Can Angular momentum travel down an electric cord?

[benj]

Do you recall the Feynman Paradox?

Here's the question. We all know that energy can travel down an electric
cord. The evidence is everywhere, The heating of irons, the pulling of
solenoids against springs, the lifting of elevators etc. And clearly when
there is linear motion with mass (mv) there is also momentum which is
conserved.

But Feynman's apparatus deals with ANGULAR MOMENTUM. It starts with
nothing moving and ends with the charged disk rotating. As he points out,
it appears that angular momentum is not conserved. But Feynman notes that
the missing angular momentum is in the EM fields about the apparatus.
Hence it IS conserved.

The question, though, is how did that angular momentum get into the
fields and did it come down the electric cord when the solenoid was given
a constant current?

[Timo Nieminen]

On Saturday, 18 May 2013 08:18:20 UTC+10, benj wrote:
> Do you recall the Feynman Paradox?

[...]

> The question, though, is how did that angular momentum get into the
> fields and did it come down the electric cord when the solenoid was given
> a constant current?

Start with the charges in place, and no current in the solenoid with no current. As you increase the current in the solenoid, there will be a torque on the charges. This is just the paradox run in reverse. 0 angular momentum in, L_f put into the field, and -L_f delivered to the rest of the world through the reaction forces resulting from the forces that act to keep the charges stationary.

Start with current in the solenoid, and charge the spheres. As you move the charge towards the spheres, there will be torque. These forces will result in -L_f transferred to the rest of the world.

But you can send angular momentum down a wire (or other type of wavguide, such as an optical fibre). Firstly, if you send energy down your waveguide, you are sending momentum as well. For most choices of origin of coordinate system, that means you have angular momentum (which will be orbital angular momentum). You can send angular momentum about an axis lying along the waveguide as well (e.g., using modes with azimuthal dependence exp(im*phi)).

[benj]

On Fri, 17 May 2013 17:34:55 -0700, Timo Nieminen wrote:

> On Saturday, 18 May 2013 08:18:20 UTC+10, benj wrote:
>> Do you recall the Feynman Paradox?
> [...]
>> The question, though, is how did that angular momentum get into the
>> fields and did it come down the electric cord when the solenoid was
>> given a constant current?
>
> Start with the charges in place, and no current in the solenoid with no
> current. As you increase the current in the solenoid, there will be a
> torque on the charges. This is just the paradox run in reverse. 0
> angular momentum in, L_f put into the field, and -L_f delivered to the
> rest of the world through the reaction forces resulting from the forces
> that act to keep the charges stationary.

OK. If you do this, then as you raise the current the charges spin. But
the experiment must start with the initial conditions that the disk is
not spinning. Therefore you reach in and grab the disk stopping it. But
you've just extracted angular momentum from the disk. And I might note
that if you run the experiment you'll find you left angular momentum in
the fields. This to me raises the interesting point (which has to do with
the question) When you raised the current in the coil, was angular
momentum conserved? What I mean is You start with zero angular momentum.
Then the disk spins which is Angular momentum. Also the fields contains
some (shown by the spin when the fields collapse). Hence one might
surmise that EQUAL amounts of angular momentum went into the fields and
the spin so that it stayed conserved at zero. Hence NO angular momentum
came in the wires to the solenoid!

> Start with current in the solenoid, and charge the spheres. As you move
> the charge towards the spheres, there will be torque. These forces will
> result in -L_f transferred to the rest of the world.

If I start with a current and no charges and then charge the spheres I
don't see where any torque comes from. If I have radial conductors
carrying charges to the spheres I don't think they'll turn.

> But you can send angular momentum down a wire (or other type of
> wavguide, such as an optical fibre). Firstly, if you send energy down
> your waveguide, you are sending momentum as well. For most choices of
> origin of coordinate system, that means you have angular momentum (which
> will be orbital angular momentum). You can send angular momentum about
> an axis lying along the waveguide as well (e.g., using modes with
> azimuthal dependence exp(im*phi)).

No question about sending energy and linear momentum down a wire. As for
angular momentum a wave guide is a good idea! I'm now wondering about
things like circular polarization etc. but for a lone wire guide, I'm not
so sure.

For example the Poynting vector for a positively charged wire carrying a
steady current is directed down the wire in space about that wire. For
angular momentum you'd have to have circulating Poynting energy as in the
Feynman example. Oddly this energy flows in the direction of the current
for a positively charge wire and opposite to the current for a negatively
charged one. It does not circulate about the wire.

[Timo Nieminen]

Yes. Again, this is just the original paradox in reverse (assuming the fields change slowly enough so that radiation can be neglected).

> > Start with current in the solenoid, and charge the spheres. As you move
> > the charge towards the spheres, there will be torque. These forces will
> > result in -L_f transferred to the rest of the world.
>
> If I start with a current and no charges and then charge the spheres I
> don't see where any torque comes from. If I have radial conductors
> carrying charges to the spheres I don't think they'll turn.

Direction of current, direction of magnetic field + right-hand-rule gives direction of force. Why don't you think this results in a torque about the solenoid axis?

> > But you can send angular momentum down a wire (or other type of
> > wavguide, such as an optical fibre). Firstly, if you send energy down
> > your waveguide, you are sending momentum as well. For most choices of
> > origin of coordinate system, that means you have angular momentum (which
> > will be orbital angular momentum). You can send angular momentum about
> > an axis lying along the waveguide as well (e.g., using modes with
> > azimuthal dependence exp(im*phi)).
>
> No question about sending energy and linear momentum down a wire. As for
> angular momentum a wave guide is a good idea! I'm now wondering about
> things like circular polarization etc. but for a lone wire guide, I'm not
> so sure.

If you can send momentum, you can send angular momentum. It's only angular momentum about an axis along the wire that presents any difficulty. This (usually, if not always) requires a multi-mode waveguide (note that a wire is just a simple waveguide).

> For example the Poynting vector for a positively charged wire carrying a
> steady current is directed down the wire in space about that wire. For
> angular momentum you'd have to have circulating Poynting energy as in the
> Feynman example. Oddly this energy flows in the direction of the current
> for a positively charge wire and opposite to the current for a negatively
> charged one. It does not circulate about the wire.

Stratton considers the problem in sections 9.15 and 9.16. In summary, there are a whole bunch of modes that carry angular momentum about the wire. For wires of realistic finite conductivity, everything except the lowest order TM mode is strongly absorbed by the wire, and this mode doesn't carry AM about the wire. For infinite conductivity (or a wire with suitable (including infinite) magnetic conductivity, like a superconductor), these modes can be practical.

Fundamentally, having circulation of energy about the wire is no problem. The optical equivalent in multi-mode fibres is easy enough (again, see Stratton). The practical optical analog in free-space is Laguerre-Gauss modes.

[benj]

On Fri, 17 May 2013 19:11:12 -0700, Timo Nieminen wrote:

> On Saturday, 18 May 2013 11:42:49 UTC+10, benj wrote:

>> Then the disk spins which is Angular momentum. Also the fields contains
>> some (shown by the spin when the fields collapse). Hence one might
>> surmise that EQUAL amounts of angular momentum went into the fields and
>> the spin so that it stayed conserved at zero. Hence NO angular momentum
>> came in the wires to the solenoid!
>
> Yes. Again, this is just the original paradox in reverse (assuming the
> fields change slowly enough so that radiation can be neglected).

Yes. So we agree that momentum distribution etc. ought to be the same
whether we go forward or run it in reverse. So does angular momentum
split keeping total angular momentum zero? (assuming no radiation)

>> If I start with a current and no charges and then charge the spheres I
>> don't see where any torque comes from. If I have radial conductors
>> carrying charges to the spheres I don't think they'll turn.
>
> Direction of current, direction of magnetic field + right-hand-rule
> gives direction of force. Why don't you think this results in a torque
> about the solenoid axis?

There is a torque but it's on the wires carrying the charge to the
spheres! So it doesn't do anything. Say you had an arrangement like a Ben
Franklin electrostatic motor where a spark jumps from the wires to the
spheres to charge them. Do they rotate? I don't think so. But as soon as
the charges put an E field in space, S circulates and the AM is now
located in space and not in the stationary charges, right? Which is
exactly the same situation we had when we spun up the charges when the
current reached its value and then we manually stopped the rotation
taking out the mechanical AM.

> If you can send momentum, you can send angular momentum. It's only
> angular momentum about an axis along the wire that presents any
> difficulty. This (usually, if not always) requires a multi-mode
> waveguide (note that a wire is just a simple waveguide).

The problem as I see it is that sending momentum implies sending energy
and that's OK. But angular momentum is momentum that is bent into a
circle. So somehow that implies geometry entering in. So a straight wire
has not circularity and that kills the possibility of angular momentum,
Feynman's thing has Poynting energy going around in a circle even with
static fields and that does seem to imply angular momentum.

>> For example the Poynting vector for a positively charged wire carrying
>> a steady current is directed down the wire in space about that wire.
>> For angular momentum you'd have to have circulating Poynting energy as
>> in the Feynman example. Oddly this energy flows in the direction of the
>> current for a positively charge wire and opposite to the current for a
>> negatively charged one. It does not circulate about the wire.
>
> Stratton considers the problem in sections 9.15 and 9.16. In summary,
> there are a whole bunch of modes that carry angular momentum about the
> wire. For wires of realistic finite conductivity, everything except the
> lowest order TM mode is strongly absorbed by the wire, and this mode
> doesn't carry AM about the wire. For infinite conductivity (or a wire
> with suitable (including infinite) magnetic conductivity, like a
> superconductor), these modes can be practical.

This is interesting. So if one regards a single wire as a waveguide
(transmission line) there are higher order modes that aren't seen in
practice with practical wires, but theoretically exist for more "perfect"
wires such as superconducting. I'll have to look into this more.

> Fundamentally, having circulation of energy about the wire is no
> problem. The optical equivalent in multi-mode fibres is easy enough
> (again, see Stratton). The practical optical analog in free-space is
> Laguerre-Gauss modes.

I have no problem with AM in optical modes, and waveguides also follow
quite reasonably, but this raises the whole problem to quite another
level as I was considering only slow moving charges and current in near-
static modes rather than high frequency radiation. (light certainly is
high frequency) Which as we said above, we ignore radiation, so then the
whole waveguide/optical modes thing becomes moot. So the whole question
then becomes just how does AM flow about EM situations going in and out
of fields and exchanging with mechanical AM in apparatus. Not so much as
RF but in slower DC static type situations.

[Poutnik]

benj posted Sat, 18 May 2013 04:17:53 GMT

> Feynman's thing has Poynting energy going around in a circle even with
> static fields and that does seem to imply angular momentum.
>

As photons carry angular momentum,
any EM interaction implies angular momentum,
even if overall change can be zero.


--
Poutnik

[Timo Nieminen]

On Saturday, 18 May 2013 14:17:53 UTC+10, benj wrote:
> On Fri, 17 May 2013 19:11:12 -0700, Timo Nieminen wrote:
>
>
>
> > On Saturday, 18 May 2013 11:42:49 UTC+10, benj wrote:
>
>
>
> >> Then the disk spins which is Angular momentum. Also the fields contains
>
> >> some (shown by the spin when the fields collapse). Hence one might
>
> >> surmise that EQUAL amounts of angular momentum went into the fields and
>
> >> the spin so that it stayed conserved at zero. Hence NO angular momentum
>
> >> came in the wires to the solenoid!
>
> >
>
> > Yes. Again, this is just the original paradox in reverse (assuming the
>
> > fields change slowly enough so that radiation can be neglected).
>
>
>
> Yes. So we agree that momentum distribution etc. ought to be the same
> whether we go forward or run it in reverse. So does angular momentum
> split keeping total angular momentum zero? (assuming no radiation)

Why not?

> >> If I start with a current and no charges and then charge the spheres I
> >> don't see where any torque comes from. If I have radial conductors
> >> carrying charges to the spheres I don't think they'll turn.
> >
> > Direction of current, direction of magnetic field + right-hand-rule
> > gives direction of force. Why don't you think this results in a torque
> > about the solenoid axis?
>
> There is a torque but it's on the wires carrying the charge to the
> spheres! So it doesn't do anything.

No, it's a torque acting on the charges as they travel to the spheres.

The surface charges on the wires will exert an opposing torque stopping the charges from gaining AM. There will be a reaction torque exerted by the charges on the wire, so the rest of the world exerts a mechanical torque to stop the wires from gaining AM. The reaction torque exerted by the wires on the rest of the world means that, in the end, the rest of the world gains the AM instead of the charges as they move to the spheres.

> Say you had an arrangement like a Ben
> Franklin electrostatic motor where a spark jumps from the wires to the
> spheres to charge them. Do they rotate? I don't think so.

Direction of motion of charges, direction of magnetic field + right-hand-rule gives direction of force. Isn't this azimuthal about the solenoid axis?

> > If you can send momentum, you can send angular momentum. It's only
> > angular momentum about an axis along the wire that presents any
> > difficulty. This (usually, if not always) requires a multi-mode
> > waveguide (note that a wire is just a simple waveguide).
>
> The problem as I see it is that sending momentum implies sending energy
> and that's OK. But angular momentum is momentum that is bent into a
> circle.

No. Consider mechanical angular momentum. Just L=rxp, with no requirements for anything to be bent in a circle. A planet orbiting the sun, and a planet in the same position travelling at the same speed in a straight line have the same angular momentum.

> Feynman's thing has Poynting energy going around in a circle even with
> static fields and that does seem to imply angular momentum.

Indeed. But we don't need anything going in a circle to have AM. We have cases where we have angular momentum about some axis with no azimuthal component of the Poynting vector (which will be spin AM).

> So the whole question
> then becomes just how does AM flow about EM situations going in and out
> of fields and exchanging with mechanical AM in apparatus. Not so much as
> RF but in slower DC static type situations.

Electric motors are fun. AM is transferred within the motor via evanescent waces (the different parts of the motor are in each other's near fields), but you still have the usual hbar per photon (or whatever higher integral amount with high-AM multipole fields). E = hbar*omega tells you that the motor should be more efficient at lower frequencies. Which we see in practice.

[Poutnik]

>
> On Saturday, 18 May 2013 14:17:53 UTC+10, benj wrote:
> > On Fri, 17 May 2013 19:11:12 -0700, Timo Nieminen wrote:
> >

> > ........................................So does angular momentum

> > split keeping total angular momentum zero? (assuming no radiation)
>

Same as linear momentum can split keeping total angular momentum zero.

E.g. a gun, a powder, tha gas and a bullet
have zero linear momentum before shooting and after it.


--
Poutnik

[Salmon Egg]

In article <e92dd2f9-c651-4b1f...@googlegroups.com>,

I usually do not bother reading posts by benj, but the problem of
angular momentum in EM fields has been on my mind for a while. There now
seems to be a flurry of papers talking about orbital and spin angular
momentum in the literature. I admit to not fully understanding the
subject. All this would be based on classical, not quantum, physics.

Even in TEM waveguide, you can have what I believe fits into spin
angular momentum. Think of a three-phase cable composed of three similar
circular and parallel conductors. The three symmetrical components,
positive, negative, and zero phase sequences would carry positive,
negative, and zero spins respectively. Think of how the electric field
vector twists down the length of the cable. Because of the low frequency
involved, such cables are not usually thought of as being waveguides.

In the more typical TE and TM waveguides, there are longitudinal
magnetic and electric fields as well. Along with transverse field
components required for a Poynting vector in the direction of overall
energy propagation. These transverse components combine with
longitudinal components to give localized momentum (Poynting vector)
traveling normal to the energy propagation. That is your orbital angular
momentum.

