voltages are not in the wire.

[rfeecs]

a piece of wire in a circuit under the influence a varying magnetic field doesn't behave as zero ohm resistor but as non-ideal voltage source.

This is the crux of the argument.

Clearly the wire was breaking the law (KVL?) by operating under the influence. 

Is there voltage "in the wires"?  There is no electric field in the wires or tangential to the surface of the wires, so physics says the answer is no.

But there is a charge separation between the ends of the wires so something is going on.

One can make a lumped model by representing this charge separation by a battery in series with the wire.  Now with a lumped model, KVL works fine.

But if we try to measure the voltage across the wire, we get zero volts.  However we can get a voltage if we position the wires differently.  If we do it just right, then we can get the numbers that make KVL work again.

I'm sure people who make a lumped model know that there is no battery in the wire.  They are just making a model.  It isn't reality.

Forget the measuring equipment and just look at the loop by itself.  Calculate the voltage along the path of the circuit.  The voltage across the wires are zero.

Choose a different path and you may get a different answer.  Because voltage depends on path in this case.

[bdunham7]

Where do we go from here?

Why here, of course!



You tap into the Lewin ring and since the rotational E-field from the flux doesn't magically end at the ring, you need to run a wire out some distance perpendicularly to any E-field, so that there is no E_ind.  Once you are far enough that all fields are negligibly low, you then connect an electroscope.  You can then connect the electroscopes to points A, B, C and D and see what results you get.  What do you expect to see?

[bsfeechannel]

It isn't "just two resistors and nothing else", it has a closed loop of wire in a varying magnetic field. It has a transformer (as has been pointed out over and over here). If the model doesn't include the transformer, it is an incomplete model.

What transformer? Replace the solenoid by a moving magnet, or by the radiation coming from a radio station. The voltages in your circuit will not add up to zero. This was not only proven theoretically but practically. It is the principle behind the reception of every radio.

And we're not discussing the model. We are discussing if it is possible to measure two different voltages at the same two points by the same or different but identical meters on a circuit subjected to an externally generated varying magnetic field.

It is. Period.

[rfclown]

It isn't "just two resistors and nothing else", it has a closed loop of wire in a varying magnetic field. It has a transformer (as has been pointed out over and over here). If the model doesn't include the transformer, it is an incomplete model.

What transformer? Replace the solenoid by a moving magnet, or by the radiation coming from a radio station. The voltages in your circuit will not add up to zero. This was not only proven theoretically but practically. It is the principle behind the reception of every radio.

And we're not discussing the model. We are discussing if it is possible to measure two different voltages at the same two points by the same or different but identical meters on a circuit subjected to an externally generated varying magnetic field.

It is. Period.

What transformer? The transformer that is there. What is a transformer? Two coils with a coupled field. It is a circuit element. If you neglect a circuit element in a circuit diagram, it is incorrect. It is no more of an error than if there was a AC (or DC) source in the loop and you neglect to put it in the circuit diagram. It is obvious that this tread isn't going to convince anyone of anything.

[jesuscf]

I'm afraid this won't make your argument stronger. You really need to take geometry and the fields into account and where the magnetic flux "flows".

The core is split, it's two toroids which both carry about half of the total flux each. His "Lewin loop" is around the center column where the two toroids join, same as the yellow reference loop, they both experience the total magnetic flux. However, when he measures the voltage across the "red" parts, his probe leads and the red wire form a loop around one of the two toroids, encompassing half of the magnetic flux. So what he actually measures is not the voltage across that red wire, it is the EMF in his measurement loop, which is, since it experiences about half of the total flux, half of the voltage induced in his reference loop.

PS: this would be a brilliant experiment: Use a split core with different permeability in each of the toroids, so that the flux is no longer evenly split. I wonder if the voltages would still add up

Yes, you are right, in 'fromjesse' experiment a loop is formed with the bottom red wire with half the magnetic flux, but a loop is also formed with the top red wire with also half the magnetic flux! As a result the induced voltages due to the half magnetic flux in the red wires cancel each other!   As for the test leads of the voltmeter, a voltage is induced in one of the leads and an opposite voltage is induced in the other lead and they cancel each other as well!  The setup ends up measuring only the effect of the central core flux in the bottom red wire.

