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Poynting vector

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Richard

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Jun 19, 2001, 5:26:14 PM6/19/01
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If you run a current through a fat wire the Poynting vector points in
towards the axis of the wire.
What does this mean?
Thanks

Richard

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Jun 19, 2001, 7:25:11 PM6/19/01
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Richard <gaz...@yahoo.com> wrote in message
news:8907bc57.01061...@posting.google.com...
I just thought of something.

Does the Poynting vector point inward because the energy of the system is
stored in the fields outside the wire and dissipated inside the wire via
Joule heating?


John Park

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Jun 20, 2001, 2:03:26 AM6/20/01
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What if the resisitivity is vanishingly small?

--John Park


Richard

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Jun 20, 2001, 8:57:27 AM6/20/01
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Thanks a lot John. Now I have no idea what's going on. So how about some
help here?


John Park <af...@FreeNet.Carleton.CA> wrote in message
news:9gpebe$k19$1...@freenet9.carleton.ca...

John

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Jun 20, 2001, 10:05:01 AM6/20/01
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No, the Poynting vector is just a peculiar thing in EM theory.
Check out Feynman in the second book of the F-lectures on physics. He calls it
"a really nuts theory".
There seems however, not to be an alternative.

Eric Prebys

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Jun 20, 2001, 11:32:23 AM6/20/01
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Then the electric field goes to zero and so does the Poynting vector.

-Eric

--
---------------------------------------------------------
* Eric Prebys, Physics Department, Princeton University *
* 609-258-4910, FAX: -6360, Email: pre...@princeton.edu *
* WWW: http://www.princeton.edu/~prebys/ *
---------------------------------------------------------

Eric Prebys

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Jun 20, 2001, 11:31:25 AM6/20/01
to

The interpretation of the Poynting vector as energy flow comes out of
manipulating Maxwell's equations around a bit to get a conservation
law, but it really has meaning only in the context of propagating
waves (ie, time varying fields). There's a nice discussion of this
very thing in Eyges "The Classical Electromagnetic Field" (In my 1972
edition
its pages 192->203).

In your particularly popular example, if you examine it closely, you'll
find that the total energy flux correctly gives you the Joule heating,
but clearly the direction is meaningless.

Personally, I've always hated this example because it totally confuses
students who might otherwise be starting to understand the material.

If you want a better example of why the Poynting vector is meaningless
for static fields, consider this...

- You have a superconducting solenoid.
- Insert two oppositely charged plates into the aperture so that
the E-field is at a right angle to the B-field
There you have it! Clearly there's no energy dissipation in this system,
but you have a very well defined Poynting vector.

David Rutherford

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Jun 20, 2001, 9:02:27 PM6/20/01
to

Sure there is, try this

http://www.softcom.net/users/der555/newtransform.pdf

or this,

http://www.softcom.net/users/der555/elecmass.pdf

An island of sanity in an ocean of nuts.

--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555

David Rutherford

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Jun 20, 2001, 9:47:28 PM6/20/01
to

Eric Prebys wrote:
>
> The interpretation of the Poynting vector as energy flow comes out of
> manipulating Maxwell's equations around a bit to get a conservation
> law,

The conservation laws are fundamentally flawed, since the momentum
of the field is incompletely described by the Poynting vector
term ExB/4pi when integrated over all space, as pointed out by Feynman
("Lectures on Physics", Vol. 2, pg. 28-2).

For the correct conservation laws, see section 18, Eq. (18.4) in

http://www.softcom.net/users/der555/newtransform.pdf

Ed Green

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Jun 20, 2001, 10:55:31 PM6/20/01
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From: "Richard" Gaz...@yahoo.com

>Thanks a lot John. Now I have no idea what's going on. So how about some
>help here?

Richard, I have heard evil rumors of the Poynting vector; it
may satisfy conservation of momentum if integrated over
any closed surface, but it may not reflect an absolute
momentum flux. So you answer may be "nothing".

By the way, before I proceed further into infamy, can
you tell me if a "fat" wire means that the fields do not
go to infinity anywhere, and the Poynting vector also
remains bounded?

I'm also not sure I would absolutely give up your idea
of resistivity... ah, I see somebody has noticed that
the E field will go to zero for a superconductor.


Ed Green

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Jun 20, 2001, 11:05:46 PM6/20/01
to
From: Eric Prebys pre...@princeton.edu

>The interpretation of the Poynting vector as energy flow comes out of
>manipulating Maxwell's equations around a bit to get a conservation
>law, but it really has meaning only in the context of propagating
>waves (ie, time varying fields). There's a nice discussion of this
>very thing in Eyges "The Classical Electromagnetic Field" (In my 1972
>edition
>its pages 192->203).
>
>In your particularly popular example, if you examine it closely, you'll
>find that the total energy flux correctly gives you the Joule heating,
>but clearly the direction is meaningless.

Is it? Just to be devil's advocate, I would say any time you have
infinities: infinitely long wire, situation in steady state for an
infinitely long time and so forth; the "causal" relation between
fields, charges and currents is obscured; it's not clear what
causes what, just that they all hang together self-consistently.

So it's not _obviously_ meaningless to me that energy is
flowing in from the field to the wire; the E-field is driving
the current which is dissipating the power, is it not?

>Personally, I've always hated this example because it totally confuses
>students who might otherwise be starting to understand the material.

Tough. ;)

>If you want a better example of why the Poynting vector is meaningless
>for static fields, consider this...
>
> - You have a superconducting solenoid.
> - Insert two oppositely charged plates into the aperture so that
> the E-field is at a right angle to the B-field
>There you have it! Clearly there's no energy dissipation in this system,
>but you have a very well defined Poynting vector.

But who said anything about energy dissipation? Presumably
the Poynting vector is divergenceless; and while it may seem
quirky to have little loops of energy or momentum flux doing
nothing it is not evidently absurd. Atoms and electrons may
have angular momentum which persists "forever", why not
the EM field?


Eric Prebys

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Jun 21, 2001, 1:32:04 AM6/21/01
to

Ed Green wrote:
>
> >If you want a better example of why the Poynting vector is meaningless
> >for static fields, consider this...
> >
> > - You have a superconducting solenoid.
> > - Insert two oppositely charged plates into the aperture so that
> > the E-field is at a right angle to the B-field
> >There you have it! Clearly there's no energy dissipation in this system,
> >but you have a very well defined Poynting vector.
>
> But who said anything about energy dissipation? Presumably

You're right there, I should have said "energy flow", but...

> the Poynting vector is divergenceless; and while it may seem

^^^^^^^^^^^^^^
No! Think again about the geometry I set up. The superconducting
solenoid sets up a more or less uniform field in its aperture.
If I insert a plate capacitor into that aperture with the E-field
at right angles to the B-field, then I've created a little region
where the Poynting vector is constant, and it's zero everywhere else.
It goes to the edge of the plates and then stops - patently NOT
divergenceless!

> quirky to have little loops of energy or momentum flux doing
> nothing it is not evidently absurd. Atoms and electrons may
> have angular momentum which persists "forever", why not
> the EM field?

-Eric

franz heymann

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Jun 21, 2001, 4:08:47 AM6/21/01
to

Eric Prebys <pre...@princeton.edu> wrote in message
news:3B3186D4...@princeton.edu...

The E field of the plates does not stop sharply at the edge of
the plate. You have to take into account its extent over all
space.

>
>
>
> > quirky to have little loops of energy or momentum flux doing
> > nothing it is not evidently absurd. Atoms and electrons may
> > have angular momentum which persists "forever", why not
> > the EM field?
>

Franz Heymann

Eric Prebys

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Jun 21, 2001, 9:23:41 AM6/21/01
to

>
> The E field of the plates does not stop sharply at the edge of
> the plate. You have to take into account its extent over all
> space.
>

You're right of course. That was really stupid of me! Actually, I
was just logging in to cancel my last message when I realized it
was bullshit. Too late, I guess. Well, that's what I get for posting
at 1AM. Of course the Poynting vector is divergenceless, since that
part just comes from Maxwell's eqns.

If I haven't lost all credibility, let me add to my argument...

I *still* claim that in that in this context, while the Poynting
vector is (I now admit) divergenceless, the interpretation as
energy flow is, at the very least, confusing.

The interpretation of the Poynting Vector as energy flow comes from
the fact that you get a differential equation that relates the
divergence of the Poynting vector to the rate of change of the
local energy contained by the fields. Clearly if local energy is
changing, then it's appropriate to interpret this as energy flow;
however, if the fields are static, this just reduces to the statement
that the Poynting Vector is divergenceless. Looking back at the
example I gave, while this is mathematically correct, I think you'll
agree that there is no real physical interpretation to the strange
circular "flow" of energy it describes, certainly not in the sense
we get when we apply it to, say, dipole radiation.

Sorry again about my last post.

-Eric


> >
> >
> >
> > > quirky to have little loops of energy or momentum flux doing
> > > nothing it is not evidently absurd. Atoms and electrons may
> > > have angular momentum which persists "forever", why not
> > > the EM field?
> >
>
> Franz Heymann

--

Richard

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Jun 21, 2001, 10:32:49 AM6/21/01
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Ed Green <null...@aol.com> wrote in message
news:20010620225531...@ng-fm1.aol.com...


Sorry Ed. I don't even know enough to tell if your question is serious or
not. I'm trying to learn this stuff on my own (just because I'm interested)
and it's pretty tough going.

Was it sergeant Clink who use to say "I know nossing..absolutely nossing"

Richard

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Jun 21, 2001, 10:32:49 AM6/21/01
to

Ed Green <null...@aol.com> wrote in message
news:20010620225531...@ng-fm1.aol.com...

Richard

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Jun 21, 2001, 10:48:50 AM6/21/01
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null...@aol.com (Ed Green) wrote in message news:<20010620225531...@ng-fm1.aol.com>...

Sorry Mr. Ed. I don't even know enough to tell if your question is serious


or not. I'm trying to learn this stuff on my own
(just because I'm interested) and it's pretty tough going.

Was it sergeant Clink who used to say "I know nossing….I know nossing"

A horse is a horse of course of course.

Ed Green

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Jun 21, 2001, 1:09:43 PM6/21/01
to
From: "Richard" Gaz...@yahoo.com

>Ed Green <null...@aol.com> wrote in message
>news:20010620225531...@ng-fm1.aol.com...

>> I have heard evil rumors of the Poynting vector; it


>> may satisfy conservation of momentum if integrated over
>> any closed surface, but it may not reflect an absolute
>> momentum flux. So you answer may be "nothing".
>>
>> By the way, before I proceed further into infamy, can
>> you tell me if a "fat" wire means that the fields do not
>> go to infinity anywhere, and the Poynting vector also
>> remains bounded?
>>
>> I'm also not sure I would absolutely give up your idea
>> of resistivity... ah, I see somebody has noticed that
>> the E field will go to zero for a superconductor.
>
>
>Sorry Ed. I don't even know enough to tell if your question is serious or
>not. I'm trying to learn this stuff on my own (just because I'm interested)
>and it's pretty tough going.

You mean am I trolling you? Oh well, I'm not, but of course
that assertion carries no weight.

What I meant is given that you added the extra adjective "fat",
to the properties of your wire, I wondered if there were a reason.
And the reason I tentatively came up with was that E and B
might remain finite for a current carrying wire of finite diameter,
but I think B would go to infinity at an infinitely thin current.

>Was it sergeant Clink who use to say "I know nossing..absolutely nossing"

Yes, I recall: cute fat N***s. But wait, Klink was the commandant.
The sergeant was Schultz.

Funny, as a kid, I never got that joke: Colonel Klink,

Ed Green

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Jun 21, 2001, 1:36:40 PM6/21/01
to
From: Eric Prebys pre...@princeton.edu

>> >If you want a better example of why the Poynting vector is meaningless
>> >for static fields, consider this...
>> >
>> > - You have a superconducting solenoid.
>> > - Insert two oppositely charged plates into the aperture so that
>> > the E-field is at a right angle to the B-field
>> >There you have it! Clearly there's no energy dissipation in this system,
>> >but you have a very well defined Poynting vector.
>>
>> But who said anything about energy dissipation? Presumably
>
>You're right there, I should have said "energy flow", but...
>
>> the Poynting vector is divergenceless; and while it may seem
> ^^^^^^^^^^^^^^
>No! Think again about the geometry I set up. The superconducting
>solenoid sets up a more or less uniform field in its aperture.
>If I insert a plate capacitor into that aperture with the E-field
>at right angles to the B-field, then I've created a little region
>where the Poynting vector is constant, and it's zero everywhere else.
>It goes to the edge of the plates and then stops - patently NOT
>divergenceless!

