particles as rotating EM fields
cherring
ohms),
this property allows free space to appear to an EM propagation as though its
occurring in a resonant series tuned circuit something akin to an infinite
wave guide
or an infinite transmission line.
An inductor in series with a capacitor passes a given frequency from one
point to another. Make this infinitely long and you have something similar
to light traveling in free space.
However, place an inductor and a capacitor in parallel and in this case we
have what is commonly called a tank circuit. The E field feeds the B field
which in turn feeds the E field and so on giving us a short term, frequency
dependent energy storage device (not perpetual, energy is lost).
The Energy is trapped at a somewhat fixed point in the circuit as opposed to
traveling down the circuit. The energy and fields are caught in a loop.
Light appears to behave for the most part as the linear example above. But
is there an equivalent tank circuit in free space? It is my belief that
particles are such animals - the ultimate tank circuit for the smallest
subset of an EM wave. EM waves are dynamic electric and magnetic field
vectors at right angles to each other (positive-north nodes and
negative-south nodes but spatial - three dimensional) with a third vector of
motion perpendicular to both of these, the Poynting vector. When all three
vectors are in balance you get linear motion.
However, disrupt one of the field vectors, say, the electric field vector in
a strong Coulomb field (the E field is out of balance with the B (or H)
field)
such as near the nucleus of an atom and the Poynting vector must change
direction to compensate or the photon must give up energy
to the field (generally in the form of motion) or both. The wave gets
reflected or scattered or even partially absorbed. But what happens in pair
production? Some of the photon's energy is converted into 2 particles of
opposite charge and spin flying off in different directions. But what is the
process? For the amount of energy that gets converted into the 2 particles,
Ephoton = 2Eparticles or,
h*v_o=2*E where h*v_o is the minimum photonic energy for conversion
Rearranging and dividing by 2*pi we can arrive at,
(1/2)*(h/(2*pi)) = E/(2*pi*v_o)
The left side of this equation is the units of spin while 2*pi*v_o is
angular frequency which I will designate as w_o, therefore,
(1/2)*(h/(2*pi)) = E/w_o eq.1
thus we have particle spin equal to the energy divide by an angular
frequency,
an angular frequency associated with the original photon frequency! It
appears as though the linear vector has become a rotating vector. It sure
has the appearance of the tank circuit mentioned above. Now, since E=m*c^2
and 2*pi*v_o = 2*pi*c/wl_o (wl_o - wave length) we can also show,
(1/2)*(h/(2*pi)) = (m*wl_o*c)/(2*pi)
and if we define wl_1/2 to be the half wavelength such that (wl_o)/2 =
wl_1/2, then,
(1/2)*(h/(2*pi)) = (m*wl_1/2*c)/pi eq. 2
In eq. 1 the energy and angular frequency are directly proportional whereas
in the eq. 2 form the mass is inversely proportional to the half-wavelength
(2 particles-1/2 the wavelength!!)! As energy goes up then angular frequency
goes up but the half-wavelength decreases (in circular harmonic steps of
course)
and mass increases - as though we now have a physical explanation of
relativistic
mass increase! Because a circular wavelength can only change in harmonic
steps
we also have quantized motion! A limit to linear velocity exists because
energy gets
absorbed as harmonic energy of ever decreasing circumference (which is also
more
point like). Stopping the particle would lead to the harmonic energy being
released
all at once, Bremsstrahlung. An explanation of mesons, they are just a
harmonic state
of the electron or positron and the neutron because now k capture would be a
transition to a harmonic state followed by something similar to a hydrogen
atom.
The split between spatial nodes of the photon would have to occur at the
zero cross
over between the spatial nodes because we have 2 equal but opposite charges
with
each taking half of the B (or H) field thereby forming the smallest possible
current loop - a
single charge rotating around itself.
Quark confinement would be explained also because you cannot subdivide a
spatial node.
Unfortunately the equations do not speak to stability: why the
electron/positron is the first stable particle pair, why mesons are not
stable, and why the proton/antiproton is the next stable pair (a very large
quantized jump).
One other note, the angular velocity would be pi*c, over 3 times the speed
of linear light!
ThomasL283
>Date: 3/30/2001 8:05 PM Pacific Daylight Time
>Message-id: <SPbx6.221544$bb.19...@news1.rdc1.tx.home.com> Wrote:
>Because free space has a non-zero reactive property (an impedance of 377
>ohms),
>this property allows free space to appear to an EM propagation as though its
>occurring in a resonant series tuned circuit something akin to an infinite
>wave guide
Yes, the math is fully developed in my book "Vector particle and Nuclear
Models" ISBN 0-9631546-8-0 available from
http://www.amazon.com
or Barnes and Noble at
>Snip
>The Energy is trapped at a somewhat fixed point in the circuit as opposed to
>traveling down the circuit. The energy and fields are caught in a loop.
Yes, the energy of the photon is alternately exchanged between the capacitance
and inductance, exactly analogous to a familair resonant circuit. This storage
conserves the energy of the photon as:
hf=0.5(ExH sin^2 theta + HxE cos^2theta)
And transports the energy of the photon, apparently with infinite Q ( no loss)
.
>Light appears to behave for the most part as the linear example above. But
>is there an equivalent tank circuit in free space?
Yes:
>It is my belief that
>particles are such animals - the ultimate tank circuit for the smallest
>subset of an EM wave.
No, the particles model as closed Poynting vectors combined into a cube like
structure that preserves the orthogonality of the energy and conserves the
linear momentum of the photon(s) in the particle spin angular momentum. The
math for these arguments are fully developed in the referenced book.
>Snip
>such as near the nucleus of an atom and the Poynting vector must change
>direction to compensate or the photon must give up energy
>to the field (generally in the form of motion) or both. The wave gets
>reflected or scattered or even partially absorbed. But what happens in pair
>production?
>Some of the photon's energy is converted into 2 particles of
>opposite charge and spin flying off in different directions. But what is the
>process?
Well the opposite charges are inferior to the near field magnetic moment
forces, and since the particles are born spinning in the same direction, their
magnetic moments oppose allowing the pair to separate. Were it not so, the
enormous attracting force between their plus and minus charges would prevent
pair production. The math is fully developed in the referenced book.
You really should get a copy of the book, since your ideas seem to be similar
to those already developed.
Regards: Tom
Thomas Lockyer http://www.best.com/~lockyer (74 and retired)
"When you can measure what you are speaking about and express it in numbers,
you know something about it." Lord Kelvin (1824-1907)
cherring
"ThomasL283" <thoma...@aol.com> wrote in message
news:20010401151413...@ng-fb1.aol.com...
cherring
It seems to me that the nuclear Coulomb field would be sufficient to
separate the particles within some impact parameter. The EM field strength
is, after all, only the strength of 2 charges separated by half the
wavelength (center to center). I would think a minimum field strength of 1e
in collision proximity would be a sufficient field strength to accomplish
this, hence, say for a nucleus with 82 protons, pair production should be
able to occur within any distance where the field strength exceeds e.