Field distribution in dielectric (optical) waveguides are even more
intricate. Also consider slow wave found on helical waveguides such as
used to interact with electron beams in traveling-wave tubes.

--

Sam

Conservatives are against Darwinism but for natural selection.
Liberals are for Darwinism but totally against any selection.

[Timo Nieminen]

On Saturday, 18 May 2013 19:08:57 UTC+10, Salmon Egg wrote:
>
> Even in TEM waveguide, you can have what I believe fits into spin
> angular momentum.

[...]

> Think of how the electric field
> vector twists down the length of the cable.

Just so.

> In the more typical TE and TM waveguides, there are longitudinal
> magnetic and electric fields as well. Along with transverse field
> components required for a Poynting vector in the direction of overall
> energy propagation. These transverse components combine with
> longitudinal components to give localized momentum (Poynting vector)
> traveling normal to the energy propagation. That is your orbital angular
> momentum.

Again, just so.

That's a good brief summary of the basics of how to carry spin and orbital AM about the propagation direction.

[benj]

On Sat, 18 May 2013 00:00:59 -0700, Timo Nieminen wrote:

> On Saturday, 18 May 2013 14:17:53 UTC+10, benj wrote:

>> There is a torque but it's on the wires carrying the charge to the
>> spheres! So it doesn't do anything.
>
> No, it's a torque acting on the charges as they travel to the spheres.

Torque acting on charges which act on wires. Ok?

> The surface charges on the wires will exert an opposing torque stopping
> the charges from gaining AM. There will be a reaction torque exerted by
> the charges on the wire, so the rest of the world exerts a mechanical
> torque to stop the wires from gaining AM. The reaction torque exerted by
> the wires on the rest of the world means that, in the end, the rest of
> the world gains the AM instead of the charges as they move to the
> spheres.

There are torques but no AM as V=0. So in reality there is no AM of
charges or of wires. The world gains nothing. the wires deliver no AM to

the spheres.

>> Say you had an arrangement like a Ben Franklin electrostatic motor
>> where a spark jumps from the wires to the spheres to charge them. Do
>> they rotate? I don't think so.
>
> Direction of motion of charges, direction of magnetic field +
> right-hand-rule gives direction of force. Isn't this azimuthal about the
> solenoid axis?

Yes. but that is when charges are trapped inside the wires. if the wire
ends just about the sphere, when the "spark" jumps charging it, it is
downward parallel to the magnetic field and hence experiences no force/
torque when it is free to move and be accelerated.

>> > If you can send momentum, you can send angular momentum. It's only
>> > angular momentum about an axis along the wire that presents any
>> > difficulty. This (usually, if not always) requires a multi-mode
>> > waveguide (note that a wire is just a simple waveguide).
>>
>> The problem as I see it is that sending momentum implies sending energy
>> and that's OK. But angular momentum is momentum that is bent into a
>> circle.
>
> No. Consider mechanical angular momentum. Just L=rxp, with no
> requirements for anything to be bent in a circle. A planet orbiting the

> sun, and a planet in the same position traveling at the same speed in a

> straight line have the same angular momentum.

Obviously the INSTANTANEOUS angular momentum of the two situations is
identical, but in one case the system of the orbiting planet has AM in
some value while the linear one does not... or does it? IF i integrate
the total angular momentum of the path of the linear planet, I get a
value for angular momentum. If I do the same thing from the other side of
the path I also get AM but negative. If I do it from a planetary frame I
get zero. Point is angular momentum (as I mentioned) depends on the
geometry of the situation. Which is why I say "bent into a circle" or in
your case part of a circle. And that raises the question is AM "real" or
just a mathematical construct?

>> Feynman's thing has Poynting energy going around in a circle even with
>> static fields and that does seem to imply angular momentum.
>
> Indeed. But we don't need anything going in a circle to have AM. We have
> cases where we have angular momentum about some axis with no azimuthal
> component of the Poynting vector (which will be spin AM).

I"m trying to avoid nuclear AM as well as radiation here. But the point
would be IF you can have AM without a circulating Poynting vector (and I
believe you can) then WHERE which is to say in which fields is the AM
stored when it is EM AM?

[benj]

On Sat, 18 May 2013 02:08:57 -0700, Salmon Egg wrote:

> I usually do not bother reading posts by benj, but the problem of
> angular momentum in EM fields has been on my mind for a while.

Just read posts by Timo and Poutnik and don't bother reading any of mine.

Simple solution.

[Timo Nieminen]

On Sunday, 19 May 2013 03:53:37 UTC+10, benj wrote:
> On Sat, 18 May 2013 00:00:59 -0700, Timo Nieminen wrote:
> > On Saturday, 18 May 2013 14:17:53 UTC+10, benj wrote:
>
> >> There is a torque but it's on the wires carrying the charge to the
> >> spheres! So it doesn't do anything.
> >
> > No, it's a torque acting on the charges as they travel to the spheres.
>
> Torque acting on charges which act on wires. Ok?

Why do you think a torque wouldn't do anything if it acts on the wires?

> > The surface charges on the wires will exert an opposing torque stopping
> > the charges from gaining AM. There will be a reaction torque exerted by
> > the charges on the wire, so the rest of the world exerts a mechanical
> > torque to stop the wires from gaining AM. The reaction torque exerted by
> > the wires on the rest of the world means that, in the end, the rest of
> > the world gains the AM instead of the charges as they move to the
> > spheres.
>
> There are torques but no AM as V=0. So in reality there is no AM of
> charges or of wires. The world gains nothing. the wires deliver no AM to
> the spheres.

V=0 is an approximation, made because the rest of the world is very massive, and has a very high moment of inertia. Just the same as in linear momentum problems where we assume that the rest of the world makes a perfect "momentum sink" without moving enough to have to include that velocity in our calculations.

Just as force is transfer of momentum, torque is transfer of AM. If you exert a torque T on the rest of the world for time t, you have transferred Tt of angular momentum. Not zero if T and t are non-zero.

> >> Say you had an arrangement like a Ben Franklin electrostatic motor
> >> where a spark jumps from the wires to the spheres to charge them. Do
> >> they rotate? I don't think so.
> >
> > Direction of motion of charges, direction of magnetic field +
> > right-hand-rule gives direction of force. Isn't this azimuthal about the
> > solenoid axis?
>
> Yes. but that is when charges are trapped inside the wires. if the wire
> ends just about the sphere, when the "spark" jumps charging it, it is
> downward parallel to the magnetic field and hence experiences no force/
> torque when it is free to move and be accelerated.

If you have gotten the charges to where they can jump downwards parallel to the magnetic field, then you have moved them (along wires, or otherwise) perpendicular to the field already.

What matters is how much angular momentum is transferred. If it's transferred via the wires whether the wires are directly connected to the spheres or not, what difference does it make?

> >> The problem as I see it is that sending momentum implies sending energy
> >> and that's OK. But angular momentum is momentum that is bent into a
> >> circle.
> >
> > No. Consider mechanical angular momentum. Just L=rxp, with no
> > requirements for anything to be bent in a circle. A planet orbiting the
> > sun, and a planet in the same position traveling at the same speed in a
> > straight line have the same angular momentum.
>
> Obviously the INSTANTANEOUS angular momentum of the two situations is
> identical, but in one case the system of the orbiting planet has AM in
> some value while the linear one does not... or does it? IF i integrate
> the total angular momentum of the path of the linear planet, I get a
> value for angular momentum. If I do the same thing from the other side of
> the path I also get AM but negative.

Yes. So? You'll notice that if you choose your original on the other side of the path, then the AM of the orbiting planet is no longer constant (since you ar now taking moments about a point outside the orbit, not about a point at the central star, and the star now exerts a torque on the planet). Where the paths coincide, the AM is still the same for both.

What do you mean by "integrate the total angular momentum of the path of the linear planet"? Why would you do such a thing? (Constant angular momentum, so if you integrate along the path, you get infinity. Useful?)

> If I do it from a planetary frame I
> get zero.

In both cases, for both the linearly-moving planet and the orbiting planet.

> Point is angular momentum (as I mentioned) depends on the
> geometry of the situation.

It depends on our choice of coordinate system.

> Which is why I say "bent into a circle" or in
> your case part of a circle.

It doesn't depend on this part of the geometry.

> And that raises the question is AM "real" or
> just a mathematical construct?

It's as real as linear momentum. Linear momentum is also a conserved quantity, and also dependent on our choice of coordinate system.

> >> Feynman's thing has Poynting energy going around in a circle even with
> >> static fields and that does seem to imply angular momentum.
>
> > Indeed. But we don't need anything going in a circle to have AM. We have
> > cases where we have angular momentum about some axis with no azimuthal
> > component of the Poynting vector (which will be spin AM).
>
> I"m trying to avoid nuclear AM as well as radiation here.

?? What nuclear AM?

> But the point
> would be IF you can have AM without a circulating Poynting vector (and I
> believe you can) then WHERE which is to say in which fields is the AM
> stored when it is EM AM?

... which brings us to the often-argued-here problem of the circularly polarised plane wave.

[benj]

On Sat, 18 May 2013 14:02:18 -0700, Timo Nieminen wrote:

>> There are torques but no AM as V=0. So in reality there is no AM of
>> charges or of wires. The world gains nothing. the wires deliver no AM
>> to the spheres.
>
> V=0 is an approximation, made because the rest of the world is very
> massive, and has a very high moment of inertia. Just the same as in
> linear momentum problems where we assume that the rest of the world
> makes a perfect "momentum sink" without moving enough to have to include
> that velocity in our calculations.
>
> Just as force is transfer of momentum, torque is transfer of AM. If you
> exert a torque T on the rest of the world for time t, you have
> transferred Tt of angular momentum. Not zero if T and t are non-zero.

Hmmm. You are right about this. The torque applied by the charges due to
the magnetic field and the current does indeed get absorbed by the
"world". But it doesn't get applied to the charges. They are stationary
afterwards. But Somehow there has to be a flow of momentum up the wires
to the charges which when charged clearly produce a momentum in the E x H
field of the charges plus the solenoid. So the question now is just what
is the "flow" of that momentum? Where did it come from and how did it
find it's way into the fields? Especially since the torque in the wires
is absorbed by the "world"?

> What matters is how much angular momentum is transferred. If it's
> transferred via the wires whether the wires are directly connected to
> the spheres or not, what difference does it make?

True but we just decided that the torque produced and the T-delta t
momentum was absorbed by the "world" and doesn't get to the charges or
their fields, right?

> What do you mean by "integrate the total angular momentum of the path of
> the linear planet"? Why would you do such a thing? (Constant angular
> momentum, so if you integrate along the path, you get infinity. Useful?)

no what I'm saying is that if I integrate v X R for the orbiting planet,
I get a constant value for it's total angular momentum which is the sum
of it's instantaneous momentum at any point in the orbit. The other
planet however has an instantaneous momentum that varies from zero (v
parallel with r) to some value and back to zero again. To So the total
momentum as it passes the observer is an integration of all instantaneous
values. Is that infinite? I doubt it's more than one orbit of the near
planet.

>> And that raises the question is AM "real" or just a mathematical
>> construct?

> It's as real as linear momentum. Linear momentum is also a conserved
> quantity, and also dependent on our choice of coordinate system.

Yes, but the conservation doesn't depend on coordinate system, does it?
The laws of physics can't depend on the coordinate system chosen.

[Timo Nieminen]

On Sunday, 19 May 2013 15:24:17 UTC+10, benj wrote:
> On Sat, 18 May 2013 14:02:18 -0700, Timo Nieminen wrote:
> >

> > [...] Just the same as in

> > linear momentum problems where we assume that the rest of the world
> > makes a perfect "momentum sink" without moving enough to have to include
> > that velocity in our calculations.
> >
> > Just as force is transfer of momentum, torque is transfer of AM. If you
> > exert a torque T on the rest of the world for time t, you have
> > transferred Tt of angular momentum. Not zero if T and t are non-zero.
>
> Hmmm. You are right about this. The torque applied by the charges due to
> the magnetic field and the current does indeed get absorbed by the
> "world". But it doesn't get applied to the charges. They are stationary
> afterwards. But Somehow there has to be a flow of momentum up the wires
> to the charges

No. The AM comes from the field (see below). Why would it need to come up the wires?

> ... which when charged clearly produce a momentum in the E x H

> field of the charges plus the solenoid. So the question now is just what
> is the "flow" of that momentum? Where did it come from and how did it
> find it's way into the fields? Especially since the torque in the wires
> is absorbed by the "world"?

See below.

> > What matters is how much angular momentum is transferred. If it's
> > transferred via the wires whether the wires are directly connected to
> > the spheres or not, what difference does it make?
>
> True but we just decided that the torque produced and the T-delta t
> momentum was absorbed by the "world" and doesn't get to the charges or
> their fields, right?

Go back and look at what exerts what torques on what. That is, what transfers what AM to what. As the charges move to the spheres, the field exerts a torque on them. That is, the field transfers angular momentum to them. But the charges will transfer an amount equal in magnitude and opposite in direction to the field. End result: if the rest of the world gains L angular momentum, the field gains -L.

> > What do you mean by "integrate the total angular momentum of the path of
> > the linear planet"? Why would you do such a thing? (Constant angular
> > momentum, so if you integrate along the path, you get infinity. Useful?)
>
> no what I'm saying is that if I integrate v X R for the orbiting planet,
> I get a constant value for it's total angular momentum which is the sum
> of it's instantaneous momentum at any point in the orbit.

Why integrate? Just find rxp = rx(mv). If we place the origin at the star, then rxp is constant. No integration needed.

> The other
> planet however has an instantaneous momentum that varies from zero (v
> parallel with r) to some value and back to zero again.

No. Just sit down and do the maths. As the angle between r and v becomes smaller, the magnitude of r becomes larger. Just as for the orbiting planet, rxp=rx(mv) is constant.

> To So the total
> momentum as it passes the observer is an integration of all instantaneous
> values. Is that infinite?

?? The trajectory is infinitely long, unless it starts and stops at some particular times. The AM is constant and non-zero - the integral will be infinite. Why would you integrate along it?

> I doubt it's more than one orbit of the near
> planet.

> >> And that raises the question is AM "real" or just a mathematical
> >> construct?
>
> > It's as real as linear momentum. Linear momentum is also a conserved
> > quantity, and also dependent on our choice of coordinate system.
>
> Yes, but the conservation doesn't depend on coordinate system, does it?

No, it doesn't. Not for linear momentum, and not for angular momentum.

> The laws of physics can't depend on the coordinate system chosen.

So we assume.

[benj]

On Sat, 18 May 2013 22:43:11 -0700, Timo Nieminen wrote:

> On Sunday, 19 May 2013 15:24:17 UTC+10, benj wrote:

> Go back and look at what exerts what torques on what. That is, what
> transfers what AM to what. As the charges move to the spheres, the field
> exerts a torque on them. That is, the field transfers angular momentum
> to them. But the charges will transfer an amount equal in magnitude and
> opposite in direction to the field. End result: if the rest of the world
> gains L angular momentum, the field gains -L.

Ok. Charges move to spheres. Magnetic field exerts torque on them. Wire
holds them back hence absorbs that torque. So you say that therefore the
AM must be coming from the magnetic field. But this can't be correct
because before we bring charges there is the field and it has ZERO AM
because there are no charges present! E x H = zero. But once charges are
there clearly now there is AM stored in the fields (both E and H) from
circulating S. Hence instead of REDUCING AM in field as you say the
opposite must be true, namely that somehow the act of bringing charges to
the spheres puts AM in the field. Charged spheres have no motion (or
torque) during all this so no AM in them until H collapses.

