The Polarized Helical Wave Photon.
Len Gaasenbeek
When a light photon travels through space it follows a helical path. It
also spins around its own axis as it travels along.
If the photon travels away from the observer and follows a clockwise helix,
it will spin around its own axis in a clockwise direction.
The photon's spin axis points forward and towards the centre line of the
helical wave such that its angle continuously changes as it travels along.
This exerts a gyroscopic force on the photon which causes it to corkscrew
through space rather than follow a straight path.
If an observer were able to see the helical photon wave from the back it
would appear as a circle to him.
So what happens if a helical photon wave squeezes through a narrow slit and
becomes polarized?
I propose that when a helical wave photon becomes polarized it begins to
follow a substantially sinusoidal rather than a helical path. The only way
it can do this is for the photon to reverse its spin each time it changes
direction.
As the polarized photon travels away from the observer, depending on its
spin direction, on the way up the photon follows a nearly squashed flat
counter-clockwise helix, whereas on its way down it follows a squashed flat
clockwise helix.
That is to say, it appears to describe a slender narrow figure eight instead
of a circle, as it travels away from the observer. The axis of the eight
will be in line with its polarization plane.
In order for the polarized helical photon wave to remain stable and
symmetrical, each photon must continue reversing its spin direction and
helical rotation (what there is left of it) each half cycle.
Once the slit forces the helical wave photon to reverse its spin by flipping
its spin axis, its resultant altered trajectory will continue to flip its
spin axis over each half cycle, as it travels away from the slit. This
causes the helical photon wave to remain polarized in line with the slit
that polarized it.
It isn't easy to describe the above phenomenon in words.
However I will be glad to elaborate on the above scenario should anyone have
any questions.
Enjoy, Len.
............................................................
Androcles
"Len Gaasenbeek" <gaas...@rideau.net> wrote in message
news:10ljaev...@corp.supernews.com...
No you are not. Both sal and myself asked you questions and you gave no
answers.
Androcles
| Enjoy, Len.
| ............................................................
|
|
|
|
Dirk Van de moortel
"Androcles" <andr...@nospamblueyonder.co.uk> wrote in message news:Vii6d.106$tm4...@text.news.blueyonder.co.uk...
[snip]
> | It isn't easy to describe the above phenomenon in words.
> | However I will be glad to elaborate on the above scenario should anyone
> | have any questions.
>
> No you are not. Both sal and myself asked you questions and you gave no
> answers.
How sweet... both sal and yourself asked questions.
Geez... what a stronzo you are.
Dirk Vdm
FrediFizzx
Ken S. Tucker
I'd like to know more a "helicity". In a recent thread, Eugene's book,
both Fredi and Bilge referred to the helicity of a photon as invariant.
I would really like to read more about that if anyone would care to
post, a bit of basic ideas about that.
TIA
Ken
Androcles
"Ken S. Tucker" <dyna...@vianet.on.ca> wrote in message
news:2202379a.04092...@posting.google.com...
Ok:
1) Freddi has pinpointed an excellent page,
http://www.physics.gla.ac.uk/Optics/projects/singlePhotonOAM/
which demonstrates the constant energy vector (the radius)
and the spin. Look carefully at the simplest plane-wave model,
I'll refer to it in a moment.
2) I have described the components of that energy at
http://www.androc1es.pwp.blueyonder.co.uk/Radio%20Wave.htm
which enables the propagation without aether.
In the plane-wave model on (1), notice there is an invisible helix
as the "bicycle wheel" moves in the direction of its axis. This
is most obvious in the models on the left of the page. Try to
visualize the B-field and E-field components the radius represents.
3) Len's helical wave stretches or shinks this invisible helix to
cause the speed of the photon to be exactly c for the observer.
Len's notion is magic, not physics, but at least he was thinking.
Unfortunately nobody can discuss Len's idea with him, it's HIS
theory and he'll not answer any questions he doesn't like.
As you can see, he'll elaborate if you agree, but he'll not do so if
you raise an objection. I object to "each photon must continue
reversing its spin direction and helical rotation" since that violates
conservation of angular momentum.
4) Consider the bicycle wheel model on page (1), imagine it
a sphere, then consider the Earth's 23.5 degree tilt with respect
to the plane of its orbit. Not shown is the plane wave rotated
through 90 degrees, aligned with the direction of motion
as would be the normal condition for a bicycle.
5) Bilge is just an arrogant idiot who will not reason, but you
can probably work that out for yourself.
Androcles.
| TIA
| Ken
Ken S. Tucker
> "Ken S. Tucker" <dyna...@vianet.on.ca> wrote in message
> news:2202379a.04092...@posting.google.com...
> | I'd like to know more a "helicity". In a recent thread, Eugene's book,
> | both Fredi and Bilge referred to the helicity of a photon as invariant.
> | I would really like to read more about that if anyone would care to
> | post, a bit of basic ideas about that.
>
> Ok:
>
> 1) Freddi has pinpointed an excellent page,
> http://www.physics.gla.ac.uk/Optics/projects/singlePhotonOAM/
> which demonstrates the constant energy vector (the radius)
> and the spin. Look carefully at the simplest plane-wave model,
> I'll refer to it in a moment.
>
> 2) I have described the components of that energy at
> http://www.androc1es.pwp.blueyonder.co.uk/Radio%20Wave.htm
> which enables the propagation without aether.
Nice site (Fredi has a nice site too), but you should
check your substitutions of x' in your disproof of SR,
very carefully.
> In the plane-wave model on (1), notice there is an invisible helix
> as the "bicycle wheel" moves in the direction of its axis. This
> is most obvious in the models on the left of the page. Try to
> visualize the B-field and E-field components the radius represents.
>
>
> 3) Len's helical wave stretches or shinks this invisible helix to
> cause the speed of the photon to be exactly c for the observer.
> Len's notion is magic, not physics, but at least he was thinking.
> Unfortunately nobody can discuss Len's idea with him, it's HIS
> theory and he'll not answer any questions he doesn't like.
> As you can see, he'll elaborate if you agree, but he'll not do so if
> you raise an objection. I object to "each photon must continue
> reversing its spin direction and helical rotation" since that violates
> conservation of angular momentum.
> 4) Consider the bicycle wheel model on page (1), imagine it
> a sphere, then consider the Earth's 23.5 degree tilt with respect
> to the plane of its orbit. Not shown is the plane wave rotated
> through 90 degrees, aligned with the direction of motion
> as would be the normal condition for a bicycle.
> 5) Bilge is just an arrogant idiot who will not reason, but you
> can probably work that out for yourself.
There's a lot of good ideas for what a photon "really" is,
but there's not a concensus I've found.
> Androcles.
Ken
sal
What's a "stronzo"?
(Besides an Italian word which Google can't seem to translate.)
--
I can be contacted through http://www.physicsinsights.org
Dirk Van de moortel
"sal" <pragm...@nospam.org> wrote in message news:pan.2004.09.29....@nospam.org...
Haven't you guessed? ;-)
Haven't you seen Amarcord?
Secondo lo Zingarelli minore:
1. Pezzo di sterco sodo, di forma cilindrica
2. Persona spregevole e tale da essere ingiuriata
Not you obviously :-)
Dirk Vdm
Androcles
"Ken S. Tucker" <dyna...@vianet.on.ca> wrote in message
news:2202379a.04092...@posting.google.com...
| "Androcles" <andr...@nospamblueyonder.co.uk> wrote in message
news:<0Xt6d.354$tm4...@text.news.blueyonder.co.uk>...
| > "Ken S. Tucker" <dyna...@vianet.on.ca> wrote in message
| > news:2202379a.04092...@posting.google.com...
|
| > | I'd like to know more a "helicity". In a recent thread, Eugene's book,
| > | both Fredi and Bilge referred to the helicity of a photon as
invariant.
| > | I would really like to read more about that if anyone would care to
| > | post, a bit of basic ideas about that.
| >
| > Ok:
| >
| > 1) Freddi has pinpointed an excellent page,
| > http://www.physics.gla.ac.uk/Optics/projects/singlePhotonOAM/
| > which demonstrates the constant energy vector (the radius)
| > and the spin. Look carefully at the simplest plane-wave model,
| > I'll refer to it in a moment.
| >
| > 2) I have described the components of that energy at
| > http://www.androc1es.pwp.blueyonder.co.uk/Radio%20Wave.htm
| > which enables the propagation without aether.
|
| Nice site (Fredi has a nice site too), but you should
| check your substitutions of x' in your disproof of SR,
| very carefully.
I have. Extremely carefully. Refer to the original paper at
http://www.fourmilab.ch/etexts/einstein/specrel/www/
It clearly says
"IF we place x' = x-vt..." and goes on to conclude
chsi = x'/sqrt(1-v^2/c^2) (chsi is the greek letter)
which many people misconstrue as
x' = (x-vt)/sqrt(1-v^2/c^2)
Obviously it makes no sense to claim
x' = x-vt and
x' = (x-vt)/sqrt(1-v^2/c^2),
UNLESS x' = 0, in which case v = 0 also, and that is trivial.
When you calculate the moving clock runs slow, remember it
runs fast on the return trip. There is an "IF" preceding the calculation
you make.
|
| > In the plane-wave model on (1), notice there is an invisible helix
| > as the "bicycle wheel" moves in the direction of its axis. This
| > is most obvious in the models on the left of the page. Try to
| > visualize the B-field and E-field components the radius represents.
| >
| >
| > 3) Len's helical wave stretches or shinks this invisible helix to
| > cause the speed of the photon to be exactly c for the observer.
| > Len's notion is magic, not physics, but at least he was thinking.
| > Unfortunately nobody can discuss Len's idea with him, it's HIS
| > theory and he'll not answer any questions he doesn't like.
| > As you can see, he'll elaborate if you agree, but he'll not do so if
| > you raise an objection. I object to "each photon must continue
| > reversing its spin direction and helical rotation" since that violates
| > conservation of angular momentum.
|
| > 4) Consider the bicycle wheel model on page (1), imagine it
| > a sphere, then consider the Earth's 23.5 degree tilt with respect
| > to the plane of its orbit. Not shown is the plane wave rotated
| > through 90 degrees, aligned with the direction of motion
| > as would be the normal condition for a bicycle.
|
| > 5) Bilge is just an arrogant idiot who will not reason, but you
| > can probably work that out for yourself.
|
| There's a lot of good ideas for what a photon "really" is,
| but there's not a concensus I've found.
Well, at least your interest in the helical nature of the beast
has been replied to, whether you accept it or not.