Oops.

I also suppose the fact that we may do some work bringing
the charged capacitor into the magnetic field doesn't matter.

I suppose edge effects are not going to rescue us either?

Hang on though; I'm not 100% convinced: introduce a
finite capacitor "edge on" in a region of constant vertical B
so that energy is ostensively flowing left-to-right across the
gap, originating in nothing and ending in nothing.

But now think about the fringe field of a finite gap; bulging
out from the sides and in fact wrapping around the edges;
so we have a kind of toroid of distorted field (assuming a
circular gap). Now it is certainly plausible if not obvious that
if we carefully consider the Poynting vector in this toroid we
will have a return energy flow along the edges exactly
balancing the putative left to right flow in the gap.

In fact, I would wager money on it. ;)

No, I'm not sure what this closed energy loop would mean;
I said I thought it widely accepted that the Poynting vector
may describe closed energy loops which were apparently
fictitious: though I would not be too fast to dismiss them
as entirely meaningless: but the real message in your
example may be that we cannot have a finite capacitor
whose field stops short at the edges. The Poynting vector
knows! :-)

Ed Green

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Jun 21, 2001, 1:38:43 PM6/21/01
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From: "franz heymann" franz....@care4free.net

>The E field of the plates does not stop sharply at the edge of
>the plate. You have to take into account its extent over all
>space.

Franz once again expresses tersely what I expressed in
many words...

Oh well, at least I didn't know the answer when I started
writing!


BillSincl

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Jun 21, 2001, 1:41:18 PM6/21/01
to
> I see somebody has noticed that
>the E field will go to zero for a superconductor.
>

In that case there is no dissipation of energy.

Ed Green

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Jun 21, 2001, 1:49:06 PM6/21/01
to
>From: Eric Prebys pre...@princeton.edu

> Of course the Poynting vector is divergenceless, since that
>part just comes from Maxwell's eqns.

>I *still* claim that in that in this context, while the Poynting


>vector is (I now admit) divergenceless, the interpretation as
>energy flow is, at the very least, confusing.
>
>The interpretation of the Poynting Vector as energy flow comes from
>the fact that you get a differential equation that relates the
>divergence of the Poynting vector to the rate of change of the
>local energy contained by the fields. Clearly if local energy is
>changing, then it's appropriate to interpret this as energy flow;
>however, if the fields are static, this just reduces to the statement
>that the Poynting Vector is divergenceless. Looking back at the
>example I gave, while this is mathematically correct, I think you'll
>agree that there is no real physical interpretation to the strange
>circular "flow" of energy it describes, certainly not in the sense
>we get when we apply it to, say, dipole radiation.

Again, I wouldn't be so sure. It may be trying to tell us
something, it may be an artifact. Sometimes things which
look Iike artifacts turn out to be more "physical" than we
thought; cf. Aharanov-Bohm effect.

Again, the question of angular momentum comes to mind:
I imagine if we have closed loops of Poynting vector we
would predict the field to have angular momentum, even
while seemingly "static".

>Sorry again about my last post.

I found it stimulating. ;)


Ed Green

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Jun 21, 2001, 2:28:25 PM6/21/01
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>From: bill...@aol.com (BillSincl)

>> I see somebody has noticed that
>>the E field will go to zero for a superconductor.
>>
>
>In that case there is no dissipation of energy.

Yes, that was the point.


Roy McCammon

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Jun 21, 2001, 2:22:42 PM6/21/01
to

That's pretty much it. Outside the wire, Poynting vector
points axially along the wire and represents energy
delivered to the load and lost to the wire. Inside the
wire the Poynting vector is directed radially inward and
represents ohmic heating of the wire. If the wire has
no resistance, then inside the wire the E field and
Poynting vector is zero and no energy is lost to heating
the wire.

--
If you are one in a million; there are 6000 people just like
you.

Opinions expressed herein are my own and may not represent those of my employer.

Roy McCammon

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Jun 21, 2001, 2:39:06 PM6/21/01
to
Eric Prebys wrote:

> I *still* claim that in that in this context, while the Poynting
> vector is (I now admit) divergenceless, the interpretation as
> energy flow is, at the very least, confusing.

perhaps you have brought in an unwarranted assumption
that is confusing you. Something like "if I cannot
see it, then its not there." Its these unwarranted
assumptions that our intuition sometimes drags in that
often lead to apparent paradoxes.

Roy McCammon

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Jun 21, 2001, 2:35:31 PM6/21/01
to
Ed Green wrote:

> Again, the question of angular momentum comes to mind:
> I imagine if we have closed loops of Poynting vector we
> would predict the field to have angular momentum, even
> while seemingly "static".

yes, that is right. In the case where the two plates
are identical and equal and opposite charge and placed
symmetrically in the solenoid, the angular momentum
is zero (there is just as much clock wise flux as
there is ccw flux). But charge them unequally or place
them off axis and there is non zero angular momentum
stored in the field. This means that as you brought
it together, you had to apply a torque.

Ace Schallger

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Jun 21, 2001, 8:32:19 PM6/21/01
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[[ This message was both posted and mailed: see
the "To," "Cc," and "Newsgroups" headers for details. ]]

In article <3B314A66...@softcom.net>, David Rutherford
<druth...@softcom.net> wrote:

> John wrote:
> >
> > No, the Poynting vector is just a peculiar thing in EM theory.
> > Check out Feynman in the second book of the F-lectures on physics. He
> > calls it "a really nuts theory".
> > There seems however, not to be an alternative.
>
> Sure there is, try this
>
> http://www.softcom.net/users/der555/newtransform.pdf
>
> or this,
>
> http://www.softcom.net/users/der555/elecmass.pdf
>
> An island of sanity in an ocean of nuts.

Well, David you arrive by a tortuous route at the notion that there are
not electric monopoles in your newtransform.pdf paper. That is a solid
conclusion but there's a simpler and more direct route which is just
considering Maxwell's equations written in terms of E and H only.

Maxwell's equations when written in terms of E and H only we have:

1) Del X H = permittivity*partial E/partial t
2) Del X E = -permeability*partial H/partial t
3) Del dot E = 0
4) Del dot H = 0

Del dot H = 0 tells us that the H loop described in the first equation
has zero divergence or that there are no sources or sinks in the H.
This is commonly interpreted to mean that there are no magnetic
monopoles or that a north or south magnetic pole or magnetic charge
cannot exist in isolation.

Then what does Del dot E tell us? In fact, it is telling us the same
thing except with repect to the E loop. It tells us that E loop
described in the second equation has zero divergence or that there are
no sources or sinks in the E. Can this be interpreted to mean that
there are no electric monopoles or that a negative or positive electric
pole or charge cannot exist in isolation? That is exactly what it
means. It really means that for every charge there must exist its
relational conjugate, that they cannot actually exist in absolute
isolation. This tells us that an E loop must consist of both a charge
and its conjugate just as Del dot H = 0 tells us that an H loop must
consist of its magnetic charge and its conjugate. So if we follow the
logic of these four equations then the first one is telling us that if
E is changing with time at some point then H has curl at that point and
can be considered as forming a small closed loop around the changing E.
Also we see that if E is changing with time that H, also, will change
even though not necessarily in the same way. The second equation is
telling us that if H is changing with time at some point then E has
curl at that point and can be considered as forming small closed loops
around the changing H lines. This tells us once again that we have
changing E which was the original hypothesis of the first equation.
But this changing E is a small distance away from the original
disturbance and we know that there is a finite amount of time from the
appearance of one changing E to the next and we can properly surmise
that the d/t relationship implied here is c, the velocity of light.
So, these equations tell us about the propagation of electromagnetic
radiation but they also tell us that electromagnetic radiation consists
of a charge and its conjugate in superposition. If we rotate the
electric flux density vector, D, pi/2 radians all in the same sense
(all clockwise or all anticlockwise) everywhere in an electric flux
loop that flux loop becomes converted to a magnetic flux loop or H
loop. If we rotate the magnetic flux density vector, B, pi/2 radians
all in the same sense (all clockwise or all anticlockwise) everywhere
in a magnetic flux loop (H loop) that flux loop becomes converted to an
electric flux loop or E loop. This tells us that an H loop and an E
loop are two different modes of the same thing. And it further tells
us that the propagation of electromagnetic radiation is the classic
example of such mode changes propagating in a straight line. But
because it tells us that electromagnetic radiation consists of a charge
and its conjugate in superposition and because we recognize that the
presence of a photon at the site of an atom causes the electron to be
repelled from the nucleus it also tells us that a photon is a time
gradient field. (since this is a clip from another post I haven't
included the part that describes the complete interaction of elementary
charges which, in fact, will be quite opposite to Coulombic
interactions if the two charges have a common de Broglie wavelength
that is equal to or greater than the interparticle distance [calculated
from a center of momentum frame].

We can conclude that any flux loop produces a monolithic gravitational
field the terminus of which is along the toroidal axis of such a flux
loop. Since a gravitational field is time gradient field then
particles that approach the terminus achieve a condition where they
will overlap in momentum space and if they are same charged they will
attractively interact and if oppositely charge they will repulsively
interact. Therefore, it is easy to predict that a gravitational field
will cause a strong charge separation effect.

There's also no doubt that you are intuitively correct about a rotation
transverse to the direction of translation but it appears to me that
you are still caught in or trapped in the old notion of a 'field' as a
continuous structure. We mustn't confuse the mathematics with the
actual physics even if the mathematics is sometimes a great aid to
calculation for engineering purposes. An electric flux loop can be
thought of almost like a current loop in that because there are
incremental directional changes then we have, D, or electric flux
density with circulation. A current loop would be self repulsive and
therefore expansive or tending to expand because everywhere at pi
radians around the loop from a given point P would be an antiparallel
current segment at P' but the flux loop has zero divergence and that
means it must act like an incompressible fluid. How does the minor
radius of a flux loop torus decrease while the major radius increases?
By an orthogonal acceleration, and that orthogonal component is what we
consider the magnetic field. We can think of the propagation of a
charged particle along its spin axis with the spin being an orthogonal
acceleration (to the translational axis) which then produces a helical
propagation. When the pitch of the helix reaches zero then there has
been a pi/2 radian rotation of the direction of propagation and an in
the case of an electric flux loop, well, it has become converted into a
magnetic flux loop and in the case of a magnetic flux loop it becomes
converted into an electric flux loop.

In other words, "Maxwell's equations are Lorentz invariant and as a
consequence of this the electric field and the magnetic field at
a given point in space at a given instant are no longer seen as
independent entities but rather as components of a single object or
tensor. This is on exactly the same footing as saying that the
coordinates of a particle have no independent meaning only the distance
from the origin is objective, since individual coordinates can be made
to vary by changing the direction of the orthogonal coordinate axes."
(Quoted from a private email from Daniel Pisello author of
'Gravitation, Electromagnetism and Quantized Charge'.)

Regards,

David Rutherford

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Jun 21, 2001, 9:30:34 PM6/21/01
to

My eyeballs hurt 3-\.

Ed Green

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Jun 22, 2001, 1:24:46 AM6/22/01
to
>From: rbmcc...@mmm.com (Roy McCammon)

>> Richard <gaz...@yahoo.com> wrote in message
>> news:8907bc57.01061...@posting.google.com...
>> > If you run a current through a fat wire the Poynting vector points in
>> > towards the axis of the wire.
>> > What does this mean?
>> > Thanks
>> >
>> I just thought of something.
>>
>> Does the Poynting vector point inward because the energy of the system is
>> stored in the fields outside the wire and dissipated inside the wire via
>> Joule heating?
>
>That's pretty much it. Outside the wire, Poynting vector
>points axially along the wire and represents energy
>delivered to the load and lost to the wire. Inside the
>wire the Poynting vector is directed radially inward and
>represents ohmic heating of the wire. If the wire has
>no resistance, then inside the wire the E field and
>Poynting vector is zero and no energy is lost to heating
>the wire.