But an examination of the experiment seems to show that your explanation
gives torque in the correct direction namely that if bringing charge in
produced torque on the "world" clockwise, the field torque will be
counter clockwise and then when the solenoid current stops it appears as
if the spheres also move counter clockwise as expected. But the question
then just what "field" stores the AM? Either q or H alone doesn't do it.
Somehow it takes both q and H to get AM stored.

So it appears as if the act of bringing charges into a magnetic field
somehow stores AM in that combo. In other words to store AM in fields in
space always requires both E and H fields be present!

> No. Just sit down and do the maths. As the angle between r and v becomes
> smaller, the magnitude of r becomes larger. Just as for the orbiting
> planet, rxp=rx(mv) is constant.

Ah so! Forgot about R being larger at distance! And also forgot about
conservation of momentum requiring rxp = constant...DUH!

[Timo Nieminen]

On Sunday, 19 May 2013 16:36:58 UTC+10, benj wrote:
> On Sat, 18 May 2013 22:43:11 -0700, Timo Nieminen wrote:
>
>
>
> > On Sunday, 19 May 2013 15:24:17 UTC+10, benj wrote:
>
>
>
> > Go back and look at what exerts what torques on what. That is, what
>
> > transfers what AM to what. As the charges move to the spheres, the field
>
> > exerts a torque on them. That is, the field transfers angular momentum
>
> > to them. But the charges will transfer an amount equal in magnitude and
>
> > opposite in direction to the field. End result: if the rest of the world
>
> > gains L angular momentum, the field gains -L.
>
>
>
> Ok. Charges move to spheres. Magnetic field exerts torque on them. Wire
> holds them back hence absorbs that torque. So you say that therefore the
> AM must be coming from the magnetic field.

Not from the magnetic field. The electromagnetic field. You can't separate this in general into the energy/momentum/AM of the field of source 1, the field of source 2, etc. Note that the total EM field = field of source 1 + field of source 2 etc.; these add linearly. But the energy is not linear in the fields, so you can't assign it to the fields of individual sources (except in special cases). Likewise, momentum and angular momentum.

> But this can't be correct
> because before we bring charges there is the field and it has ZERO AM
> because there are no charges present! E x H = zero.

No problem at all. Just like linear momentum, angular momentum is a vector quantity. Start with zero, give +L to thing 1, field ends up with -L. L-L=0, as started with. A mechanical analog for linear momentum, the recoil of a firing gun, has been mentioned up-thread.

> But once charges are
> there clearly now there is AM stored in the fields (both E and H) from
> circulating S. Hence instead of REDUCING AM in field as you say the
> opposite must be true, namely that somehow the act of bringing charges to
> the spheres puts AM in the field. Charged spheres have no motion (or
> torque) during all this so no AM in them until H collapses.

Doesn't matter that the spheres don't move. What matters is that the charge moves (to get to them). What about torque on the charge as it moves? Total torque is zero. A non-zero torque is exerted by the field, and an equal and opposite torque is exerted by the wire on the charge. Likewise, a torque is exerted on the wire by the charge, and an equal and opposite torque is exerted on the wire by the rest of the world. The only things that have non-zero total torque acting on them as the charge moves in are the EM field and the rest of the world; thus these are the things that exchange AM (through the various intermediaries).

> But an examination of the experiment seems to show that your explanation
> gives torque in the correct direction namely that if bringing charge in
> produced torque on the "world" clockwise, the field torque will be
> counter clockwise and then when the solenoid current stops it appears as
> if the spheres also move counter clockwise as expected.

Doing this, the Feynman paradox reversed followed by the original paradox, shows that you store AM in the field. You put it in, and get it back out. We know that AM is conserved, so this should be unsurprising.

> But the question
> then just what "field" stores the AM? Either q or H alone doesn't do it.
> Somehow it takes both q and H to get AM stored.
>
> So it appears as if the act of bringing charges into a magnetic field
> somehow stores AM in that combo. In other words to store AM in fields in
> space always requires both E and H fields be present!

And what does it need to store linear momentum in an EM field? If you can store linear momentum, you can store angular momentum.

[benj]

On Sun, 19 May 2013 00:43:05 -0700, Timo Nieminen wrote:


> Not from the magnetic field. The electromagnetic field. You can't
> separate this in general into the energy/momentum/AM of the field of
> source 1, the field of source 2, etc. Note that the total EM field =
> field of source 1 + field of source 2 etc.; these add linearly. But the
> energy is not linear in the fields, so you can't assign it to the fields
> of individual sources (except in special cases). Likewise, momentum and
> angular momentum.

> And what does it need to store linear momentum in an EM field? If you
> can store linear momentum, you can store angular momentum.

This is really the problem, isn't it? Where exactly is the angular
momentum stored in fields? Note that energy can be stored in just an E or
H field alone. But angular momentum requires BOTH an electric AND
magnetic field. You have just sort of thrown the whole mess into a box
and called it the "electromagnetic field" which is certainly true, but it
avoids any explanation of what exactly acts on what etc.

Feynman says that Angular momentum is stored in the circulation of
Poynting energy, but one can show this isn't correct. For example, rework
Feynman's paradox slightly: Put a charged hoop inside a long solenoid.
Start the current and it rotates. OK. We've got the E field from the
charge and the B field from the solenoid. That gives a circulation of
energy around the hoop. All is well.

But now change that and put the hoop OUTSIDE the solenoid. Oops. Now
there is H inside the coil but no charge and q outside the coil but no H!
Hence no Poynting vector and no circulating energy! Yet the hoop still
rotates! So where is that angular momentum stored?

[Timo Nieminen]

On Monday, 20 May 2013 14:37:43 UTC+10, benj wrote:
> On Sun, 19 May 2013 00:43:05 -0700, Timo Nieminen wrote:
>
> > And what does it need to store linear momentum in an EM field? If you
> > can store linear momentum, you can store angular momentum.
>
> This is really the problem, isn't it? Where exactly is the angular
> momentum stored in fields? Note that energy can be stored in just an E or
> H field alone. But angular momentum requires BOTH an electric AND
> magnetic field.

Problem? Why is it a problem? From observation, the EM field can store energy, linear momentum, and angular momentum. Why single out AM as a mystery? We don't know any more about how the field stores energy or linear momentum.

Combine (a) Coulomb's law, (b) retardation, and (c) conservation of energy, momentum, and angular momentum, and the EM field behaves as it must.

> You have just sort of thrown the whole mess into a box
> and called it the "electromagnetic field" which is certainly true, but it
> avoids any explanation of what exactly acts on what etc.

What acts on what is described by the Lorentz force law, or by the Maxwell stress tensor if you prefer. Not an explanation, but a working model. We can test the quantitative models; much harder to test stories that we put behind them.

> But now change that and put the hoop OUTSIDE the solenoid.

You mean like Feynman's version?

> Oops. Now
> there is H inside the coil but no charge and q outside the coil but no H!
> Hence no Poynting vector and no circulating energy!

You mean an infinitely long solenoid (unlike Feynman's version)?

> Yet the hoop still
> rotates!

Does it? Show!

> So where is that angular momentum stored?

First, show that there is angular momentum. Then worry about this part.

[Larry Harson]

Don't forget the mechanical momentum in the current!

Feynman himself suggests his paradox is resolved by taking into
account the momentum in the field, but how much momentum is accounted
for by the current in the solenoid?

Regards, larry.

[Larry Harson]

On May 18, 3:11 am, Timo Nieminen <t...@physics.uq.edu.au> wrote:

> Stratton considers the problem in sections 9.15 and 9.16. In summary, there are a whole bunch of modes that carry angular momentum about the wire. For wires of realistic finite conductivity, everything except the lowest order TM mode is strongly absorbed by the wire, and this mode doesn't carry AM about the wire. For infinite conductivity (or a wire with suitable (including infinite) magnetic conductivity, like a superconductor), these modes can be practical.

I wonder what the history of this problem is?

And here was me thinking all these years it was Feynman's perceptive
genius that first discovered this paradox.

Regards, larry

[Salmon Egg]

In article
<bfb54fb3-34ae-41d7...@k3g2000vbn.googlegroups.com>,

I do not think Feynman, as good as he was, was first notice angular
momentum in the electromagnetic field. In Goldstein's Classical
Mechanic, extension is made to include electromagnetism in the
lagrangian. Vector potential, usually notated by a bold A enters into
this lagrangian, I do not remember the details. If the lagrangian does
not vary with a generalized coordinate, the conjugate momentum is
conserved. I am fairly sure that that extension was well known before
Goldstein's book became a standard. Many quantum mechanical hamiltonians
involve the use of vector potential when magnetic effects were studied.

[benj]

On Sun, 19 May 2013 21:57:46 -0700, Timo Nieminen wrote:

> On Monday, 20 May 2013 14:37:43 UTC+10, benj wrote:
>> On Sun, 19 May 2013 00:43:05 -0700, Timo Nieminen wrote:

>> This is really the problem, isn't it? Where exactly is the angular
>> momentum stored in fields? Note that energy can be stored in just an E
>> or H field alone. But angular momentum requires BOTH an electric AND
>> magnetic field.
>
> Problem? Why is it a problem? From observation, the EM field can store
> energy, linear momentum, and angular momentum. Why single out AM as a
> mystery? We don't know any more about how the field stores energy or
> linear momentum.

Right. So long as Aristotle said it, that should be enough. Science
really doesn't investigate "problems"! Science is there to tell everybody
how much you do know! Come on Timo.

> Combine (a) Coulomb's law, (b) retardation, and (c) conservation of
> energy, momentum, and angular momentum, and the EM field behaves as it
> must.

Sure. Why? What are the details? You do know what science does, right?

>> You have just sort of thrown the whole mess into a box and called it
>> the "electromagnetic field" which is certainly true, but it avoids any
>> explanation of what exactly acts on what etc.
>
> What acts on what is described by the Lorentz force law, or by the
> Maxwell stress tensor if you prefer. Not an explanation, but a working
> model. We can test the quantitative models; much harder to test stories
> that we put behind them.

What we are discussing here is indeed a "working model". That's the whole
point!

>> But now change that and put the hoop OUTSIDE the solenoid.
>
> You mean like Feynman's version?

No.

>> Oops. Now there is H inside the coil but no charge and q outside the
>> coil but no H! Hence no Poynting vector and no circulating energy!
>
> You mean an infinitely long solenoid (unlike Feynman's version)?

Yes. (or large toroid if you wish)

>> Yet the hoop still rotates!
>
> Does it? Show!

Transformer.

[benj]

On Mon, 20 May 2013 06:11:24 -0700, Larry Harson wrote:

> On May 17, 11:18 pm, benj <b...@iwaynet.net> wrote:
>> Do you recall the Feynman Paradox?
>>
>> Here's the question. We all know that energy can travel down an
>> electric cord. The evidence is everywhere, The heating of irons, the
>> pulling of solenoids against springs, the lifting of elevators etc. And
>> clearly when there is linear motion with mass (mv) there is also
>> momentum which is conserved.
>>
>> But Feynman's apparatus deals with ANGULAR MOMENTUM. It starts with
>> nothing moving and ends with the charged disk rotating. As he points
>> out,
>> it appears that angular momentum is not conserved. But Feynman notes
>> that the missing angular momentum is in the EM fields about the
>> apparatus. Hence it IS conserved.
>>
>> The question, though, is how did that angular momentum get into the
>> fields and did it come down the electric cord when the solenoid was
>> given a constant current?

Let me point out at this point that Timo put his finger on the excellent
point that the angular momentum does NOT come down the cord when the
magnet was powered up because we imagined a case where the magnet is
powered when the spheres are uncharged. Since there is no charge there
can be no angular momentum! (it takes charge AND magnetic fields) Then we
charge the spheres. But now there is charge (coming down the charging
wires) plus magnetic field (from the solenoid) The current due to the
movement of those charges Timo pointed out creates forces which as all
wires are radial creates a torque which creates an angular momentum. That
angular momentum as Timo noted is absorbed by "the world". And to
conserve angular momentum an equal but opposite amount must somehow go to
be stored in space or as we say in the "fields". That portion stored in
the fields is what comes back out and rotates the disk when the current
stops again conserving angular momentum.

The very interesting point Timo came up with here is that in this
situation, the angular momentum is NOT a result of the magnet charged
with current. But rather flows into the system as a result of charging
the spheres and draining away the mechanical half. The same idea would be
if you charged the coil with spheres already charged and let the disk
spin. And then reached in and stopped it. Your HAND would then represent
draining off mechanical angular momentum to "the world" and only the
stored field momentum would be left again giving the usual final result.
But in this case the angular momentum WAS a result of the current
charging the solenoid! But again total angular momentum in the wire is
zero.

But note that in either case angular momentum does NOT come "down the
wires" because while there is indeed angular momentum involved there are
two halves (mechanical and field) that are equal and opposite hence the
TOTAL angular momentum coming down the wires in any case is ZERO! In any
case it is remarkable that energy with zero angular momentum in these
cases, sort of like "pair production", can split itself into to two equal
and opposite pieces that can be dealt with separately.

> Don't forget the mechanical momentum in the current!
>
> Feynman himself suggests his paradox is resolved by taking into account
> the momentum in the field, but how much momentum is accounted for by the
> current in the solenoid?
>
> Regards, larry.

Virtually none because the drift velocity of the electrons is about
walking speed! Might be different in an electron beam.

[Timo Nieminen]

On Monday, 20 May 2013 23:11:24 UTC+10, Larry Harson wrote:
>
> Don't forget the mechanical momentum in the current!

Negligible. There will also be momentum due to transport of energy along the wire, if the coil has resistance, or if there is a resistive load.

Drift velocity of about 1mm/s, about 10^19 electrons per mm per amp, so 10^22 electrons moving at 10^-3 m/s for a 1 metre long wire, and 10^-30 kg/electron makes 10^-11 N.s of momentum.

If we have a 1 ohm resistive load, with 1A, we need to deliver 1W. The momentum flux through a cross-section will be 1W/c = 10^-9. Momentum along a 1m wire will be this divided by c, for a total of 10^-17 N.s.

[Timo Nieminen]

On Tuesday, 21 May 2013 04:46:14 UTC+10, benj wrote:
> On Sun, 19 May 2013 21:57:46 -0700, Timo Nieminen wrote:
> > On Monday, 20 May 2013 14:37:43 UTC+10, benj wrote:
> >> On Sun, 19 May 2013 00:43:05 -0700, Timo Nieminen wrote:
>
> >> This is really the problem, isn't it? Where exactly is the angular
> >> momentum stored in fields? Note that energy can be stored in just an E
> >> or H field alone. But angular momentum requires BOTH an electric AND
> >> magnetic field.
> >
> > Problem? Why is it a problem? From observation, the EM field can store
> > energy, linear momentum, and angular momentum. Why single out AM as a
> > mystery? We don't know any more about how the field stores energy or
> > linear momentum.
>
> Right. So long as Aristotle said it, that should be enough. Science
> really doesn't investigate "problems"! Science is there to tell everybody
> how much you do know! Come on Timo.

Ah, back to your usual strawmen!

No, what I said is that the storage of energy and linear momentum is equally mysterious. Go back two steps, and ponder on the mysteries of the storage of energy in the EM field. If you think there is no mystery there, that the questions are all answered, then you should have no difficulty with AM.