There may never be. I don't know what an electron really is either.
Whatever the matter (mass) it is made of, it is of no known element.
The electron can be diffracted.
All we can say of the photon is what we know of radio waves.
Fools like Bilge, Andersen and moortel (well, he's just a puppy
that wants to bark with the rest of the hounds) insist the electric
and magnetic fields are in phase, denying conservation of energy.
They get this crazy notion from Maxwell, who believed in aether.
Maxwell's equations do not deny a 90 degree phase shift between
the E and B fields, as they seem to think.
| > Androcles.
| Ken
sal
[ ... ]
>
> All we can say of the photon is what we know of radio waves. Fools like
> Bilge, Andersen and moortel (well, he's just a puppy that wants to bark
> with the rest of the hounds) insist the electric and magnetic fields are
> in phase, denying conservation of energy. They get this crazy notion from
> Maxwell, who believed in aether. Maxwell's equations do not deny a 90
> degree phase shift between the E and B fields, as they seem to think.
Actually, for a traveling plane wave, this appears to be a consequence of
Faraday's law: curl(E) = -@B/@t. Just by eyeballing the E and B fields,
the curl of E will peak at the zero-crossings, when the spatial rate of
change is the highest, so that's also when @B/@t must be highest. And
@B/@t is naturally highest when B is crossing zero ... and that implies
that E and B cross zero at the same points in space. And that, in turn,
implies they're in phase.
That wasn't a rigorous proof, of course, but it can be proved, too, and
it can also be shown by Appealing to Authority. Jackson's too rich for my
blood so I won't quote from it, but Griffiths, in his extremely readable
introductory text on electrodynamics, covers this point on page 378.
Quote: "Evidently, E and B are in phase, and mutually perpendicular...".
Evidently, the energy isn't evenly distributed in space along the wave.
But it's all zipping along with the wave anyway (The Poynting vector, S =
ExB, points in the direction of the wave, of course -- or it better,
unless somebody (God?) got a sign wrong!). (Note that if the E and B
fields were out of phase, S would point the _wrong_ _way_ half the time,
which would imply the energy wasn't flowing with the wave.)
The situation for a standing wave is presumably rather different.
Androcles
| On Wed, 29 Sep 2004 22:31:33 +0000, Androcles wrote:
| [ ... ]
| >
| > All we can say of the photon is what we know of radio waves. Fools like
| > Bilge, Andersen and moortel (well, he's just a puppy that wants to bark
| > with the rest of the hounds) insist the electric and magnetic fields
are
| > in phase, denying conservation of energy. They get this crazy notion
from
| > Maxwell, who believed in aether. Maxwell's equations do not deny a 90
| > degree phase shift between the E and B fields, as they seem to think.
|
| Actually, for a traveling plane wave, this appears to be a consequence of
| Faraday's law: curl(E) = -@B/@t.
| Just by eyeballing the E and B fields,
Ugh...
| the curl of E will peak at the zero-crossings,
| when the spatial rate of
| change is the highest, so that's also when @B/@t must be highest.
"It was deduced in 1862 by James Clerk Maxwell that the converse happens:
that a changing E-field produces a magnetic field. Put the two together,
that a change in one gives the other and vice-versa, and you might wonder
where it all ends. In fact it ends with a dance between the two forms of
energy in which E and M continually rub shoulders within what is called an
electromagnetic wave - of which light, radio and X-rays are all examples.
Maxwell expresses Faraday's law in a more general vector field equation -
curl E = -@B/@t"
http://www.ee.surrey.ac.uk/Workshop/advice/coils/faraday.html
Possibly you are confused by the term "curl". This is when the field
has reached its maximum value and is "curling" back on itself (changing
direction), and the magnetic field is in the middle of its div(e) as it
crosses
zero. Curl is when the field has its lowest rate of change. The slope at
the top of a sine wave is zero.
See http://www.androc1es.pwp.blueyonder.co.uk/Radio%20Wave.htm
| And
| @B/@t is naturally highest when B is crossing zero ...
| and that implies
| that E and B cross zero at the same points in space.
No, it doesn't imply that at all. If both fields were zero at some
point in space, there would be no energy at that point unless it
were stored in some other form, such as a compressed aether.
| And that, in turn,
| implies they're in phase.
No, it implies E and B are phase-shifted by 90 degrees.
"In fact it ends with a dance between the two forms of
energy in which E and M continually rub shoulders within what is called an
electromagnetic wave - of which light, radio and X-rays are all examples. "
|
| That wasn't a rigorous proof, of course,
It wasn't a proof at all.
| but it can be proved, too, and
| it can also be shown by Appealing to Authority.
I'll quite happily appeal to the authority of Maxwell, if that is what it
takes.
"a changing E-field produces a magnetic field" - Maxwell, Faraday,
Observation.
Observation is the greatest authority of all.
| Jackson's too rich for my
| blood so I won't quote from it, but Griffiths, in his extremely readable
| introductory text on electrodynamics, covers this point on page 378.
| Quote: "Evidently, E and B are in phase, and mutually perpendicular...".
Evidently, Griffiths has no idea what he's talking about, because evidently
(by which I mean the result of experiment, and not "obviously" as you mean
it) the B and E fields are out of phase, as ANY practicing
electrical/electronic engineer will tell you.
I will, however, agree as to them being mutually perpendicular.
|
| Evidently, the energy isn't evenly distributed in space along the wave.
That's the same as saying it isn't evidently conserved, and evidently it is.
When a water wave propagates on a pond, at any given instant the height of a
molecule of water represents potential energy and its fall or rise
represents
kinetic energy. At mean pond level, the kinetic energy is at its greatest
and
the potential energy at zero. At its greatest height, the potential energy
is at
its greatest and the kinetic energy is zero. This applies to every molecule
of water taking part in the ripple. EVIDENTLY the energy is distributed
evenly at any instant.
| But it's all zipping along with the wave anyway (The Poynting vector, S =
| ExB, points in the direction of the wave, of course -- or it better,
| unless somebody (God?) got a sign wrong!). (Note that if the E and B
| fields were out of phase, S would point the _wrong_ _way_ half the time,
| which would imply the energy wasn't flowing with the wave.)
|
| The situation for a standing wave is presumably rather different.
In other words the Poynting vector is now for two waves travelling in
opposite directions and we have a node or series of nodes.
If you travel alongside a photon it will have the same energy, relative
to you, as travelling alongside another car. NONE. That doesn't mean
the wheels are not turning.
Androcles
|
| --
| I can be contacted through http://www.physicsinsights.org, but I am
| unable to listen to reason.
Dirk Van de moortel
"Androcles" <andr...@nospamblueyonder.co.uk> wrote in message news:O8V6d.181$xb....@text.news.blueyonder.co.uk...
What does a mathematical no-brain like you know about
someone being confused about the term "curl"?
Keep your mouth shut - you stink like hell.
Dirk Vdm
sal
>
> "sal" <pragm...@nospam.org> wrote in message
> news:pan.2004.09.30...@nospam.org...
> | On Wed, 29 Sep 2004 22:31:33 +0000, Androcles wrote: [ ... ]
> | >
> | > All we can say of the photon is what we know of radio waves. Fools
> | > like Bilge, Andersen and moortel (well, he's just a puppy that wants
> | > to bark with the rest of the hounds) insist the electric and magnetic
> | > fields
> are
> | > in phase, denying conservation of energy. They get this crazy notion
> from
> | > Maxwell, who believed in aether. Maxwell's equations do not deny a 90
> | > degree phase shift between the E and B fields, as they seem to think.
> |
> | Actually, for a traveling plane wave, this appears to be a consequence
> | of Faraday's law: curl(E) = -@B/@t.
>
> | Just by eyeballing the E and B fields,
> Ugh...
Well, I can picture it just fine. And I understand curl well
enough to be able to judge roughly what it'll be just by looking at a
vector field. What's so strange about that?
> | the curl of E will peak at the zero-crossings, when the spatial rate
> | of change is the highest, so that's also when @B/@t must be highest.
>
> "It was deduced in 1862 by James Clerk Maxwell that the converse
> happens: that a changing E-field produces a magnetic field. Put the two
> together, that a change in one gives the other and vice-versa, and you
> might wonder where it all ends. In fact it ends with a dance between the
> two forms of energy in which E and M continually rub shoulders within
> what is called an electromagnetic wave - of which light, radio and
> X-rays are all examples. Maxwell expresses Faraday's law in a more
> general vector field equation - curl E = -@B/@t"
> http://www.ee.surrey.ac.uk/Workshop/advice/coils/faraday.html
>
> Possibly you are confused by the term "curl".
No, I am not confused by the term "curl". Del X V is closely related to
the exterior derivative of a vector, and it is also a _derivative_. At
the peak and trough of the wave, the field is locally unidirectional, the
magnitude is not changing, and the curl -- along with the time derivative
-- is zero.
You may not be able to picture the curl of a sine wave in 3-space
directly, but you ought to be able to imagine integrating the field around
a small square path. Make the square lie in the same plane as the wave.
If you take the integral at the point where the wave is constant -- max or
min -- it will be zero. If you take it where the field is changing, the
"up" and "down" parts of the path won't cancel. And that integral is the
magnitude of the curl. If you look at the resulting magnitude, laid on a
vector which is perpendicular to the area enclosed by the loop, you'll
find it's parallel to the B field, and pointing in the same direction as
-@B/@t .... but only if the B field is changing in step with the E field
at that point. If B is at a peak while E is changing rapidly the two
won't match.
Since curl(B) = @E/@t in vacuum, you can run the same argument from either
field to the other, of course.
> This is when the field has
> reached its maximum value and is "curling" back on itself (changing
> direction),
No, it's not. Sorry. Curl is a derivative -- a _first_ derivative, not
a _second_ derivative -- and when you're looking at a maximum or minimum
in a (locally) unidirectional vector field, as you are near the peak or
trough of a traveling plane sine wave, curl is necessarily zero (as I
already pointed out).
Perhaps you are confused by the fact that CURL of E is maxed out at the
point where the MAGNITUDE of E is at a minimum?
Here, let's look at the actual value of curl(E) in this case. Assume it's
a plane wave, traveling along the X axis, with E directed along the Y
axis. Since the X and Z components of E are identically zero, and the Y
component varies only with X (it's independent of Z), the only nonzero
term in curl(E) in this case is
@Ey/@x, directed in the Z direction
and that has maximum magnitude when E=0, and it's zero when E's magnitude
is at its maximum. Similarly, B is changing most rapidly when B=0. So,
those two points -- B=0 and E=0 -- must occur at the same place in space.