I had always thought of the curled Poynting vectors for each particle as
being opposite (opposite spins) but I do believe your correct. They would
both be in the same direction. If the positive electric field is more than
doubled in relation to the north magnetic field (the Coulomb field adding to
the EM positive field) the Poynting vector would turn away from the electric
field in order to compensate. But the negative field would be negated and
hence the Poynting vector would turn inward toward the missing field and,
hence, in the same direction as the positive particle. However, the Coulomb
field would definitely send the positive particle off away from the nucleus
due to the like-field repulsion whereas the electron would turn toward the
Coulomb field and sail off in this direction curving around the nucleus much
as we do with a sling-shot of a satellite around a planet.
ThomasL283
>Date: 4/5/2001 11:37 PM Pacific Daylight Time
>Message-id: <Sudz6.247867$bb.20...@news1.rdc1.tx.home.com>wrote;
>"ThomasL283" <thoma...@aol.com> wrote in message
>news:20010401151413...@ng-fb1.aol.com...
>> >"cherring" cherr...@home.com
>> >Date: 3/30/2001 8:05 PM Pacific Daylight Time
>> >Message-id: <SPbx6.221544$bb.1903106@news1.r
>dc1.tx.home.com> Wrote:
>>
> But what happens in
>pair
>> >production?
>> >Some of the photon's energy is converted into 2 particles of
>> >opposite charge and spin flying off in different directions. But what is
>the
>> >process?
>> Well the opposite charges are inferior to the near field magnetic moment
>> forces, and since the particles are born spinning in the same direction,
>their
>> magnetic moments oppose allowing the pair to separate.
>>Were it not so, the
>> enormous attracting force between their plus and minus charges would
>prevent
>> pair production. The math is fully developed in the referenced book.
>It seems to me that the nuclear Coulomb field would be sufficient to
>separate the particles within some impact parameter. The EM field strength
>is, after all, only the strength of 2 charges separated by half the
>wavelength (center to center).
The force in Newton between electron and positron charges (e) is:
Fe= (e^2)/(4pi Eo m^2)
The force in Newton between the electron and positron's Bohr magneton (Ub) is:
Fu=(2 Uo Ub Ub)/(2pi m^4)
Where Eo is the permittivity and Uo is the permeability of the vacuum, e is the
fundamental charge and (m) is the separation in meters.
If you combine equations above for a separation (m) you get:
m=sqr((4 Uo Eo Ub Ub)/(e^2)) which equals exactly lambda bar for the electron
as their separation distance at birth.
I view the pair production as forming particles out of a single photon
fireball, created by the photon scattering in matter. The scattering turns the
Poynting vectors back on themselves and forms a spinning fireball. The spin
creates charge and magnetic moments that then separate the pair, per the above
newly created EM forces.
BTW, it is shown (on page 36 of my book) that a spinning EM particle developes
exactly the fundamental charge
(e) and spin angular momentum of (1/2 h bar) regardless of it's size. This
supports the spinning EM cube geometry and explains why (e) and (1/2 h bar) are
common characteristics for subatomic particles with desparate masses. For
example the muon and proton.
cherring
No disagreement here but the Coulomb field of the nucleus had to play a more
direct role and by virtual of like and unlike fields must continue to play a
role in the eventual direction the 2 particles take.
> BTW, it is shown (on page 36 of my book) that a spinning EM particle
developes
> exactly the fundamental charge
> (e) and spin angular momentum of (1/2 h bar) regardless of it's size.
This
> supports the spinning EM cube geometry and explains why (e) and (1/2 h
bar) are
> common characteristics for subatomic particles with desparate masses.
For
> example the muon and proton.
More to say later but don't have the time right now.
cherring
"cherring" <cherr...@home.com> wrote in message
news:otMz6.254842$bb.20...@news1.rdc1.tx.home.com...
I agree with your statement that energy must be conserved, that
S=0.5[(EcrossH)] implies a zero between positive-north and negative-south
excursions. I have always had a problem with this also. There is nothing
with which to pass the energy forward. Your correction looks like it
accomplishes this. However, your equation...
S=0.5[(sin omega t E)*(sin omega t H) + (cos omega t H)*(cos omega t E)]
does not mean what you say - an axial flow of the E & H fields at the null.
Since for any vector A and B, A cross B = - B cross A, your left side can
become...
- (cos omega t E)*(cos omega t H)
Which implies a set of opposite fields, hence, our typical interpretation,
positive & negative electric fields and north & south magnetic fields. But,
your use of the cosine function serves to shift this latter part by 90
degrees thus forming the negative-south fields as the positive-north fields
collapse. In other words when the positive-north fields are at a maximum the
negative-south fields are zero and vice-versa. After a maximum is reached,
the collapsing fields would generate the opposing fields and so on down the
chain. The energy would be transferred via a chaining mechanism.
You show the correct diagram on page 11 & 16 but not on page 13. Shift the
forward nodes in this diagram back 90 degrees as the equation implies and
there is no need to posulate axial flow for E & H.
cherring
The old equation in your book would produce vectors that look something like
these below. The vertical and horizontal represent instantaneous E & H
fields while the dots represent displacement along the Poynting vector
(Ignore the disproportions in the two vectors. They are assumed to be
equal). It is easy to see that you are correct, that energy is not
conserved. Also, notice that at the null there is zero instantaneous
displacement so the speed would be zero as well! Certainly not a constant at
that moment in time.
|
|
| |
| |
------. |
----. |
--.
.--
| .----
| .------
| |
| |
|
|
But with the change you propose not only is the energy conserved but the 2
vector pairs of positive-north and negative-south always work together to
maintain a constant displacement vector over time. This now gives a physical
reason for why c is a constant (besides that it is (e_0*u_o)*1/2).
|
|
|
| |
------. |
|
----.--
| |
|
---.---
|
| |
--.----
|
| .------
| |
|
|
|
I think you can drop the 0.5 in your the equation because there is no need
to average S. At the max of one one set of vectors pairs S is max while for
the other set S is zero and then vise versa 180 degrees later. Anywhere in
between they add to always equal Smax - constant displacement with time! But
then again it may be needed for other reasons.
"cherring" <cherr...@home.com> wrote in message
news:DP2A6.258902$bb.20...@news1.rdc1.tx.home.com...
cherring
"cherring" <cherr...@home.com> wrote in message
news:_PrA6.264754$bb.20...@news1.rdc1.tx.home.com...
ThomasL283
>Date: 4/9/2001 4:44 PM Pacific Daylight Time
>Message-id: <_PrA6.264754$bb.20...@news1.rdc1.tx.home.com> Wrote:
>> I agree with your statement that energy must be conserved, that
>> S=0.5[(EcrossH)] implies a zero between positive-north and negative-south
>> excursions. I have always had a problem with this also. There is nothing
>> with which to pass the energy forward.
Yes the Fig 2.4 page 12 is typical of standard text traveling wave diagram.
>Your correction looks like it
>> accomplishes this. However, your equation...
>>
>> S=0.5[(sin omega t E)*(sin omega t H) + (cos omega t H)*(cos omega t E)]
>>
>> does not mean what you say -
>an axial flow of the E & H fields at the
>null.
>> Since for any vector A and B, A cross B = - B cross A, your left side can
>> become...
>>
>> - (cos omega t E)*(cos omega t H)
>> - (cos omega t E)*(cos omega t H)
>>
>> Which implies a set of opposite fields, hence, our typical interpretation,
>> positive & negative electric fields and north & south magnetic fields.