I'm confused now. I thought Richard was asserting the Poynting
vector pointed inward _outside_ the wire, which would mean
the E-field was axially along the wire, which made sense to me.

Now, why would the E-field point radially outward, as your
version would seem to require? The wire is not charged, it's
just conducting current, isn't it? Or is charge necessarily
piled up by virtue of distributed capacitance?

I had not even thought about power delivered to the load; but
now I am concerned that the current returning from the load
is just the same as the current going to the load! So surely
the Poynting vector is just the same on both branches of the
circuit, and the power is coming back.... err....

Eureka!

The positive branch of the circuit indeed has a net positive
charge piled up, and the negative branch a net negative
charge piled up, and _both_ branches have a component
of the Poynting vector pointing at the load.

That's cute.


Ed Green

unread,
Jun 22, 2001, 1:49:52 AM6/22/01
to
From: rbmcc...@mmm.com (Roy McCammon)

>Ed Green wrote:
>
>> Again, the question of angular momentum comes to mind:
>> I imagine if we have closed loops of Poynting vector we
>> would predict the field to have angular momentum, even
>> while seemingly "static".
>
>yes, that is right. In the case where the two plates
>are identical and equal and opposite charge and placed
>symmetrically in the solenoid, the angular momentum
>is zero (there is just as much clock wise flux as
>there is ccw flux). But charge them unequally or place
>them off axis and there is non zero angular momentum
>stored in the field. This means that as you brought
>it together, you had to apply a torque.

That is cute also. I see why there is no net angular momentum
in the totally symmetric case: we have a kind of flux "fountain"
wrapping back along both sides of the plates. I think I can
see what is going on in the totally assymetric capacitor -- a
single charged plate. We then would have flux (meaning
momentum flux) circulating around the plate, AFAICT.

That's a fairly strange conclusion about the field. It's not
just that we apply a torque in setting things up, but that
if we measure the torque on plate and solenoid they
don't cancel, as if we were spinning something up. Can
this be demonstrated? Just to take the other side of
devil's advocacy, are you sure it's possible to set up
situations with J stored in a static field, or does nature
cleverly make attempts to isolate net angular field
momentum like trying to isolate a magnetic monopole
by bisecting a bar magnet?

If it is real then we have less reason to be weirded out by
microscopic versions of angular momentum like spin; we
can construct a dissipationless macroscopic version.

franz heymann

unread,
Jun 21, 2001, 2:47:41 PM6/21/01
to

Ed Green <null...@aol.com> wrote in message
news:20010621133640...@ng-xc1.aol.com...

I won't take you on. I don't bet when it is a cert that I am on
to a loser :-).


>
> No, I'm not sure what this closed energy loop would mean;
> I said I thought it widely accepted that the Poynting vector
> may describe closed energy loops which were apparently
> fictitious: though I would not be too fast to dismiss them
> as entirely meaningless: but the real message in your
> example may be that we cannot have a finite capacitor
> whose field stops short at the edges. The Poynting vector
> knows! :-)

Franz Heymann


franz heymann

unread,
Jun 21, 2001, 2:16:45 PM6/21/01
to

Eric Prebys <pre...@princeton.edu> wrote in message
news:3B31F55D...@princeton.edu...

>
>
> >
> > The E field of the plates does not stop sharply at the edge
of
> > the plate. You have to take into account its extent over all
> > space.
> >
>
> You're right of course. That was really stupid of me! Actually,
I
> was just logging in to cancel my last message when I realized
it
> was bullshit. Too late, I guess. Well, that's what I get for
posting
> at 1AM. Of course the Poynting vector is divergenceless, since
that
> part just comes from Maxwell's eqns.
>
> If I haven't lost all credibility, let me add to my argument...
>
> I *still* claim that in that in this context, while the
Poynting
> vector is (I now admit) divergenceless, the interpretation as
> energy flow is, at the very least, confusing.

I agree. I don't know what the significance is of energy
circulating endlessly around loops, as Poynting says it does when
you have independent static fields E and B occupying the same
space.

>
> The interpretation of the Poynting Vector as energy flow comes
from
> the fact that you get a differential equation that relates the
> divergence of the Poynting vector to the rate of change of the
> local energy contained by the fields. Clearly if local energy
is
> changing, then it's appropriate to interpret this as energy
flow;
> however, if the fields are static, this just reduces to the
statement
> that the Poynting Vector is divergenceless. Looking back at
the
> example I gave, while this is mathematically correct, I think
you'll
> agree that there is no real physical interpretation to the
strange
> circular "flow" of energy it describes, certainly not in the
sense
> we get when we apply it to, say, dipole radiation.
>
> Sorry again about my last post.

I make worse mistakes than you do.

Roy McCammon

unread,
Jun 22, 2001, 10:30:40 AM6/22/01
to
Ed Green wrote:
>
> >From: rbmcc...@mmm.com (Roy McCammon)

You're doing good now. You need a couple of other pieces
of information. The first is that the speed of EM waves
in a good conductor is very slow, about like a brisk walk.
The second is that Snell's law is operative. The ratio of
speed outside the conductor to the speed inside the
conductor is so large that if we apply Snell's law, the
wave entering the conductor from the outside will always
be bent to perpendicular. So, visualize the EM wave
coming from the source toward the load traveling almost
axially to the wire ( or visualize it hitting the wire at
a very shallow angle). At the point where it impinges on
the wire, it makes an almost right angle turn and dives
into the wire, where it is dissipated as heat and where
it gives rise to an axial E field inside the wire ( which
gives rise to the current that gives rise to the B field
which is part of the external EM wave which is why a wire
can guide an EM wave from a source to a load.)

The best way to visualize is to be the EM wave. You
are running along the top of a big conductor. Your
head is your E field vector pointing radially away
from the wire. Your right arm is your B field vector
pointing circumferentially around the wire. Your
nose is your Poynting vector. You run along until an
electron grabs your foot. You pivot on your foot by
ninety degrees and end up propagating face down into
the wire. Your E field vector is now pointed along
the wire, your Poynting vector points into the wire,
and you are not going nearly as fast as you were.
Your axial momentum is transferred to the electron
that grabbed your foot which contributes to the wire
current.

Richard

unread,
Jun 22, 2001, 3:38:54 PM6/22/01
to

Roy McCammon <rbmcc...@mmm.com> wrote in message
news:3B335690...@mmm.com...

> You're doing good now. You need a couple of other pieces
> of information. The first is that the speed of EM waves
> in a good conductor is very slow,

<snip>

Now I'm very confused. If the speed of EM waves in a good conductor is very
slow, then what's the big deal about the recent experiments that have
succeeded in the slowing of the speed of light

Ed Green

unread,
Jun 22, 2001, 9:09:12 PM6/22/01
to
From: "Richard" Gaz...@yahoo.com

ITIYM experiments which have slowed the speed of light glacially.

I've heard it repeated that there is something extremely slow
connected with a conductor, but generally reported as the
carrier speed; which of course may also be true.

OTOH the speed of an EM wave bound to a conductor as an
"external wave guide" is a substantial fraction of the speed of
light, for what that's worth.

Ed Green

unread,
Jun 22, 2001, 9:21:16 PM6/22/01
to
From: rbmcc...@mmm.com (Roy McCammon)

I am speechless.

And I suppose the depth of the shallow indentation we
make in the conductor when we fall face downward
is the skin depth.

Bennett Standeven

unread,
Jun 22, 2001, 10:57:06 PM6/22/01
to
"Richard" <Gaz...@yahoo.com> wrote in message news:<w7NY6.30$56.1...@e420r-atl3.usenetserver.com>...

They slowed the group velocity in the recent experiments, rather than
the phase velocity. That might make a difference.

Old Man

unread,
Jun 23, 2001, 1:00:38 AM6/23/01
to

franz heymann <franz....@care4free.net> wrote in message
news:3b32f129$0$15022$cc9e...@news.dial.pipex.com...

>
> Eric Prebys <pre...@princeton.edu> wrote in message
> news:3B31F55D...@princeton.edu...
> >
> >
> > >
> > > The E field of the plates does not stop sharply at the edge
> of
> > > the plate. You have to take into account its extent over all
> > > space.
> > >
> >
> > You're right of course. That was really stupid of me! Actually,
> I
> > was just logging in to cancel my last message when I realized
> it
> > was bullshit. Too late, I guess. Well, that's what I get for
> posting
> > at 1AM. Of course the Poynting vector is divergenceless, since
> that
> > part just comes from Maxwell's eqns.
> >
> > If I haven't lost all credibility, let me add to my argument...
> >
> > I *still* claim that in that in this context, while the
> Poynting
> > vector is (I now admit) divergenceless, the interpretation as
> > energy flow is, at the very least, confusing.
>
> I agree. I don't know what the significance is of energy
> circulating endlessly around loops, as Poynting says it does when
> you have independent static fields E and B occupying the same
> space.
>
> Franz Heymann

I am convinced otherwise. One must consider the forces and torque's
required to assemble the system. In this way, the Poynting vector is just
expressing conservation of energy and angular momentum. [Old Man]


Roy McCammon

unread,
Jun 23, 2001, 4:08:53 AM6/23/01
to
Ed Green wrote:

> I am speechless.

Enlightenment has different effects on different people.

> And I suppose the depth of the shallow indentation we
> make in the conductor when we fall face downward
> is the skin depth.

Well it doesn't account for the dependence on wavelength,
but its not entirely unlike that.

Richard

unread,
Jun 23, 2001, 11:59:04 AM6/23/01
to

Ed Green <null...@aol.com> wrote in message
news:20010622210912...@ng-fx1.aol.com...

>
Well, I'll be damned.

I calculated the propagation speed of a 1 MHz radio wave in copper and it
came out to 400m/s. I guess we learn something new every day.... IF we visit
sci.physics.


franz heymann

unread,
Jun 23, 2001, 3:45:24 AM6/23/01
to

Roy McCammon <rbmcc...@mmm.com> wrote in message
news:3B335690...@mmm.com...

The only EM *waves* which can exist in a good conductor are
evanescent waves near the surface. They do not "propagate" in
the usual sense of the word. I think you must be referring to
the drift speed of electrons, which is something different.

Franz Heymann


franz heymann

unread,
Jun 23, 2001, 4:10:37 AM6/23/01
to

Old Man <o...@bg.net> wrote in message
news:TdVY6.1053$io1....@newsfeed.slurp.net...

To make the matter more tractable, how about forgetting about an
extremely complicated thing like a pair of finite conductors in a
locally uniform magnetic field.
The simplest situation I can visualise is that of a point charge
sitting at the centre of a circular loop carrying a current. The
Poynting field consists of a set of coaxial circles. If you
bring the charge in along the axis of the circle from infinity to
the centre of the circle, there is never any force or torque
exerted on the charge. and yet the Poynting field is established.
Where do you go from here?

Franz Heymann


franz heymann

unread,
Jun 23, 2001, 2:15:28 PM6/23/01
to

Richard <Gaz...@yahoo.com> wrote in message
news:XZ2Z6.2089$f6.6...@e420r-atl2.usenetserver.com...

A 1 mHz wave will not propagate in copper at all, except as an
evanescent wave.
The copper acts essentially as a reflector.

Franz Heymann


Ed Green

unread,
Jun 23, 2001, 2:50:41 PM6/23/01
to
From: "Richard" Gaz...@yahoo.com

>Ed Green <null...@aol.com> wrote in message
>news:20010622210912...@ng-fx1.aol.com...

>> I've heard it repeated that there is something extremely slow


>> connected with a conductor, but generally reported as the
>> carrier speed; which of course may also be true.
>>
>> OTOH the speed of an EM wave bound to a conductor as an
>> "external wave guide" is a substantial fraction of the speed of
>> light, for what that's worth.
>
>>
>Well, I'll be damned.
>
>I calculated the propagation speed of a 1 MHz radio wave in
>copper and it came out to 400m/s.

How did you do that?

Serious question for a dummy (me).

> I guess we learn something
> new every day.... IF we visit
>sci.physics.