[ ... ]
>
> | but it can be proved, too, and
> | it can also be shown by Appealing to Authority.
>
> I'll quite happily appeal to the authority of Maxwell, if that is what it
> takes.
> "a changing E-field produces a magnetic field" - Maxwell, Faraday,
> Observation.
> Observation is the greatest authority of all.
So? Your quote from Maxwell given here does not address this question
directly, and in any case you started out by saying Maxwell himself said
the fields were in phase. You said:
> | > ... [they believe] the electric and magnetic fields are in phase,
> | > denying conservation of energy. They get this crazy notion from
> | > Maxwell ...
You started by saying Maxwell was all wet, and now you want to use him to
bolster your argument?
>
> | Jackson's too rich for my
> | blood so I won't quote from it, but Griffiths, in his extremely
> | readable introductory text on electrodynamics, covers this point on
> | page 378. Quote: "Evidently, E and B are in phase, and mutually
> | perpendicular...".
>
> Evidently, Griffiths has no idea what he's talking about,
Wrong.
You're not just dismissing relativity here, you know...
And energy _IS_ conserved -- it's just not evenly distributed throughout
space. What's the problem with that? The energy distribution in the
plane wave is like beads on a string, and they're all moving through space
in tandem. The wave carries the energy along with it -- isn't that what
you'd expect, after all?
And yes, I snipped a bunch, from the point where you denied the validity
of a totally standard E&M text on down.
Address the fact that curl(E) in a plane wave goes as @Ey/@x, and the
magnitude of that derivative reaches a maximum when E=0, and then we can
take it from there.
Androcles
| At
| the peak and trough of the wave, the field is locally unidirectional,
Do you mean along the X-axis, Y-axis, Z-axis, sqrt(ax^2+by^2+cz^2),
or the T-axis?
If sqrt(ax^2+by^2+cz^2),what are the values of a,b, and c?
Is gravity unidirectional? If so, I doubt Henri, you or I can agree on the
direction, except "down". It isn't East, West, North or South, is it?
I'll agree for the T- axis, and not just locally, either.
the magnitude is not changing, and the curl -- along with the time
derivative
| -- is zero.
Magnitude of what? The E-field? Or the curl of the E-field?
|
| You may not be able to picture the curl of a sine wave in 3-space
| directly, but you ought to be able to imagine integrating the field around
| a small square path.
Integrate a sine wave? Of course I can, it's -cos().
That integrates to -sin, which integrates to cos, which integrates to sin.
Shoot, you have only to look at the slope of the sine wave on the
T-axis to see it is one = cos(0) -- err..locally.
| Make the square lie in the same plane as the wave.
A square with sides magnitude and time? Ok.
| If you take the integral at the point where the wave is constant -- max or
| min -- it will be zero.
Yeah... so? The E-field isn't changing, so the B-field is zero. That's what
Faraday and Maxwell are saying.
| If you take it where the field is changing,
Which is at maximum when the E-Field is zero...
| the
| "up" and "down" parts of the path won't cancel.
What path? I don't know what you are talking about. When did
paths come into it? We are talking about fields, not paths. This
is about physics, not a walk in the countryside. The only thing
that crosses an E-Field is a B-field, not a path.
|And that integral is the
| magnitude of the curl.
Yeah, so? The magnitude of the curl of the E-field
is the magnitude of the B-field.
curl E = -@B/@t.
| If you look at the resulting magnitude, laid on a
| vector which is perpendicular to the area enclosed by the loop,
What loop? You are continually introducing new terms.
I can't follow you. Squares with time on one side, paths, loops,
it's hopeless. Next you'll be telling me the E and B fields are
in phase, which is known by experiment to be wrong.
you'll
| find it's parallel to the B field, and pointing in the same direction as
| -@B/@t .... but only if the B field is changing in step with the E field
| at that point. If B is at a peak while E is changing rapidly the two
| won't match.
|
| Since curl(B) = @E/@t in vacuum, you can run the same argument from either
| field to the other, of course.
And your argument will be just as invalid.
When the B-field is at it greatest rate of change, the spark plug in your
car engine fires. We get that when the current in the primary winding
of the coil is interrupted and the energy stored in the B-field collapses.
Connect a couple of wires across the switch of an old solenoid doorbell
and you'll soon find out the jolt you'll get from the 3V-9V battery that
operates it. Heck, I was doing that stuff when I was 12 years old,
shocking myself and my kid sister. Understanding it came easily to me
later when I gained the math.
|
| > This is when the field has
| > reached its maximum value and is "curling" back on itself (changing
| > direction),
|
| No, it's not. Sorry.
I'm done arguing. Obviously you have never played with electricity
as a child. I did.
The E and B fields of a photon are 90 degrees phase shifted, whether
you believe it or not.
You obviously have a reading comprehension problem. Nowhere did I say
"Maxwell claims the E and B fields are in phase and is all wet".
What I actually quoted was
http://www.ee.surrey.ac.uk/Workshop/advice/coils/faraday.html
Look at your own take on what I said: " [they believe] ".
If you want to argue that the E and B fields are in phase, take it
up with Faraday.
I'm done.
Androcles.
sal
on this topic, I had no idea whether the E and B fields were in or out of
phase in EM radiation. After I saw your post, I went and looked at what
Griffiths had to say about it, and I thought about how the curl of a
traveling plane wave must behave. As so often happens when I reply to
your posts, I learned something from it. (Whether it was what you wanted
me to learn, well, that's a separate question...)
On Thu, 30 Sep 2004 19:17:59 +0000, Androcles wrote:
>
> "sal" <pragm...@nospam.org> wrote in message
> news:pan.2004.09.30....@nospam.org... | On Thu, 30 Sep 2004
> 14:56:46 +0000, Androcles wrote: |
> |
> | > "sal" <pragm...@nospam.org> wrote in message | >
> news:pan.2004.09.30...@nospam.org...
> | > On Wed, 29 Sep 2004 22:31:33 +0000, Androcles wrote:
[ ... ]
[The wrapping got a little shattered here -- oops]
> | > | > All we can say of the photon is what we know of radio waves.
> Fools | > | > like Bilge, Andersen and moortel (well, he's just a puppy
> that | > | > wants to bark with the rest of the hounds) insist the
> electric | > | > and
> magnetic
> | > | > fields
> | > are
> | > | > in phase, denying conservation of energy. They get this crazy |
> > | > notion
> | > from
> | > | > Maxwell, who believed in aether. Maxwell's equations do not deny
> a 90
> | > | > degree phase shift between the E and B fields, as they seem to
> think.
[sal]
> | >Actually, for a traveling plane wave, this appears to be a | > |
> consequence of Faraday's law: curl(E) = -@B/@t. | >
> | > | Just by eyeballing the E and B fields
[A]
| > Ugh...
[sal]
> | Well, I can picture it just fine. And I understand curl well enough
> to | be able to judge roughly what it'll be just by looking at a vector
> | field. What's so strange about that? |
> | > | the curl of E will peak at the zero-crossings, when the spatial
> rate | > | of change is the highest, so that's also when @B/@t must be
> highest.
[A]
> | > "It was deduced in 1862 by James Clerk Maxwell that the converse | >
> happens: that a changing E-field produces a magnetic field. Put the | >
> two together, that a change in one gives the other and vice-versa, and |
> > you might wonder where it all ends. In fact it ends with a dance | >
> between the two forms of energy in which E and M continually rub | >
> shoulders within what is called an electromagnetic wave - of which | >
> light, radio and X-rays are all examples. Maxwell expresses Faraday's |
> > law in a more general vector field equation - curl E = -@B/@t" | >
> http://www.ee.surrey.ac.uk/Workshop/advice/coils/faraday.html
> | > Possibly you are confused by the term "curl".
[sal]
> | No, I am not confused by the term "curl". Del X V is closely related
> to | the exterior derivative of a vector, and it is also a _derivative_.
[From here down the wrapping is OK, I think, and who said what should be
reasonably clear.]
[A]
> Good.
[SAL]
> | At
> | the peak and trough of the wave, the field is locally unidirectional,
[A]
> Do you mean along the X-axis, Y-axis, Z-axis, sqrt(ax^2+by^2+cz^2), or
> the T-axis?
We're talking about ordinary 3-space here, and EM waves which propagate
through that space. Beyond that initial assumption, it doesn't really
matter for that particular assertion what direction a plane wave is
traveling in -- at the locations where the E field's magnitude is at a
peak, the E field is at a plateau, and the partial derivatives with
respect to all spatial dimensions and the partial with respect to time of
that field are all zero. And for a plane wave, at the points where the E
field magnitude is at a peak, the E field is not changing direction -- it
is "locally unidirectional", which is to say at all points close to the
peak the field points in the same direction.
I actually thought that was all pretty apparent.
> Is gravity unidirectional? If so, I doubt Henri, you or I can agree on
> the direction, except "down". It isn't East, West, North or South, is
> it? I'll agree for the T- axis, and not just locally, either.
>
> the magnitude is not changing, and the curl -- along with the time
> derivative -- is zero.
>
> Magnitude of what? The E-field? Or the curl of the E-field?
Magnitude of the E field, as I again thought was obvious.
At the peaks in the E field, the magnitude of the E field is momentarily
unchanging.
Since the magnitude and direction are momentarily unchanging, the curl of
the E field is necessarily zero at those points.
This should, again, be obvious.
[sal]
> | You may not be able to picture the curl of a sine wave in 3-space
> | directly, but you ought to be able to imagine integrating the field
> | around a small square path.
[A]
> Integrate a sine wave?
No, that's not what I meant. Did you not recognize a simple application
of Stokes' theorem here? I thought you would realize what I was getting
at, and didn't go into great detail as a result.
Stokes' theorem in 3 dimensions, stated briefly, in English: Take the
line integral around a closed path of the value V*F where V is the tangent
to the path and F is the value of a vector field at that point on the
path. "*" in this case represents ordinary dot product. That integral is
equal to the integral of curl(F) taken over a surface bounded by the path;
at each point on the surface we actually use the value curl(F)*N where N
is the normal vector to the surface, and "*" again represents dot product.
We're in 3-space here, and the closed path is in _space_ -- it doesn't
extend into the time dimension.