[rfeecs]

One can make a lumped model by representing this charge separation by a battery in series with the wire.  Now with a lumped model, KVL works fine.

In my mind where this modeling method breaks down badly is in the case of the uniform resistive loop.  Here the loop is just a high resistance wire, for example.

If the loop is uniform, then there is no static charge anywhere in the loop.  So where to put the battery?

You could just stick one battery at one spot in the loop.  To me, this is a half decent approximation.  It works everywhere except the section of the loop including the battery.

But people are modeling this as lots of little batteries alternating with lots of little resistors.  So the voltage through the loop goes up through the batteries and down through the resistors all around the loop.  So the voltage between any two points on the loop is zero.

At the least this model is utterly confusing.  The voltage between two points is zero?  But what happened to ohm's law?  V=IR?  What about the power?  The resistor is going to get hot, like an induction heater isn't it?  I thought P=(I squared)R, or (V squared)/R.  But there's zero voltage.  So where's the power coming from?  (I know, the batteries).  Not a very good model.

[bdunham7]

But people are modeling this as lots of little batteries alternating with lots of little resistors.  So the voltage through the loop goes up through the batteries and down through the resistors all around the loop.  So the voltage between any two points on the loop is zero.

At the least this model is utterly confusing.  The voltage between two points is zero?  But what happened to ohm's law?  V=IR?  What about the power?  The resistor is going to get hot, like an induction heater isn't it?  I thought P=(I squared)R, or (V squared)/R.  But there's zero voltage.  So where's the power coming from?  (I know, the batteries).  Not a very good model.

The problem with trying to model something you don't understand with elements that you do understand and think are equivalent is that if you don't understand it in the first place, how can you know that your model is actually equivalent?

And this issue doesn't just apply to the 'KVLers', although most of them seem messy and unclear, which is why I've devoted my arguments to the issue of what fundamental disagreement causes the two camps to differ.  Unfortunately I'm figuring this out as I go with a pretty rusty education and a bit of spare time. 

[rfclown]

One can make a lumped model by representing this charge separation by a battery in series with the wire.  Now with a lumped model, KVL works fine.

In my mind where this modeling method breaks down badly is in the case of the uniform resistive loop.  Here the loop is just a high resistance wire, for example.

If the loop is uniform, then there is no static charge anywhere in the loop.  So where to put the battery?

You could just stick one battery at one spot in the loop.  To me, this is a half decent approximation.  It works everywhere except the section of the loop including the battery.

But people are modeling this as lots of little batteries alternating with lots of little resistors.  So the voltage through the loop goes up through the batteries and down through the resistors all around the loop.  So the voltage between any two points on the loop is zero.

At the least this model is utterly confusing.  The voltage between two points is zero?  But what happened to ohm's law?  V=IR?  What about the power?  The resistor is going to get hot, like an induction heater isn't it?  I thought P=(I squared)R, or (V squared)/R.  But there's zero voltage.  So where's the power coming from?  (I know, the batteries).  Not a very good model.

This to me is one of the few posts in this thread to invoke some thinking, not just a reaction. What is being induced in the coupled circuit is a current, not a voltage. A better model might be a current source. You can't distrubute that along the wire length like you can a voltage source, but it removes the problem mentioned here with using the distributed voltage source model.

[bdunham7]

What is being induced in the coupled circuit is a current, not a voltage.

Really?  You have a rotational E-field that exerts a force on the charge.  There's no guarantee that they'll move and if there is another counterforce, they won't.  A current source implies that they move no matter what.

[jesuscf]

What transformer? The transformer that is there. What is a transformer? Two coils with a coupled field. It is a circuit element. If you neglect a circuit element in a circuit diagram, it is incorrect. It is no more of an error than if there was a AC (or DC) source in the loop and you neglect to put it in the circuit diagram. It is obvious that this tread isn't going to convince anyone of anything.

This is a nice summary of the state of this tread.

[rfclown]

What is being induced in the coupled circuit is a current, not a voltage.

Really?  You have a rotational E-field that exerts a force on the charge.  There's no guarantee that they'll move and if there is another counterforce, they won't.  A current source implies that they move no matter what.