You have no idea how welcome such a sentiment is to a
jaded and crackpot warn sci.physicist.


Richard

unread,
Jun 23, 2001, 3:28:06 PM6/23/01
to

Ed Green <null...@aol.com> wrote in message
news:20010623145041...@ng-ca1.aol.com...
Well, this is from Griffiths which is a first year text so take it for what
it's worth.

Propagation velocity = wavelength times frequency
= (4x10^-4)*10^6 = 400m/s

Wavelength = 2pi*sqrt(2/omega*sigma*mu_0)

It's a little more convoluted than this, but you can find it all in chapter
9.4 problem 9.18 of the third edition.

However if Franz Heymann says it's wrong then it's probably wrong.

Roy McCammon

unread,
Jun 23, 2001, 7:15:37 PM6/23/01
to
franz heymann wrote:
.

> To make the matter more tractable, how about forgetting about an
> extremely complicated thing like a pair of finite conductors in a
> locally uniform magnetic field.
> The simplest situation I can visualise is that of a point charge
> sitting at the centre of a circular loop carrying a current. The
> Poynting field consists of a set of coaxial circles. If you
> bring the charge in along the axis of the circle from infinity to
> the centre of the circle, there is never any force or torque
> exerted on the charge. and yet the Poynting field is established.
> Where do you go from here?

You will find that if the flux is circulating clockwise
inside the ring, it will circulate counter clockwise
outside the ring. I haven't done the calculation but
I suspect it all sums to zero.

Roy McCammon

unread,
Jun 23, 2001, 7:29:06 PM6/23/01
to

I get 414 m/s.

Roy McCammon

unread,
Jun 23, 2001, 7:34:05 PM6/23/01
to
franz heymann wrote:
.
> A 1 mHz wave will not propagate in copper at all, except as an
> evanescent wave.
> The copper acts essentially as a reflector.

In an evanescent wave, E and H have a phase difference of
90 degrees. In a good conductor the phase difference is
very close to 45 degrees. But in any case, there is a
substantial impedance mismatch between air and copper.

Old Man

unread,
Jun 23, 2001, 8:04:30 PM6/23/01
to

franz heymann <franz....@care4free.net> wrote in message
news:3b34d534$0$15029$cc9e...@news.dial.pipex.com...

I think that, with the charge exactly at the center, the field angular
momentum inside of the current loop cancels that outside of the loop, to
yield zero net field angular momentum. An application of torque is required
to place the charge at any off-axis position, however, and in this case, the
final net field circulation will reflect that effort with a non-zero value.
Feynman illustrated this problem with a permanent dipole magnet, and,
without comment, sticks the charge, in an off axis position, to the side off
the magnet. Decades latter, another skillful slight-of-hand artist, named
Franz Heymann, sticks, without comment, the charge in the exact center of a
current loop!!! [Old Man]

franz heymann

unread,
Jun 23, 2001, 5:27:02 PM6/23/01
to

Richard <dic...@yahoo.com> wrote in message
news:U16Z6.2352$f6.1...@e420r-atl2.usenetserver.com...

The reason why it is wrong is that you took no account of the
conductivity of the copper.

Franz Heymann

franz heymann

unread,
Jun 24, 2001, 4:01:13 AM6/24/01
to

Roy McCammon <rbmcc...@ieee.org> wrote in message
news:3B352768...@ieee.org...

Mea culpa! I misused the term "evanescent". In my mind was the
fact that the wave propagates *inside* the copper with an
exponential decay.

Franz Heymann


franz heymann

unread,
Jun 24, 2001, 4:05:47 AM6/24/01
to

Roy McCammon <rbmcc...@ieee.org> wrote in message
news:3B352314...@ieee.org...

Indeed that is so. (I havent checked the summing to zero). But
there is still a region in which energy appears to be circulating
in a clockwise direction and one in which it is circulating in
the opposite direction.

Franz Heymann


franz heymann

unread,
Jun 24, 2001, 4:29:33 AM6/24/01
to

Old Man <o...@bg.net> wrote in message
news:e_9Z6.1569$n%1.5...@newsfeed.slurp.net...

Roy has mentioned the same point. Which of the three of us is
going to do the calculation? In view of the conundrum I mention
lower down, I would not be all that certain that the total
circulating *energy* would be zero.

> An application of torque is required
> to place the charge at any off-axis position, however, and in
this case, the
> final net field circulation will reflect that effort with a
non-zero value.
> Feynman illustrated this problem with a permanent dipole
magnet, and,
> without comment, sticks the charge, in an off axis position, to
the side off
> the magnet. Decades latter, another skillful slight-of-hand
artist, named
> Franz Heymann, sticks, without comment, the charge in the exact
center of a
> current loop!!! [Old Man]

OK. Let us make the problem even more interesting :-

Replace the current loop with a true magnetic dipole and repeat
my experiment. There are then no considerations like "inside",
"outside" or "off-centre" All of the Poynting vector circulates
in the same direction. This is not a hypothetical situation at
all, since a free electron is de facto a point charge associated
with a concomitant magnetic dipole field.

Where does this lead us to? Infinite circulating energy. The
story is of course related to the infinite stationary (?) energy
associated with a point charge or a point dipole, but even if you
integrate the the Poynting vector only from some cut-off
outwards, you get a non zero value.

Franz Heymann

Ed Green

unread,
Jun 24, 2001, 9:27:49 AM6/24/01
to
From: "franz heymann" franz....@care4free.net

<propagation of EM waves in metal>

>The reason why it is wrong is that you took no account of the
>conductivity of the copper.

My first "explanation" for the idea that it might be meaningful
to assign some velocity to EM waves _inside_ copper,
whereas my even earlier reaction was that EM waves cannot
exist inside a conductor, was that the simple prohibition
failed to take account of the _finite_ conductivity of copper.

I then, yes, e'en I, looked up a form for the speed of light,
and discovered: speed = 1/sqrt(eps*mu)

I then began to wonder how one would discover these properties
for copper. I suppose they are frequency dependent?

I suppose if a wave is very strongly damped then the speed
given by the above relation may not mean much; we may
not be able to follow a crest whose phase velocity is given
by 1/sqrt(eps*mu). On the other hand, I presume given
any losses in a medium a solution to Maxwell's equations is
given by the product of a traveling sinuisoid times an
exponential envelope, and the phase velocity of the
sinusoid would be given by this expression, and in this
sense we can always calculate the speed of propagation
at least formally.

Does any of this correspond with reality?

In counterpoint to my ignorance I can only offer the defense
that I come to praise Maxwell's theory, not to bury it.

With regard to circulating energy -- I can't resist -- I would
suggest that a theory which gives results in accord with
observation but also contains some more or less strange
structure apparently not connected with observation may
occasion raised eyebrows, but cannot be rejected as
"unphysical" simply on that account; particularly if there
is no apparent way to keep the observable outcomes while
eliminating the unobserved structure. Who knows, what
seems strange might even be real.

If static fields can store angular momentum then there
should be a lecture demonstration: we bring some
configuration of charge and current together and measure
the torques. If we are clever we manage to do this in
such a way that the torques on the contraints do not
cancel. We then disassemble the charge and current
configuration and demonstrate we recover the torque defect.

This would be analagous to David Rutherford's favorite
thought experiment purporting to show whatever it is
purporting to show, and showing to those who do not
wish to overturn electromagnetic theory that as seen
from some reference frames two charges constrained
to be in uniform rectilinear motion forever transfer an
impulse to the field as they cross and then recover this
deposit. In this thought experiment the constraints
keeping the charges steady on course and speed would
supply the forces needed to do so, and in these reference
frames with the "momentum defect" the time integral of
the forces would show this, just as in this second thought
experiment the time integral of the torques supplied by
the constraints would show an angular momentum defect.


Roy McCammon

unread,
Jun 24, 2001, 11:09:07 AM6/24/01
to
franz heymann wrote:
.

>
> Replace the current loop with a true magnetic dipole and repeat
> my experiment. There are then no considerations like "inside",
> "outside" or "off-centre" All of the Poynting vector circulates
> in the same direction. This is not a hypothetical situation at
> all, since a free electron is de facto a point charge associated
> with a concomitant magnetic dipole field.

lets build our dipole out of two monopoles. Put a monopole
of strength +1 at +1 on the z axis and mirror it with a monopole
of strength -1 at -1 on the z axis. Let the electric charge
approach from infinity along the x axis. As it approaches,
it cuts lines of flux. It experiences a force in the y direction.
If you are going to keep it on the x axis you have to apply
a counter acting torque, all the way to the origin. The
non zero net circulating flux reflects this fact.

Richard

unread,
Jun 24, 2001, 11:48:16 AM6/24/01
to

franz heymann <franz....@care4free.net> wrote in message
news:3b3595a8$0$15022$cc9e...@news.dial.pipex.com...


But I did. I used the conductivity to find the wavelength (.4mm).

Roy McCammon

unread,
Jun 24, 2001, 11:58:28 AM6/24/01
to
Ed Green wrote:

> I then, yes, e'en I, looked up a form for the speed of light,
> and discovered: speed = 1/sqrt(eps*mu)

there's more to the story (isn't there always). A more
general expression for phase velocity is

Vp = real part (w/k) where w = omega = frequency in rad/sec
k = wave number

in a good low loss dielectric, k = 2pi/L where L = wavelength
so Vp is just the familiar w*L/2pi which reduces to 1/sqrt(eu)
where e = epsilon = permittivity , u = mu = permeability

But, in a good conductor k is complex with real part and
imaginary part almost equal and equal to sqrt (wus/2)
where s = sigma = conductivity

What does the imaginary part of k mean? It simply
means that the wave is attenuated as it travels.

So, in a conductor, the wavelength is much shorter
and the wave is rapidly attenuated (absorbed) as it
propagates into the conductor.

> I then began to wonder how one would discover these properties
> for copper. I suppose they are frequency dependent?

In general they are. This leads to the phenomena called
dispersion.


>
> I suppose if a wave is very strongly damped then the speed
> given by the above relation may not mean much;

the phase velocity still enters into matching the boundary
conditions between the external wave and the internal wave.
This is why the internal wave propagates into the
conductor at almost a normal angle.

> we may
> not be able to follow a crest whose phase velocity is given
> by 1/sqrt(eps*mu).

we can still follow zero crossings

> On the other hand, I presume given
> any losses in a medium a solution to Maxwell's equations is
> given by the product of a traveling sinuisoid times an
> exponential envelope, and the phase velocity of the
> sinusoid would be given by this expression, and in this
> sense we can always calculate the speed of propagation
> at least formally.

yup

> Does any of this correspond with reality?

double yup, you're getting the hang of it.

> With regard to circulating energy -- I can't resist -- I would
> suggest that a theory which gives results in accord with
> observation but also contains some more or less strange
> structure apparently not connected with observation may
> occasion raised eyebrows, but cannot be rejected as
> "unphysical" simply on that account; particularly if there
> is no apparent way to keep the observable outcomes while
> eliminating the unobserved structure. Who knows, what
> seems strange might even be real.

right

> If static fields can store angular momentum then there
> should be a lecture demonstration: we bring some
> configuration of charge and current together and measure
> the torques. If we are clever we manage to do this in
> such a way that the torques on the contraints do not
> cancel. We then disassemble the charge and current
> configuration and demonstrate we recover the torque defect.

Feynman has a good write up on this in volume 2 of
his lectures. The experiment is demonstratable in principal,
but I'm not sure if the magnitude of the effect is large
enough to observe in a lecture demo. The upshot is
that if angular momentum is to be conserved, it has to
be stored somewhere, and assuming that it is stored in
the field and running the calculations exactly accounts
for the amount that must be stored. Not that that makes
it so, of course. A pragmatist would say it works and
gives exactly correct answers so whether it is true or
not, the world is as if it were true and I'll just
take it as true. The skeptic might say that maybe
god (or nature) puts the stored angular momentum in
a flash in his hip pocket and then gives it back when its
needed. Since experiment isn't going to decide, you wind
up believing whatever makes you comfortable.

> David Rutherford's favorite thought experiment

I don't know it

Old Man

unread,
Jun 24, 2001, 2:48:33 PM6/24/01
to

Roy McCammon <rbmcc...@ieee.org> wrote in message
news:3B36028B...@ieee.org...