If you find that energy in EM fields is still a Great Mystery, then start there, where the details are a little simpler. Why pick the case where you will make the largest number of elementary mistakes?

> > Combine (a) Coulomb's law, (b) retardation, and (c) conservation of
> > energy, momentum, and angular momentum, and the EM field behaves as it
> > must.
>
> Sure. Why? What are the details? You do know what science does, right?

Why? We don't know. We start with observation, not deriving the laws of physics from logic alone. We observe electric forces, we don't predict them ab initio.

Retardation follows as a consequence of the principle of relativity, in the framework of SR or GR. The principle of relativity is a basic principle ("the laws of physics are independent of human inventions such as coordinate systems"), but alone is insufficient - retardation does not follow from Galileian relativity.

We do have excellent quantitative models of the details, with some ambiguities, since the conservation laws are only certain in their integral forms (leaving the densities ambiguous). Note well: we have persistent ambiguities even in our quantitative model. These appear to be beyond experimental resolution. (Finding a case where the conservation laws fail might be a great help, but alas, such has not been found yet.)

> >> But now change that and put the hoop OUTSIDE the solenoid.
> >
> > You mean like Feynman's version?
>
> No.

> >> Oops. Now there is H inside the coil but no charge and q outside the
> >> coil but no H! Hence no Poynting vector and no circulating energy!
> >
> > You mean an infinitely long solenoid (unlike Feynman's version)?
>
> Yes. (or large toroid if you wish)
>
> >> Yet the hoop still rotates!
> >
> > Does it? Show!
>
> Transformer.

That doesn't show it.

Go ahead, show it!

[Salmon Egg]

In article <SalmonEgg-741EC...@news80.forteinc.com>,

Salmon Egg <Salm...@sbcglobal.net> wrote:

> I do not think Feynman, as good as he was, was first notice angular
> momentum in the electromagnetic field. In Goldstein's Classical
> Mechanic, extension is made to include electromagnetism in the
> lagrangian. Vector potential, usually notated by a bold A enters into
> this lagrangian, I do not remember the details. If the lagrangian does
> not vary with a generalized coordinate, the conjugate momentum is
> conserved. I am fairly sure that that extension was well known before
> Goldstein's book became a standard. Many quantum mechanical hamiltonians
> involve the use of vector potential when magnetic effects were studied.

To refresh my memory, I looked at the Wikipedia article on the
Hamiltonian formulation of classical mechanics. It is obvious from the
formulation that if a variable is missing from the hamiltonian, the time
rate change of the conjugate momentum is zero. That is conservation of
any kind of momentum in a nutshell.

As an example. consider the motion of a planet about a heavy sun. The
hamiltonian is the sum of the kinetic energy and potential energy
expressed in generalized coordinates and their conjugate momenta.
H = T(angular momentum) + V(radius).
Because the angular position of the planet does not appear in the
hamiltonian, angular momentum is conserved. Because the radius does
appear in the hamiltonian, radial momentum is not conserved.

[Timo Nieminen]

The earliest discussion I know of angular momentum of static fields is by J. J. Thomson; see pg 25 in Electricity and Matter, http://archive.org/details/electricityandma00thomiala

Stratton covers higher-order modes in waveguides, including the conducting wire. He doesn't connect these to angular momentum. However, that such modes relate to angular momentum is old stuff.

[benj]

On Mon, 20 May 2013 14:38:39 -0700, Timo Nieminen wrote:

> On Tuesday, 21 May 2013 04:46:14 UTC+10, benj wrote:

>> Right. So long as Aristotle said it, that should be enough. Science
>> really doesn't investigate "problems"! Science is there to tell
>> everybody how much you do know! Come on Timo.

> Ah, back to your usual strawmen!

Just can't go very long without lapsing into your fantasy worlds can you?

> No, what I said is that the storage of energy and linear momentum is
> equally mysterious. Go back two steps, and ponder on the mysteries of
> the storage of energy in the EM field. If you think there is no mystery
> there, that the questions are all answered, then you should have no
> difficulty with AM.

Pure fantasy! Nobody questioned what you said nor even implied there was
no mystery. You always do this! You take someone's position and always
say they said the opposite of what they really said and then chop the
strawman to pieces. (note how you are accusing ME of using the strawman
that in fact you yourself are using here!)

> If you find that energy in EM fields is still a Great Mystery, then
> start there, where the details are a little simpler. Why pick the case
> where you will make the largest number of elementary mistakes?

I get it! If there are mysteries and questions in science they cannot be
consider as they interest people but instead MUST be taken in the order
that YOU have deemed will be productive. Why not pick an area that is
complex over one that is simple? The complex one will doubtless have the
greatest need for understanding. No?

> Why? We don't know. We start with observation, not deriving the laws of
> physics from logic alone. We observe electric forces, we don't predict
> them ab initio.
>
> Retardation follows as a consequence of the principle of relativity, in
> the framework of SR or GR. The principle of relativity is a basic
> principle ("the laws of physics are independent of human inventions such
> as coordinate systems"), but alone is insufficient - retardation does
> not follow from Galileian relativity.

Retardation does not "follow" from relativity. You have the cart before
the ass again. First you say it correctly and then in the next sentence
you turn it all backwards! Retardation is OBSERVED. Relativity followed
from those observations. Retardation of course therefore also doesn't
follow from Galilean relativity which simply states that a state of
motion at a constant velocity in a straight line cannot be distinguished
from a state of rest.

Why am I discussing relativity with you? I NEVER enter the "who is
smarter than Einstein" contests on the INTERNET.

> We do have excellent quantitative models of the details, with some
> ambiguities, since the conservation laws are only certain in their
> integral forms (leaving the densities ambiguous). Note well: we have
> persistent ambiguities even in our quantitative model. These appear to
> be beyond experimental resolution. (Finding a case where the
> conservation laws fail might be a great help, but alas, such has not
> been found yet.)

Ah yes, that which is not known by you can NEVER be known by anybody. Got
it.

>> >> Yet the hoop still rotates!
>> >
>> > Does it? Show!
>>
>> Transformer.
>
> That doesn't show it.
>
> Go ahead, show it!

Not going to try to explain emf to you.

I see you've lapsed into your usual craziness again.
That was a nice discussion for a while! We'll have to try it again
sometime!

[Timo Nieminen]

On Tuesday, 21 May 2013 10:21:18 UTC+10, benj wrote:
> On Mon, 20 May 2013 14:38:39 -0700, Timo Nieminen wrote:
> > On Tuesday, 21 May 2013 04:46:14 UTC+10, benj wrote:
>
> >> Right. So long as Aristotle said it, that should be enough. Science
> >> really doesn't investigate "problems"! Science is there to tell
> >> everybody how much you do know! Come on Timo.
>
> > Ah, back to your usual strawmen!
>
> Just can't go very long without lapsing into your fantasy worlds can you?

Your strawman, your lies, and you personal attacks are plain to see, not fantasy. You had choices: you could have avoided resorting to strawmen in the first place. You could have stopped. Instead you chose to add lies and personal attacks to your initial strawman. At least you are consistent. But why choose to be immoral?

> > If you find that energy in EM fields is still a Great Mystery, then
> > start there, where the details are a little simpler. Why pick the case
> > where you will make the largest number of elementary mistakes?
>
> I get it! If there are mysteries and questions in science they cannot be
> consider as they interest people but instead MUST be taken in the order
> that YOU have deemed will be productive. Why not pick an area that is
> complex over one that is simple? The complex one will doubtless have the
> greatest need for understanding. No?

You might make fewer errors considering the simple case. Why pick the case

where you will make the largest number of elementary mistakes?

Why would the complex case have the greatest need for understanding? Let alone that it is doubtless. You claim "doubtless": show it.

> > Why? We don't know. We start with observation, not deriving the laws of
> > physics from logic alone. We observe electric forces, we don't predict
> > them ab initio.
> >
> > Retardation follows as a consequence of the principle of relativity, in
> > the framework of SR or GR. The principle of relativity is a basic
> > principle ("the laws of physics are independent of human inventions such
> > as coordinate systems"), but alone is insufficient - retardation does
> > not follow from Galileian relativity.
>
> Retardation does not "follow" from relativity.

Wrong. You can start with the PoR and SR, and retardation immediately follows.

> You have the cart before
> the ass again. First you say it correctly and then in the next sentence
> you turn it all backwards! Retardation is OBSERVED. Relativity followed
> from those observations.

You're the one who has been whining that it isn't enough to observe that things happen (such as the storage of energy, momentum, and AM in EM fields); you say we NEED to know WHY!

And now you whine about using general principles to say why. Make up you mind! Do we want to know why or not?

Yes, our general principles are based on observation. But note that we don't need to use retardation to obtain SR; we can use a different set of observations, and use SR to predict retardation. Meanwhile, we can observe retardation, and note that the observation agrees with the prediction from principles.

You have some kind of problem with trying to compare theoretical predictions and experiments? Plenty of stuff is both predicted and experimentally observed! Seems a very strange thing to whine about!

> >> >> Yet the hoop still rotates!
> >> >
> >> > Does it? Show!
> >>
> >> Transformer.
> >
> > That doesn't show it.
> >
> > Go ahead, show it!
>
> Not going to try to explain emf to you.

So, back to correcting your elementary errors!

You claim (a) there is no torque on the charges as you move them in, and (b) no torque on the charges as you take the current in the solenoid from zero to the steady current, and (c) the hoop rotates after the current is turned off.

Start with (c). Line integral of E.dl = area integral of dB/dt. So, for a given radius r, we have E = constant/(2*pi*r). If we have a charge at some distance from the solenoid, we have force qE and torque rxqE = q*constant/(2*pi). Note that this torque is independent of the distance from the solenoid. Unlike the short coil considered by Feynman, you can't start "outside" the field of the solenoid - you get the same torque no matter how far away you have the charges.

If you begin with a short solenoid, and the charges close to the solenoid, and increase the length of the solenoid to infinity, you push the external magnetic field out towards infinity. Essentially, this is the same as Feynman's version with the current raised from zero with the charges in place. Making the unphysical assumption of an infinitely long solenoid hides that this is the same case.

If you could have been bothered trying it, you should have seen this yourself.

With a toroidal coil, which is not an unphysical case, you can start with the charges outside the field of the coil. You can't avoid affecting the coil as you move the charges in to make a ring around the coil. Note that the when the charges are in place, you have non-zero ExB.

[benj]

On Mon, 20 May 2013 18:19:46 -0700, Timo Nieminen wrote:

> On Tuesday, 21 May 2013 10:21:18 UTC+10, benj wrote:

<snip babbling fantasy and lies>

>> Retardation does not "follow" from relativity.
>
> Wrong. You can start with the PoR and SR, and retardation immediately
> follows.

PoR? Physician of Record? Partially Optimal Routing? Pastrami on Rye?
Plane of Rotation? So once again mathematical fantasy is more real than
reality to you.

> You're the one who has been whining that it isn't enough to observe that
> things happen (such as the storage of energy, momentum, and AM in EM
> fields); you say we NEED to know WHY!
>
> And now you whine about using general principles to say why. Make up you
> mind! Do we want to know why or not?

Nobody is whining about "general principles" the whine is about creating
a more detailed model.

> Yes, our general principles are based on observation. But note that we
> don't need to use retardation to obtain SR; we can use a different set
> of observations, and use SR to predict retardation. Meanwhile, we can
> observe retardation, and note that the observation agrees with the
> prediction from principles.

Retardation is simple. Action at a distance turned out to be bullshit.
(even though many still believe in it today) What does that have to do
with relativity? Retardation says actions take time to travel from A to
B. SR says those actions can't travel faster than c, but that really has
no relevance here.

> You have some kind of problem with trying to compare theoretical
> predictions and experiments? Plenty of stuff is both predicted and
> experimentally observed! Seems a very strange thing to whine about!
>
>> >> >> Yet the hoop still rotates!

>> > Go ahead, show it!
>>
>> Not going to try to explain emf to you.
>
> So, back to correcting your elementary errors!
>
> You claim (a) there is no torque on the charges as you move them in, and
> (b) no torque on the charges as you take the current in the solenoid
> from zero to the steady current, and (c) the hoop rotates after the
> current is turned off.

All lies.
a) I claimed there was no MOMENTUM, not no torque. I said v = 0. You
pointed out that wasn't true. So that is no longer an error. It is
corrected.

b) I NEVER made such a claim. PURE fantasy on your part.

c) Another spurious claim. It's the whole point of Feynman's "paradox".
What do you smoke to make up this fantasy? Although there is pretty fuzzy
thinking here since you don't say if you mean hoop rotates starting from
stopped as in Feynman or slows to stop from b) if we just send current to
zero without changing initial conditions. You don't seem to know what you
are talking about.

> Start with (c). Line integral of E.dl = area integral of dB/dt. So, for
> a given radius r, we have E = constant/(2*pi*r). If we have a charge at
> some distance from the solenoid, we have force qE and torque rxqE =
> q*constant/(2*pi). Note that this torque is independent of the distance
> from the solenoid. Unlike the short coil considered by Feynman, you
> can't start "outside" the field of the solenoid - you get the same
> torque no matter how far away you have the charges.

Ah, so now you've finally decided the charges outside the solenoid DO
experience a torque and hence rotate. Good for you. Glad we've corrected
your errors. I knew I wouldn't have to explain emf to you.

> If you begin with a short solenoid, and the charges close to the
> solenoid, and increase the length of the solenoid to infinity, you push
> the external magnetic field out towards infinity. Essentially, this is
> the same as Feynman's version with the current raised from zero with the
> charges in place. Making the unphysical assumption of an infinitely long
> solenoid hides that this is the same case.

Doesn't have to be "infinite" as much as you think math is more real then
reality. Just long enough it keeps end effects far enough away and just
choose two cases: Hoop just inside the solenoid and just outside. Anyway,
large toroid eliminates the "infinite" problem.

> If you could have been bothered trying it, you should have seen this
> yourself.

Bwahahaha! Hey, Bunky, *I* said there was torque you denied it! Now you
have it all reversed again saying I took YOUR position and you took mine!
Cute.

> With a toroidal coil, which is not an unphysical case, you can start
> with the charges outside the field of the coil. You can't avoid
> affecting the coil as you move the charges in to make a ring around the
> coil. Note that the when the charges are in place, you have non-zero
> ExB.

How is ExB non zero when charges are in place? No B outside. No charges
inside.

Do it the other way. Charges in place and coil current zero. Start
current. Hoop rotates. Stop it by hand (mechanical A. momentum removed).
Now stop current. A. Mom left in fields transfers to charges and rotates
hoop! Only there are NO fields to store that AM so just where does it
come from? Magic?

[Timo Nieminen]

On Tuesday, May 21, 2013 12:41:14 PM UTC+10, benj wrote:
> On Mon, 20 May 2013 18:19:46 -0700, Timo Nieminen wrote:
>
> >> Retardation does not "follow" from relativity.
> >
> > Wrong. You can start with the PoR and SR, and retardation immediately
> > follows.
>
> PoR? Physician of Record? Partially Optimal Routing? Pastrami on Rye?
> Plane of Rotation?

As in the immediately preceding discussion, "principle of relativity".

> So once again mathematical fantasy is more real than
> reality to you.

You think that the laws of physics do depend on human choice of coordinate systems? The PoR says it doesn't. If that is fantasy ...