You may be right, I was just replying off the top of my head, but a current source model doesn't imply current. If I draw a current course with nothing attached to it, it is assumed that no current flows. If you get into a debate over that, it will be simillar to this thread.

[jesuscf]

At the least this model is utterly confusing.  The voltage between two points is zero?  But what happened to ohm's law?  V=IR?  What about the power?  The resistor is going to get hot, like an induction heater isn't it?  I thought P=(I squared)R, or (V squared)/R.  But there's zero voltage.  So where's the power coming from?  (I know, the batteries).  Not a very good model.

There is a very nice video explaining all of this from Bob DuHamel (RSD Academy) that I watched after following a link in a YouTube comment.  The video is not listed and I can not find it know.  As soon as I find it, I'll post the link here.

In an unbroken loop of wire under the influence of a varying magnetic field, each segment of arbitrary length of wire behaves as a voltage source in series with a resistance.  The induced voltage in that segment is consumed by the resistance in that segment.  When we add them up, the net contribution is zero.  That is why if we measure voltage (correctly) across two arbitrary points of the loop of the unbroken wire the voltage we get is zero.  If we open the loop, now the generated voltage at the terminals of the open loop doesn't match the consumed voltage (the current through the resistors is zero) and we get a net voltage output.

The same happens if you add a resistor between the open terminals: the net voltage drop inside the loop doesn't match the voltage generated and we get a net voltage across the external resistor.  If you make a closed loop out of resistors of the same value instead of wire, you'll observe exactly the same behavior!  The net voltage generated in each resistor is consumed by the voltage drop in the resistor and the net effect is zero.  In both cases, (wire loop and resistor loop) KVL always holds. 

Your observation of the loop getting hot is correct!  That is why is not a good idea to short circuit the secondary of a transformer!

EDIT: Here is the video I mentioned above.

[bdunham7]

You may be right, I was just replying off the top of my head, but a current source model doesn't imply current. If I draw a current course with nothing attached to it, it is assumed that no current flows. If you get into a debate over that, it will be simillar to this thread.

As we both know, a real-life current source will have a maximum compliance voltage above which it is no longer a current source.  However, if you are using an ideal current source in a theoretical model, the voltage goes to infinity and your model breaks, unless you are modeling an electric arc.  No assumptions should be made about a broken model, you need to revise it so that works or it is meaningless. 

There's a lot of broken models in this discussion, and even more messy ones.

[bsfeechannel]

What transformer? The transformer that is there. What is a transformer? Two coils with a coupled field. It is a circuit element. If you neglect a circuit element in a circuit diagram, it is incorrect. It is no more of an error than if there was a AC (or DC) source in the loop and you neglect to put it in the circuit diagram.

A transformer is a lumped component. It is a black box with three, four or more terminals. We know that we excite the primary and we get voltages on the secondary.

But, what is really going on inside it? Do the wires generate voltage? If they do, why is not possible to measure that voltage? Or better, why do we get an undefined measurement? If the wires don't generate voltage, what is in fact generating this voltage?

It is obvious that this tread isn't going to convince anyone of anything.

I haven't seen a single person that understood that the voltages are not in the wires and were convinced that they were. But seen lots who realized where the misconception is, changed sides and never came back.

[bsfeechannel]

Why here, of course!



You tap into the Lewin ring and since the rotational E-field from the flux doesn't magically end at the ring, you need to run a wire out some distance perpendicularly to any E-field, so that there is no E_ind.  Once you are far enough that all fields are negligibly low, you then connect an electroscope.  You can then connect the electroscopes to points A, B, C and D and see what results you get.  What do you expect to see?

Are you trying to measure the contribution of the displacement charges to the Coulomb field inside the wire?

[bdunham7]

Are you trying to measure the contribution of the displacement charges to the field inside the wire?

Sigh.  That is not my ultimate intent nor am I sure that this will do exactly that--although perhaps it boils down to the same thing.  This is not the last step.  So, do I see varying electroscope readings at A,B,C and D or are they all zero?

[jesuscf]

The core is split, it's two toroids which both carry about half of the total flux each. His "Lewin loop" is around the center column where the two toroids join, same as the yellow reference loop, they both experience the total magnetic flux. However, when he measures the voltage across the "red" parts, his probe leads and the red wire form a loop around one of the two toroids, encompassing half of the magnetic flux. So what he actually measures is not the voltage across that red wire, it is the EMF in his measurement loop, which is, since it experiences about half of the total flux, half of the voltage induced in his reference loop.