Right. Franz and I are old hands at describing the motion of an electron in
a magnetic field between two appositely charged magnetic monopoles, just as
you have set up here. For some interesting results see

http://www.iw.net/~jakoepke

[Old Man]


franz heymann

unread,
Jun 24, 2001, 4:42:10 PM6/24/01
to

Ed Green <null...@aol.com> wrote in message
news:20010624092749...@ng-fx1.aol.com...

Harnwell, "Principles of Elecrticity and Electromagnetism", gives
a very full account of the propagation of EM waves in conductors.
It turns out that the phase velocity is just omega*skindepth.
The latter is determined by the magnetic permeability and the
conductivity of the conductor and the frequency.

>
> In counterpoint to my ignorance I can only offer the defense
> that I come to praise Maxwell's theory, not to bury it.
>
> With regard to circulating energy -- I can't resist -- I would
> suggest that a theory which gives results in accord with
> observation but also contains some more or less strange
> structure apparently not connected with observation may
> occasion raised eyebrows, but cannot be rejected as
> "unphysical" simply on that account; particularly if there
> is no apparent way to keep the observable outcomes while
> eliminating the unobserved structure. Who knows, what
> seems strange might even be real.
>
> If static fields can store angular momentum then there
> should be a lecture demonstration: we bring some
> configuration of charge and current together and measure
> the torques. If we are clever we manage to do this in
> such a way that the torques on the contraints do not
> cancel. We then disassemble the charge and current
> configuration and demonstrate we recover the torque defect.

I am seriously bothered about using the concept of the Poynting
vector at all in the context of coexisting static fields. Has
anybody seen a derivation of the significance of this vector for
overlapping, independent, static E and B fields? It is usually
introduced only in the context of EM waves. I have a strong
feeling that in the case of static fields, all that can actually
be deduced is that the PV integrated over a closed surface is
zero, and that nothing can be said about its significance as far
as an open surface element is concerned.

>
> This would be analagous to David Rutherford's favorite
> thought experiment purporting to show whatever it is
> purporting to show, and showing to those who do not
> wish to overturn electromagnetic theory that as seen
> from some reference frames two charges constrained
> to be in uniform rectilinear motion forever transfer an
> impulse to the field as they cross and then recover this
> deposit. In this thought experiment the constraints
> keeping the charges steady on course and speed would
> supply the forces needed to do so, and in these reference
> frames with the "momentum defect" the time integral of
> the forces would show this, just as in this second thought
> experiment the time integral of the torques supplied by
> the constraints would show an angular momentum defect.

Franz Heymann


franz heymann

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Jun 24, 2001, 4:44:35 PM6/24/01
to

Roy McCammon <rbmcc...@ieee.org> wrote in message
news:3B360E1C...@ieee.org...

Thanks for a good commentary.

Franz Heymann


franz heymann

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Jun 24, 2001, 4:47:06 PM6/24/01
to

Richard <dic...@yahoo.com> wrote in message
news:NVnZ6.6449$g4.1...@e420r-atl2.usenetserver.com...

My apologies. Once again I got my knickers in a twist.

Franz Heymann

franz heymann

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Jun 24, 2001, 5:09:31 PM6/24/01
to

Roy McCammon <rbmcc...@ieee.org> wrote in message
news:3B36028B...@ieee.org...

Now repeat the experiment, this time bringing the charge from
infinity along the z axis. Same final state, but no force
anywhere.along the path.

However, what I really had in mind was a *true* dipole, pole
strength tending to infinity and separation tending to zero, or
current tending to infinity and loop diameter tending to zero.

I have another problem with your experiment. The magnetic field
after the charge has been introduced is identical with that
before its introduction. So in what way does it reflect the fact
that the charge has been introduced?

The more I think about it, the more wary I become of the use of
the PV in the context of static fields. I veer more and more
towards the idea that all you can say in these circumstances is
that the integral of BxE over a closed surface is zero, and you
can say nothing about its behaviour over an open surface element.
What am I missing out on?

Franz Heymann


Roy McCammon

unread,
Jun 24, 2001, 10:08:44 PM6/24/01
to
franz heymann wrote:

> Now repeat the experiment, this time bringing the charge from
> infinity along the z axis. Same final state, but no force
> anywhere.along the path.

As long as the point electron is coming down the z axis, there is
no torque. But I'm unwilling to push the point electron through
the point monopole because of the infinities, so I get very close
to the monopole and go around it and then on down the z axis to
the origin. As long as I'm moving along the z axis, there is no
torque, but that little trip around the monopole was through a
region of immense B field and there was strong torque. Again I
suspect that the properly computed path integrals will yield the
exact result of assuming Poynting is momentum.

> However, what I really had in mind was a *true* dipole, pole
> strength tending to infinity and separation tending to zero, or
> current tending to infinity and loop diameter tending to zero.

I suspect that math works out the same for any finite strength of
monopoles with non-zero separation but I don't know if the limits
exist when you force a point electron into a point dipole.

> I have another problem with your experiment. The magnetic field
> after the charge has been introduced is identical with that
> before its introduction. So in what way does it reflect the fact
> that the charge has been introduced?

I'm not sure what you are saying, but I note that there is
only one field and that field has a magnetic component and
an electric component and that the total field before bringing
the electron in from infinity is different from afterwards.


> The more I think about it, the more wary I become of the use of
> the PV in the context of static fields. I veer more and more
> towards the idea that all you can say in these circumstances is
> that the integral of BxE over a closed surface is zero, and you
> can say nothing about its behaviour over an open surface element.
> What am I missing out on?

Just that computing the angular momentum of the fields tells
you how much ang mo you will get if you let the fields collapse.

Think of this, EM fields transport and store energy. It would
be odd if it could transport ang mo without being able to
store it. We need look no further than an electric motor
to conclude that EM fields can transport ang mo.

Ed Green

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Jun 24, 2001, 10:49:02 PM6/24/01
to
From: "franz heymann" franz....@care4free.net

>Roy McCammon <rbmcc...@mmm.com> wrote in message
>news:3B335690...@mmm.com...

>> You're doing good now. You need a couple of other pieces


>> of information. The first is that the speed of EM waves
>> in a good conductor is very slow, about like a brisk walk.
>
>The only EM *waves* which can exist in a good conductor are
>evanescent waves near the surface. They do not "propagate" in
>the usual sense of the word. I think you must be referring to
>the drift speed of electrons, which is something different.

Thought of the same thing, but Roy McCammon makes a
strong impression on me as a man who knows what he is
talking about. Therefore I would offer more than even money
that he in fact knows what he is talking about here, and
his remarks have a sensible interpretation.


Ed Green

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Jun 24, 2001, 11:05:22 PM6/24/01
to
>From: "franz heymann" franz....@care4free.net

>A 1 mHz wave will not propagate in copper at all, except as an
>evanescent wave.
>The copper acts essentially as a reflector.

I apologize if you have already answered this elsewhere, but allow
me to engage in my professional specialty: educated guesswork. ;)

First all all, copper is a good conductor but not a perfect
conductor. That means current flow to null out internal
fields takes finite time and fields can penetrate "evanescently".

Second of all, it seems reasonable to me that EM waves can
penetrate lossy media. Take an extreme example, an
insulator with a conductivity of 10^{-umptiscratz} mho/cm
(umptiscratz is a large number sometimes used in military
applications).

When the conductivity leaves "zero", are EM waves
completetely extinguished? That is a rhetorical question.

[Note: I am worried about "lossy", since I think this may
not only include conductivity, but also polarization hysterisis,
if there is such a thing, or resonance, which might be the
same thing, or maybe some yet other description.]

Anyway, given that a finite conductance, however small,
does not immediately extinguish EM waves but presumably
merely attenuates them leaving a defined phase velocity,
then also presumably we could formally follow this phase
velocity down to a region where good conductors attentuate
EM waves, in a technical sense, "damned fast".

I am encouraged in my belief in Roy McCammon's essential
technical grasp of the situation by his reference to Snell's
law and his detailed consideration of the Poynting vector.
His is not the writing of a man likely to mistake drift velocity
for phase velocity. If he has made a technical error, it must
be much more subtle. ;)

Presumably complex impedences and dielectric constants
and permeabilities and all that make all this work out to a tau.


Ed Green

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Jun 24, 2001, 11:07:24 PM6/24/01
to
From: "franz heymann" franz....@care4free.net

Would one of you learned gentlemen like to outline the difference
between an "evanescent" wave and one merely showing an
exponential decay, and, for optional credit, relate that to
the difference in phases differences?


Ed Green

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Jun 24, 2001, 11:17:36 PM6/24/01
to
From: "franz heymann" franz....@care4free.net

>Roy McCammon <rbmcc...@ieee.org> wrote in message
>news:3B360E1C...@ieee.org...

<...>

>Thanks for a good commentary.

Seconded with enthusiasm.

Hey! They must not give out those ieee.org addresses for
nothing. ;) I'd take a second course in EM taught by him
any day.


Ed Green

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Jun 24, 2001, 11:35:43 PM6/24/01
to
From: "franz heymann" franz....@care4free.net

>I am seriously bothered about using the concept of the Poynting
>vector at all in the context of coexisting static fields. Has
>anybody seen a derivation of the significance of this vector for
>overlapping, independent, static E and B fields? It is usually
>introduced only in the context of EM waves. I have a strong
>feeling that in the case of static fields, all that can actually
>be deduced is that the PV integrated over a closed surface is
>zero, and that nothing can be said about its significance as far
>as an open surface element is concerned.

For what my opinion is worth, I am comfortable with the idea
of dynamic fields which don't appear to be "doing" anything;
as I said, merely a macroscopic extension of spin and other
expressions of intrinsic microscopic angular momentum:
why should microscopic systems have all the fun?

The emphasis on angular momentum would result from the
distressing propensity of spatially limited systems having
net linear momentum to leave us and go elsewhere; but
the ideas and treatment would be very similar.


Ed Green

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Jun 24, 2001, 11:48:35 PM6/24/01
to
From: "franz heymann" franz....@care4free.net

>Replace the current loop with a true magnetic dipole and repeat
>my experiment. There are then no considerations like "inside",
>"outside" or "off-centre" All of the Poynting vector circulates
>in the same direction. This is not a hypothetical situation at
>all, since a free electron is de facto a point charge associated
>with a concomitant magnetic dipole field.
>
>Where does this lead us to? Infinite circulating energy. The
>story is of course related to the infinite stationary (?) energy
>associated with a point charge or a point dipole, but even if you
>integrate the the Poynting vector only from some cut-off
>outwards, you get a non zero value.

In the same spirit that we might attempt to represent the
energy or mass of the electron as stored in the external
field, integrated to some cut-off value to represent ignorance,
as a professor of mine once put it, we could imagine the
angular momentum of the circulating Poynting vector
integrated to some cutoff as representing the spin angular
momentum, also stored in the external field.

The particle can now pack its bags and go on vacation, leaving
its external fields to represent its effects to the outside world.

There may be some self-consistency problems in picking
the cutoff as there appear to be for the field energy, but the
spirit of the thing is the same, and may provide a strong hint
up to some break-down of the theory near the core what
is going on.


Roy McCammon

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Jun 25, 2001, 12:14:53 AM6/25/01
to
Ed Green wrote:

> Would one of you learned gentlemen like to outline the difference
> between an "evanescent" wave and one merely showing an
> exponential decay, and, for optional credit, relate that to
> the difference in phases differences?

I must admit that I cannot cite an authoritative reference that says
an evanescent field is such and such; all I can find is examples that
have been labeled evanescent. There could be more than one correct
definition in general use, however, every case that I have encountered
has these features: The time phase of the E and H fields everywhere
are at 90 degrees difference from each other, and the time phase the
field is not a function of position (everywhere the same). The
amplitude usually attenuates with distance. The fact that E and H
are at 90 degrees tells you the field cannot transport net energy
(averaged over one cycle) but energy can slosh back and forth. The
field inside the copper does transport energy into the copper where
it is absorbed as heat. E and H inside the copper are at 45 degrees
(over a significant frequency range) so you might say that the wave
is half evanescent. But the essence of an evanescent wave is that
it only stores energy and does not transport it, hence my reluctance
to characterize the wave inside the copper as evanescent.