If the wave is traveling along the X axis, with the E field pointing along
the Y axis, then the path I was suggesting integrating around would have
two sides parallel to the X axis, and two sides parallel to the Y axis. If
it starts at (x1,y1,z1), then the path might be described as
(x1,y1,z1) -> (x1+delta,y1,z1)
-> (x1+delta,y1+delta,z1)
-> (x1,y1+delta,z1)
-> (x1,y1,z1)
Again, I didn't go into detail because I thought you would see immediately
what I was getting at.
[A]
> Of course I can, it's -cos(). That integrates to -sin, which integrates
> to cos, which integrates to sin. Shoot, you have only to look at the
> slope of the sine wave on the T-axis to see it is one = cos(0) --
> err..locally.
>
>
>
> | Make the square lie in the same plane as the wave. A square with sides
> magnitude and time? Ok.
No, the path lies entirely in space, with no time extension, as I already
said.
[sal]
> | If you take the integral at the point where the wave is constant --
> max or min -- it will be zero.
[A]
> Yeah... so? The E-field isn't changing, so the B-field is zero. That's
> what Faraday and Maxwell are saying.
WRONG !!!
The E field isn't changing, so the CURL of the B-field is zero, _NOT_ the
MAGNITUDE. And at that point, the B field also isn't changing, so the
CURL of the E-field is also zero.
The magnitude of the B field (or E field) may or may not be zero where its
curl is zero! In this case, it is decidedly not zero at the point where it
has zero curl.
> | If you take it where the field is changing,
>
> Which is at maximum when the E-Field is zero...
>
> | the
> | "up" and "down" parts of the path won't cancel.
>
> What path? I don't know what you are talking about.
In the application of Stoke's theorem, above, I talked about a line
integral around a square .... a _path_ which had four straight segments
joined by 90 degree bends. That's the _path_ I was talking about.
Again, I didn't fill in all the details because I thought you would
immediately see what I was talking about.
> |And that integral is the
> | magnitude of the curl.
> Yeah, so? The magnitude of the curl of the E-field
> is the magnitude of the B-field.
> curl E = -@B/@t.
And it's also the value you get integrating the E field around a closed
path -- a closed curve -- a loop -- whatever you want to call it. (Oh, of
course you need to multiply the curl by the area to get the actual value
of the line integral.)
> | If you look at the resulting magnitude, laid on a vector which is
> | perpendicular to the area enclosed by the loop,
>
> What loop?
loop == path == closed curve == that square I talked about integrating
around.
That's the "loop" in question.
All of these terms are commonly used in talking about Stokes' theorem and
line integrals; I was not being intentionally obscure.
> You are continually introducing new terms. I can't follow you. Squares
> with time on one side, paths, loops, it's hopeless. Next you'll be
> telling me the E and B fields are in phase, which is known by experiment
> to be wrong.
I'd be interested in hearing about such experiments, because as far as I
can see, for a plane wave E and B have got to be in phase. They're tilted
90 degrees (in space) with respect to each other, of course.
> | you'll find it's parallel to the B field, and pointing in the same
> | direction as -@B/@t .... but only if the B field is changing in step
> | with the E field at that point. If B is at a peak while E is changing
> | rapidly the two won't match.
> |
> | Since curl(B) = @E/@t in vacuum, you can run the same argument from
> | either field to the other, of course.
>
> And your argument will be just as invalid. When the B-field is at it
> greatest rate of change, the spark plug in your car engine fires.
My spark plugs were never fired by an ignition system that used radio
waves propagating in free space. They used waves propagating down a
waveguide, commonly called a "spark plug wire", which is something far
more complicated than a vacuum and I would not claim I can see anything
about how the E and B fields relate in that situation just by looking at a
picture.
> We get
> that when the current in the primary winding of the coil is interrupted
> and the energy stored in the B-field collapses.
Yes, I'm quite aware of the effect. In fact, while I was digging in the
attic for an antenna to attach to the VCR so we can tape the debate
tonight, I ran across an old bit of circuitry I built for fun some time
back: It has a couple 9V batteries on it, an oscillator and a voltage
multiplier, and it fires occasional pulses into an ignition coil to make
sparks. It never got as much bang as I would have liked, though, because
it operated by discharging a capacitor into the coil; with that
arrangement, the voltage across the primary is limited by the voltage in
the capacitor when it fires. When we _interrupt_ the current using, say,
breaker points, the voltage can go a great deal higher than that. In
fact, when breaking the circuit, it could easily go high enough to blow
the little mosfet I was using to trigger it into little bits, which is one
reason I didn't do it that way. Another was my current source, which was
pretty wimpy; in the usual operating mode ignition coils want something on
the order of an amp flowing through them when the circuit breaks.
> Connect a couple of
> wires across the switch of an old solenoid doorbell and you'll soon find
> out the jolt you'll get from the 3V-9V battery that operates it. Heck, I
> was doing that stuff when I was 12 years old, shocking myself and my kid
> sister. Understanding it came easily to me later when I gained the math.
>
>
>
>
> | > This is when the field has
> | > reached its maximum value and is "curling" back on itself (changing
> | > direction),
> |
> | No, it's not. Sorry.
>
> I'm done arguing. Obviously you have never played with electricity as a
> child.
You constantly forget: I'm not a physics guy, I'm a math guy.
You've just described curl as a second derivative, and it's not... and
math guys do typically know something about curl, after all.
And you stopped just one strategic paragraph before I stated the formula
for curl(E) for a plane wave. You can check this, if you like; once
again, to quote,
> | Here, let's look at the actual value of curl(E) in this case. Assume
> | it's a plane wave, traveling along the X axis, with E directed along
> | the Y axis. Since the X and Z components of E are identically zero,
> | and the Y component varies only with X (it's independent of Z), the
> | only nonzero term in curl(E) in this case is
> |
> | @Ey/@x, directed in the Z direction
> |
> | and that has maximum magnitude when E=0, and it's zero when E's
> | magnitude is at its maximum. Similarly, B is changing most rapidly
> | when B=0. So, those two points -- B=0 and E=0 -- must occur at the
> | same place in space.
Do you have some objection to the claim that curl(E) = @Ey/@x, in the Z
direction, in that case? Since Ey = k*sin(a*X + b*t), for some constants
a and b and k, the Z component of curl(E) = a*k*cos(a*X + b*t), which most
certainly maxes out when E == 0.
Androcles
Which is when the magnitude of the B field is momentarily zero but IS
changing.
In other words, the E and B fields are phase shifted by a quarter circle, as
I thought was obvious and EVIDENT.
|
| Since the magnitude and direction are momentarily unchanging, the curl of
| the E field is necessarily zero at those points.
|
| This should, again, be obvious.
What you are omitting to state is the domain of the function.
In the time domain, the magnitude (codomain or image) of the E field is
sine(t)
and the magnitude of the B field is cos(t).
|
| [sal]
| > | You may not be able to picture the curl of a sine wave in 3-space
| > | directly, but you ought to be able to imagine integrating the field
| > | around a small square path.
|
| [A]
| > Integrate a sine wave?
|
| No, that's not what I meant. Did you not recognize a simple application
| of Stokes' theorem here? I thought you would realize what I was getting
| at, and didn't go into great detail as a result.
|
| Stokes' theorem in 3 dimensions, stated briefly, in English: Take the
| line integral around a closed path of the value V*F where V is the tangent
| to the path and F is the value of a vector field at that point on the
| path. "*" in this case represents ordinary dot product. That integral is
| equal to the integral of curl(F) taken over a surface bounded by the path;
| at each point on the surface we actually use the value curl(F)*N where N
| is the normal vector to the surface, and "*" again represents dot product.
|
| We're in 3-space here, and the closed path is in _space_ -- it doesn't
| extend into the time dimension.
Now you are getting there. However, we live in a universe where time passes
and the E and B field MAGNITUDES do not occur at the same place AT THE SAME
TIME. If you travel with the wave, as a dragonfly might fly along with a
water wave, the MAGNITUDE NEVER CHANGES. However, the dragonfly can choose
the trough or the crest to hover above. Now we have effectively eliminated
time in our study.
|
| If the wave is traveling along the X axis, with the E field pointing along
| the Y axis, then the path I was suggesting integrating around would have
| two sides parallel to the X axis, and two sides parallel to the Y axis. If
| it starts at (x1,y1,z1), then the path might be described as
|
| (x1,y1,z1) -> (x1+delta,y1,z1)
| -> (x1+delta,y1+delta,z1)
| -> (x1,y1+delta,z1)
| -> (x1,y1,z1)
|
| Again, I didn't go into detail because I thought you would see immediately
| what I was getting at.
What you were getting at is that the B and E fields are in phase, and they
are not.
|
| [A]
| > Of course I can, it's -cos(). That integrates to -sin, which integrates
| > to cos, which integrates to sin. Shoot, you have only to look at the
| > slope of the sine wave on the T-axis to see it is one = cos(0) --
| > err..locally.
| >
| >
| >
| > | Make the square lie in the same plane as the wave. A square with sides
| > magnitude and time? Ok.
|
| No, the path lies entirely in space, with no time extension, as I already
| said.
What you are discussing is in the form y = f(x), not y = f(t).
curl(E) = -d@B/@t <-- see that '@t'?
REPEAT... can you see @t??
|
| [sal]
| > | If you take the integral at the point where the wave is constant --
| > max or min -- it will be zero.
|
| [A]
| > Yeah... so? The E-field isn't changing, so the B-field is zero. That's
| > what Faraday and Maxwell are saying.
|
| WRONG !!!
|
| The E field isn't changing, so the CURL of the B-field is zero, _NOT_ the
| MAGNITUDE.
If the E field isn't changing, the magnitude of the B field is constant.
I was incomplete when I said it was zero, I meant a constant zero when the E
field was at a maximum. So yes, what I said was wrong, but not "WRONG !!!"
That condition arises as you fly along with the wave, as the dragonfly
might.
Maybe that is where you relativists come to the ridiculous conclusion that
time stands still if you travel at the speed of light.
And at that point, the B field also isn't changing, so the
| CURL of the E-field is also zero.
|
| The magnitude of the B field (or E field) may or may not be zero where its
| curl is zero! In this case, it is decidedly not zero at the point where it
| has zero curl.
It has zero curl when it is constant. Is that better?
|
|
| > | If you take it where the field is changing,
| >
| > Which is at maximum when the E-Field is zero...
| >
| > | the
| > | "up" and "down" parts of the path won't cancel.
| >
| > What path? I don't know what you are talking about.
|
| In the application of Stoke's theorem, above, I talked about a line
| integral around a square .... a _path_ which had four straight segments
| joined by 90 degree bends. That's the _path_ I was talking about.
|
| Again, I didn't fill in all the details because I thought you would
| immediately see what I was talking about.