Exactly. What he had was NOT Lewin's setup. The meters that were measuring the voltages across the wires were forming additional loops around half of the magnetic field in the opposite direction. No wonder he got negative voltages.

bsfeechannel, in how many loops is the voltmeter circuit in the 'fromjesse' experiment as shown in the figure below?  (Tip: the answer is in a previous post)

[jesuscf]

"A toroidal vacuum chamber encircles the core of a large magnet. The magnetic field is produced by pulsed coils; the magnetic flux inside the radius of the vacuum chamber changes with time. Increasing flux generates an azimuthal electric field which accelerates electrons in the chamber.
In the absence of an air gap, there is little magnetic flux outside the core."

Emphasis mine, this is what I was referring to. It's the electric field the electrons interact with.

But it's not independent from the changing magnetic field. The way I see it is they are two manifestations of the same phenomenon. (Like in the propagation of an EM wave, it's not that the changing B field causes a changing E field that causes... ). But anyway, we can consider this just nitpicking.
It appears to me that you now understand Faraday's law in full.

I cannot help imagining Lewin stroking his hands with satisfaction while, in a black hood and cloak, reads your exchange with jesuscf and mutters "Well done, my apprentice... Hihihihi.... Now, strike him down with your path integral!"

May the nonconservative force be with you.

Really?  In the attached diagram courtesy of bsfeechannel, in how may loops is the voltmeter?  Is there some EMF cancellation going on?  Here is a tip: if you draw a circuit and KVL doesn't add up, then you are missing something in the circuit!

[thinkfat]

The core is split, it's two toroids which both carry about half of the total flux each. His "Lewin loop" is around the center column where the two toroids join, same as the yellow reference loop, they both experience the total magnetic flux. However, when he measures the voltage across the "red" parts, his probe leads and the red wire form a loop around one of the two toroids, encompassing half of the magnetic flux. So what he actually measures is not the voltage across that red wire, it is the EMF in his measurement loop, which is, since it experiences about half of the total flux, half of the voltage induced in his reference loop.

Exactly. What he had was NOT Lewin's setup. The meters that were measuring the voltages across the wires were forming additional loops around half of the magnetic field in the opposite direction. No wonder he got negative voltages.

bsfeechannel, in how many loops is the voltmeter circuit in the 'fromjesse' experiment as shown in the figure below?  (Tip: the answer is in a previous post)

In one loop only. No matter how you see it. Especially if you argue that the wire is a voltage source - the wire will "enforce" a voltage across itself and anything happening in any other part of the circuit will not matter.

There is no other path going through the volt meter, and if the magnetic flux wasn't evenly distributed between the two toroids, you would even measure different voltages "across" each of the red wires.

[bsfeechannel]

Sigh.  That is not my ultimate intent nor am I sure that this will do exactly that--although perhaps it boils down to the same thing.  This is not the last step.  So, do I see varying electroscope readings at A,B,C and D or are they all zero?

I'm not sure, so I'll bet they will show different readings.

[Sredni]

IT DEPENDS ON THE PATH

Of course it does. 

But then why do you ask, and i quote, "Is there potential across the ends or not?" without specifying the path?
Anyway, my yesterday answer was focused on the the long straight conductor through the center of the torus. For some reason (well, I had your sphere machine in mind) I thought you were talking about that all along and that the long U shaped path was just another question at the end (I even skipped it as non related). So, there has been a misunderstanding but what I wrote in my previous answer still applies. It is nevertheless clear that with that U shape the "potential across" meant the voltage on a path joining them in the proximity of the terminals. Sincere apologies for this misunderstanding on my part.

I nevertheless confirm that you should not use the word potential and in general - and even in this case because there are paths that can be run around the core.

The pictures I had drawn yesterday are mainly centered on the partial turn that according to some represent the 'elementary battery', but at this point I might as well post them. The second one was about the nearly complete but still partial turn and that is basically the same as your U shaped turn.
What I wrote in the previous answer still apply - but I would have worded it in a different manner. Can I blame the tequila?