The most common example of an evanescent EM field is the field outside
an air-glass interface when "total" internal reflection of the light
inside the glass occurs. If the angle is shallow enough there is
"complete" internal reflection, but there is still EM field out in
the air. It is reactive and falls off in amplitude very rapidly as
you move out from the interface. No energy is lost by this field
(hence complete reflection), but the external field does influence
the phase of the reflected field. In effect, the light wouldn't
know to fully internally reflect if it didn't know there was air on
the other side and it wouldn't know that if it didn't have that
evanescent field out there on the other side feeling the media
constants over there.

Another example occurs in the vicinity of an electrically
small orifice in a cavity. Like the mess on the front of a
microwave oven.

Roy McCammon

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Jun 25, 2001, 12:27:25 AM6/25/01
to
Ed Green wrote:

> They must not give out those ieee.org addresses for
> nothing.

You got to pay your dues if you want to play the blues
(or have an ieee.org address)

franz heymann

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Jun 25, 2001, 5:29:58 AM6/25/01
to

Ed Green <null...@aol.com> wrote in message
news:20010624230724...@ng-mb1.aol.com...

There is confusion in my various optics and EM books on the
precise meaning of "evanescent waves". I now tend to think that
the term may actually be applied to any situation where, at a
boundary between two media, the conditions are such as to give
rise to a wave with an exponential decay in the second medium.

Franz Heymann


franz heymann

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Jun 25, 2001, 5:58:24 AM6/25/01
to

Ed Green <null...@aol.com> wrote in message
news:20010624224902...@ng-mb1.aol.com...

Yes

Franz Heymann

franz heymann

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Jun 25, 2001, 6:19:38 AM6/25/01
to

Roy McCammon <rbmcc...@ieee.org> wrote in message
news:3B369D22...@ieee.org...

> franz heymann wrote:
>
> > Now repeat the experiment, this time bringing the charge from
> > infinity along the z axis. Same final state, but no force
> > anywhere.along the path.
>
> As long as the point electron is coming down the z axis, there
is
> no torque. But I'm unwilling to push the point electron
through
> the point monopole because of the infinities, so I get very
close
> to the monopole and go around it and then on down the z axis to
> the origin. As long as I'm moving along the z axis, there is
no
> torque, but that little trip around the monopole was through a
> region of immense B field and there was strong torque. Again I
> suspect that the properly computed path integrals will yield
the
> exact result of assuming Poynting is momentum.

But the answer would depend on whether you negotiated the pole to
the left or to the right


>
> > However, what I really had in mind was a *true* dipole, pole
> > strength tending to infinity and separation tending to zero,
or
> > current tending to infinity and loop diameter tending to
zero.
>
> I suspect that math works out the same for any finite strength
of
> monopoles with non-zero separation but I don't know if the
limits
> exist when you force a point electron into a point dipole.

But that is exactly what an electron is. To the best of our
knowledge, it is a point particle possessing both charge and
magnetic moment.

>
> > I have another problem with your experiment. The magnetic
field
> > after the charge has been introduced is identical with that
> > before its introduction. So in what way does it reflect the
fact
> > that the charge has been introduced?
>
> I'm not sure what you are saying, but I note that there is
> only one field and that field has a magnetic component and
> an electric component and that the total field before bringing
> the electron in from infinity is different from afterwards.

OK. I will think more about this one.

>
>
> > The more I think about it, the more wary I become of the use
of
> > the PV in the context of static fields. I veer more and more
> > towards the idea that all you can say in these circumstances
is
> > that the integral of BxE over a closed surface is zero, and
you
> > can say nothing about its behaviour over an open surface
element.
> > What am I missing out on?
>
> Just that computing the angular momentum of the fields tells
> you how much ang mo you will get if you let the fields
collapse.

Yes, but the field nevertheless consists of two non interacting
static parts The electric field has no associated angular
momentum, whereas the magnetic field does have angular momentum.
Which brings me back to the point where I said I must think some
more. If there is any angular momentum in the field, it arises
only from the magnetic field, which did not change when the
particle was introduced.

>
> Think of this, EM fields transport and store energy. It would
> be odd if it could transport ang mo without being able to
> store it. We need look no further than an electric motor
> to conclude that EM fields can transport ang mo.

Not EM fields in this case, but purely M fields. E fields do not
have angular momentum.

It gets trickier every time I think about it.
Have you ever seen a derivation which shows that energy flow per
square cm per unit time is related to the Poynting vector for two
static overlapping fields? I only know that the integral of the
Poynting vector over a closed surface is zero under these
circumstances.

Franz Heymann


franz heymann

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Jun 25, 2001, 6:26:33 AM6/25/01
to

Ed Green <null...@aol.com> wrote in message
news:20010624234835...@ng-mb1.aol.com...

Interesting. I have just belatedly remembered that there is a
relationship between the spin and the magnetic moment of an
electron. I am a little pressed for time, and will comment on it
somewhat later. (If someone else does not do so in the meantime!)

Franz Heymann

Ed Green

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Jun 25, 2001, 9:20:25 AM6/25/01
to
From: Roy McCammon rbmcc...@ieee.org

>Ed Green wrote:
>
>> Would one of you learned gentlemen like to outline the difference
>> between an "evanescent" wave and one merely showing an
>> exponential decay, and, for optional credit, relate that to
>> the difference in phases differences?
>
>I must admit that I cannot cite an authoritative reference that says
>an evanescent field is such and such; all I can find is examples that
>have been labeled evanescent.

<...>

Thanks! My guess was going to be that "evanescent" fields
decayed rapidly in time, and some other kind rapidly in space,
but that just shows how dangerous it is to try to reason from
the words used.

>Another example occurs in the vicinity of an electrically
>small orifice in a cavity. Like the mess on the front of a
>microwave oven.

Yeah, I know. I've been meaning to clean that. :-)

Sorry, couldn't resist.


Roy McCammon

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Jun 25, 2001, 9:35:05 AM6/25/01
to

yes, I intended to say "mesh" but these fingers seem to
have a certain amount of autonomy.

--
If you are one in a million; there are 6000 people just like
you.

Opinions expressed herein are my own and may not represent those of my employer.

Roy McCammon

unread,
Jun 25, 2001, 9:46:01 AM6/25/01
to
franz heymann wrote:
>
> I now tend to think that
> the term may actually be applied to any situation where, at a
> boundary between two media, the conditions are such as to give
> rise to a wave with an exponential decay in the second medium.

I just don't think that is special enough to warrant
a special name like "evanescent". Any traveling wave
in a lossy linear media will attenuate exponentially.
That is to say, that it looses a certain percentage for
each unit length of travel. Also stated as so many
dB per meter.

Roy McCammon

unread,
Jun 25, 2001, 10:09:22 AM6/25/01
to
Ed Green wrote:

> First all all, copper is a good conductor but not a perfect
> conductor. That means current flow to null out internal
> fields takes finite time

Something on the order of atto seconds or maybe fempto
seconds. As long as you stay below about 1 THz, you
can neglect the "relaxation time".

> When the conductivity leaves "zero", are EM waves
> completetely extinguished?

yes, in the limit. Go far enough and you can
get them as small as you want.


> [Note: I am worried about "lossy", since I think this may
> not only include conductivity, but also polarization hysterisis,

yes, there is dielectric loss. In typical transmission
lines, it is usually less than the loss due to finite
conductivity, but it may significant.


> Anyway, given that a finite conductance, however small,
> does not immediately extinguish EM waves but presumably
> merely attenuates them leaving a defined phase velocity,
> then also presumably we could formally follow this phase
> velocity down to a region where good conductors attentuate
> EM waves, in a technical sense, "damned fast".

yes, on the order of dB's per wavelength (which are
much shorter than in free space)

> Presumably complex impedences and dielectric constants
> and permeabilities and all that make all this work out
> to a tau.

yup (depending on what you mean by "to a tau")

Alan Boswell

unread,
Jun 25, 2001, 10:47:03 AM6/25/01
to

Roy McCammon wrote:

> I must admit that I cannot cite an authoritative reference that says > an evanescent field is such and such; all I can find is examples that
> have been labeled evanescent. There could be more than one correct
> definition in general use, however, every case that I have encountered
> has these features: The time phase of the E and H fields everywhere
> are at 90 degrees difference from each other, and the time phase the
> field is not a function of position (everywhere the same). The
> amplitude usually attenuates with distance. The fact that E and H
> are at 90 degrees tells you the field cannot transport net energy
> (averaged over one cycle) but energy can slosh back and forth. The
> field inside the copper does transport energy into the copper where
> it is absorbed as heat. E and H inside the copper are at 45 degrees
> (over a significant frequency range) so you might say that the wave
> is half evanescent. But the essence of an evanescent wave is that
> it only stores energy and does not transport it, hence my reluctance
> to characterize the wave inside the copper as evanescent.
>

Roy
I agree with you. There are two separate conditions here - one with a
completely real propagation constant, fields in time quadrature, no
energy transport, the other with equal real and imaginary parts to the
propagation constant and some energy transport. The first condition is
one that all waveguide modes go into when the frequency drops below
cutoff, so it has a useful meaning in waveguide technology, and it is
what people commonly call "evanescent" in my opinion. If it's applied
equally to anything with an exponential decay, I don't see the need for
a special word.
Alan

Richard

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Jun 25, 2001, 11:37:54 AM6/25/01
to

Ed Green <null...@aol.com> wrote in message
news:20010624233543...@ng-mb1.aol.com...

Please excuse me as I don't know if this thread is even where I should be
asking this question.

Griffiths analyzes a coaxial cable with a battery at one end and a resistor
at the other end.

"The momentum in the fields is (a bunch of letters). This is an astonishing
result. The cable is not moving, and the fields are static and yet we are
asked to believe that there is momentum in the system......In this case it
turns out that there is hidden mechanical momentum associated with the flow
of current, and this exactly cancels the momentum in the fields....It is
actually a relativistic effect."

Does this statement by Griffiths correctly address the meaning of the PV in
static fields?


Roy McCammon

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Jun 25, 2001, 11:38:08 AM6/25/01
to
franz heymann wrote:
>

> But the answer would depend on whether you negotiated the pole to
> the left or to the right

The B field in any xy plane is radial. You get the
same torque no matter which side you go around.

> But that is exactly what an electron is. To the best
> of our knowledge, it is a point particle possessing
> both charge and magnetic moment.

No argument, but I don't know how to handle the infinities,
so I'll have to settle for an arbitrarily close approach
and hope that the limit exists.

> Yes, but the field nevertheless consists of two non interacting

> static parts.

That's like saying that if a penny is lying motionless
opn a table then the face doesn't interact with the rim.

> The electric field has no associated angular momentum,
> whereas the magnetic field does have angular momentum.
> Which brings me back to the point where I said I must think some
> more. If there is any angular momentum in the field, it arises
> only from the magnetic field, which did not change when the
> particle was introduced.

I think you may be confusing curl with angular
momentum.

> > Think of this, EM fields transport and store energy. It would
> > be odd if it could transport ang mo without being able to
> > store it. We need look no further than an electric motor
> > to conclude that EM fields can transport ang mo.
>
> Not EM fields in this case, but purely M fields. E fields do not
> have angular momentum.

There is only one field (broken record endlessly repeats).

Lets try an analogy. You have no trouble accepting
that static fields store energy. That's because you've
been told that over and over and perhaps have made the
calculations and see that calculating the field energy
gives you the same answer as path integrals of force
needed to assemble the charge configuration. You can't
see that energy while it is being stored. It's not
doing anything, but you can get it back, as is easily
demonstrated in a classroom. The idea of static
fields storing angular momentum is analogous, except
that the effect is too small to be casually demonstrated.

> It gets trickier every time I think about it.


> Have you ever seen a derivation which shows that energy flow per
> square cm per unit time is related to the Poynting vector for two
> static overlapping fields?

Feynman discusses this in vol. 2 of "Lectures..."

My recollection is that as long as you stick to classical
physics, you cannot derive that the actual energy flow
must be identical to the Poynting vector, but rather
you can show that that interpretation is consistent
with experiment and the various conservation laws.