Because the vectors are exactly always at 90 degrees the idea of a right handed
screw and cross product is not applicable. That analysis can only be applied to
waveforms that change phase in relation to each other. The lateral E and H are
and remain in a sine-sine relationship, in a traveling wave.
>But,
>> your use of the cosine function
>serves to shift this latter part by 90
>> degrees thus forming the negative-south fields as the positive-north
>fields
>> collapse. In other words when the positive-north fields are at a maximum
>the
>> negative-south fields are zero and
>vice-versa. After a maximum is reached,
>> the collapsing fields would generate the opposing fields and so on down
>the
>> chain. The energy would be transferred via a chaining mechanism.
>>
Yes, the thing to realize is that we are talking about TWO conjugate waveforms
in this model.
>> You show the correct diagram on page 11 & 16 but not on page 13. Shift the
>> forward nodes in this diagram back 90 degrees as the equation implies and
>> there is no need to posulate axial flow for E & H.
I tried to show that the lateral E and H cannot be in sine-sine and be mutually
dependent and inseperable, they must be in a sine-cosine as shown on page 11.
Page 11 further shows that the rates of change of the sine and cosine are
exactly c at the RMS point (circled on the diagram). The diagram on page 16
shows that the photon model velocity (c) is an RMS value.
On page 13 the traveling wave was pictured as being the symbiosis of TWO
conjugate resonances.
S=0.5[(sin omega t E)*(sin omega t H) + (cos omega t H)*(cos omega t E)]
This serves to create an axial Poynting (S) vector that combines cos H from one
and cos E from the other, in the normal time rate of change (sine-cosine) from
the lateral E an H. This looking at the Poynting vector as a dynamic cycling
vector is a brand new way. The resulting axial Poynting vector gives
translation at an RMS velocity of (c) as shown on page 13, Fig 2.6, inch worm
style. But Fig 2.4 page 12 does not have this mechanism for storing a
constant energy. Only the proposed Fig 2.6 page 13 schematic can conserve the
energy over all time by resonantly stroring the energy alternately between
lateral and axial development.
>But with the change you propose not only is the energy conserved but the 2
>vector pairs of positive-north and negative-south always work together to
>maintain a constant displacement vector over time. This now gives a physical
>reason for why c is a constant (besides that it is (e_0*u_o)*1/2).
> Snip a nice diagram.
Yes, but as was shown, this is an RMS velocity as the photon cycles.
>I think you can drop the 0.5 in your the equation because there is no need
>to average S. At the max of one one set of vectors pairs S is max while for
>the other set S is zero and then vise versa 180 degrees later. Anywhere in
>between they add to always equal Smax - constant displacement with
No, I first questioned that also, but look at the graph on page 14, Fig.2.8.
This is a MathCad produced graph of the equations on page 14 Fig.2.7. In the
case of Maxwell, the energy cannot be negative, so the 0.5 serves to cut off
the peaks and fill in the valleys to average the energy over the whole cycle
(and hide the fact that energy is not conserved at theta 180 or 360 on the
cycle). On the other hand, notice the graph of the VPP energy of the equation
Fig 2.7 (and repeated above) gives a constant value over all time (heavy line
on the graph).
cherring
Wait a mintue, where was my brain on this??? Your using squared sine and
cosine functions.
S=0.5[E*H*sin^2 + H*E*cos^2] if this is just plain old multiplication then
= 0.5*E*H[sin^2 + cos^2]=0.5*E*H ?
> This serves to create an axial Poynting (S) vector that combines cos H
from one
> and cos E from the other, in the normal time rate of change (sine-cosine)
from
> the lateral E an H. This looking at the Poynting vector as a dynamic
cycling
> vector is a brand new way. The resulting axial Poynting vector gives
By any law of vectors you can't have an axial Poynting vector created from
axial E&H vectors!
Write a wave equation for the diagram I gave - 2 EH pairs, one phase shifted
by 90 degrees. Work from there.
ThomasL283
>Date: 4/10/2001 10:05 PM Pacific Daylight Time
>Message-id: <OCRA6.271987$bb.21...@news1.rdc1.tx.home.com> Wrote:
>>"ThomasL283" <thoma...@aol.com> wrote in message
>>news:20010410002032...@ng-fn1.aol.com...
>Snip
>> This serves to create an axial Poynting (S) vector that combines cos H
>from one
>> and cos E from the other, in the normal time rate of change (sine-cosine)
>from
>> the lateral E an H. This looking at the Poynting vector as a dynamic
>cycling
>> vector is a brand new way. The resulting axial Poynting vector gives
>By any law of vectors you can't have an axial Poynting vector created from
>axial E&H vectors!
Yes, you are correct. This model makes the axial vector real, whilst the
Poynting vector just denotes the transport of power across an area. The
proposed new model is showing the Poynting vector in a brand new way, as an
axial compound vector of E and H, so as to conserve the energy. (sin^2 theta
+ cos^2 theta = 1 over all time)
>
>Write a wave equation for the diagram I gave - 2 EH pairs, one phase shifted
>by 90 degrees. Work from there.
The diagram you gave did not have a mechanism for translation. Photon
translation is the reason for postulating the axial vector as a real
combination of cosine E and H. The theory is that this axial development
produces the velocity of light by physical translation. The photon model
stitches out the dimensions of space and time, resonantly.
If you read on in the book, the photon model allows developing boundary
conditions, that produce perfect spinning EM field structures, by using the
calculus of related rates (page 19) and the fact that E and H are the same
length at exactly (c) velocity, of increase (decrease) growth (the rms point
of Fig 2.3 page 11).
The model allows showing a reason for the electron and positron pair being
created at their one specific (exact) energy. (page 25, 26)
But the best proof that particles are rotating EM field cubes is in given on
pages 34, 35 and 36. The rotating EM cubes, regardless of their size, are
shown to have exactly the fundamental charge (e) and spin angular momentum
(1/2 h bar). The magic, that makes this so, is in the volume and current loop
area, from the geometry of the spinning EM cube. The model predicted EM cube
edge lengths is lambda bar, making the spinning volume (lambda cubed / 16
pi^2) and current loops (lambda squared / 4pi).
Note that any cube volume and attendant current loop areas gives exactly the
same charge and spin.
cherring
> >"cherring" cherr...@home.com
> >Date: 4/10/2001 10:05 PM Pacific Daylight Time
> >Message-id: <OCRA6.271987$bb.21...@news1.rdc1.tx.home.com> Wrote:
>
> >>"ThomasL283" <thoma...@aol.com> wrote in message
> >>news:20010410002032...@ng-fn1.aol.com...
>
> >Snip
>
> >> This serves to create an axial Poynting (S) vector that combines cos H
> >from one
> >> and cos E from the other, in the normal time rate of change
(sine-cosine)
> >from
> >> the lateral E an H. This looking at the Poynting vector as a dynamic
> >cycling
> >> vector is a brand new way. The resulting axial Poynting vector gives
>
> >By any law of vectors you can't have an axial Poynting vector created
from
> >axial E&H vectors!