Ok, here is the partial turn and some paths


https://i.postimg.cc/jqJ7FwKh/screenshot.png

and here is still another partial turn that runs almost - but not - completely around the core. You can consider this a representation of your long U coil.


https://i.postimg.cc/QN5Fq2k4/screenshot-2.png

As you can see it all depends on the path. And the need for a closed (mathematical, non necessarily material) path is that it's the only kind of path that allows me to apply Faraday's law to get the answer. We already discussed that the local form of Maxwell equations doesn't really change this, because they are partial differential equations that requiire boundary conditions. (Also look up at the definition of curl and divergence and you will see that surfaces and boundary paths, volumes and boundary surfaces are built in into the equations.)

Now, let me answer your question about the U coil. You ask about the electrometers at the end and I have seen the experiment performed with FET electrometers. They put them near an outlet and they sense the (alternating) accumulation of charge at the terminals. The core is in the transformer on a pole near the house in the US, or even a mile away in a big transformer that delivers power to whole city blocks in some parts of the rest of the world. So, I'd say that with a linearly increasing excitation you will see the leaf open - for as long as you can ramp up the current in the primary.

What I am perplexed about is your fixation with 'where there are no fields'. If the charges are there, there is at least their field, so you can't do away with it. If there are no fields to push and keep them there, then the charges are no reason to accumulate there.

To put this in context, I was comparing three phenomena- a conductor in a static, irrotational conservative field, a conductor in a rotational non-conservative field derived from dB/dt and a conductor moving across magnetic lines as in a generator, where the charges experience the Lorentz force.  So in the first case, I think we all agree with your assertion.  Apparently, and don't attribute this assertion to me, the third case does result in a potential gradient over the length of the conductor.  Is that right or wrong?  And if it is right, then how do you differentiate the effect of the Lorentz force from the local Eind force in the second example?

Well, yesterday I also did a couple of pictures for the straight rod near a variable magnetic flux region. They are no different from the ones above, I only separated the paths that give zero voltage, like these


https://i.postimg.cc/cJ2JRhKG/screenshot-4.png

from the ones that, by linking the emf, give a voltage different from zero (and a multiple of the EMF)


https://i.postimg.cc/1zk9nzT9/screenshot-5.png

The second picture is the reason I wrote - adding in the edit: IN GENERAL

"and this is again (EDIT: in general) wrong for the reason above."

The paths above show that in general VBA can be nonzero.
Immediately after I added (emphasy added now):

"But, if you consider a region of space that contains all of your piece of conductor and none of the magnetic flux, like a cylinder around the rod in the middle of the torus, in that region of space all paths you can imagine will never enclose the magnetic flux region. in this sense you can consider it as equipotential."

And this is basically the situation depicted in the former picture. If I consider paths that do not run around the variable magnetic region. I can consider - AND PLEASE NOTE THAT I AM PUTTING IT IN QUOTES - the conductor as 'equipotential', meaning that the voltage (difference) between any two points of the conductor (and in particular on the surface) is zero. This is apparently indistinguishable from the conductor polarized by immersion in the electrostatic field generated by two charged plates.

I believe that there could be a distinction, tho. It's in the nature - I would say 'the shape'  - of the field. In both cases, if you stay in your 'safe space' or 'magnetic free bubble' you can consider the conductor 'equipotential' (THE QUOTES! THE QUOTES!!!), but I seriously doubt that you could manage to make the field lines (and presumably the distribution of charge on the surface) the same in both cases (maybe a few lines will be reasonable close but then the fields will differ). What I mean is that the shape of the field lines coming out of charged plates will be straight if the plates are parallel, with the field magnitude constant, or hyperbolic shaped if the plates are at an angle, but you probably won't be able to replicate both the shape of the lines and the way the field increase or decrease on large portions of space. (I am not 100% positive.)
This in turn means that even the distorted field will be different.

But from the point of view of an observer of a polarized body that is in a room next to the room where a giant solenoid is being powered, the voltage along any path he can devise inside the room will be zero.
Does this answer your question?

I am not going to touch the motional case even with a ten foot pole. We are already wasting too much space and time to address the Lewin ring and we would fall down a relativistic rabbit hole with no end. I suggest you look up Purcell for an introduction to that kind of stuff.