> I only know that the integral of the
> Poynting vector over a closed surface is zero under these
> circumstances.

yes. In effect, nothing is heating up and
no energy or momentum gets away.

Alan Boswell

unread,
Jun 25, 2001, 12:12:47 PM6/25/01
to

Roy McCammon wrote:
>
perhaps you have brought in an unwarranted assumption
> that is confusing you. Something like "if I cannot
> see it, then its not there." Its these unwarranted
> assumptions that our intuition sometimes drags in that
> often lead to apparent paradoxes.
>

Roy
The charged particle next to a bar magnet had a good airing on the EM NG
last year. It's conceptually similat to the solenoid/capacitor case
above, and it's another case where there is a supposedly endless flow of
circulating energy described by the Poynting flux, which is nonzero at
all points where the electric and magnetic fields are not tangential.
Last year you said the circulating energy arises from energy introduced
when the particle was brought up to the magnet (I think), but I have had
trouble with that concept. The total energy can be found by integrating
the E and H field distributions, and seems to be independent of the
distance between the particle and the magnet.
Eyges says that the flow of energy is a useful concept only when used as
a vague shorthand description, and that when one tries to identify and
localise the energy flow too closely, the dilemmas arise.
It seems to me that when there is a flow of EM energy, a Poynting vector
description can always be found, but that when there is a Poynting
vector there is not always necessarily an energy flow. In other words a
Poynting vector is a necessary but not a sufficient condition for a flow
of energy - do you agree or not?
Alan

Ace Schallger

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Jun 25, 2001, 1:29:17 PM6/25/01
to
In article <20010624092749...@ng-fx1.aol.com>, Ed Green
<null...@aol.com> wrote:

> If static fields can store angular momentum then there
> should be a lecture demonstration: we bring some
> configuration of charge and current together and measure
> the torques. If we are clever we manage to do this in
> such a way that the torques on the contraints do not
> cancel. We then disassemble the charge and current
> configuration and demonstrate we recover the torque defect.

Take two Leyden Jars with plastic inserts. Charge them up using as
small Van de graaff generator or Wimhurst Machine. Carefully separate
each Leyden Jar assembly into its three components 1) outer can, 2)
inner can and 3) plastic seperator. Ground all metal cans together
and then ground them to a metal water pipe or other good electrical
ground. Ground the plastic inserts as well in the same manner.

Carefully re-assemble the Leyden Jars.

Dare you lick your thumb and forefinger and then touch them across any
outer and inner can? Try it. Tell me what it was that knocks you on
your butt if you touch left hand and right hand to inner and outer cans
of any Leyden Jar that has been charged, disassembled as above and
grounded all together and then reassembled?

Could it have been torsion or strain stored in the plastic insert?

Ace Schallger

Roy McCammon

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Jun 25, 2001, 2:09:58 PM6/25/01
to
Alan Boswell wrote:

> Last year you said the circulating energy arises from energy introduced
> when the particle was brought up to the magnet (I think), but I have had
> trouble with that concept.

Hard to remember what I said last year without the
complete context, but I think it would be better
said that the circulating energy can be interpreted
as angular momentum which is introduced by torques
you must apply as you bring the configuration together.

> The total energy can be found by integrating the E
> and H field distributions, and seems to be independent of the
> distance between the particle and the magnet.

Yes, total energy does not change, but angular
momentum does.


> Eyges says that the flow of energy is a useful concept only when used as
> a vague shorthand description, and that when one tries to identify and
> localise the energy flow too closely, the dilemmas arise.
> It seems to me that when there is a flow of EM energy, a Poynting vector
> description can always be found, but that when there is a Poynting
> vector there is not always necessarily an energy flow. In other words a
> Poynting vector is a necessary but not a sufficient condition for a flow
> of energy - do you agree or not?
> Alan

Experiments to date are not sensitive enough to weigh the
energy while its in transit. There is no direct evidence
about where it is when in transit. All we know is where
it winds up when it is dissipated (or does something else
that we can measure). So, we cannot say that a non-zero
Poynting vector is a sufficient condition that there
is energy flow. It is sufficient to conclude from where
the energy is being sourced and to where it is being dumped
and how hot the wire becomes. As a pragmatist, I find
the assumption that Poynting is identical with energy
flow to be useful under almost conditions for the prediction
of macroscopic behavior (all save where QED must be used).

Roy McCammon

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Jun 25, 2001, 2:18:53 PM6/25/01
to
Richard wrote:

> Griffiths analyzes a coaxial cable with a battery at one end and a resistor
> at the other end.
>
> "The momentum in the fields is (a bunch of letters). This is an astonishing
> result. The cable is not moving, and the fields are static and yet we are
> asked to believe that there is momentum in the system......In this case it
> turns out that there is hidden mechanical momentum associated with the flow
> of current, and this exactly cancels the momentum in the fields....It is
> actually a relativistic effect."

I don't have Griffiths, but what I would expect is
once the circuit was energized there would be a
reaction that would see the source and load pushed
away from each other ( dP/dt ) which would be
countered by mechanical stress (stretching) in the
wires or other mechanical mountings.

Steve Harris

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Jun 25, 2001, 3:18:46 PM6/25/01
to

Ace Schallger wrote in message <250620011029047511%phys...@att.net>...


The zap is stored as an electric field across the plastic. You can ground it
at a single point (or several) and it doesn't come off because it doesn't
conduct over the
surface to the contact point. You can WIPE it off both sides with a grounded
conductor, and then have no problem (no electricity is seen when you
reassemble).
Or, you can melt it and recast, and similarly get no charge on assembly.
But until
you get rid of this surface charge on the plastic, which you can feel as
static when you wipe it, any reassembly of the Leyden jar will leave with
a fully-capable charged capacitor.


franz heymann

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Jun 25, 2001, 12:39:34 PM6/25/01
to

franz heymann <franz....@care4free.net> wrote in message
news:3b37124a$0$12241$cc9e...@news.dial.pipex.com...

And, as Alan has just reminded us, the waves below the cutoff
freq. of a waveguide are also called "evanescent"
It appears that the term is actually used to refer to all waves
which have an exponential part in the distance factor.

Franz Heymann


franz heymann

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Jun 25, 2001, 2:03:34 PM6/25/01
to

Richard <dic...@yahoo.com> wrote in message
news:3SIZ6.16637$g4.4...@e420r-atl2.usenetserver.com...

The PV in the case of a steady current in a coaxial cable,
integrated over a surface transverse to the axis of the cable in
fact turns out to be VI, which is the power transmitted past that
cross section of the cable. In saying this, I neglect the
resistivity of the wires. If this is taken into account, it
turns out that the E field has a small longitudinal component at
the surface of the wire. If you integrate the PV over a small
cylindrical surface touching the conductor, you find that the
power transported from the field to the wire is just the I^2R
loss in the wire. So in this case the PV does its job perfectly
sensibly.

However, my problem is this: In the coax example, the E field
and the B field are truly concomitant, in the sense that one of
them could not exist without the presence of the other. In fact,
the fields are just those of a single zero frequency wave
propagated in the dielectric medium. Here I expect the PV to do
its job, as indeed it does.

But now place the whole cable inside a strong externally supplied
magnetic field. The PV in the cable is now something totally
different and yet neither the power transmitted nor the Joule
losses are affected.

Franz Heymann


franz heymann

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Jun 25, 2001, 2:33:17 PM6/25/01
to

Roy McCammon <rbmcc...@mmm.com> wrote in message
news:3B375AE0...@mmm.com...

> franz heymann wrote:
> >
>
> > But the answer would depend on whether you negotiated the
pole to
> > the left or to the right
>
> The B field in any xy plane is radial. You get the
> same torque no matter which side you go around.

Touche

>
> > But that is exactly what an electron is. To the best
> > of our knowledge, it is a point particle possessing
> > both charge and magnetic moment.
>
> No argument, but I don't know how to handle the infinities,
> so I'll have to settle for an arbitrarily close approach
> and hope that the limit exists.
>
> > Yes, but the field nevertheless consists of two non
interacting
> > static parts.
>
> That's like saying that if a penny is lying motionless
> opn a table then the face doesn't interact with the rim.

The analogy escapes me.

>
> > The electric field has no associated angular momentum,
> > whereas the magnetic field does have angular momentum.
> > Which brings me back to the point where I said I must think
some
> > more. If there is any angular momentum in the field, it
arises
> > only from the magnetic field, which did not change when the
> > particle was introduced.
>
> I think you may be confusing curl with angular
> momentum.

I think you are right.

>
> > > Think of this, EM fields transport and store energy. It
would
> > > be odd if it could transport ang mo without being able to
> > > store it. We need look no further than an electric motor
> > > to conclude that EM fields can transport ang mo.
> >
> > Not EM fields in this case, but purely M fields. E fields do
not
> > have angular momentum.
>
> There is only one field (broken record endlessly repeats).

Are you saying that under all circumstances, even when E and B
arise from different sources, a coordinate system can alway be
found such that there is only an E field or only a B field
present? This is not a leading question. It is a genuine one.

>
> Lets try an analogy. You have no trouble accepting
> that static fields store energy. That's because you've
> been told that over and over and perhaps have made the
> calculations and see that calculating the field energy
> gives you the same answer as path integrals of force
> needed to assemble the charge configuration. You can't
> see that energy while it is being stored. It's not
> doing anything, but you can get it back, as is easily
> demonstrated in a classroom.

Up to here I am totally happy.

The idea of static
> fields storing angular momentum is analogous, except
> that the effect is too small to be casually demonstrated.

OK as far as it goes, but I have never yet seen an expression for
the angular momentum density, proven for the case of two
independent electic and magnetic fields. As I have said before,
all I am aware of in this context is effectively that divExB = 0.

>
> > It gets trickier every time I think about it.
>
>
> > Have you ever seen a derivation which shows that energy flow
per
> > square cm per unit time is related to the Poynting vector for
two
> > static overlapping fields?
>
> Feynman discusses this in vol. 2 of "Lectures..."
>
> My recollection is that as long as you stick to classical
> physics, you cannot derive that the actual energy flow
> must be identical to the Poynting vector, but rather
> you can show that that interpretation is consistent
> with experiment and the various conservation laws.

Thanks for the reference. I am due for a good read.


>
> > I only know that the integral of the
> > Poynting vector over a closed surface is zero under these
> > circumstances.
>
> yes. In effect, nothing is heating up and
> no energy or momentum gets away.

The discussion so far has been most stimulating. I have to do
some reading and cogitation before I can usefully contribute any
more.

Franz Heymann

Ace Schallger

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Jun 25, 2001, 7:09:54 PM6/25/01
to
In article <9h82rn$fjm$1...@slb6.atl.mindspring.net>, Steve Harris
<sbha...@ix.netcom.com> wrote:

> The zap is stored as an electric field across the plastic. You can ground it
> at a single point (or several) and it doesn't come off because it doesn't
> conduct over the
> surface to the contact point. You can WIPE it off both sides with a grounded
> conductor, and then have no problem (no electricity is seen when you
> reassemble).
> Or, you can melt it and recast, and similarly get no charge on assembly.
> But until
> you get rid of this surface charge on the plastic, which you can feel as
> static when you wipe it, any reassembly of the Leyden jar will leave with
> a fully-capable charged capacitor.

Well, the fact that isn't a 'surface charge' as you suppose, should be
a big clue. Something tells me you have no experience doing any of
these things.

Ace Schallger

Richard

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Jun 25, 2001, 7:56:52 PM6/25/01
to

Roy McCammon <rbmcc...@mmm.com> wrote in message
news:3B37808D...@mmm.com...

> Richard wrote:
>
> > Griffiths analyzes a coaxial cable with a battery at one end and a
resistor
> > at the other end.
> >
> > "The momentum in the fields is (a bunch of letters). This is an
astonishing
> > result. The cable is not moving, and the fields are static and yet we
are
> > asked to believe that there is momentum in the system......In this case
it
> > turns out that there is hidden mechanical momentum associated with the
flow
> > of current, and this exactly cancels the momentum in the fields....It is
> > actually a relativistic effect."
>
> I don't have Griffiths, but what I would expect is
> once the circuit was energized there would be a
> reaction that would see the source and load pushed
> away from each other ( dP/dt ) which would be
> countered by mechanical stress (stretching) in the
> wires or other mechanical mountings.
>
> --
Oops, I neglected to mention that there's a line charge Lambda on the inner
conductor. So I guess what he is saying is that two static fields *can*
carry momentum which is offset by some kind of relativistic effect having to
do with the current.