>
> Yes, you are correct. This model makes the axial vector real, whilst the
> Poynting vector just denotes the transport of power across an area. The
> proposed new model is showing the Poynting vector in a brand new way, as
an
> axial compound vector of E and H, so as to conserve the energy. (sin^2
theta
> + cos^2 theta = 1 over all time)
You can't postulate this out of thin air. The Poynting vector has a basis in
matrix math. Vector A cross vector B can be shown in matrix math to yield
vector C, orthogonal to A & B.
> >
> >Write a wave equation for the diagram I gave - 2 EH pairs, one phase
shifted
> >by 90 degrees. Work from there.
>
> The diagram you gave did not have a mechanism for translation. Photon
> translation is the reason for postulating the axial vector as a real
> combination of cosine E and H. The theory is that this axial development
> produces the velocity of light by physical translation. The photon model
> stitches out the dimensions of space and time, resonantly.
>
> If you read on in the book, the photon model allows developing boundary
> conditions, that produce perfect spinning EM field structures,
So does this -- (1/2)*(h/(2*pi)) = E/w_o -- where E is energy and w_o is
the angular frequency associated with a photon of frequency v such that it
is the h*v energy of the photon for which electron- positron pair production
occurs. This is directly derivable from E= 2*m_o*c^2, E= h*v, w_o=2*pi*v, &
v*wl=c (wl is wavelength). You simply picked the point at which this occurs,
when the rate of change of all three vectors are the same.
You also stated in your book that you propose a cubic shape in order to
maintain orthogonality between the three vectors E, H & S as if the cube is
the only shape that can accomplish this, which is not correct (E here is the
electric field intensity). Since there is no physical containment in space
to force a shape, as in a waveguide, the most likely shape is a curvature of
some kind which is what we see in the macro-world, fields of circular shape
in free space, i.e., for a current loop the magnetic field is not cubic but
circular and even for an open face capacitor that is square the field lines
protrude to produce a cylinder shape. Hence if there is no physical
structure at all then some sort of curvature is the most likely.
>by using the
> calculus of related rates (page 19) and the fact that E and H are the same
> length at exactly (c) velocity, of increase (decrease) growth (the rms
point
> of Fig 2.3 page 11).
> The model allows showing a reason for the electron and positron pair
being
> created at their one specific (exact) energy. (page 25, 26)
So would the above.
> But the best proof that particles are rotating EM field cubes is in given
on
> pages 34, 35 and 36. The rotating EM cubes, regardless of their size, are
> shown to have exactly the fundamental charge (e) and spin angular
momentum
> (1/2 h bar).
As would the above.
>The magic, that makes this so, is in the volume and current loop
> area, from the geometry of the spinning EM cube. The model predicted EM
cube
> edge lengths is lambda bar, making the spinning volume (lambda cubed / 16
> pi^2) and current loops (lambda squared / 4pi).
What is proved is that the properties of particles are intrinsic in the
photon. Once particles are produced in dimensional space its easy to get the
correct values because fundamental geometric shapes are easy to relate to
each other. I got the spin value directly from the wavelength which is not a
shape!
ThomasL283
>Date: 4/12/2001 8:13 PM Pacific Daylight Time
>Message-id: <H9uB6.277780$bb.21...@news1.rdc1.tx.home.com> Wrote:
>Snip<
>> >By any law of vectors you can't have an axial Poynting vector created
>from
>> >axial E&H vectors!
>>
>> Yes, you are correct. This model makes the axial vector real, whilst the
>> Poynting vector just denotes the transport of power across an area. The
>> proposed new model is showing the
>>Poynting vector in a brand new way, as
>>an
>> axial compound vector of E and H, so as to conserve the energy. (sin^2
>>theta
>> + cos^2 theta = 1 over all time)
>You can't postulate this out of thin air. The Poynting vector has a basis in
>matrix math. Vector A cross vector B can be shown in matrix math to yield
>vector C, orthogonal to A & B.
The purpose of the model was to explain a traveling wave of EM energy having an
(in phase) E and H that somehow translated at (c). The Poynting vector
describes the (energy / area x time) but does not show the mechanism for energy
flow. You are right that maths show vector C orthogonal to A X B, but only
by definition, not by model.
>Snip<
>You also stated in your book that you propose a cubic shape in order to
>maintain orthogonality between the three vectors E, H & S as if the cube is
>the only shape that can accomplish this, which is not correct (E here is the
>electric field intensity). Since there is no physical containment in space
>to force a shape, as in a waveguide, the most likely shape is a curvature of
>some kind which is what we see in the macro-world, fields of circular shape
>in free space, i.e., for a current loop the magnetic field is not cubic but
>circular and even for an open face capacitor that is square the field lines
>protrude to produce a cylinder shape. Hence if there is no physical
>structure at all then some sort of
>curvature is the most likely.
>
The model is a closed particle constructed of photons. The photon is modeled
as a poynting vector,
The photon is a boson that ordinarily should not react with another boson
(photon).
There is , however, one unique energy where this is violated. Experimentally,
a boson connects to boson at the photo- production energy of the electron and
positron.
An effort was made to show why this unique energy has the value it does, on
pages 25, 26. In essence it is theorized that eqaul E and H field energy only
occupy the same length at the energy for pair production, without defining the
value. This, coupled with the rates of change being equal, at that equal
energy, allows the photons to connect to each other in a perpetual spinning
dance.
The mass (energy) for e+e- production, is thus the same here or on alpha
century.
>What is proved is that the properties of >particles are intrinsic in the
>photon. Once particles are produced in >dimensional space its easy to get the
>correct values because fundamental >geometric shapes are easy to relate to
>each other. I got the spin value directly >from the wavelength which is not a
>shape!
Yes, the fundamental constants are all related by a number of precise
mathematical expressions.
See page 21. this model automatically gives us the structures for the leptons,
by simply combining the photon model in all possible ways. See page 28, the
given cube geometry maintains all related electron fundamental constants in
their proper ratios.
Thus, unlike other theories, these model structures do not have to be "reverse
engineered", rather, they are given to us blindly and automatically, by the
model rules. (Nature has to operate automatically, so these models parallel
nature)
cherring
An AC current in a loop of wire is a vector model and the fields are by no
means cubic.
ThomasL283
>Date: 4/16/2001 10:55 AM Pacific Daylight Time
>Message-id: <vmGC6.6315$JI6....@news1.rdc1.tx.home.com> Wrote:
>Snip<
>> The model is a closed particle constructed of photons. The photon is
>modeled
>> as a Poynting vector,
>
>An AC current in a loop of wire is a vector model and the fields are by no
>means cubic.
Yes, the model has to create the current loops from the spinning cube
structure.
As you can see from page 28, Figure 4.5 the front and back cube faces have
current loop areas (L2) whose circumferences are described by the eight cube
corners.
The Bohr magneton has the dimensions of Ampere times area (A x m^2) and these
model determined two loop areas L2 give the Bohr from Ub=( L2 x e x c)/ lambda.
Where lambda is the electrons known Compton wavelength, e is the fundamental
charge and c is the velocity of light. Thus L2 tends to verify the spinning EM
cube structure.
As was mentioned earlier, one can hang ALL of the electrons numbers of the
geometry of the spinning cube of EM field energy.