Anyway, what does that have to do with the point? If you start with the PoR and SR, it is straighforward to deduce that we have retardation of EM effects. That is a fact. Whether or not it is mathematical is irrelevant.

> > You're the one who has been whining that it isn't enough to observe that
> > things happen (such as the storage of energy, momentum, and AM in EM
> > fields); you say we NEED to know WHY!
> >
> > And now you whine about using general principles to say why. Make up you
> > mind! Do we want to know why or not?
>
> Nobody is whining about "general principles"

You did.

> the whine is about creating
> a more detailed model.

What kind of more detailed model?

Note that we don't even have experimentally resolvable unambiguous answers for what energy, momentum and AM densities are in terms of the fields. What kind of "detailed model" do you want to put as a back-story to the fields that will tell us anything useful?

> > Yes, our general principles are based on observation. But note that we
> > don't need to use retardation to obtain SR; we can use a different set
> > of observations, and use SR to predict retardation. Meanwhile, we can
> > observe retardation, and note that the observation agrees with the
> > prediction from principles.
>
> Retardation is simple. Action at a distance turned out to be bullshit.
> (even though many still believe in it today) What does that have to do
> with relativity? Retardation says actions take time to travel from A to
> B. SR says those actions can't travel faster than c, but that really has
> no relevance here.

So SR says there must be retardation, but SR has no relevance to retardation?

> >> >> >> Yet the hoop still rotates!
> >> > Go ahead, show it!
> >>
> >> Not going to try to explain emf to you.
> >
> > So, back to correcting your elementary errors!
> >
> > You claim (a) there is no torque on the charges as you move them in, and
> > (b) no torque on the charges as you take the current in the solenoid
> > from zero to the steady current, and (c) the hoop rotates after the
> > current is turned off.
>
> All lies.
> a) I claimed there was no MOMENTUM, not no torque. I said v = 0.

Is there torque? If there is torque, you transfer AM. If you start with everything having zero AM, and a torque acts, then your total AM will be zero, but some things will have non-zero AM.

> You
> pointed out that wasn't true. So that is no longer an error. It is
> corrected.

For the original Feynman version, yes. The infinite solenoid case gives an important difference: the torque on the charges is independent of how far away they are.

> b) I NEVER made such a claim. PURE fantasy on your part.

Then where is the mystery? If you don't have (a) and (b), then everything adds up and there is no problem at all, if you are careful how you take limits as the coil is made infinitely long.

> c) Another spurious claim.

Oh? You don't claim that the hoop rotates?

> It's the whole point of Feynman's "paradox".

But you're not claiming it is the case here?

> Doesn't have to be "infinite" as much as you think math is more real then
> reality.

Can't keep away from personal attacks, can you? Why do you choose to bluster, lie, and attack? Nobody forces you to be immoral, so why choose to be?

In any case, YOU are the one claiming magic results from the maths of infinite coils. Do not transfer YOUR "math is more real than reality" that you are basing YOUR argument on to others.

> Just long enough it keeps end effects far enough away and just
> choose two cases: Hoop just inside the solenoid and just outside.

So, for a solenoid of cross-section A, the flux though the hoop is BA. The flux outside the hoop is -BA. Plenty of B field outside the loop. Electric field of the loop, B outside the loop: azimuthal ExB. Axial rx(ExB). Where is the problem?

Make the solenoid longer and longer, none of that changes. Take the limit in a physically meaningful was as the solenoid becomes infinitely long, and still none of that changes. If you naively take the infinitely long solenoid as your starting point, and then proceed, you have a problem.

> Anyway,
> large toroid eliminates the "infinite" problem.

And also gives us a region with ExB that is azimuthal about the hoop axis. So where is the problem?

> > With a toroidal coil, which is not an unphysical case, you can start
> > with the charges outside the field of the coil. You can't avoid
> > affecting the coil as you move the charges in to make a ring around the
> > coil. Note that the when the charges are in place, you have non-zero
> > ExB.
>
> How is ExB non zero when charges are in place? No B outside. No charges
> inside.

You think that you can't have non-zero E where the charge is zero? We have div(D)=rho, thus div(D)=0 where there is no charge. Not the same as D=0.

For the hoop to rotate, the toroidal coil and the hoop must pass through each other's inner holes. AM stored in the part of the toroid within the hoop, in the plane of the hoop, is zero. What about the rest of the toroid? Consider the part of the toroid in the plane of the hoop, outside the hoop. What is the field of the hoop there?

> Do it the other way. Charges in place and coil current zero. Start
> current. Hoop rotates. Stop it by hand (mechanical A. momentum removed).
> Now stop current. A. Mom left in fields transfers to charges and rotates
> hoop! Only there are NO fields to store that AM so just where does it
> come from? Magic?

Which coil? The unphysical infinite coil? Again, you're assuming that your maths is more real than reality. Start with a long coil, find where the AM is, and then see how the AM stored in the fields changes as you take the limit of infinite length. The region of space where the AM is stored is "pushed" out towards infinity, but the stored AM in that external field stays constant.

So where is the problem?

[Salmon Egg]

In article <b430db88-608e-4d64...@googlegroups.com>,

Timo Nieminen <ti...@physics.uq.edu.au> wrote:

> Your strawman, your lies, and you personal attacks are plain to see, not
> fantasy. You had choices: you could have avoided resorting to strawmen in the
> first place. You could have stopped. Instead you chose to add lies and
> personal attacks to your initial strawman. At least you are consistent. But
> why choose to be immoral?

Now you know why I do not bother reading posts by the subject poster. My
only contact is when third parties are unfortunate enough to respond.

[Larry Harson]

On 21 May, 01:05, Timo Nieminen <t...@physics.uq.edu.au> wrote:
> On Monday, 20 May 2013 23:20:53 UTC+10, Larry Harson  wrote:
> > On May 18, 3:11 am, Timo Nieminen <t...@physics.uq.edu.au> wrote:
>
> > > Stratton considers the problem in sections 9.15 and 9.16. In summary, there are a whole bunch of modes that carry angular momentum about the wire. For wires of realistic finite conductivity, everything except the lowest order TM mode is strongly absorbed by the wire, and this mode doesn't carry AM about the wire. For infinite conductivity (or a wire with suitable (including infinite) magnetic conductivity, like a superconductor), these modes can be practical.
>
> > I wonder what the history of this problem is?
>
> > And here was me thinking all these years it was Feynman's perceptive
> > genius that first discovered this paradox.
>

> The earliest discussion I know of angular momentum of static fields is by J. J. Thomson; see pg 25 in Electricity and Matter,http://archive.org/details/electricityandma00thomiala

>
> Stratton covers higher-order modes in waveguides, including the conducting wire. He doesn't connect these to angular momentum. However, that such modes relate to angular momentum is old stuff.

I'm surprised it's as late as 1905, since Poynting published his paper
on the transfer of energy in 1884:

Poynting, J. H. (1884). "On the Transfer of Energy in the
Electromagnetic Field". Philosophical Transactions of the Royal
Society of London

The obvious thought would have been to consider the transfer of linear
and angular momentum in the electromagnetic field.

Maxwell mentions "electrokinetic momentum" starting on page 207 of his
second volume, but doesn't appear to my limited competence, to have
anything to do with electromagnetic momentum of the field as I
understand it.

Regards, Larry.

Regards, Larry.

[Salmon Egg]

In article
<d83404e9-9c7b-4458...@g9g2000vbl.googlegroups.com>,

Larry Harson <larry...@softhome.net> wrote:

> I'm surprised it's as late as 1905, since Poynting published his paper
> on the transfer of energy in 1884:
>
> Poynting, J. H. (1884). "On the Transfer of Energy in the
> Electromagnetic Field". Philosophical Transactions of the Royal
> Society of London
>
> The obvious thought would have been to consider the transfer of linear
> and angular momentum in the electromagnetic field.
>
> Maxwell mentions "electrokinetic momentum" starting on page 207 of his
> second volume, but doesn't appear to my limited competence, to have
> anything to do with electromagnetic momentum of the field as I
> understand it.

It took Einstein and the E = m * c^2 to put the whole concept together.

My learning came by using the equations for infinite TEM plane waves
using the formulation from "Waves and Fields in Modern Radio" by Ramo
and Whinnery. Send a plane wave onto a semi-infinite block of dielectric
(vacuum) with a bit of conductivity �. For small �, the reflection at
the boundary --> 0. When the magnetic force on the current gets
integrated along the direction of propagation, the result gives the rate
of linear momentum transfer, As � --> infinity, the wave gets totally
reflected and the momentum rate transfer gets doubled.

It is a bit more complicated for angular momentum, and I have not tried
to work it through. My guess is that the attenuation of E and H requires
there to be a longitudinal component so as to require div E and div H to
be zero in a charge free volume.

[benj]

On Tue, 21 May 2013 03:37:46 -0700, Salmon Egg wrote:

> In article <b430db88-608e-4d64...@googlegroups.com>,
> Timo Nieminen <ti...@physics.uq.edu.au> wrote:
>
>> Your strawman, your lies, and you personal attacks are plain to see,
>> not fantasy. You had choices: you could have avoided resorting to
>> strawmen in the first place. You could have stopped. Instead you chose
>> to add lies and personal attacks to your initial strawman. At least you
>> are consistent. But why choose to be immoral?
>
> Now you know why I do not bother reading posts by the subject poster. My
> only contact is when third parties are unfortunate enough to respond.

Thanks for the laughs guys. Bye.

[benj]

On Mon, 20 May 2013 21:08:07 -0700, Timo Nieminen wrote:

> Can't keep away from personal attacks, can you? Why do you choose to
> bluster, lie, and attack? Nobody forces you to be immoral, so why choose
> to be?

Can't keep away from personal attacks, can you? Why do you choose to
bluster, lie, and attack? Nobody forces you to be immoral, so why choose
to be?

> In any case, YOU are the one claiming magic results from the maths of
> infinite coils. Do not transfer YOUR "math is more real than reality"
> that you are basing YOUR argument on to others.

Why do you make up stories and then claim they are what the other person
is saying? Is this how you always won on the debating team? We've already
said a toroid does the same job. Why do you keep insisting on an
"infinite" coil. No such thing exists (which you've already said). So why
are you harping and harping on it. It's irrelevant.

>> Just long enough it keeps end effects far enough away and just choose
>> two cases: Hoop just inside the solenoid and just outside.
>
> So, for a solenoid of cross-section A, the flux though the hoop is BA.
> The flux outside the hoop is -BA. Plenty of B field outside the loop.
> Electric field of the loop, B outside the loop: azimuthal ExB. Axial
> rx(ExB). Where is the problem?

There is NO steady state magnetic flux outside a toroid. Give it up
already!

> Make the solenoid longer and longer, none of that changes. Take the
> limit in a physically meaningful was as the solenoid becomes infinitely
> long, and still none of that changes. If you naively take the infinitely
> long solenoid as your starting point, and then proceed, you have a
> problem.

Solenoid Schmolenoid! We chucked that problem back with Toroid. Although
your description is not really valid. A long solenoid will act pretty
much as a toroid and not as your fantasy. But who cares? If you want to
get picky, then say toroid and shut up.

>> Anyway,
>> large toroid eliminates the "infinite" problem.

Yes. So why the big harangue on long solenoids?

> And also gives us a region with ExB that is azimuthal about the hoop
> axis. So where is the problem?

Where? There is NO such region if the charge is outside the coil. (Coil
is grounded) What's the static E field inside a closed conducting
container containing no charge?

>> > With a toroidal coil, which is not an unphysical case, you can start
>> > with the charges outside the field of the coil. You can't avoid
>> > affecting the coil as you move the charges in to make a ring around
>> > the coil. Note that the when the charges are in place, you have
>> > non-zero ExB.

I'm still waiting for you to show me just where that non-zero ExB is
located...

>> How is ExB non zero when charges are in place? No B outside. No charges
>> inside.

> You think that you can't have non-zero E where the charge is zero? We
> have div(D)=rho, thus div(D)=0 where there is no charge. Not the same as
> D=0.

Gauss's law and symmetry. QED. E=0 inside coil.

> For the hoop to rotate, the toroidal coil and the hoop must pass through
> each other's inner holes. AM stored in the part of the toroid within the
> hoop, in the plane of the hoop, is zero. What about the rest of the
> toroid? Consider the part of the toroid in the plane of the hoop,
> outside the hoop. What is the field of the hoop there?

Static B is zero everywhere outside the toroid. Period. Everybody knows
that.

>> Do it the other way. Charges in place and coil current zero. Start
>> current. Hoop rotates. Stop it by hand (mechanical A. momentum
>> removed). Now stop current. A. Mom left in fields transfers to charges
>> and rotates hoop! Only there are NO fields to store that AM so just
>> where does it come from? Magic?

> Which coil? The unphysical infinite coil? Again, you're assuming that
> your maths is more real than reality. Start with a long coil, find where
> the AM is, and then see how the AM stored in the fields changes as you
> take the limit of infinite length. The region of space where the AM is
> stored is "pushed" out towards infinity, but the stored AM in that
> external field stays constant.

Will you please shove your infinite solenoid where the sun don't shine? I
LOVE the way you accuse me of having "maths more real than reality" right
after I pointed out it's what you were doing. I am NOT the person fixated
on infinite solenoids here. My suggestion of a toroid solved that way
back. Give it up. Are you saying the hoop "pushes" the AM around to the
"bottom" of the toroid? Didn't we just say the static B field is zero
EVERYWHERE outside the Toroid?

> So where is the problem?

Where's the B (eef)?

[Timo Nieminen]

On Wednesday, 22 May 2013 08:06:31 UTC+10, benj wrote:
> On Mon, 20 May 2013 21:08:07 -0700, Timo Nieminen wrote:
>
> Yes. So why the big harangue on long solenoids?

You introduced the infinite solenoid. You waved your hands about the "magic" effects we can naively get from an infinite solenoid. The long solenoid shows how to do it properly.

> > And also gives us a region with ExB that is azimuthal about the hoop
> > axis. So where is the problem?
>
> Where? There is NO such region if the charge is outside the coil. (Coil
> is grounded) What's the static E field inside a closed conducting
> container containing no charge?

You want to treat the toroid as a closed conductor? In that case, you won't have an induced E field outside. So then the hoop doesn't rotate.

If the induced field is not blocked by the toroidal coil as a closed conductor, then the E field of the charges isn't blocked.

For some fun, consider the case of a toroidal coil as a closed conductor, with the charges just inside. Put them close enough the the inside surface of the toroid so that the volume over which you get a net azimuthal ExB is negligible. What happens when you turn down the voltage driving the current in the toroid?

Be careful with idealisations! Especially when you deduce magic consequences! Again, you are treating (naive) maths as more real than reality.

(If you want to be correct, if it's grounded, it won't have zero charge. Just leave off the "containing zero charge".)

> > You think that you can't have non-zero E where the charge is zero? We
> > have div(D)=rho, thus div(D)=0 where there is no charge. Not the same as
> > D=0.
>
> Gauss's law and symmetry. QED. E=0 inside coil.

Not QED. Think again about that "symmetry".

> Will you please shove your infinite solenoid where the sun don't shine? I
> LOVE the way you accuse me of having "maths more real than reality" right
> after I pointed out it's what you were doing. I am NOT the person fixated
> on infinite solenoids here.

You introduced it. If it isn't relevant, if it isn't useful to the discussion, why did you do so? Your claimed magic only happens with the infinite toroid, and only if you don't bother takes the limits in a physically meaningful way.

> My suggestion of a toroid solved that way
> back.

No, it doesn't.