Ed Green

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Jun 25, 2001, 9:13:33 PM6/25/01
to
From: rbmcc...@mmm.com (Roy McCammon)

>Ed Green wrote:
>
>> First all all, copper is a good conductor but not a perfect
>> conductor. That means current flow to null out internal
>> fields takes finite time
>
>Something on the order of atto seconds or maybe fempto
>seconds. As long as you stay below about 1 THz, you
>can neglect the "relaxation time".

Well... and by the way, feel free to stop any time you find
the questions tiresome... I have trouble understanding how
anything like an EM wave can pentrate copper at frequencies
much less than that. You have an E field, the electrons
redistribute themselves with relaxation time constants
much shorter than the period of the E field, you have no
E field, ergo, no wave at all. Or do I have something wrong?

Surely if we adopt the approximation "well, the relaxation
time is essentially instantaneous" we have thrown out
these teeny tiny eency weency field penetrations also?

>> When the conductivity leaves "zero", are EM waves
>> completetely extinguished?
>
>yes, in the limit. Go far enough and you can
>get them as small as you want.

Well of course. I really meant "excluded", and I was reacting
to Franz's various comments to the effect that there is no EM
wave in a conductor, by asking a rhetorical question; if we have
a very slightly lossy medium we would not hesitate to say we have a wave
penetrating the
medium even if it is eventually extinguished, so presumably,
following this along to good conductors, there is no point at
which we have zero penetration; the extinction just happens
over shorter distances.

I'm not sure "evanescent" is being misused: in the pure case
you say we have a little exponentially attentuated standing wave;
here we have an admixture of exponentially attentuated
travelling wave.

>> [Note: I am worried about "lossy", since I think this may
>> not only include conductivity, but also polarization hysterisis,
>
>yes, there is dielectric loss. In typical transmission
>lines, it is usually less than the loss due to finite
>conductivity, but it may significant.
>
>> Anyway, given that a finite conductance, however small,
>> does not immediately extinguish EM waves but presumably
>> merely attenuates them leaving a defined phase velocity,
>> then also presumably we could formally follow this phase
>> velocity down to a region where good conductors attentuate
>> EM waves, in a technical sense, "damned fast".
>
>yes, on the order of dB's per wavelength (which are
>much shorter than in free space)
>
>> Presumably complex impedences and dielectric constants
>> and permeabilities and all that make all this work out
>> to a tau.
>
>yup (depending on what you mean by "to a tau")

You pretty much covered this in your other response to me
where you said that the more eneral expression for phase
velocity was Re(w/k), IIRC. I meant I presume it is one
of those situations where if we make everything in sight a
complex quantity things magically works out with the
correct phase shifts and decay constants and all that. :)


Ed Green

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Jun 25, 2001, 9:40:44 PM6/25/01
to
>From: "Richard" dic...@yahoo.com

>Ed Green <null...@aol.com> wrote in message

>> For what my opinion is worth, I am comfortable with the idea


>> of dynamic fields which don't appear to be "doing" anything;
>> as I said, merely a macroscopic extension of spin and other
>> expressions of intrinsic microscopic angular momentum:
>> why should microscopic systems have all the fun?
>>
>> The emphasis on angular momentum would result from the
>> distressing propensity of spatially limited systems having
>> net linear momentum to leave us and go elsewhere; but
>> the ideas and treatment would be very similar.
>>
>
>Please excuse me as I don't know if this thread is even where I should be
>asking this question.

Seems darn appropriate to me.

>Griffiths analyzes a coaxial cable with a battery at one end and a resistor
>at the other end.
>
>"The momentum in the fields is (a bunch of letters). This is an astonishing
>result. The cable is not moving, and the fields are static and yet we are
>asked to believe that there is momentum in the system......In this case it
>turns out that there is hidden mechanical momentum associated with the flow
>of current, and this exactly cancels the momentum in the fields....It is
>actually a relativistic effect."
>
>Does this statement by Griffiths correctly address the meaning of the PV in
>static fields?

I am made nervous, to adopt the infinitely correct style of Franz
and Old Man :-), by the words "it is a relativistic effect". This
sounds suspiciously like handwaving. _Anything_ involving
the description of EM fields alone is necessarily a "relativistic
effect", since the fields have Lorentz invariance built in 'em.

On the other hand, I see Griffith's point. Given my hasty
education in some aspects of Poynting vectors in this thread
I think the Poynting vector in the case of a coaxial cable
carrying DC current to a load would be largely confined to
the space between the central and outer conductors and
point at the load. Now, we have been thinking of this in
terms of energy flux, but the Poynting vector also represents
momentum flux, with some c or other thrown in.

What Griffiths notes is that momentum is flowing from the
battery to the load, and yet the load is not flying away!
How can this be?

Well, momentum flux is just another name for "stress", and
evidently there are some mechanical stresses in the wire
which just cancel this mysterious momentum flux. Phew. ;)

Now... sorry, I can't help going all philosophical on you,
I'm not sure what "mechanical" means in this case, since
aren't _all_ forces and stresses in matter electrical in
origin? Evidentally it means some forces arising from
the lattice and the charge carriers not reflected in the
"macroscopic" EM description.

It could be as stupid as the force resulting from the retardation
of the drifting electrons by collisions with the lattice; though
I'm not sure why this would not cancel for the two legs.
Maybe it has something to do with the "turning" of the
current in the load; the image comes to mind of turning
a distant water wheel with a fire hose; the stream carries
both energy and momentum, the turning of the stream
at the wheel creates a force, and there must be a
balancing force holding the wheel down if it is not to
be blown away.

This appearence of extra stresses need to balance the EM
description is reminiscent of the Poincare stresses I just
happened to be reading about, needed to correctly account
for the momentum of a moving electron, at least its field
momentum.

Correctly address? I don't know. Stirs the pot some more
maybe. ;) Now you have heard from the peanut gallery,
let's hear from the experts...


Roy McCammon

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Jun 25, 2001, 11:25:22 PM6/25/01
to
Ed Green wrote:

>
> From: rbmcc...@mmm.com (Roy McCammon)
.
> >Something on the order of atto seconds or maybe fempto
> >seconds. As long as you stay below about 1 THz, you
> >can neglect the "relaxation time".
>
> I have trouble understanding how
> anything like an EM wave can pentrate copper at frequencies
> much less than that. You have an E field, the electrons
> redistribute themselves with relaxation time constants
> much shorter than the period of the E field, you have no
> E field, ergo, no wave at all.

The relaxation time is the time between the establishment
of the E field and the establishment of the current density,
i.e. how long you wait before the equation J = sigma * E
is sufficiently close.

But the conductivity is finite. The current density is
may be trying to cancel the E field, but the finite
conductivity holds it back.

> Surely if we adopt the approximation "well, the relaxation
> time is essentially instantaneous" we have thrown out
> these teeny tiny eency weency field penetrations also?

no, we would just be saying that J = sigma * E is
instantaneously correct.

> Well of course. I really meant "excluded", and I was reacting
> to Franz's various comments to the effect that there is no EM
> wave in a conductor, by asking a rhetorical question; if we have
> a very slightly lossy medium we would not hesitate to say we have a wave
> penetrating the
> medium even if it is eventually extinguished, so presumably,
> following this along to good conductors, there is no point at
> which we have zero penetration; the extinction just happens
> over shorter distances.

si

> I'm not sure "evanescent" is being misused: in the pure case
> you say we have a little exponentially attentuated standing wave;
> here we have an admixture of exponentially attentuated
> travelling wave.

But as I said earlier, all propagating waves in lossy linear
material see an exponential attenuation in the large scale.
Again I feel the essence of evanescent is that is does not
propagate and no energy gets away.


> You pretty much covered this in your other response to me
> where you said that the more eneral expression for phase
> velocity was Re(w/k), IIRC. I meant I presume it is one
> of those situations where if we make everything in sight a
> complex quantity things magically works out with the
> correct phase shifts and decay constants and all that. :)

that's exactly right.

Ed Green

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Jun 25, 2001, 11:32:50 PM6/25/01
to
From: "Richard" dic...@yahoo.com

>Oops, I neglected to mention that there's a line charge Lambda on the inner
>conductor. So I guess what he is saying is that two static fields *can*
>carry momentum which is offset by some kind of relativistic effect having to
>do with the current.

See my other comments on "relativistic effect". It is presumptuous
for an amateur to say a professional is in effect blowing smoke,
but occasionally I will try just a little presumption!

Ed Green

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Jun 25, 2001, 11:38:29 PM6/25/01
to
From: "franz heymann" franz....@care4free.net

>The PV in the case of a steady current in a coaxial cable,
>integrated over a surface transverse to the axis of the cable in
>fact turns out to be VI, which is the power transmitted past that
>cross section of the cable. In saying this, I neglect the
>resistivity of the wires. If this is taken into account, it
>turns out that the E field has a small longitudinal component at
>the surface of the wire. If you integrate the PV over a small
>cylindrical surface touching the conductor, you find that the
>power transported from the field to the wire is just the I^2R
>loss in the wire. So in this case the PV does its job perfectly
>sensibly.
>
>However, my problem is this: In the coax example, the E field
>and the B field are truly concomitant, in the sense that one of
>them could not exist without the presence of the other. In fact,
>the fields are just those of a single zero frequency wave
>propagated in the dielectric medium. Here I expect the PV to do
>its job, as indeed it does.
>
>But now place the whole cable inside a strong externally supplied
>magnetic field. The PV in the cable is now something totally
>different and yet neither the power transmitted nor the Joule
>losses are affected.

Well, all of these thought experiments involving charges and
monopoles and suspicious Poynting vectors I think amount
to the same thought experiment: we form a system where
the Poynting vector has a non-zero curl; then we ask "does
this mean anything"? The question would seem to have
physical content, because we should be able to analyze
the construction of the situation and point to the unrequited
torque now stored in the field. Well, maybe at least in
situations where the field has net angular momentum.

In situations where we have circulating momentum but
no net angular momentum, I guess we just have to say
"huh, that's interesting; it's self-consitent, I wonder if
it is real", and await further developments.


Ed Green

unread,
Jun 25, 2001, 11:42:03 PM6/25/01
to
>From: Ace Schallger phys...@att.net

>Take two Leyden Jars with plastic inserts. Charge them up using as
>small Van de graaff generator or Wimhurst Machine. Carefully separate
>each Leyden Jar assembly into its three components 1) outer can, 2)
>inner can and 3) plastic seperator. Ground all metal cans together
>and then ground them to a metal water pipe or other good electrical
>ground. Ground the plastic inserts as well in the same manner.
>
>Carefully re-assemble the Leyden Jars.
>
>Dare you lick your thumb and forefinger and then touch them across any
>outer and inner can? Try it. Tell me what it was that knocks you on
>your butt if you touch left hand and right hand to inner and outer cans
>of any Leyden Jar that has been charged, disassembled as above and
>grounded all together and then reassembled?
>
>Could it have been torsion or strain stored in the plastic insert?

No, the one time I thought I understood this I think it was
static surface charge on the plastic. Essentially immobilized
on the plastic, it induces mobile polarization charge in the metal,
which in turn is free to knock you on your butt, leaving the surface
charge in the plastic neutralized by an intimate layer of surface
charge on the plastic/metal contact.


Alan Boswell

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Jun 26, 2001, 4:10:50 AM6/26/01
to

Roy McCammon wrote:
> yes, on the order of dB's per wavelength (which are
> much shorter than in free space)

54.5dBs per wavelength (exp(-2pi)).
Alan

Ed Green

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Jun 26, 2001, 9:39:51 AM6/26/01
to
From: Alan Boswell alan.b...@baesystems.com

>I agree with you. There are two separate conditions here - one with a
>completely real propagation constant, fields in time quadrature, no
>energy transport, the other with equal real and imaginary parts to the
>propagation constant and some energy transport.

Is "propagation constant" just another name for "k", the
wavenumber, or are they merely good friends?

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