Well, I didn't. As far as I could see, your square had sides
magnitude and time. A path around that would imply time runs backwards.
So you weren't just wrong, but WRONG!!!
|
|
|
| > |And that integral is the
| > | magnitude of the curl.
| > Yeah, so? The magnitude of the curl of the E-field
| > is the magnitude of the B-field.
| > curl E = -@B/@t.
|
| And it's also the value you get integrating the E field around a closed
| path -- a closed curve -- a loop -- whatever you want to call it. (Oh, of
| course you need to multiply the curl by the area to get the actual value
| of the line integral.)
|
|
| > | If you look at the resulting magnitude, laid on a vector which is
| > | perpendicular to the area enclosed by the loop,
| >
| > What loop?
|
| loop == path == closed curve == that square I talked about integrating
| around.
|
| That's the "loop" in question.
|
| All of these terms are commonly used in talking about Stokes' theorem and
| line integrals; I was not being intentionally obscure.
|
|
| > You are continually introducing new terms. I can't follow you. Squares
| > with time on one side, paths, loops, it's hopeless. Next you'll be
| > telling me the E and B fields are in phase, which is known by experiment
| > to be wrong.
|
| I'd be interested in hearing about such experiments, because as far as I
| can see, for a plane wave E and B have got to be in phase. They're tilted
| 90 degrees (in space) with respect to each other, of course.
Do it yourself. All you need are some resistors, a solenoid, a switch, a
nine volt
battery and a voltmeter, all of which you'll get from Radio Shack. Connect
everything except the voltmeter in series and make a closed path, then
connect
the voltmeter in parallel with the solenoid. Oh, and you need a paperclip or
nail to detect when the B-field is present. Don't buy a digital voltmeter,
they react too slowly, get a cheaper swinging needle type. Now play with it,
turning the switch on and off. With a high value resistor, you'll see the
voltage rise slowly. Turn the switch off and watch the needle kick hard the
opposite way. Now play with reversing the battery terminals and different
resistor values.
|
| > | you'll find it's parallel to the B field, and pointing in the same
| > | direction as -@B/@t .... but only if the B field is changing in step
| > | with the E field at that point. If B is at a peak while E is changing
| > | rapidly the two won't match.
| > |
| > | Since curl(B) = @E/@t in vacuum, you can run the same argument from
| > | either field to the other, of course.
| >
| > And your argument will be just as invalid. When the B-field is at it
| > greatest rate of change, the spark plug in your car engine fires.
|
| My spark plugs were never fired by an ignition system that used radio
| waves propagating in free space. They used waves propagating down a
| waveguide, commonly called a "spark plug wire", which is something far
| more complicated than a vacuum and I would not claim I can see anything
| about how the E and B fields relate in that situation just by looking at a
| picture.
That's because you don't experiment, you look at mathematics and hope
to understand Nature from other people.
| > We get
| > that when the current in the primary winding of the coil is interrupted
| > and the energy stored in the B-field collapses.
|
| Yes, I'm quite aware of the effect. In fact, while I was digging in the
| attic for an antenna to attach to the VCR so we can tape the debate
| tonight, I ran across an old bit of circuitry I built for fun some time
| back: It has a couple 9V batteries on it, an oscillator and a voltage
| multiplier, and it fires occasional pulses into an ignition coil to make
| sparks. It never got as much bang as I would have liked, though, because
| it operated by discharging a capacitor into the coil; with that
| arrangement, the voltage across the primary is limited by the voltage in
| the capacitor when it fires. When we _interrupt_ the current using, say,
| breaker points, the voltage can go a great deal higher than that. In
| fact, when breaking the circuit, it could easily go high enough to blow
| the little mosfet I was using to trigger it into little bits, which is one
| reason I didn't do it that way. Another was my current source, which was
| pretty wimpy; in the usual operating mode ignition coils want something on
| the order of an amp flowing through them when the circuit breaks.
Well, if you know that much, why cant you understand the E field and
B field are phase shifted?
|
| > Connect a couple of
| > wires across the switch of an old solenoid doorbell and you'll soon find
| > out the jolt you'll get from the 3V-9V battery that operates it. Heck, I
| > was doing that stuff when I was 12 years old, shocking myself and my kid
| > sister. Understanding it came easily to me later when I gained the math.
| >
| >
| >
| >
| > | > This is when the field has
| > | > reached its maximum value and is "curling" back on itself (changing
| > | > direction),
| > |
| > | No, it's not. Sorry.
| >
| > I'm done arguing. Obviously you have never played with electricity as a
| > child.
|
| You constantly forget: I'm not a physics guy, I'm a math guy.
|
| You've just described curl as a second derivative, and it's not... and
| math guys do typically know something about curl, after all.
|
| And you stopped just one strategic paragraph before I stated the formula
| for curl(E) for a plane wave.
Yeah, well, I was busy so I dropped it. How does it it feel to have that
done to you, as you have done so many times to me?
You can check this, if you like; once
| again, to quote,
|
| > | Here, let's look at the actual value of curl(E) in this case. Assume
| > | it's a plane wave, traveling along the X axis, with E directed along
| > | the Y axis. Since the X and Z components of E are identically zero,
| > | and the Y component varies only with X (it's independent of Z), the
| > | only nonzero term in curl(E) in this case is
| > |
| > | @Ey/@x, directed in the Z direction
| > |
| > | and that has maximum magnitude when E=0, and it's zero when E's
| > | magnitude is at its maximum. Similarly, B is changing most rapidly
| > | when B=0. So, those two points -- B=0 and E=0 -- must occur at the
| > | same place in space.
Well good grief, obviously if B=E=0, there is no energy to consider and an
infinite number of places in space where that is true. Sheesh, I'm talking
about
the ripples on a pond and you are talking about a flat surface.
Paul B. Andersen
"Androcles" <andr...@nospamblueyonder.co.uk> skrev i melding news:uu77d.657$xb....@text.news.blueyonder.co.uk...
>
> "sal" <pragm...@nospam.org> wrote in message
> news:pan.2004.09.30....@nospam.org...
> | Androcles, perhaps I should mention that before I read your earlier post
> | on this topic, I had no idea whether the E and B fields were in or out of
> | phase in EM radiation. After I saw your post, I went and looked at what
> | Griffiths had to say about it, and I thought about how the curl of a
> | traveling plane wave must behave. As so often happens when I reply to
> | your posts, I learned something from it. (Whether it was what you wanted
> | me to learn, well, that's a separate question...)
> |
> |
> | On Thu, 30 Sep 2004 19:17:59 +0000, Androcles wrote:
> | >
> | > "sal" <pragm...@nospam.org> wrote in message
> | > news:pan.2004.09.30....@nospam.org... | On Thu, 30 Sep 2004
> | > 14:56:46 +0000, Androcles wrote: |
> | > |
> | > | > "sal" <pragm...@nospam.org> wrote in message | >
> | > news:pan.2004.09.30...@nospam.org...
> | > | > On Wed, 29 Sep 2004 22:31:33 +0000, Androcles wrote:
> | [ ... ]
> | > | > | > All we can say of the photon is what we know of radio waves.
> | > | > | > Fools like Bilge, Andersen and moortel (well, he's just a puppy
> | > | > | > that wants to bark with the rest of the hounds) insist the
> | > | > | > electric and magnetic fields are
> | > | > | > in phase, denying conservation of energy. They get this crazy
> | > | > | > notion from
> | > | > | > Maxwell, who believed in aether.
Indeed.
So you will have to include Maxwell in the list of fools
believing that the fields are in phase.
> | > | > | > Maxwell's equations
> | > | > | > do not denya 90
> | > | > | > degree phase shift between the E and B fields, as they seem to
> | > | > | > think.
Androcles, I challenge you to prove that the wave:
Ex(z,t) = Eo*sin(wt - kz)
Hy(z,t) = Eo/Z*cos(wt - kz)
is a possible solution of Maxwell's equations.
I don't expect an answer, of course,
because I know this is way beyond your abilites.
What is evident to Androcles is usually wrong.
This is no exception.
> | Since the magnitude and direction are momentarily unchanging, the curl of
> | the E field is necessarily zero at those points.
> |
> | This should, again, be obvious.
>
> What you are omitting to state is the domain of the function.
> In the time domain, the magnitude (codomain or image) of the E field is
> sine(t)
> and the magnitude of the B field is cos(t).
Wrong.
> | [sal]
> | > | You may not be able to picture the curl of a sine wave in 3-space
> | > | directly, but you ought to be able to imagine integrating the field
> | > | around a small square path.
> |
> | [A]
> | > Integrate a sine wave?
> |
> | No, that's not what I meant. Did you not recognize a simple application
> | of Stokes' theorem here? I thought you would realize what I was getting
> | at, and didn't go into great detail as a result.
> |
> | Stokes' theorem in 3 dimensions, stated briefly, in English: Take the
> | line integral around a closed path of the value V*F where V is the tangent
> | to the path and F is the value of a vector field at that point on the
> | path. "*" in this case represents ordinary dot product. That integral is
> | equal to the integral of curl(F) taken over a surface bounded by the path;
> | at each point on the surface we actually use the value curl(F)*N where N
> | is the normal vector to the surface, and "*" again represents dot product.
> |
> | We're in 3-space here, and the closed path is in _space_ -- it doesn't
> | extend into the time dimension.
>
> Now you are getting there. However, we live in a universe where time passes
> and the E and B field MAGNITUDES do not occur at the same place AT THE SAME
> TIME. If you travel with the wave, as a dragonfly might fly along with a
> water wave, the MAGNITUDE NEVER CHANGES. However, the dragonfly can choose
> the trough or the crest to hover above. Now we have effectively eliminated
> time in our study.
Let us do a bit of math, and see what Maxwell say.
Let's consider a sinusoidal, linearly polarized, plane wave
propagating along the z-axis, and with the E field
in the x direction and the H field in y direction.
The E and H fields are in phase, and the relation between
their amplitudes is given by the impedance of space Z.