Either the field from the charges gets inside the toroid, or there is no induced E field outside the toroid.

[benj]

On Tue, 21 May 2013 15:51:20 -0700, Timo Nieminen wrote:

> On Wednesday, 22 May 2013 08:06:31 UTC+10, benj wrote:
>> On Mon, 20 May 2013 21:08:07 -0700, Timo Nieminen wrote:
>>
>> Yes. So why the big harangue on long solenoids?
>
> You introduced the infinite solenoid. You waved your hands about the
> "magic" effects we can naively get from an infinite solenoid. The long
> solenoid shows how to do it properly.

You introduced the infinite solenoid. You waved your hands about the
"magic" effects we can naively get from an infinite solenoid. The long
solenoid shows how to do it properly.

(I'm not going to bother to go back and read your posts to you but I
could., Short memory here.)

> You want to treat the toroid as a closed conductor? In that case, you
> won't have an induced E field outside. So then the hoop doesn't rotate.

Not true. Note the coil is wound like laminations of transformer.
Assuming low resistance, (or we said superconducting) there is no or
little potential from one end to other. Hence acts as Faraday shield.
Does NOT however shield the induced emf. Care to rethink this?

> If the induced field is not blocked by the toroidal coil as a closed
> conductor, then the E field of the charges isn't blocked.

And the icing on the cake is that if the coil is superconducting then it
it reflects then internal B field as well without resorting to Toroidal
arguments.

> For some fun, consider the case of a toroidal coil as a closed
> conductor, with the charges just inside. Put them close enough the the
> inside surface of the toroid so that the volume over which you get a net
> azimuthal ExB is negligible. What happens when you turn down the voltage
> driving the current in the toroid?

Obviously still turns, Because volume isn't "zero" it's just small. And
as the hoop gets close to wall B stays same but E gets very large to
compensate for smaller volume. Right?

> Be careful with idealisations! Especially when you deduce magic
> consequences! Again, you are treating (naive) maths as more real than
> reality.

Be careful with idealisations! Especially when you deduce magic
consequences! Again, you are treating (naive) maths as more real than
reality.

> (If you want to be correct, if it's grounded, it won't have zero charge.
> Just leave off the "containing zero charge".)

Are you saying that inside a hollow conductor containing no charge that
somehow there will be charge inside? You know that's not right. What are
you trying to say?

>> > You think that you can't have non-zero E where the charge is zero? We
>> > have div(D)=rho, thus div(D)=0 where there is no charge. Not the same
>> > as D=0.
>>
>> Gauss's law and symmetry. QED. E=0 inside coil.
>
> Not QED. Think again about that "symmetry".

You think about it.

>> Will you please shove your infinite solenoid where the sun don't shine?
>> I LOVE the way you accuse me of having "maths more real than reality"
>> right after I pointed out it's what you were doing. I am NOT the person
>> fixated on infinite solenoids here.

> You introduced it. If it isn't relevant, if it isn't useful to the
> discussion, why did you do so? Your claimed magic only happens with the
> infinite toroid, and only if you don't bother takes the limits in a
> physically meaningful way.

I did NOT introduce it. YOU introduced it! (Did NOT! Did too! etc.) I
introduced the "long" solenoid as an approximation to think about and you
jumped on it and made it infinite over some detail (unable to see the big
picture here) and started shoving Angular Momentum to infinity! Just
shove the whole thing! If you won't accept an approximate argument then
stick with the toroidal one.

>> My suggestion of a toroid solved that way back.
>
> No, it doesn't.
>
> Either the field from the charges gets inside the toroid, or there is no
> induced E field outside the toroid.

I tell you that this statement is false. AM requires both a static E and
H field to be stored, but emf only needs a changing flux to be induced
all around (inside and outside) the toroid. NO static E field inside (or
outside) the coil needed at all. Our charged hoop charges are really just
test charges showing qE forces, in other words measuring the EMF field
which provides the torque and moves the hoop.

[Timo Nieminen]

On Wednesday, May 22, 2013 12:29:08 PM UTC+10, benj wrote:
> On Tue, 21 May 2013 15:51:20 -0700, Timo Nieminen wrote:
> > On Wednesday, 22 May 2013 08:06:31 UTC+10, benj wrote:
> >> On Mon, 20 May 2013 21:08:07 -0700, Timo Nieminen wrote:
> >>
> >> Yes. So why the big harangue on long solenoids?
> >
> > You introduced the infinite solenoid. You waved your hands about the
> > "magic" effects we can naively get from an infinite solenoid. The long
> > solenoid shows how to do it properly.
>
> You introduced the infinite solenoid. You waved your hands about the
> "magic" effects we can naively get from an infinite solenoid. The long
> solenoid shows how to do it properly.

No, you introduced it. To quote you: "Now there is H inside the coil but no charge and q outside the coil but no H!". If it isn't infinite, you don't have zero H outside.

Why lie about such things? You introduced it, and made a big deal about magic effects that result from the unphysical assumptions.

> > For some fun, consider the case of a toroidal coil as a closed
> > conductor, with the charges just inside. Put them close enough the the
> > inside surface of the toroid so that the volume over which you get a net
> > azimuthal ExB is negligible. What happens when you turn down the voltage
> > driving the current in the toroid?

> Obviously still turns, Because volume isn't "zero" it's just small.

Obviously? Note well the wording of the question: "What happens when you turn down the voltage driving the current in the toroid?", not "What happens when you reduce B inside the toroid?"

Sometimes it is worth trying to do the actual calculation, not just say "obviously ...".

> And
> as the hoop gets close to wall B stays same but E gets very large to
> compensate for smaller volume. Right?

Wrong. Why does the E field of a charged hoop care about an exterior wall (in the space inside the wall)?

> > (If you want to be correct, if it's grounded, it won't have zero charge.
> > Just leave off the "containing zero charge".)
>
> Are you saying that inside a hollow conductor containing no charge that
> somehow there will be charge inside? You know that's not right. What are
> you trying to say?

This point isn't relevant to the main discussion, just a minor correction. Put a grounded conductor near some charge, and it will be charged, not neutral. Won't give you any internal field, so not a big deal.

> > Either the field from the charges gets inside the toroid, or there is no
> > induced E field outside the toroid.
>
> I tell you that this statement is false. AM requires both a static E and
> H field to be stored, but emf only needs a changing flux to be induced
> all around (inside and outside) the toroid. NO static E field inside (or
> outside) the coil needed at all. Our charged hoop charges are really just
> test charges showing qE forces, in other words measuring the EMF field
> which provides the torque and moves the hoop.

Take a region of magnetic field surrounded by an infinite conducting tube of cross-sectional area A, B along the tube axis. Flux through tube is BA.

If the tube is a good conductor, E in the tube wall is zero (or close enough to it). So, if we integrate E.dl around a loop around the tube inside the tube wall, we have integral(E.dl)=0 (or close enough to it). So dB/dt=0.

If you were to try to change the field inside, you would get an induced current on the inner surface of the tube which will keep the internal B field constant.

We can look at the external E field. We have, within the tube wall, E=0. So, the azimuthal and axial components of E just outside the tube are zero as well. Outside the tube, we have curl(E)=0, dE/dz=0, dE/d(phi)=0. From symmetry, E_z=0. So d(r*E_phi)/dr=0, or E_phi=constant/r. Just outside the wall, E_phi=0, so the constant is zero, and E_phi=0 everywhere outside the conducting tube.

Feel free to calculate what happens for a toroidal tube of finite conductivity.

[Timo Nieminen]

On Wednesday, May 22, 2013 6:40:48 AM UTC+10, Salmon Egg wrote:
>
> My learning came by using the equations for infinite TEM plane waves
> using the formulation from "Waves and Fields in Modern Radio" by Ramo
> and Whinnery. Send a plane wave onto a semi-infinite block of dielectric

> (vacuum) with a bit of conductivity ß. For small ß, the reflection at

> the boundary --> 0. When the magnetic force on the current gets
> integrated along the direction of propagation, the result gives the rate

> of linear momentum transfer, As ß --> infinity, the wave gets totally

> reflected and the momentum rate transfer gets doubled.
>
> It is a bit more complicated for angular momentum, and I have not tried
> to work it through. My guess is that the attenuation of E and H requires
> there to be a longitudinal component so as to require div E and div H to
> be zero in a charge free volume.

For the infinite TEM plane wave and angular momentum, your wave is circularly polarised. No problems with div(E) or div(H) since dE_x/dx=0 etc. since it's an infinite plane wave.

For a circularly polarised plane wave in a non-absorbing isotropic medium, you have an induced dipole moment per unit volume that is in the same direction as E. So no torque. Note that we have a polarisation that is synchronous with the E field, and a polarisation current that is a quarter-wave out-of-phase.

But if we have some conductivity, not only do we have a polarisation current that is out-of-phase, we also have a conduction current that is in-phase with E. This gives us an effective polarisation that lags behind E. So, in an absorbing medium, we have some angle between the instantaneous polarisation and the instantaneous E field. The field exerts a torque to align the dipole moment with the field, so we get a torque per unit volume acting on the medium. Integrate into the half space, and you get the total transferred angular momentum per unit area, which will be I/omega, for irradiance I and angular frequency omega.

[Salmon Egg]

In article <2bd442bb-9d03-4e20...@googlegroups.com>,

Timo Nieminen <ti...@physics.uq.edu.au> wrote:

> For the infinite TEM plane wave and angular momentum, your wave is circularly
> polarised. No problems with div(E) or div(H) since dE_x/dx=0 etc. since it's
> an infinite plane wave.

As I said, I have not looked at this problem in any detail.

Nevertheless, even in the absence of a longitudinal field component.
current density vectors in a slightly conductive medium will be rotated
to different angles along the propagation direction. As we know, wires
carrying current have magnetic torques that tend to line up currents to
flow in the same direction.

With that knowledge, I ought to be able to integrated the differential
couples down the propagation path to get the torque. When I get around
to it.

[Larry Harson]

On May 17, 11:18 pm, benj <b...@iwaynet.net> wrote:
> Do you recall the Feynman Paradox?
>
> Here's the question. We all know that energy can travel down an electric
> cord. The evidence is everywhere, The heating of irons, the pulling of
> solenoids against springs, the lifting of elevators etc. And clearly when
> there is linear motion with mass (mv) there is also momentum which is
> conserved.
>
> But Feynman's apparatus deals with ANGULAR MOMENTUM. It starts with
> nothing moving and ends with the charged disk rotating. As he points out,
> it appears that angular momentum is not conserved. But Feynman notes that
> the missing angular momentum is in the EM fields about the apparatus.
> Hence it IS conserved.
>
> The question, though, is how did that angular momentum get into the
> fields and did it come down the electric cord when the solenoid was given
> a constant current?

Benj, you majored in electrical engineering rather than physics,
right?

In which case it's highly likely you haven't come across the general
definition of mechanical angular momentum as:

AM = rxp

The definition of angular electromagnetic momentum then follows using
electromagnetic momentum:

EAM = epsilon_0 integral r x (ExB) dv

Regards, Larry.

[benj]

On Tue, 21 May 2013 21:57:15 -0700, Timo Nieminen wrote:

> On Wednesday, May 22, 2013 12:29:08 PM UTC+10, benj wrote:

>> You introduced the infinite solenoid. You waved your hands about the
>> "magic" effects we can naively get from an infinite solenoid. The long
>> solenoid shows how to do it properly.
>
> No, you introduced it. To quote you: "Now there is H inside the coil but
> no charge and q outside the coil but no H!". If it isn't infinite, you
> don't have zero H outside.
>
> Why lie about such things? You introduced it, and made a big deal about
> magic effects that result from the unphysical assumptions.

No lie. I introduced the "long" solenoid which is approximately true to
above. You made it "infinite" in an attempt to make it exactly true.
<shrug>

>> Obviously still turns, Because volume isn't "zero" it's just small.
>
> Obviously? Note well the wording of the question: "What happens when you
> turn down the voltage driving the current in the toroid?", not "What
> happens when you reduce B inside the toroid?"

Here, lets try to get you to be scientifically exact. When you say "turn
down" do you mean what happens WHILE I'm turning down the drive or AFTER
I've turned it down?

Of course the electric field driving the charges is created NOT by the B
fields but by it's derivative! So if you mean "after" then he answer is
nothing happens as the emf depends on the rate of change of B and not
it's value. If you mean "while" then "obviously" an E field is created
and the charges turn.

> Sometimes it is worth trying to do the actual calculation, not just say
> "obviously ...".

Obvious to me. See if the calculations don't agree.

>> And as the hoop gets close to wall B stays same but E gets very large
>> to compensate for smaller volume. Right?
>
> Wrong. Why does the E field of a charged hoop care about an exterior
> wall (in the space inside the wall)?

How does the magnitude of the E field from charges vary with distance
from them?

> Take a region of magnetic field surrounded by an infinite conducting
> tube of cross-sectional area A, B along the tube axis. Flux through tube
> is BA.

This is not our situation! We have a coil (Faraday shield) not a
conducting tube. Not the same!

> If the tube is a good conductor, E in the tube wall is zero (or close
> enough to it). So, if we integrate E.dl around a loop around the tube
> inside the tube wall, we have integral(E.dl)=0 (or close enough to it).
> So dB/dt=0.
>
> If you were to try to change the field inside, you would get an induced
> current on the inner surface of the tube which will keep the internal B
> field constant.

What you say is true for your problem. But in OUR problem what happens
when you try to change the field is the E field appears in the coil and
thus appears to us as INDUCTANCE of the coil, which is doubtless handled
in some manner by our coil power supply. So yes, these effects are there,
but it is possible to change the magnetic field in spite of inductance,
though not instantaneously.

> We can look at the external E field. We have, within the tube wall, E=0.
> So, the azimuthal and axial components of E just outside the tube are
> zero as well. Outside the tube, we have curl(E)=0, dE/dz=0, dE/d(phi)=0.
> From symmetry, E_z=0. So d(r*E_phi)/dr=0, or E_phi=constant/r. Just
> outside the wall, E_phi=0, so the constant is zero, and E_phi=0
> everywhere outside the conducting tube.

> Feel free to calculate what happens for a toroidal tube of finite
> conductivity.

Again, we do not have a conducting tube. Are you saying that transformers
can't work when there is a Faraday shield between the primary and
secondary? I don't think so. What you are saying is that a shorted turn
kills the transformer action. Which is true. (within limits) But that is
not our situation.

[Timo Nieminen]

On Thursday, 23 May 2013 02:37:06 UTC+10, benj wrote:
> On Tue, 21 May 2013 21:57:15 -0700, Timo Nieminen wrote:
> > On Wednesday, May 22, 2013 12:29:08 PM UTC+10, benj wrote:
>
> >> You introduced the infinite solenoid. You waved your hands about the
> >> "magic" effects we can naively get from an infinite solenoid. The long
> >> solenoid shows how to do it properly.
> >
> > No, you introduced it. To quote you: "Now there is H inside the coil but
> > no charge and q outside the coil but no H!". If it isn't infinite, you
> > don't have zero H outside.
> >
> > Why lie about such things? You introduced it, and made a big deal about
> > magic effects that result from the unphysical assumptions.
>
> No lie. I introduced the "long" solenoid which is approximately true to
> above. You made it "infinite" in an attempt to make it exactly true.

The long solenoid doesn't have zero field outside. Flux inside is BA, flux outside is -BA. So, you were just wrong then, not over-idealising.