But more than that, ANY sized cube of spinning EM fields has exactly the
fundamental charge (e) and spin angular momentum (1/2h bar). See the math on
page 34,35,36 and again with the proton model core particle on page 58.
So, you can see why this model seems to be the correct one, no matter how
counter intuitive certain premises may be. The numbers all work out, and the
answers are given by the model geometry, exactly.
There is not another model that can make the claim of complete agreement with
the fundamental constants!
Thomas Lockyer http://www.best.com/~lockyer (74 and retired)
The book ISBN 0963154680 at http://www.fatbrain,com
ThomasL283
had a typo.
The book is also stocked at
Use ISBN 0963154680 if you are interested in optaining a copy.
Thomas Lockyer http://www.best.com/~lockyer (74 and retired)
The book ISBN 0963154680 at http://www.fatbrain.com
cherring
> >"cherring" cherr...@home.com
> >Date: 4/16/2001 10:55 AM Pacific Daylight Time
> >Message-id: <vmGC6.6315$JI6....@news1.rdc1.tx.home.com> Wrote:
> >Snip<
>
> >> The model is a closed particle constructed of photons. The photon is
> >modeled
> >> as a Poynting vector,
> >
> >An AC current in a loop of wire is a vector model and the fields are by
no
> >means cubic.
>
> Yes, the model has to create the current loops from the spinning cube
> structure.
Again, space doesn't support spinning cubes. The macro-world proves what
space supports when no physical structure exists.
Jim Carr
in article <_PrA6.264754$bb.20...@news1.rdc1.tx.home.com>:
}
<... much obviously snipped by Thomas ...>
}
} "cherring" <cherr...@home.com> wrote in message
} news:DP2A6.258902$bb.20...@news1.rdc1.tx.home.com...
} > I agree with your statement that energy must be conserved, that
} > S=0.5[(EcrossH)] implies a zero between positive-north and negative-south
} > excursions. I have always had a problem with this also. There is nothing
} > with which to pass the energy forward.
In article <20010410002032...@ng-fn1.aol.com>
thoma...@aol.com (ThomasL283) writes:
>
>Yes the Fig 2.4 page 12 is typical of standard text traveling wave diagram.
Are you pointing out the obvious error made by "cherring" in assuming
that a traveling wave does not travel and that Maxwell's equations
do not lead to a traveling wave when making that observation about
the instantaneous value of S?
} > Which implies a set of opposite fields, hence, our typical interpretation,
} > positive & negative electric fields and north & south magnetic fields.
>Because the vectors are exactly always at 90 degrees the idea of a right handed
>screw and cross product is not applicable.
You have that exactly wrong concerning classical E+M.
--
James Carr <j...@scri.fsu.edu> http://www.scri.fsu.edu/~jac/
Dopeler Effect: The tendency of stupid ideas to seem smarter when they
come at you rapidly. (anon source via e-chain-letter)
ThomasL283
>Date: 4/19/2001 6:49 PM Pacific Daylight Time
>Message-id: <7BMD6.22117$JI6.1...@news1.rdc1.tx.home.com> wrote:
>snip<
>> >An AC current in a loop of wire is a vector model and the fields are by
>no
>> >means cubic.
>> Yes, the model has to create the current loops from the spinning cube
>> structure.
>Again, space doesn't support spinning cubes. The macro-world proves what
>space supports when no physical structure exists.
The model does show that the cube particle spin directly results from the
momentum of the Poynting vector directions going in the same direction in front
and back cube faces. See page 28.
Is the electron really a spinning cube? I don't know. All that can be said for
sure is that the model supports all of the electron's fundamental physical
constants, gives the structures for the neutrinos and scales to all composite
particles in the particle zoo.
Nice thing about vector cube particles, you can combine the electron and
neutrino model vectors, member by member and the resultants scale to the
proton, neutron, pion and muon sturctures.
The model gives the mass of the proton to within 3 ppm and the mass of the
neutron to within 400 ppb of the CODATA.
More than that, the neutron model actually REQUIRES the decay electron and
neutrino and gives their mass contribution to the proton model, from the
scaling. See Chapters 6, 7 and 8.
Regards: Tom:
Thomas Lockyer http://www.best.com/~lockyer (74 and retired)
The book ISBN 0963154680 at http://www.fatbrain.com
ThomasL283
>Date: 4/19/2001 8:49 PM Pacific Daylight Time
>Message-id: <9bobko$4p$1...@news.fsu.edu>wrote;
>
> <... much obviously snipped by Thomas ...>
It would help with our discussions if you get a copy of the book (Vector
Particles And Nuclear Models). See signature info below. I think you will find
the book interesting, (esp. with your excellent experimental background in
nuclear reactions)
>} "cherring" <cherr...@home.com> wrote in message
>} news:DP2A6.258902$bb.20...@news1.rdc1.tx.home.com...
>} > I agree with your statement that energy must be conserved, that
>} > S=0.5[(EcrossH)] implies a zero
>between positive-north and negative-south
>} > excursions. I have always had a problem with this also. There is nothing
>} > with which to pass the energy forward.
>In article <20010410002032...@ng-fn1.aol.com>
>thoma...@aol.com (ThomasL283) writes:
>>
>>Yes the Fig 2.4 page 12 is typical of standard text traveling wave diagram.
>
> Are you pointing out the obvious error made by "cherring" in assuming
> that a traveling wave does not travel and that Maxwell's equations
> do not lead to a traveling wave when making that observation about
> the instantaneous value of S?
I think the point, we both agreed upon, is that the Poynting vector has a
problem in that it also describes a static field, and thus does not show how
the energy translates, in any case.
(Others have discussed this same fact. See "The Classical Electromagnetic
Field" Eyges Addison Wesley 1972, page 195-203)
>Snip<
>>Because the vectors are exactly always at 90 degrees the idea of a right
>handed
>>screw and cross product is not applicable.
> You have that exactly wrong concerning classical E+M.
>
The point I was trying to make, Jim, is that when the angle is exactly 90
degrees (as it is in this case) then the vector cross product becomes a simple
product of E times H.
Daniel McCarty
>product of E times H.
>
angle between E and H is 90, the magnitude is EH and the direction is
perpendicular to both. These magnitudes are themselves sinusoidal scalar
functions of time and position.
Danny, PhD education MS physics MSE electrical engineering.
cherring
E = hv...
Take E/v=h and multiply by 1/2 then divide by 2*pi and knowing that for the
conversion energy v = v_o and that E/2 equals E_o, the energy mass of either
the electron or positron, one can easily show...
E_o/(2*pi*v_o) = (1/2)[h/(2*pi)] eq 1 (the right side is spin
units)
Further, since 2*pi*v_o is angular frequency w_o, we get...
E_o/w_o = (1/2)[h/(2*pi)] eq2 (simpliest form)
Which means that half the energy from the photon that can create an
electron-positron pair, roughly 1.022 Mev, divide by the angular frequency
equivalent to the linear frequency of the photon is particle spin, a
rotating vector
but circular.
Going back to the left side of eq 1 and knowing that v_o = c/wl_o where wl_o
is the photon wavelength and that (wl_o)/2 = wl_1/2, half wavelength and
that
E_o = m_o*c^2...