Ex(z,t) = Eo*sin(wt - kz)
Ey(z,t) = 0
Ez(z,t) = 0
Hz (z,t) = 0
Hy(z,t) = Eo/Z*sin(wt - kz)
Hz(z,t) = 0
u is permeability, e is permittivity of space
c is speed of light c^2 = e/u
Z = impedance of space = sqrt(e/u) = u*c
k is wave number = 2pi/lamda = w/c
Let's see if this is a solution of Maxwells equations.
curl E = - u*dH/dt
curl H = e*dE/dt
Since E has only an x-component, and H has
only an y-component, and all changes are in
the z-direction, these equations can be simplified to:
dEx/dz = -u*dHy/dt
dHy/dz = -e*dEx/dt
dEx/dz = -k*Eo*cos(wt-kz) = -(w/c)*Eo*cos(wt-kz)
-u*dHy/dt = w*(-u)*(Eo/(u*c))cos(wt-kz) = -(w/c)*Eo*cos(wt-kz)
The first equation is fulfilled
dHy/dz = -k*(Eo/(u*c))*cos(wt-kz)= -w*e*Eo*cos(wt-kz)
-e*dEx/dt = w*(-e)*Eo*cos(wt-kz) = -w*e*Eo*cos(wt-kz)
The second equation is fullfilled.
The wave above is a solution of Maxwell's equations.
QED
Actually, the simple fact that the impedance of free space
is resistive (real) suffice to prove that the fields must be in phase.
But what about the energy? Is it conserved?
The power density (or intensity, power per area)
in the wave is E X H. This vector is in the z-direction,
the power is flowing in the z-direction.
P(x,t) = (Eo^2/Z)*(sin(wt-kz))^2
The energy is not distributed evenly along the wave.
It is moving along in chuks.
It is zero whenever (wt -kz) = n*pi, n integer.
Not very remarkable, and it has of course nothing
with conservation of energy to do.
That an incandescent lamp flicker when powered
by a/c does not mean that energy isn't conserved.
Why should energy have to be evenly
distributed to be conserved? :-)
A strange idea!
But Androcles will never learn, of course.
He will die will all his misconceptions intact.
Won't you, Androcles?
Paul
sal
>
> "sal" <pragm...@nospam.org> wrote in message
> news:pan.2004.09.30....@nospam.org...
> |
> | On Thu, 30 Sep 2004 19:17:59 +0000, Androcles wrote:
> | >
> | > "sal" <pragm...@nospam.org> wrote in message
[snip stuff we're no longer discussing]
> | >
> | > the magnitude is not changing, and the curl -- along with the time
> | > derivative -- is zero.
> | >
> | > Magnitude of what? The E-field? Or the curl of the E-field?
> |
> | Magnitude of the E field, as I again thought was obvious.
> |
> | At the peaks in the E field, the magnitude of the E field is momentarily
> | unchanging.
>
> Which is when the magnitude of the B field is momentarily zero but IS
> changing. In other words, the E and B fields are phase shifted by a
> quarter circle, as I thought was obvious and EVIDENT.
It seems like it should be that way, but, since curl(E)=-@B/@t, it's
not. Consider:
(Notation in these posts is awkward, unfortunately, but the time has come
to write some vectors regardless.)
By definition of the curl, we have, at one particular instant (NB -- this
mess is supposed to be a 3-vector, shown as three components):
curl(E) = (@Ez/@y - @Ey/@z, @Ex/@z - @Ez/@x, @Ey/@x - @Ex/@y)
This is a property of the instantaneous _spatial_ distribution of the E
field. The time domain doesn't enter into it.
The Maxwell equation known as Faraday's law connects this with B (and with
the time domain):
curl(E) = -@B/@t
For a plane wave traveling along the X axis, with the E field parallel to
the Y axis, and the B field parallel to the Z axis, we have:
Ey = k * sin(a*X + b*t) (for some constants k, a, b)
and Ex = Ez = 0.
Then,
@Ey/@x = k*a*cos(a*X + b*t)
@Ey/@t = k*b*cos(a*X + b*t)
and all the other partials of components of E are zero.
Then from the definition of curl(E), given above, we have
curl(E) = (0, 0, k*a*cos(a*X + b*t))
By Faraday's law, therefore, we have
-@B/@t = (0, 0, k*a*cos(a*X + b*t))
or, in other words,
k*a*cos(a*X + b*t) = (-@B/@t)_z
and the X and Y components of @B/@t are zero.
But integrating that over time, at a single point in space, we find
(k*a/b)*sin(a*X + b*t) = (B(t))_z
and comparing with the original value for E, we find that Bz = Ey * (a/b)
And those are in phase.
This is messy and awkward, and my original "intuitive" argument contained
the same freight but was, on the whole, easier to grasp (or so I thought,
anyway). The statement I just gave here has the advantage of avoiding the
use of Stokes' theorem, of course.
> | Since the magnitude and direction are momentarily unchanging, the curl
> | of the E field is necessarily zero at those points.
> |
> | This should, again, be obvious.
>
> What you are omitting to state is the domain of the function. In the
> time domain, the magnitude (codomain or image) of the E field is sine(t)
> and the magnitude of the B field is cos(t).
But see above -- Maxwell's equations lead inevitably to the conclusion
that if magnitude(E) = sin(t) then magnitude(B) = sin(t), too (up to a
constant multiplier)
> | [sal]
> | > | You may not be able to picture the curl of a sine wave in 3-space
> | > | directly, but you ought to be able to imagine integrating the
> | > | field around a small square path.
> |
> | [A]
> | > Integrate a sine wave?
> |
> | No, that's not what I meant. Did you not recognize a simple
> | application of Stokes' theorem here? I thought you would realize what
> | I was getting at, and didn't go into great detail as a result.
> |
> | Stokes' theorem in 3 dimensions, stated briefly, in English: Take the
> | line integral around a closed path of the value V*F where V is the
> | tangent to the path and F is the value of a vector field at that point
> | on the path. "*" in this case represents ordinary dot product. That
> | integral is equal to the integral of curl(F) taken over a surface
> | bounded by the path; at each point on the surface we actually use the
> | value curl(F)*N where N is the normal vector to the surface, and "*"
> | again represents dot product.
> |
> | We're in 3-space here, and the closed path is in _space_ -- it doesn't
> | extend into the time dimension.
>
> Now you are getting there. However, we live in a universe where time
> passes and the E and B field MAGNITUDES do not occur at the same place
> AT THE SAME TIME.
But as it stands, that is just an assertion. It's a plausible one, but
still just an assertion -- and when we compute the curl of E (in SPACE,
at one moment in time) and compare it with the time derivative of B at
that same moment, we find that it's not correct, as I did, up above.
> If you travel with the wave, as a dragonfly might fly
> along with a water wave, the MAGNITUDE NEVER CHANGES. However, the
> dragonfly can choose the trough or the crest to hover above. Now we have
> effectively eliminated time in our study.
Yes, and if you do this, Maxwell's equations fall down the stairs and
break into little pieces. For the (spatial) curl of E and B is still
nonzero almost everywhere, but the partial of each of E and B with respect
to time is now identically zero.
The form in which we currently employ Maxwell's equations requires that C
be invariant -- in other words, special relativity is wired into them. If
you want to allow the dragonfly to fly along at C you need to patch the
equations somehow.
> | If the wave is traveling along the X axis, with the E field pointing
> | along the Y axis, then the path I was suggesting integrating around
> | would have two sides parallel to the X axis, and two sides parallel to
> | the Y axis. If it starts at (x1,y1,z1), then the path might be
> | described as
> |
> | (x1,y1,z1) -> (x1+delta,y1,z1)
> | -> (x1+delta,y1+delta,z1)
> | -> (x1,y1+delta,z1)
> | -> (x1,y1,z1)
> |
> | Again, I didn't go into detail because I thought you would see
> | immediately what I was getting at.
>
> What you were getting at is that the B and E fields are in phase, and
> they are not.
Well, the equations sure seem to say that they are...
> | [A]
> | > Of course I can, it's -cos(). That integrates to -sin, which
> | > integrates to cos, which integrates to sin. Shoot, you have only to
> | > look at the slope of the sine wave on the T-axis to see it is one =
> | > cos(0) -- err..locally.
> | >
> | >
> | >
> | > | Make the square lie in the same plane as the wave. A square with
> | > | sides
> | > magnitude and time? Ok.
> |
> | No, the path lies entirely in space, with no time extension, as I
> | already said.
>
> What you are discussing is in the form y = f(x), not y = f(t). curl(E) =
> -d@B/@t <-- see that '@t'? REPEAT... can you see @t??
But that's not the point -- or perhaps it's exactly the point. See my
little monograph at the start of this post -- the curl is defined in
space, for values of the fields chosen at a single instant of time. It's
Maxwell's equations -- and in particular, the one called Faraday's law --
which ties together the space and time domains here.
IIW curl(E) = -@B/@t is not the definition of curl(E) -- rather, it's an
assertion about how curl(E) relates to the B field.
> | [sal]
> | > | If you take the integral at the point where the wave is constant
> | > | --
> | > max or min -- it will be zero.
> |
> | [A]
> | > Yeah... so? The E-field isn't changing, so the B-field is zero.
> | > That's what Faraday and Maxwell are saying.
> |
> | WRONG !!!
> |
> | The E field isn't changing, so the CURL of the B-field is zero, _NOT_
> | the MAGNITUDE.
>
> If the E field isn't changing, the magnitude of the B field is constant.
:-) :-) :-)
If the E field isn't changing, then E must be at an extremum. Right?
If B has constant magnitude, B must be at an extremum. Right?
So if E unchanging => B constant, then they hit their extrema together.
Still right?
So, they're in phase!
> I was incomplete when I said it was zero, I meant a constant zero
Whoops -- constant, yes. Constant _zero_, no, because that would imply
there's no wave.
> when
> the E field was at a maximum. So yes, what I said was wrong, but not
> "WRONG !!!" That condition arises as you fly along with the wave, as the
> dragonfly might.
_AND_ it arises at every peak and trough in the plane wave.
And consequently, the peaks and troughs of E and B must be coincident.
And that was my point, and that's why I went overboard with the
exclamation points and the capital letters on the word "wrong".
> Maybe that is where you relativists come to the ridiculous conclusion
> that time stands still if you travel at the speed of light.
Yeah, as I said, Maxwell's equations, as currently formulated, incorporate
SR. You can't ditch SR without making some change to Maxwell's equations.
> And at that point, the B field also isn't changing, so the
> | CURL of the E-field is also zero.
> |
> | The magnitude of the B field (or E field) may or may not be zero where
> | its curl is zero! In this case, it is decidedly not zero at the point
> | where it has zero curl.
>
> It has zero curl when it is constant. Is that better?
Yes, much! (As I already said, at excessive length...)
Perhaps you meant the voltmeter should go across the resistor?