> >> Obviously still turns, Because volume isn't "zero" it's just small.
> >
> > Obviously? Note well the wording of the question: "What happens when you
> > turn down the voltage driving the current in the toroid?", not "What
> > happens when you reduce B inside the toroid?"
>
> Here, lets try to get you to be scientifically exact. When you say "turn
> down" do you mean what happens WHILE I'm turning down the drive or AFTER
> I've turned it down?

It's very simple. You specified a good conductor, and even offered the alternative of a superconductor. You said it keeps all external E fields out.

Basically, you have imagined some B field energy in a closed box. It can't go anywhere. If you turn the driving voltage off, it provides it's own to sustain the current to maintain the B field. Infinite inductance.

Could it be that a closed good/super conductor is a very poor model of a toroidal solenoid? Try a realistic model instead.

> Of course the electric field driving the charges is created NOT by the B
> fields but by it's derivative! So if you mean "after" then he answer is
> nothing happens as the emf depends on the rate of change of B and not
> it's value. If you mean "while" then "obviously" an E field is created
> and the charges turn.

"Obviously" again. Do the calculation.

> > Sometimes it is worth trying to do the actual calculation, not just say
> > "obviously ...".
>
> Obvious to me. See if the calculations don't agree.

Mine don't. You're not provided any. All you've provided so far has been errors and unphysical models and "obviously". Can you offer something of more substance?

> >> And as the hoop gets close to wall B stays same but E gets very large
> >> to compensate for smaller volume. Right?
> >
> > Wrong. Why does the E field of a charged hoop care about an exterior
> > wall (in the space inside the wall)?
>
> How does the magnitude of the E field from charges vary with distance
> from them?

If the charged hoop is close to wall, and is expanded so that the distance to the wall is halved, the field does not double as required by your assumption. Again, sometimes doing the calculation is better than declaring "obvisouly" and assuming the desired result.

For the exact result for a thing ring,
http://unapologetic.wordpress.com/2012/01/10/charged-rings-and-planes/

> > Take a region of magnetic field surrounded by an infinite conducting
> > tube of cross-sectional area A, B along the tube axis. Flux through tube
> > is BA.
>
> This is not our situation! We have a coil (Faraday shield) not a
> conducting tube. Not the same!

So why talk about a closed conducting container?

If you were thinking of something different, provide the mathematical model and solution.

> > If the tube is a good conductor, E in the tube wall is zero (or close
> > enough to it). So, if we integrate E.dl around a loop around the tube
> > inside the tube wall, we have integral(E.dl)=0 (or close enough to it).
> > So dB/dt=0.
> >
> > If you were to try to change the field inside, you would get an induced
> > current on the inner surface of the tube which will keep the internal B
> > field constant.
>
> What you say is true for your problem.

Your problem. You introduced the "closed conductor" that keeps external E fields out. I'm just pointing out that it also keeps internal induced E fields in.

> But in OUR problem what happens
> when you try to change the field is the E field appears in the coil and
> thus appears to us as INDUCTANCE of the coil, which is doubtless handled
> in some manner by our coil power supply. So yes, these effects are there,
> but it is possible to change the magnetic field in spite of inductance,
> though not instantaneously.

Ah yes, we can have the power supply magically deal with the infinite inductance you specified. From the combination of two unphysical idealisations, you wonder at the magic that happens. Maths is more real than reality!

Why not stick with a more physically plausible case?

> > We can look at the external E field. We have, within the tube wall, E=0.
> > So, the azimuthal and axial components of E just outside the tube are
> > zero as well. Outside the tube, we have curl(E)=0, dE/dz=0, dE/d(phi)=0.
> > From symmetry, E_z=0. So d(r*E_phi)/dr=0, or E_phi=constant/r. Just
> > outside the wall, E_phi=0, so the constant is zero, and E_phi=0
> > everywhere outside the conducting tube.
>
> > Feel free to calculate what happens for a toroidal tube of finite
> > conductivity.
>
> Again, we do not have a conducting tube.

So what do we have? If it isn't a closed conducting tube, why did you bother with "What's the static E field inside a closed conducting container [...]?" If we don't have a closed conducting container, isn't that just an irrelevant red herring? But you were basing your argument on it!

You claimed that the solenoid stops external E fields from getting in, but magically lets the internal induced E field out.

> Are you saying that transformers
> can't work when there is a Faraday shield between the primary and
> secondary? I don't think so.

Are you saying Faraday shields only work one-way? I don't think so.

[Larry Harson]

On May 22, 10:02 pm, Timo Nieminen <t...@physics.uq.edu.au> wrote:
> On Thursday, 23 May 2013 02:37:06 UTC+10, benj  wrote:

> For the exact result for a thing ring,http://unapologetic.wordpress.com/2012/01/10/charged-rings-and-planes/
>

What a fantastic looking blog on the application of mathematics to EM
problems.

Thanks Benj for starting the thread, and Timo for the fascinating
replies.

Regards, Larry.

[benj]

On Wed, 22 May 2013 14:02:14 -0700, Timo Nieminen wrote:

> On Thursday, 23 May 2013 02:37:06 UTC+10, benj wrote:
>> On Tue, 21 May 2013 21:57:15 -0700, Timo Nieminen wrote:
>> > On Wednesday, May 22, 2013 12:29:08 PM UTC+10, benj wrote:

> The long solenoid doesn't have zero field outside. Flux inside is BA,
> flux outside is -BA. So, you were just wrong then, not over-idealising.

Your case only matters if you are integrating over all space. If you are
close to the solenoid fields are minimal if you aren't near the ends.
Fields creating forces on charges are NEAR the solenoid. Sure it's not
"ideal" but it makes the point. You are over-idealizing. A toroid
idealizes the situation but really works little different from a long
solenoid in any practical case. Practical isn't what you are doing.

> It's very simple. You specified a good conductor, and even offered the
> alternative of a superconductor. You said it keeps all external E fields
> out.
>
> Basically, you have imagined some B field energy in a closed box. It
> can't go anywhere. If you turn the driving voltage off, it provides it's
> own to sustain the current to maintain the B field. Infinite inductance.

Yes, it's what happens when you have a superconducting coil. But you seem
to have forgotten that part of the experiment is to let the coil heat up
so it loses it's superconducting nature. Then the Energy in the B fields
gets out and inductance is no longer "infinite".

> Could it be that a closed good/super conductor is a very poor model of a
> toroidal solenoid? Try a realistic model instead.

There's nothing wrong with this model. It's a well-understood device.

>> Of course the electric field driving the charges is created NOT by the
>> B fields but by it's derivative! So if you mean "after" then he answer
>> is nothing happens as the emf depends on the rate of change of B and
>> not it's value. If you mean "while" then "obviously" an E field is
>> created and the charges turn.
>
> "Obviously" again. Do the calculation.

Let's see. I think even someone working on a GED like me can do this...
(invoke symmetry)
emf = Pi d E (d = diameter of hoop) Hence E = emf/(Pi d)
Tangential Force on charge = q E so F = (q emf)/(pi d)
torque = F x R so T = (q emf (d/2) )/(pi d) = q emf/2 pi

Since you'd rather calculate than think, here's an exercise for the
interested student.

In the above case with either a "long" solenoid or toroid of n turns per
unit length with a cylinder bearing a total charge q and of mass m either
just inside or just outside the coil find the final angular velocity of
the charged cylinder after the current is switched on and goes from zero
to a final value I?


Teacher's Answerbook: w = - mu (qnI)/2m (unit vector in axis direction)
Note minus sign means rotation is opposite to direction of current.

Good luck!

>> > Sometimes it is worth trying to do the actual calculation, not just
>> > say "obviously ...".
>>
>> Obvious to me. See if the calculations don't agree.
>
> Mine don't. You're not provided any. All you've provided so far has been
> errors and unphysical models and "obviously". Can you offer something of
> more substance?
>
>> >> And as the hoop gets close to wall B stays same but E gets very
>> >> large to compensate for smaller volume. Right?
>> >
>> > Wrong. Why does the E field of a charged hoop care about an exterior
>> > wall (in the space inside the wall)?
>>
>> How does the magnitude of the E field from charges vary with distance
>> from them?
>
> If the charged hoop is close to wall, and is expanded so that the
> distance to the wall is halved, the field does not double as required by
> your assumption. Again, sometimes doing the calculation is better than
> declaring "obvisouly" and assuming the desired result.
>
> For the exact result for a thing ring,
> http://unapologetic.wordpress.com/2012/01/10/charged-rings-and-planes/

Interesting result. Only calculation is for Z direction not radial.
Radial result is same as Coulomb's law for distances greater than R.
Anyway, if the field between the wall and hoop (lets say cylinder shall
we?) does not represent increased fields then the angular momentum must
be also stored in fields inside the cylinder. The point being that the
angular momentum stored in the E x B fields cannot depend on the diameter
of the charged cylinder... or does it? AM stored in fields is same
whether hoop turns or not, right?

>> This is not our situation! We have a coil (Faraday shield) not a
>> conducting tube. Not the same!
>
> So why talk about a closed conducting container?

because you were saying the electrostatic fields from the external
charges finds it's way inside the coil. This is not true for THOSE
fields. Electrokinetic fields (emf) however do appear about all current
elements.

> If you were thinking of something different, provide the mathematical
> model and solution.

Actually the point is to FIND the mathematical model and solution.

> So what do we have? If it isn't a closed conducting tube, why did you
> bother with "What's the static E field inside a closed conducting
> container [...]?" If we don't have a closed conducting container, isn't
> that just an irrelevant red herring? But you were basing your argument
> on it!

You've got me cornered. Here's what I'm trying to do. I want to try to
set up a situation where there is clearly stored angular momentum yet no
place where there is both static B and static E fields needed to store
that momentum. Note that it is STATIC fields that store momentum the
induced fields are all zero for this.

Thus, if we have a cylinder inside the coil clearly there is both static
E (from charges) and static B (from steady current in coil) in there and
the Poynting energy is circulating. (radial E and axis B) Hence A M can
be stored. But if the cylinder is outside the coil now B is all (mostly)
inside the coil and E is (mostly) outside. If some E leaks in that is
irrelevant because the torque does not change significantly between the
two positions. Hence there is a problem.

Where this is trying to go is that while we know that the amount of
linear momentum stored in a region of space is the integral of E x H over
that volume. But one can show that the amount of linear momentum stored
in space is given by G = q A where A is the magnetic vector potential.
Since B is zero outside the coil that means A is a constant. So where
this is headed is that while B is zero outside, A is not. There is plenty
of A out where the charge, q, is located. In fact, the value of A just
outside the coil is the same as A just inside it. Furthermore, the
momentum is in the direction of A which is circulating about the coil and
hence represents stored Angular Momentum!

This is what I was hoping you'd come up with without any prodding by me.
But nevertheless the exact relationships aren't completely clear here.

> You claimed that the solenoid stops external E fields from getting in,
> but magically lets the internal induced E field out.

Correct. Two different E fields with different properties!

>> Are you saying that transformers can't work when there is a Faraday
>> shield between the primary and secondary? I don't think so.
>
> Are you saying Faraday shields only work one-way? I don't think so.

I'm saying they only work for one type of E field. But I'm trying NOT to
get off on that tangent now. Transformers clearly work whether the
secondary is inside or outside the primary. Our charged "hoop" is really
just an exotic secondary winding.

[benj]

On Wed, 22 May 2013 09:08:47 -0700, Larry Harson wrote:

> On May 17, 11:18 pm, benj <b...@iwaynet.net> wrote:
>> Do you recall the Feynman Paradox?

> Benj, you majored in electrical engineering rather than physics, right?

I have always kept a foot in each camp. My largest shortcoming is that I
am not much of an expert mathematician although if forced I can plow my
way through any needed math. Timo obviously is a talented mathematician.
This is why a "lively discussion" with him can so productive.

> In which case it's highly likely you haven't come across the general
> definition of mechanical angular momentum as:
>
> AM = rxp

You think engineers don't have to study rotational kinematics? Wake up
call!

> The definition of angular electromagnetic momentum then follows using
> electromagnetic momentum:
>
> EAM = epsilon_0 integral r x (ExB) dv

Yes, I know this. but of course the devil is in the details. In other
words just how does all this work which is indicated in the subject
asking if angular momentum can travel down a wire. Evidence so far is
that it can't. Instead zero angular momentum splits into actual
mechanical rotation and an equal opposite part stored in the fields.
Isn't that interesting?

You will find the conversation is headed toward a discussion of G = qA (G
= stored momentum and A = magnetic vector potential)

But thanks.

[Timo Nieminen]

On Thursday, May 23, 2013 1:49:19 PM UTC+10, benj wrote:
> On Wed, 22 May 2013 14:02:14 -0700, Timo Nieminen wrote:
> > On Thursday, 23 May 2013 02:37:06 UTC+10, benj wrote:
> >> On Tue, 21 May 2013 21:57:15 -0700, Timo Nieminen wrote:
> >> > On Wednesday, May 22, 2013 12:29:08 PM UTC+10, benj wrote:
>
> > The long solenoid doesn't have zero field outside. Flux inside is BA,
> > flux outside is -BA. So, you were just wrong then, not over-idealising.
>
> Your case only matters if you are integrating over all space. If you are
> close to the solenoid fields are minimal if you aren't near the ends.

What is important is the integral of rx(ExH) over all space. That means we need to integrate over all space.

> Fields creating forces on charges are NEAR the solenoid. Sure it's not
> "ideal" but it makes the point. You are over-idealizing. A toroid
> idealizes the situation but really works little different from a long
> solenoid in any practical case. Practical isn't what you are doing.

"Practical" means that we can't ignore the regions of space that contribute to the integral of rx(ExH), and then claim that "Wow! The integral of rx(ExH) is zero!"

> > It's very simple. You specified a good conductor, and even offered the
> > alternative of a superconductor. You said it keeps all external E fields
> > out.
> >
> > Basically, you have imagined some B field energy in a closed box. It
> > can't go anywhere. If you turn the driving voltage off, it provides it's
> > own to sustain the current to maintain the B field. Infinite inductance.
>
> Yes, it's what happens when you have a superconducting coil. But you seem
> to have forgotten that part of the experiment is to let the coil heat up
> so it loses it's superconducting nature. Then the Energy in the B fields
> gets out and inductance is no longer "infinite".

You can have some fun with a model like this. For simplicity, start with a superconducting tube with axial B, turn off the superconductivity. Calculate what happens to the field angular momentum over time, and the charged hoop AM over time. Note that once the tube isn't superconducting any more, the B field leaks out. Only when the B field leaks out past the charged hoop does the flux through the hoop change. Consider the AM stored in the field outside the hoop (which has, by that time, non-zero B).

> > Could it be that a closed good/super conductor is a very poor model of a
> > toroidal solenoid? Try a realistic model instead.
>
> There's nothing wrong with this model. It's a well-understood device.

A well-understood device that doesn't let the induced field out?

> > "Obviously" again. Do the calculation.
>
> Let's see. I think even someone working on a GED like me can do this...
> (invoke symmetry)
> emf = Pi d E (d = diameter of hoop) Hence E = emf/(Pi d)
> Tangential Force on charge = q E so F = (q emf)/(pi d)
> torque = F x R so T = (q emf (d/2) )/(pi d) = q emf/2 pi

OK, so ignoring the conducting walls. Magic conducting walls that only work one-way? Real metal boxes don't work that way.