E_o/(2*pi*v_o) = (m_o*c^2)/ {2*pi*[c/(2*wl_1/2)]} = (m_o*wl_1/2*c)/pi
hence...
(m_o*c)*(wl_1/2)/pi = (1/2)[h/(2*pi)] eq 3. (the mass and half
wavelength form)
Which means that a form of momentum related to the photon, m_o*c, times a
conversion factor from meters to radians, wl_1/2/pi, equals particle spin,
angular momentum.
Eq. 2 in terms of the fine structure constant, given by a =
[(2*pi)(e_o)^2]/(h*c) where e_o is the fundumental electric charge, we
have...
E_o/w_o = [(e_o)^2]/(2*a*c)
or
e_o = [(2*a*c*E_o)/(w_o)]^(1/2) = [(m_o)*(c^2)*(wl_o)*(a)]/(pi)
Note: wl_o = (1/2)*(Compton wavelength)
Besides a rotating system of two particles, something equivalent to a toroid
should be stable yet have no charge or magnetic moment, just an electric
moment. Then there would be the electric field equivalent of a toroid which
should also be stable with no charge, just a magnetic moment. These are
candidates for neutrinos in an analagous fashion to your postulated cubic
particles (and I emphaizes the postulated!).
Magnetic pole particles are postualted but have never been observed. Indeed
from one of Dirac's equations, g_o = (1/2)*[(h*c)/(2*pi*e_o)] where g_o
would have to be a fundumental magnetic charge and e_o is the fundumental
electric charge.
"ThomasL283" <thoma...@aol.com> wrote in message
news:20010420132547...@ng-cj1.aol.com...
ThomasL283
>Date: 4/21/2001 7:07 PM Pacific Daylight Time
>Message-id: <20010421220744...@ng-fy1.aol.com> Wrote:
>>he vector cross product becomes a simple
>>product of E times H.
>>
>Am I reading this right? The vector cross product is a VECTOR. When the
>angle between E and H is 90, the magnitude is EH and the direction is
>perpendicular to both. These magnitudes are themselves sinusoidal scalar
>functions of time and position.
Yes that is understood, but it is not stated by the Poynting vector; S = 0.5(E
x H)
The angle is the included angle, between the lateral E and H and sine theta of
90 degrees is 1.0 for a traveling wave, which means the simple product of E
times H is equivalent to the vector cross product at 90 degrees, as you state.
The Poynting vector is calculated at the peak or rms value for E and H and
gives the Volt Amps per meter squared in a surface.
The point is that static fields also are given in the same units E = V/m and H
= A/m and do not travel. The Poynting vector just shows the direction of a
traveling wave for alternating traveling waves, by DEFINITION without showing a
mechanism for producing that motion.
The new definition proposed creates the mechanism for travel by a time rate of
change between lateral and axial. This would serve to conserve the energy,
over all time and provides a mechanism for motion, using the well known trig
identity;
S = 0.5 ((E x H sin^2 theta) + ( H x E cos^2 theta))
Hrere there is no doubt that we are talking time rate of change in an
alternating current, because theta is a phase angle between E and H and H and E
of the traveling wave. The mechanism for creation of travel, by this new
(dynamic) Poynting vector, is given by this math, but not by the static S = 0.5
(E x H )
Its all in the book, "Vector Particle and Nuclear Models" which you should get
to aid in these discussions. (See Sig info below)
ThomasL283
>Date: 4/22/2001 2:54 PM Pacific Daylight Time
>Message-id: <hqIE6.32289$JI6.1...@news1.rdc1.tx.home.com> Wrote:
>And you can't do the same thing with (rhetorical question)....
>
> E = hv...
>
>Take E/v=h and multiply by 1/2 then divide by 2*pi and knowing that for the
>conversion energy v = v_o and that E/2 equals E_o, the energy mass of either
>snip<
Yes, you can take E = hv and multiply by 1/2 and divide by 2*pi but you are
just begging the result.
The trouble with math alone, without a model, is that it does not add anything
to our understanding. At the fundamental level, all physical constants are
related mathemetically, so no surprize that you get these equations to work.
OTOH, I find that the geometry of the spinning cube of EM energy "form factor"
leads to a number of brand new ideas.
As was stated before, one can relate all of the electrons numbers by the
geometry of the electron model. (See Chapter 4)
More than that, the model shows the source of particle spin, and the
structural difference between the electron and positron. (see page 21, 23, and
Chapter 4)
More than that the model shows the structures for the neutrinos. (see page 21
and Chapter 5)
More than that, the form factor can be scaled to the mass of the proton and
neutron.
In fact, one can start with just the dimensionless fine structure constant (a)
and get BOTH the dimensionless mass ratios of the proton and neutron. (see
Chapter 7, pages 54, 55)
More than that, the proton and neutron models form factors allow directly
calculating the binding energy, for any nucleus, using well known EM equations
for the electric and magnetic energy. (see Chapter 13)
cherring
photon of frequency f and wavelength wl (I hope it looks right). Let the
pluses
represent thepositive E-field vector and its concentration in space, the
minus
representthe negative E-field vector and its concentration in space, the o's
represent the B-field vector coming out of the screen and the x's represent
the B-field vector going into the screen (note: a B-field is always a closed
loop and, therefore, is the mechanism for forward energy transport). The
arrows, >, represent the Poynting vector (S) and its direction and they are
placed at the null locations between the positive-north and negative-south
as markers for those null points. This gives us a typical photon energy
transport mechanism.
++++
+++++++
++++++++++
XXXXXXXXXX>OOOOOOOOOO>
---------------
-------------
------
Without any outside fields to cause interference the cross product between
the E and B fields guarantee a forward right angle direction to both E and B
and explains photon momentum and inertia.
When the wavelength, wl, is very large compared to the molecular and atomic
scale, the photon's leading E-B field. This will continue for wavelengths
that are smaller then l as stated in the previous sentence but which are
still greater than the nuclear scale.
Now consider that the wavelength, wl, of the photon is on the order of the
nuclear scale and shorter. When a photon enters a radial Coulomb field of a
nucleus (hard to show above but easy to imagine), interactions are not
instantaneous, being limited by c, which allows the photon to penetrate some
distance before any interaction takes place. For a lateral or an oblique
penetration the nuclear radial E-field vectors add vectorally with the
photon E-field vectors. When the radial field strength is small compared to
the photon E-field strength, then little or no affect will occur to the S
vector. The photon will "hit" a peak radial field strength in one monment
and then in the next will pass out of this peak area and there will be
little or no affect on S, a minor bobble at most. On a closer approach
Comptom scattering will occur.
However, when the radial E-field is of strength equal to or greater than the
photon E-field strength, the new resultant vectors will force S to follow a
curve. Both photon E-field vectors will be affect independently. There
should exist an instance in time and a critical curvature in which it is
easier to form 2 static E-fields with magnetic moments than to re-establish
the original photon. This process will be further help because of
like-charge repulsion and opposite-charge attraction with the radial Coulomb
field. You can reason a similar process for the magnetic moment, so it
participates also.
I do agree with you that this can only occur for the critical wavelength,
wl_c (or possibly half wavelength), where the photon's E & B field's rates
of
change are equal to c, the rate of forward propagation along the S vector.