With no capacitors in the circuit, and the voltmeter on the _coil_
(aka solenoid), when you turn power on the voltage will immediately rise
to 9V (across the coil) and then will _fall_ exponentially as current
increases in the coil. If the coil has small resistance compared to the
resistor, the voltage on the coil will drop to a small value as the
voltage on the resistor rises to (almost) 9V.
For the coil, we have V = L*(dI/dt) (or something like that) -- right?
So initially, dI/dt is large, and V on the coil is at its maximum; as the
current rises to its steady state value, dI/dt drops to zero and so does
the current across the coil.
> Turn the switch off and
> watch the needle kick hard the opposite way.
Yes. To see this you need the meter across the coil, as you said (or
across the switch) -- there won't be any show at the resistor, where the
voltage will just drop.
All that energy in the B field comes back with a bang (literally
-- that little spark inside the switch makes a little "bang" ... unless it
was a big coil, then it makes a big "bang").
> Now play with reversing the
> battery terminals and different resistor values.
Perhaps I'm being dense, but I don't see how to conclude anything about
the fields in a traveling plane wave from this setup.
> | > | you'll find it's parallel to the B field, and pointing in the same
> | > | direction as -@B/@t .... but only if the B field is changing in
> | > | step with the E field at that point. If B is at a peak while E is
> | > | changing rapidly the two won't match.
> | > |
> | > | Since curl(B) = @E/@t in vacuum, you can run the same argument
> | > | from either field to the other, of course.
> | >
> | > And your argument will be just as invalid. When the B-field is at it
> | > greatest rate of change, the spark plug in your car engine fires.
> |
> | My spark plugs were never fired by an ignition system that used radio
> | waves propagating in free space. They used waves propagating down a
> | waveguide, commonly called a "spark plug wire", which is something far
> | more complicated than a vacuum and I would not claim I can see
> | anything about how the E and B fields relate in that situation just by
> | looking at a picture.
>
> That's because you don't experiment, you look at mathematics and hope to
> understand Nature from other people.
Hmmm? Would _you_ claim to be able to say what the fields in a waveguide
look like just by drawing a picture?
Tell me about the E and B fields in a piece of spark plug wire leading to
a spark plug a moment after the points break. If it's a copper wire
inside, the fields travel down the wire at c/N, where N is the refractive
index of the insulation on the wire. But which way is the E field
oriented in the wave? Which way is the B field oriented? What do the
waves look like at the interface between the insulation, with N=10 (give
or take a bit) and the air, with N=1 (roughly!)? How does this change if
you use carbon-impregnated string for the conductor instead of copper?
I don't know about you, but I don't know the answers to these questions,
and I don't see how to find them _experimentally_ without using some very
fancy equipment -- that wave travels at roughly 30,000 km/sec (about C/10)
which is 'way faster than my paperclip can turn to show where the B field
is.
>
> | > We get
> | > that when the current in the primary winding of the coil is
> | > interrupted and the energy stored in the B-field collapses.
> |
> | Yes, I'm quite aware of the effect. In fact, while I was digging in
> | the attic for an antenna to attach to the VCR so we can tape the
> | debate tonight, I ran across an old bit of circuitry I built for fun
> | some time back: It has a couple 9V batteries on it, an oscillator and
> | a voltage multiplier, and it fires occasional pulses into an ignition
> | coil to make sparks. It never got as much bang as I would have liked,
> | though, because it operated by discharging a capacitor into the coil;
> | with that arrangement, the voltage across the primary is limited by
> | the voltage in the capacitor when it fires. When we _interrupt_ the
> | current using, say, breaker points, the voltage can go a great deal
> | higher than that. In fact, when breaking the circuit, it could easily
> | go high enough to blow the little mosfet I was using to trigger it
> | into little bits, which is one reason I didn't do it that way. Another
> | was my current source, which was pretty wimpy; in the usual operating
> | mode ignition coils want something on the order of an amp flowing
> | through them when the circuit breaks.
>
> Well, if you know that much, why cant you understand the E field and B
> field are phase shifted?
Because all I can experiment with is low-speed stuff that takes place at
a particular, single location. I can't see what happens when the energy
hightails it off in a traveling wave! My (old, creaky) scope might be
fast enough (if it still works at all) but setting up something to
actually sense the E and B fields in a radio wave as it goes by, and let
me actually _see_ which way each is pointing and how they're changing, is
something I couldn't begin to do with the equipment I have here.
Paperclips and analog voltmeters, both of which I have, are not
adequate for this task!
> | > Connect a couple of
> | > wires across the switch of an old solenoid doorbell and you'll soon
> | > find out the jolt you'll get from the 3V-9V battery that operates
> | > it. Heck, I was doing that stuff when I was 12 years old, shocking
> | > myself and my kid sister. Understanding it came easily to me later
> | > when I gained the math.
> | >
> | >
> | >
> | >
> | > | > This is when the field has
> | > | > reached its maximum value and is "curling" back on itself
> | > | > (changing direction),
> | > |
> | > | No, it's not. Sorry.
> | >
> | > I'm done arguing. Obviously you have never played with electricity
> | > as a child.
> |
> | You constantly forget: I'm not a physics guy, I'm a math guy.
> |
> | You've just described curl as a second derivative, and it's not... and
> | math guys do typically know something about curl, after all.
> |
> | And you stopped just one strategic paragraph before I stated the
> | formula for curl(E) for a plane wave.
>
> Yeah, well, I was busy so I dropped it. How does it it feel to have
> that done to you, as you have done so many times to me?
No prob, just so long as you don't start telling me how many degrees you
have when I point out that you misunderstood something. ;-)
> You can check this, if you like; once
> | again, to quote,
> |
> | > | Here, let's look at the actual value of curl(E) in this case.
> | > | Assume it's a plane wave, traveling along the X axis, with E
> | > | directed along the Y axis. Since the X and Z components of E are
> | > | identically zero, and the Y component varies only with X (it's
> | > | independent of Z), the only nonzero term in curl(E) in this case
> | > | is
> | > |
> | > | @Ey/@x, directed in the Z direction
> | > |
> | > | and that has maximum magnitude when E=0, and it's zero when E's
> | > | magnitude is at its maximum. Similarly, B is changing most rapidly
> | > | when B=0. So, those two points -- B=0 and E=0 -- must occur at
> | > | the same place in space.
> Well good grief, obviously if B=E=0, there is no energy to consider and
> an infinite number of places in space where that is true. Sheesh, I'm
> talking about
> the ripples on a pond and you are talking about a flat surface.
No, again, see the stuff back at the top -- I'm talking about a traveling
wave, and the places where E=B=0 at a particular instant are just the
zero crossings. There are an infinite number of them (if the wave train
is infinitely long) but there are lots of non-zero points, too.
sal
> With no capacitors in the circuit, and the voltmeter on the _coil_ (aka
> solenoid), when you turn power on the voltage will immediately rise to 9V
> (across the coil) and then will _fall_ exponentially as current increases
> in the coil. If the coil has small resistance compared to the resistor,
> the voltage on the coil will drop to a small value as the voltage on the
> resistor rises to (almost) 9V.
>
> For the coil, we have V = L*(dI/dt) (or something like that) -- right? So
> initially, dI/dt is large, and V on the coil is at its maximum; as the
> current rises to its steady state value, dI/dt drops to zero and so does
> the current across the coil.
Oops I meant the "voltage" across the coil drops to zero.
Dirk Van de moortel
> on this topic, I had no idea whether the E and B fields were in or out of
> phase in EM radiation. After I saw your post, I went and looked at what
> Griffiths had to say about it, and I thought about how the curl of a
> traveling plane wave must behave. As so often happens when I reply to
> your posts, I learned something from it. (Whether it was what you wanted
> me to learn, well, that's a separate question...)
This is what our genious has to say about E and B:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/EMWaves.html
from what I learned -again- what a stupid stronzo he actually is ;-)
Dirk Vdm
Androcles
| On Fri, 01 Oct 2004 07:15:38 +0000, Androcles wrote:
|
| >
| > "sal" <pragm...@nospam.org> wrote in message
| > news:pan.2004.09.30....@nospam.org...
| > |
| > | On Thu, 30 Sep 2004 19:17:59 +0000, Androcles wrote:
| > | >
| > | > "sal" <pragm...@nospam.org> wrote in message
|
| [snip stuff we're no longer discussing]
|
| > | >
| > | > the magnitude is not changing, and the curl -- along with the time
| > | > derivative -- is zero.
| > | >
| > | > Magnitude of what? The E-field? Or the curl of the E-field?
| > |
| > | Magnitude of the E field, as I again thought was obvious.
| > |
| > | At the peaks in the E field, the magnitude of the E field is
momentarily
| > | unchanging.
| >
| > Which is when the magnitude of the B field is momentarily zero but IS
| > changing. In other words, the E and B fields are phase shifted by a
| > quarter circle, as I thought was obvious and EVIDENT.
|
| It seems like it should be that way, but, since curl(E)=-@B/@t, it's
| not. Consider:
Nope. I'm not considering it at all. I'm done considering. Evidence reigns,
but I'll let you rant on and make the occasional comment for the benefit of
any lurkers that may be interested.
|
| (Notation in these posts is awkward, unfortunately, but the time has come
| to write some vectors regardless.)
|
| By definition of the curl, we have, at one particular instant (NB -- this
| mess is supposed to be a 3-vector, shown as three components):
|
| curl(E) = (@Ez/@y - @Ey/@z, @Ex/@z - @Ez/@x, @Ey/@x - @Ex/@y)
There isn't a 't' in there.
Nothing to consider.
| This is a property of the instantaneous _spatial_ distribution of the E
| field. The time domain doesn't enter into it.
So what are you trying to consider? I've never disputed the E field is at
right angles to the B-field. Heck, it has to be for a DC solenoid.
Fleming's right and left hand rules:
http://www.tiscali.co.uk/reference/encyclopaedia/hutchinson/m0016614.html
In my teenage years and early twenties I rewound electric motors for a
living.
Ever heard of a four-parallel delta-collected three-phase 440V 50 hp
squirrel cage induction motor? Just being able to say that got me hired
instantly :-)
|
| The Maxwell equation known as Faraday's law connects this with B (and with
| the time domain):
|
| curl(E) = -@B/@t
|
| For a plane wave traveling along the X axis, with the E field parallel to
| the Y axis, and the B field parallel to the Z axis, we have:
|
| Ey = k * sin(a*X + b*t) (for some constants k, a, b)
|
| and Ex = Ez = 0.