> >> >> And as the hoop gets close to wall B stays same but E gets very
> >> >> large to compensate for smaller volume. Right?
> >> >
> >> > Wrong. Why does the E field of a charged hoop care about an exterior
> >> > wall (in the space inside the wall)?
> >>
> >> How does the magnitude of the E field from charges vary with distance
> >> from them?
> >
> > If the charged hoop is close to wall, and is expanded so that the
> > distance to the wall is halved, the field does not double as required by
> > your assumption. Again, sometimes doing the calculation is better than
> > declaring "obvisouly" and assuming the desired result.
> >
> > For the exact result for a thing ring,
> > http://unapologetic.wordpress.com/2012/01/10/charged-rings-and-planes/
>
> Interesting result. Only calculation is for Z direction not radial.

Both are given. Read again.

> Radial result is same as Coulomb's law for distances greater than R.

No. Same as Coulomb's law for distances much greater than R.

> Anyway, if the field between the wall and hoop (lets say cylinder shall
> we?) does not represent increased fields then the angular momentum must
> be also stored in fields inside the cylinder. The point being that the
> angular momentum stored in the E x B fields cannot depend on the diameter
> of the charged cylinder... or does it? AM stored in fields is same
> whether hoop turns or not, right?

Why would AM be stored inside the cylinder? Zero E field inside.

For a long solenoid, it's obvious what happens: when the hoop is near the wall, the AM is stored in the E and B field outside the solenoid. As we shrink the cylinder, we store more AM inside the solenoid, between the solenoid and cylinder. But this AM has opposite direction to that stored outside, so reduces the total stored AM. Makes sense, if you calculate the torque exerted on the cylinder as B->0.

For a charged toroidal tube in a toroidal solenoid, what happens? Is it just equivalent to a curled-up version of the infinite superconducting tube above? (Faraday would have said "yes"!)

> >> This is not our situation! We have a coil (Faraday shield) not a
> >> conducting tube. Not the same!
> >
> > So why talk about a closed conducting container?
>
> because you were saying the electrostatic fields from the external
> charges finds it's way inside the coil. This is not true for THOSE
> fields. Electrokinetic fields (emf) however do appear about all current
> elements.

Don't both obey the same field equations? The same boundary conditions?

If you think static versus non-static matters, calculate for a tube/toroid of finite conductivity.

> > If you were thinking of something different, provide the mathematical
> > model and solution.
>
> Actually the point is to FIND the mathematical model and solution.

If you don't have a model, where do you get your "obviously" from?

> > So what do we have? If it isn't a closed conducting tube, why did you
> > bother with "What's the static E field inside a closed conducting
> > container [...]?" If we don't have a closed conducting container, isn't
> > that just an irrelevant red herring? But you were basing your argument
> > on it!
>
> You've got me cornered. Here's what I'm trying to do. I want to try to
> set up a situation where there is clearly stored angular momentum yet no
> place where there is both static B and static E fields needed to store
> that momentum. Note that it is STATIC fields that store momentum the
> induced fields are all zero for this.

By all means, try to find such a case. But it's useless if it's just based on elementary error.

> Thus, if we have a cylinder inside the coil clearly there is both static
> E (from charges) and static B (from steady current in coil) in there and
> the Poynting energy is circulating. (radial E and axis B) Hence A M can
> be stored. But if the cylinder is outside the coil now B is all (mostly)
> inside the coil and E is (mostly) outside. If some E leaks in that is
> irrelevant because the torque does not change significantly between the
> two positions. Hence there is a problem.

The AM stored with the cylinder inside the coil is in the wrong direction. The larger the gap between coil and cylinder, the more wrong-direction AM is stored, and the less AM transferred to the cylinder. Again, it can be worthwhile doing the calculation, rather than just assuming the result because "obviously".

Do you clearly have stored AM at the start? We can start with zero AM, and finish with L mechanical AM and -L field AM without having stored AM at the start. What you need is a case with the integral of rx(ExH) over all space being zero both before and after, and have a change in mechanical AM in the process.

> Where this is trying to go is that while we know that the amount of
> linear momentum stored in a region of space is the integral of E x H over
> that volume. But one can show that the amount of linear momentum stored
> in space is given by G = q A where A is the magnetic vector potential.

No, not in general. We can say that field momentum = qA for a point charge (so mv + qA is constant). Under what conditions is this correct? How well does this work for multiple charges? (Multiple charges shows that rho*A isn't the field momentum density, since that would require magic teleporting momentum, which might be seen as undesirable.)

We can construct cases with rho=0 everywhere (and thus rho*A being zero) but with non-zero field momentum (which will require the field energy to be moving, so generally we will have radiation). What works in static/quasi-static cases does not necessarily work in every case.

> Since B is zero outside the coil that means A is a constant. So where
> this is headed is that while B is zero outside, A is not. There is plenty
> of A out where the charge, q, is located. In fact, the value of A just
> outside the coil is the same as A just inside it. Furthermore, the
> momentum is in the direction of A which is circulating about the coil and
> hence represents stored Angular Momentum!
>
> This is what I was hoping you'd come up with without any prodding by me.

Something that (a) doesn't work in all cases, and (b) require magic teleporting AM (and magic teleporting linear momentum)? Why would I come up with that?

One runs into similar problems with saying that the electromagnetic energy density is rho*(scalar potential). It can be convenient to say that the field energy is q*(scalar potential) in electrostatic/quasi-static problems. But why try to elevate a mathematical convenience into a general truth when we already know cases where it breaks?

> But nevertheless the exact relationships aren't completely clear here.

> > You claimed that the solenoid stops external E fields from getting in,
> > but magically lets the internal induced E field out.
>
> Correct. Two different E fields with different properties!

Two fields which satisfy the same field equations and follow the same constitutive relations (and thus have the same boundary conditions).

[Larry Harson]

On May 23, 5:20 am, benj <b...@iwaynet.net> wrote:
> On Wed, 22 May 2013 09:08:47 -0700, Larry Harson wrote:
> > On May 17, 11:18 pm, benj <b...@iwaynet.net> wrote:
> >> Do you recall the Feynman Paradox?
> > Benj, you majored in electrical engineering rather than physics, right?
>
> I have always kept a foot in each camp. My largest shortcoming is that I
> am not much of an expert mathematician although if forced I can plow my
> way through any needed math. Timo obviously is a talented mathematician.
> This is why a "lively discussion" with him can so productive.
>
> > In which case it's highly likely you haven't come across the general
> > definition of mechanical angular momentum as:
>
> > AM = rxp
>
> You think engineers don't have to study rotational kinematics? Wake up
> call!

Most electrical engineers come across angular momentum defined in a
way analogous to linear momentum:

AM = wI, where w is angular velocity and I moment of inertial.

> > The definition of angular electromagnetic momentum then follows using
> > electromagnetic momentum:
>
> > EAM = epsilon_0 integral r x (ExB) dv
>
> Yes, I know this. but of course the devil is in the details. In other
> words just how does all this work which is indicated in the subject
> asking if angular momentum can travel down a wire. Evidence so far is
> that it can't.

>Instead zero angular momentum splits into actual
> mechanical rotation and an equal opposite part stored in the fields.
> Isn't that interesting?

Linear momentum does the same thing. There's all sorts of marvelous
"paradoxes" where action doesn't equal reaction for two moving
charges, which implies a net force on the centre of mass of the
system. The change in the linear momentum of the fields creates an
opposing force:

European Journal of Physics Volume 20 Number 1
Oleg D Jefimenko 1999 Eur. J. Phys. 20 39

A relativistic paradox seemingly violating conservation of
momentum
law in electromagnetic systems

[benj]

Actually, you have homed in on what is going on here... good for you!

I am currently writing a paper that will eventually appear on one of my
Websites that is a review of classical Electromagnetics including
momentum and energy flows based upon many of the ideas and calculations
of Jefimenko and others. The idea is to develop a cogent review that
covers the high points. These would only be the ideas of classical
electrodynamics and will not cover his ideas on gravity or relativity.

For the most part things are going quite well, but I was interested in
seeing if I had made any major mistakes in my thinking and also to get a
hint just what the response of traditional physics would be to such ideas
(I expect it NOT to be positive). For that reason I attempted to bounce a
few things off of sci.physics to see what happened. Timo responded but Jos
did not.

Nevertheless, I feel that quite a few points were nicely clarified. I
continue to work on this.

[Neil B]

Benj, would you consider including my "paradox" in your discussion/s?
Or is "my" paradox that same one you note? One way to imagine why my
paradox is a challenge, is that it allows a center of mass to be
displaced "without cost" - IOW, turn on the current and there is a
push in a direction, let the device travel, then turn the current off
and it stops - sitting there in a displaced position (hence, extra
orbital AM, "L", relative to a moving observer.) Note, these devices
are not "photon rockets" which require 300 MW per Newton of force.
BTW, send me a message or I will to you, we should stay in touch.
Cheers,

[Neil B]

On Jun 17, 6:32 am, Neil B <tyrannogen...@gmail.com> wrote:
> On May 24, 2:26 pm, benj <b...@iwaynet.net> wrote:
>
>
>
> > On Fri, 24 May 2013 10:14:00 -0700, Larry Harson wrote:
> > > On May 23, 5:20 am, benj <b...@iwaynet.net> wrote:
> > >>Instead zero angular momentum splits into actual
> > >> mechanical rotation and an equal opposite part stored in the fields.
> > >> Isn't that interesting?
>
> > > Linear momentum does the same thing. There's all sorts of marvelous
> > > "paradoxes" where action doesn't equal reaction for two moving charges,
> > > which implies a net force on the centre of mass of the system. The
> > > change in the linear momentum of the fields creates an opposing force:
>
> > > European Journal of Physics Volume 20 Number 1 Oleg D Jefimenko 1999
> > > Eur. J. Phys. 20 39
>
> > > A relativistic paradox seemingly violating conservation of
> > > momentum law in electromagnetic systems
>
> > Actually, you have homed in on what is going on here... good for you!
>

<snip>

>
> > Nevertheless, I feel that quite a few points were nicely clarified. I
> > continue to work on this.
>
> Benj, would you consider including my "paradox" in your discussion/s?
> Or is "my" paradox that same one you note? One way to imagine why my
> paradox is a challenge, is that it allows a center of mass to be
> displaced "without cost" - IOW, turn on the current and there is a
> push in a direction, let the device travel, then turn the current off
> and it stops - sitting there in a displaced position (hence, extra
> orbital AM, "L", relative to a moving observer.) Note, these devices
> are not "photon rockets" which require 300 MW per Newton of force.
> BTW, send me a message or I will to you, we should stay in touch.
> Cheers,

No, not the same - see this description by another author. Have you
seen anything like that "linear FDP" before? BTW some symbols pasted
out weird. Note also, in my paradox the field momentum leads to
further problems, rather than solving all of them:


LETTERS AND COMMENTS The vector potential of a moving charge in the
Coulomb gauge
by Jean-jacques Labarthe , 1999


Abstract

Abstract. In a recent paper, Jefimenko presented a ‘paradox ’ and
resolved it by a lengthy calculation using the vector potential in the
Lorenz gauge. This calculation can be avoided by using the vector
potential AC in the Coulomb gauge. In this Comment, the vector
potential AC created by a point charge in uniform motion is
determined, though it is not explicitly needed for the resolution of
the ‘paradox’. 1. The ‘paradox’ Let us recall the ‘paradox ’ [1].
Consider a system of two identical point charges q, moving on an axis
with opposite velocities. The electromagnetic forces that the two
charges exert upon each other are equal in magnitude and opposite: the
total momentum is constant. Consider then the same system, in a
reference frame where charge 1 is at rest at the origin and charge 2
is moving with constant velocity on an axis through the origin. Let
F12 (F21) be the electromagnetic force that charge 1 exerts on charge
2 (charge 2 on charge 1). Then F12 + F21 ̸ = 0, with the paradoxical
result that momentum conservation seems to be violated. The resolution
of the paradox is that the electromagnetic field contains momentum.
∫Defining the interaction momentum G = D × B dV, where D is the
electric field created by charge 1 and B is the magnetic field created
by charge 2, momentum conservation † is expressed as F12 + F21 + dG dt

[Neil B]

[benj]

On Mon, 17 Jun 2013 04:38:51 -0700, Neil B wrote:
> On Jun 17, 6:32 am, Neil B <tyrannogen...@gmail.com> wrote:
>> On May 24, 2:26 pm, benj <b...@iwaynet.net> wrote:

>> > Nevertheless, I feel that quite a few points were nicely clarified. I
>> > continue to work on this.
>>
>> Benj, would you consider including my "paradox" in your discussion/s?
>> Or is "my" paradox that same one you note? One way to imagine why my
>> paradox is a challenge, is that it allows a center of mass to be
>> displaced "without cost" - IOW, turn on the current and there is a push
>> in a direction, let the device travel, then turn the current off and it
>> stops - sitting there in a displaced position (hence, extra orbital AM,
>> "L", relative to a moving observer.) Note, these devices are not
>> "photon rockets" which require 300 MW per Newton of force. BTW, send me
>> a message or I will to you, we should stay in touch. Cheers,
>
> No, not the same - see this description by another author. Have you seen
> anything like that "linear FDP" before? BTW some symbols pasted out
> weird. Note also, in my paradox the field momentum leads to further
> problems, rather than solving all of them:

My original purpose was to clarify some of the EM ideas of Jefimenko and
then use that understanding to go through an understanding of just what
is going on in the FDP. The plan was (is) to re-work the experiment from
his version to one with a charged cylinder either just inside or just
outside a solenoid. The idea being that when Feynman said "You could
actually calculate this", it actually could relatively easily be
calculated. [I also like Timo's magnetic dipole approximation to
calculate the field from a small single turn loop] It was going well
until it was bounced off discussions here. A fine mess this has turned
into.

Not only is the problem your version (which in mine would become two
stripes of charge on opposite sides of the cylinder causing torque (but
not force) to be zero) but also the McDonald paper added yet another way
to calculate field momentum, namely the integral PHi J method. And of
course there is the problem that I can't access the Furry paper to see
how he derived that. And now below you've indicated another paper this
time apparently using the electric vector potential. I'm now guessing
that there is an expression for EM_mom using the scalar magnetic
potential as well. All of them seem incompatible with each other! There
are serious problems as seen in McDonald saying that magnetic fields from
superconducting currents can't store momentum! Clearly that flies right
in the face of the Feynman "hands-off no battery modification" and simply
can't be true. Why McDonald said that points out the incompatibilities
I'm talking about.

Point is that now I've begun questioning all this field momentum thing!
Some clear understanding of it is sorely needed by all concerned (me
included). Especially when you start reading McDonald where he is
explaining "hidden" momentum from mass loss of batteries!!!

So I now ask can LINEAR momentum go down a wire? Consider your device. I
raise the current and the two charge yoke shoots off into space carrying
energy and momentum. Well, energy down a wire is no surprise, but
conservation says that momentum must be balanced somewhere. We wave hands
and say it's in the EM fields. But it can't because the charge moved away
carrying kinetic momentum with it and it takes BOTH E and B to store
momentum! And even worse I see no reaction back upon the coil for this
acceleration! Usually momentum is conserved by the usual case of say a
spring pushing a BB. The BB gets forward momentum and the reaction gives
an equal momentum to the spring and gun. So what mechanism gives reaction
on the coil when your charges shoot off? One might say well, the moving
charges are currents and that creates a magnetic field. True, but I don't
see the geometry here producing a reaction force. Which is indeed why
Feynman made the original point of momentum stored in fields!

Anyway, that's the mess you made. Thanks for the reference below, I'll
have to look up it and some of the others if I can deal with the password
problem.

Benj