However, what is happening is that a linear plain wave of fields
E & B is undergoing a convolution with the particle's E & B spherical field.
"ThomasL283" <thoma...@aol.com> wrote in message
news:20010424121352...@ng-fu1.aol.com...
Jim Carr
|
| "cherring" cherr...@home.com wrote
| in article <_PrA6.264754$bb.20...@news1.rdc1.tx.home.com>:
| }
|
In article <20010421194838...@ng-cm1.aol.com>
thoma...@aol.com (ThomasL283) writes:
> It would help with our discussions if you get a copy of the book (Vector
>Particles And Nuclear Models).
I don't see how that would keep you from snipping large parts of
a discussion you don't want to address.
| } "cherring" <cherr...@home.com> wrote in message
| } news:DP2A6.258902$bb.20...@news1.rdc1.tx.home.com...
| } > I agree with your statement that energy must be conserved, that
| } > S=0.5[(EcrossH)] implies a zero between positive-north and negative-south
| } > excursions. I have always had a problem with this also. There is nothing
| } > with which to pass the energy forward.
|
| In article <20010410002032...@ng-fn1.aol.com>
| thoma...@aol.com (ThomasL283) writes:
| >Yes the Fig 2.4 page 12 is typical of standard text traveling wave diagram.
|
| Are you pointing out the obvious error made by "cherring" in assuming
| that a traveling wave does not travel and that Maxwell's equations
| do not lead to a traveling wave when making that observation about
| the instantaneous value of S?
In article <20010421194838...@ng-cm1.aol.com>
thoma...@aol.com (ThomasL283) writes:
>
>I think the point, we both agreed upon, is that the Poynting vector has a
>problem in that it also describes a static field, and thus does not show how
>the energy translates, in any case.
I can see how you might both agree on something that is wrong,
but I don't see why you think that this makes it correct.
The Poynting vector does not *describe* a field, it is calculated
from it. Big difference. It is a vector, so it is its dot product
with an area that gives an energy flux. Details like this matter.
That it does not show how a field that is not traveling travels is
a rather pointless objection because a static field is not traveling
so there is no traveling to show.
The Poynting vector does show how energy is carried by a traveling
electromagnetic wave. That there are zeroes only shows you that
what is meaningful for most applications is the time average of S.
> (Others have discussed this same fact. See "The Classical Electromagnetic
>Field" Eyges Addison Wesley 1972, page 195-203)
I don't have that book, so I can't tell what you are mistaken about,
but perhaps that concerns the fact that it is the divergence of S
that tells you about d/dt of the energy density, the observable.
This is related to the details I pointed out above.
| } > Which implies a set of opposite fields, hence, our typical interpretation,
| } > positive & negative electric fields and north & south magnetic fields.
|
| >Because the vectors are exactly always at 90 degrees the idea of a right
| >handed screw and cross product is not applicable.
|
| You have that exactly wrong concerning classical E+M.
>The point I was trying to make, Jim, is that when the angle is exactly 90
>degrees (as it is in this case) then the vector cross product becomes a
>simple product of E times H.
Then make that point, rather than make a statement that is nonsense.
That would save me the time of pointing out your first error so I can
point out your new one -- which ignores the fact that the vector cross
product still has a direction in that case. I think your "correction"
shows that you still don't understand this point.
ThomasL283
wrote in:
Message-id: <9csce1$h3o$1...@news.fsu.edu>
>j...@dirac.csit.fsu.edu (Jim Carr) wrote in <9bobko$4p$1...@news.fsu.edu>:
|
>| "cherring" cherr...@home.com wrote
>| in article <_PrA6.264754$bb.20...@news1.rdc1.tx.home.com>:
>| }
>|
>| <... much obviously snipped by Thomas ...>
>In article <20010421194838...@ng-cm1.aol.com>
>thoma...@aol.com (ThomasL283) writes:
>> It would help with our discussions if you get a copy of the book (Vector
>>Particles And Nuclear Models).
> I don't see how that would keep you from snipping large parts of
> a discussion you don't want to address.
It was useful, because Cherring did get a copy of the book. If you read
further in the thread, we were able to exchange ideas by referring to the pages
in the book, as they related to his premises, without having to clutter the
thread with hard to read ASCII copy.
>| } "cherring" <cherr...@home.com> wrote in message
>| } news:DP2A6.258902$bb.20...@news1.rdc1.tx.home.com...
>| } > I agree with your statement that energy must be conserved, that
>| } > S=0.5[(EcrossH)] implies a zero between positive-north and
negative-south
>| } > excursions. I have always had a problem with this also. There is nothing
>| } > with which to pass the energy forward.
|
>| In article <20010410002032...@ng-fn1.aol.com>
>| thoma...@aol.com (ThomasL283) writes:
>| >Yes the Fig 2.4 page 12 is typical of standard text traveling wave diagram.
|
>| Are you pointing out the obvious error made by "cherring" in assuming
>| that a traveling wave does not travel and that Maxwell's equations
>| do not lead to a traveling wave when making that observation about
>| the instantaneous value of S?
>In article <20010421194838...@ng-cm1.aol.com>
>thoma...@aol.com (ThomasL283) writes:
>
>>I think the point, we both agreed upon, is that the Poynting vector has a
>>problem in that it also describes a static field, and thus does not show how
>>the energy translates, in any case.
>I can see how you might both agree on something that is wrong,
>but I don't see why you think that this makes it correct.
The Poynting vector has no mechanism for motion, other than by definition, was
the point I think we both agreed upon.
> The Poynting vector does not *describe* a field, it is calculated
> from it. Big difference. It is a vector, so it is its dot product
> with an area that gives an energy flux. Details like this matter.
No argument. It merely describes the energy flux density. E=V/m times H=A/m
equals VA/m^2 which is the power density of the EM field.
>That it does not show how a field that is not traveling travels is
>a rather pointless objection because a static field is not traveling
>so there is no traveling to show.
Exactly, the Poynting vector Is the same in both cases. There is nothing to
distinguish traveling from static in S=0.5 (E X H)
>The Poynting vector does show how energy is carried by a traveling
>electromagnetic wave. That there are zeroes only shows you that
>what is meaningful for most applications is the time average of S.
The fine point is that the Poynting vector gives the value and direction, but
not a mechanism for travel. The new proposed "dynamic" Poynting vector is
defined as S=0.5((E sin theta)(H sin theta) + (H cos theta)(E cos theta)) which
does show a mechanism for the conservation of energy and also for the motion of
the EM field energy density.
>> (Others have discussed this same fact. See "The Classical Electromagnetic
>>Field" Eyges Addison Wesley 1972, page 195-203)
>I don't have that book, so I can't tell what you are mistaken about,
>but perhaps that concerns the fact that it is the divergence of S
>that tells you about d/dt of the energy density, the observable.
>This is related to the details I pointed out above.
>| } > Which implies a set of opposite fields, hence, our typical
interpretation,
>| } > positive & negative electric fields and north & south magnetic fields.
|
>| >Because the vectors are exactly always at 90 degrees the idea of a right
>| >handed screw and cross product is not applicable.
|
>| You have that exactly wrong concerning classical E+M.