Photons do not have to travel.
|
| Then,
|
| @Ey/@x = k*a*cos(a*X + b*t)
|
| @Ey/@t = k*b*cos(a*X + b*t)
|
| and all the other partials of components of E are zero.
|
| Then from the definition of curl(E), given above, we have
|
| curl(E) = (0, 0, k*a*cos(a*X + b*t))
|
| By Faraday's law, therefore, we have
|
| -@B/@t = (0, 0, k*a*cos(a*X + b*t))
|
| or, in other words,
|
| k*a*cos(a*X + b*t) = (-@B/@t)_z
|
| and the X and Y components of @B/@t are zero.
|
| But integrating that over time, at a single point in space, we find
|
| (k*a/b)*sin(a*X + b*t) = (B(t))_z
|
| and comparing with the original value for E, we find that Bz = Ey * (a/b)
|
| And those are in phase.
|
| This is messy and awkward, and my original "intuitive" argument contained
| the same freight but was, on the whole, easier to grasp (or so I thought,
| anyway). The statement I just gave here has the advantage of avoiding the
| use of Stokes' theorem, of course.
If B=E=0, the energy is zero also. It can only be non-zero if
Energy = hf.sqrt([B.cos(t)^2]+ [E.sin(t)^2])
I'm not interested in your crackpot theory of the spontaneous creation of
energy, and neither is the US patent office, so don't bother them with it
either. There are no perpetual motion machines.
I'd point out your mistake for you, as I have with Einstein's
―[tau(0,0,0,t)+tau(0,0,0,t+x''/(c-v)+x''/(c+v))] = tau(x'',0,0,t+x''/(c+v)),
but since you simply ignored that you'll do the same if I do it for you
here,
so it isn't worth my time looking for it. Find it yourself.
|
| > | Since the magnitude and direction are momentarily unchanging, the curl
| > | of the E field is necessarily zero at those points.
| > |
| > | This should, again, be obvious.
| >
| > What you are omitting to state is the domain of the function. In the
| > time domain, the magnitude (codomain or image) of the E field is sine(t)
| > and the magnitude of the B field is cos(t).
|
| But see above -- Maxwell's equations lead inevitably to the conclusion
| that if magnitude(E) = sin(t) then magnitude(B) = sin(t), too (up to a
| constant multiplier)
|
Then either Maxwell's equations are wrong or you do not understand them.
If B=E=0, the energy is zero also. It can only be non-zero if
Energy = hf.sqrt([B.cos(t)^2]+ [E.sin(t)^2])
I'm not interested in your crackpot theory of the spontaneous creation of
energy, and neither is the US patent office, not even from Maxwell. There
are no perpetual motion machines.
|
| > | [sal]
| > | > | You may not be able to picture the curl of a sine wave in 3-space
| > | > | directly, but you ought to be able to imagine integrating the
| > | > | field around a small square path.
| > |
| > | [A]
| > | > Integrate a sine wave?
| > |
| > | No, that's not what I meant. Did you not recognize a simple
| > | application of Stokes' theorem here? I thought you would realize what
| > | I was getting at, and didn't go into great detail as a result.
| > |
| > | Stokes' theorem in 3 dimensions, stated briefly, in English: Take the
| > | line integral around a closed path of the value V*F where V is the
| > | tangent to the path and F is the value of a vector field at that point
| > | on the path. "*" in this case represents ordinary dot product. That
| > | integral is equal to the integral of curl(F) taken over a surface
| > | bounded by the path; at each point on the surface we actually use the
| > | value curl(F)*N where N is the normal vector to the surface, and "*"
| > | again represents dot product.
| > |
| > | We're in 3-space here, and the closed path is in _space_ -- it doesn't
| > | extend into the time dimension.
| >
| > Now you are getting there. However, we live in a universe where time
| > passes and the E and B field MAGNITUDES do not occur at the same place
| > AT THE SAME TIME.
|
| But as it stands, that is just an assertion.
If B=E=0, the energy is zero also. If you consider that an assertion, so be
it.
It's a plausible one, but
| still just an assertion -- and when we compute the curl of E (in SPACE,
| at one moment in time) and compare it with the time derivative of B at
| that same moment, we find that it's not correct, as I did, up above.
Take it up with Maxwell.
|
| > If you travel with the wave, as a dragonfly might fly
| > along with a water wave, the MAGNITUDE NEVER CHANGES. However, the
| > dragonfly can choose the trough or the crest to hover above. Now we have
| > effectively eliminated time in our study.
|
| Yes, and if you do this, Maxwell's equations fall down the stairs and
| break into little pieces.
Tough. The dragonfly can move in a "square", backwards and forwards
from leading crest to trailing crest, thereby moving forwards and backwards
in time. Without time there is no energy.
For the (spatial) curl of E and B is still
| nonzero almost everywhere, but the partial of each of E and B with respect
| to time is now identically zero.
So the energy is zero. So what? its just an assertion.
|
| The form in which we currently employ Maxwell's equations requires that C
| be invariant -- in other words, special relativity is wired into them. If
| you want to allow the dragonfly to fly along at C you need to patch the
| equations somehow.
Not me. It wasn't my turn. I've patched
―[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v))
to
(t1-t)/(t2-t)*[tau(-vt,0,0,t)+tau(-vt,0,0,t+x'/V+x'/V)] = tau(x',0,0,t+x'/V)
where V = (c+v)/(1+v/c),
and I can't get the Lorentz transforms out of it.
You DO agree that (t1-t)/(t2-t) = ―, right?
| > | If the wave is traveling along the X axis, with the E field pointing
| > | along the Y axis, then the path I was suggesting integrating around
| > | would have two sides parallel to the X axis, and two sides parallel to
| > | the Y axis. If it starts at (x1,y1,z1), then the path might be
| > | described as
| > |
| > | (x1,y1,z1) -> (x1+delta,y1,z1)
| > | -> (x1+delta,y1+delta,z1)
| > | -> (x1,y1+delta,z1)
| > | -> (x1,y1,z1)
| > |
| > | Again, I didn't go into detail because I thought you would see
| > | immediately what I was getting at.
| >
| > What you were getting at is that the B and E fields are in phase, and
| > they are not.
|
| Well, the equations sure seem to say that they are..
Take it up with Maxwell. I KNOW the E and B fields are phase shifted 90
degrees, just as I know the E- field is shifted 90 degrees from the B-field
spatially.
|
|
| > | [A]
| > | > Of course I can, it's -cos(). That integrates to -sin, which
| > | > integrates to cos, which integrates to sin. Shoot, you have only to
| > | > look at the slope of the sine wave on the T-axis to see it is one =
| > | > cos(0) -- err..locally.
| > | >
| > | >
| > | >
| > | > | Make the square lie in the same plane as the wave. A square with
| > | > | sides
| > | > magnitude and time? Ok.
| > |
| > | No, the path lies entirely in space, with no time extension, as I
| > | already said.
| >
| > What you are discussing is in the form y = f(x), not y = f(t). curl(E) =
| > -d@B/@t <-- see that '@t'? REPEAT... can you see @t??
|
| But that's not the point -- or perhaps it's exactly the point. See my
| little monograph at the start of this post -- the curl is defined in
| space, for values of the fields chosen at a single instant of time.
Exactly. It only becomes a function of time if the wave moves, and all
motion is relative. I said the E and B fields are phase shifted.
That's what you challenged. Jeez... defending a Ph.D. thesis was never this
hard. You can't argue about their spatial orientation to deny it, and I'm
standing by my assertion that energy is conserved.
| It's
| Maxwell's equations -- and in particular, the one called Faraday's law --
| which ties together the space and time domains here.
|
| IIW curl(E) = -@B/@t is not the definition of curl(E) -- rather, it's an
| assertion about how curl(E) relates to the B field.
Well done.
Another way of putting it: time and space are independent quantities, but
motion is a relation between them.
|
|
| > | [sal]
| > | > | If you take the integral at the point where the wave is constant
| > | > | --
| > | > max or min -- it will be zero.
| > |
| > | [A]
| > | > Yeah... so? The E-field isn't changing, so the B-field is zero.
| > | > That's what Faraday and Maxwell are saying.
| > |
| > | WRONG !!!
| > |
| > | The E field isn't changing, so the CURL of the B-field is zero, _NOT_
| > | the MAGNITUDE.
| >
| > If the E field isn't changing, the magnitude of the B field is constant.
|
| :-) :-) :-)
|
| If the E field isn't changing, then E must be at an extremum. Right?
Agreed, since change is a function of time, by definition.
|
| If B has constant magnitude, B must be at an extremum. Right?
Wrong. Constant magnitude is a function of time also.
The constant magnitude of B is zero when E is at an extremum.
Where DO you get this perverted logic from?
| So if E unchanging => B constant, then they hit their extrema together.
| Still right?
No.
|
| So, they're in phase!
Nonsense.
|
|
| > I was incomplete when I said it was zero, I meant a constant zero
|
| Whoops -- constant, yes. Constant _zero_, no, because that would imply
| there's no wave.
There IS no wave at any instant of time.
The dragonfly hovering over a crest sees a series of bumps and hollows, not
a wave. Waves require a change in amplitude as a function of time, by
definition. We can effectively freeze time with the dragonfly model, and
then you can discuss the curl of E in terms of B.
| > when
| > the E field was at a maximum. So yes, what I said was wrong, but not
| > "WRONG !!!" That condition arises as you fly along with the wave, as the
| > dragonfly might.
|
| _AND_ it arises at every peak and trough in the plane wave.
| And consequently, the peaks and troughs of E and B must be coincident.
No. The molecule of water at the crest has potential energy. The velocity
(momentum, energy) of the falling molecule has Kinetic energy. This is
simply a spring and mass model.
| And that was my point, and that's why I went overboard with the
| exclamation points and the capital letters on the word "wrong".
You have not made your point.|
|
| > Maybe that is where you relativists come to the ridiculous conclusion
| > that time stands still if you travel at the speed of light.
|
| Yeah, as I said, Maxwell's equations, as currently formulated, incorporate
| SR. You can't ditch SR without making some change to Maxwell's equations.
|
As I said above, it's not my turn.
However, I accept curl(E) = -@B/@t
Contradiction? Not at all. As you said yourself,
"the curl is defined in space".
This equation merely ties it to the rate of change of B, which is when
B is at zero and when the E field is at an extremum.
Anyway, I've run out of time. It's Friday night and its time to chase down
naughty and willing women, and I know where I might find some.
Androcles