>>The point I was trying to make, Jim, is that when the angle is exactly 90
>>degrees (as it is in this case) then the vector cross product becomes a
>>simple product of E times H.
> Then make that point, rather than make a statement that is nonsense.
> That would save me the time of pointing out your first error so I can
> point out your new one -- which ignores the fact that the vector cross
> product still has a direction in that case. I think your "correction"
> shows that you still don't understand this point.
Hey, I'm a retired Engineer, you are a Physicist, so we do speak the same
language, but, perhaps, with a slightly different dialect.
Jim Carr
... followups to sci.physics for discussion of basic E+M ...
roac...@aol.comnet (Daniel McCarty) wrote
in article <20010421220744...@ng-fy1.aol.com>:
}
} >he vector cross product becomes a simple
} >product of E times H.
}
} Am I reading this right? The vector cross product is a VECTOR. When the
} angle between E and H is 90, the magnitude is EH and the direction is
} perpendicular to both. These magnitudes are themselves sinusoidal scalar
} functions of time and position.
In article <20010423115047...@ng-ci1.aol.com>
thoma...@aol.com (ThomasL283) writes:
>
>Yes that is understood, but it is not stated by the Poynting vector;
>S = 0.5(E x H)
Yes it is, because S, E, and H are all vectors and "x" is the
vector cross product.
>The angle is the included angle, between the lateral E and H and sine theta of
>90 degrees is 1.0 for a traveling wave, ...
Just as he said, so you support his observation.
> ... which means the simple product of E
>times H is equivalent to the vector cross product at 90 degrees, as you state.
It does not mean that, nor is that what he stated.
>The Poynting vector is calculated at the peak or rms value for E and H and
>gives the Volt Amps per meter squared in a surface.
Your statement is self-contradicatory as well as contradicting
what you wrote above about S not being a vector.
{\vec S} is calculated from {\vec E} and {\vec H}. It is S_{avg} that
is related to the rms values of the fields and you take the integral of
{\vec S} dotted into {d{\vec A}} over the area to the last quantity
you mention.
<... snip comment addressed in another article ...>
>The Poynting vector just shows the direction of a
>traveling wave for alternating traveling waves, by DEFINITION without
>showing a mechanism for producing that motion.
No one says it does. The "mechanism" is seen in Maxwell's equations.
>The new definition proposed ...
There is no reason to change the definition of {\vec S}.
You should express your new theory in terms of standard terms
used by all physicists, after you understand those terms and
how they are defined and used.
<... snip "theory" that should be discussed in s.p.particle ...>
Jim Carr
Your newsreader is broken, Thomas, or you need to learn to use it.
j...@dirac.csit.fsu.edu (Jim Carr) wrote in <9csce1$h3o$1...@news.fsu.edu>:
|
| j...@dirac.csit.fsu.edu (Jim Carr) wrote in <9bobko$4p$1...@news.fsu.edu>:
| |
| | "cherring" cherr...@home.com wrote
| | in article <_PrA6.264754$bb.20...@news1.rdc1.tx.home.com>:
| | }
| |
| | <... much obviously snipped by Thomas ...>
|
| In article <20010421194838...@ng-cm1.aol.com>
| thoma...@aol.com (ThomasL283) writes:
| > It would help with our discussions if you get a copy of the book (Vector
| >Particles And Nuclear Models).
|
| I don't see how that would keep you from snipping large parts of
| a discussion you don't want to address.
In article <20010504135601...@ng-fx1.aol.com>
thoma...@aol.com (ThomasL283) writes:
>
>It was useful, because Cherring did get a copy of the book.
Nonsequitur. Someone getting your book will not help you face
reality rather than gloss over whatever you do not understand
or do understand but don't want to deal with because it is in
conflict with your fantasy theory.
| | } "cherring" <cherr...@home.com> wrote in message
| | } news:DP2A6.258902$bb.20...@news1.rdc1.tx.home.com...
| | } > I agree with your statement that energy must be conserved, that
| | } > S=0.5[(EcrossH)] implies a zero between positive-north and negative-south
| | } > excursions. I have always had a problem with this also. There is nothing
| | } > with which to pass the energy forward.
| |
| | In article <20010410002032...@ng-fn1.aol.com>
| | thoma...@aol.com (ThomasL283) writes:
| | >Yes the Fig 2.4 page 12 is typical of standard text traveling wave diagram.
| |
| | Are you pointing out the obvious error made by "cherring" in assuming
| | that a traveling wave does not travel and that Maxwell's equations
| | do not lead to a traveling wave when making that observation about
| | the instantaneous value of S?
|
| In article <20010421194838...@ng-cm1.aol.com>
| thoma...@aol.com (ThomasL283) writes:
| >I think the point, we both agreed upon, is that the Poynting vector has a
| >problem in that it also describes a static field, and thus does not show how
| >the energy translates, in any case.
|
| I can see how you might both agree on something that is wrong,
| but I don't see why you think that this makes it correct.
>The Poynting vector has no mechanism for motion, other than by definition,
>was the point I think we both agreed upon.
The blind leading the blind, it seems, particularly if you read the
rest of my article before writing that reply -- as you should have.
| The Poynting vector does not *describe* a field, it is calculated
| from it. Big difference. It is a vector, so it is its dot product
| with an area that gives an energy flux. Details like this matter.
>No argument. It merely describes the energy flux density.
Then why the nonsense about "no mechanism for motion"?
| That it does not show how a field that is not traveling travels is
| a rather pointless objection because a static field is not traveling
| so there is no traveling to show.
>Exactly, the Poynting vector Is the same in both cases. There is nothing
>to distinguish traveling from static in S=0.5 (E X H)
You must not have read the next part of my reply. You quote it
below, but don't reply to it, leaving unclear if you got it.
| The Poynting vector does show how energy is carried by a traveling
| electromagnetic wave. That there are zeroes only shows you that
| what is meaningful for most applications is the time average of S.
>The fine point is that the Poynting vector gives the value and direction,
>but not a mechanism for travel.
It isn't supposed to. The "fine point" is that Maxwell's equations
give you the mechanism for travel, as I pointed out in the article
you are replying to but seem not to have read. You do know that we
are talking about E+M, right?
| > (Others have discussed this same fact. See "The Classical Electromagnetic
| >Field" Eyges Addison Wesley 1972, page 195-203)
|
| I don't have that book, so I can't tell what you are mistaken about,
| but perhaps that concerns the fact that it is the divergence of S
| that tells you about d/dt of the energy density, the observable.
| This is related to the details I pointed out above.
Did you understand this, Thomas?
| | } > Which implies a set of opposite fields, hence, our typical interpretation,
| | } > positive & negative electric fields and north & south magnetic fields.
| |
| | >Because the vectors are exactly always at 90 degrees the idea of a right
| | >handed screw and cross product is not applicable.
| |
| | You have that exactly wrong concerning classical E+M.
|
| >The point I was trying to make, Jim, is that when the angle is exactly 90
| >degrees (as it is in this case) then the vector cross product becomes a
| >simple product of E times H.
|
| Then make that point, rather than make a statement that is nonsense.
| That would save me the time of pointing out your first error so I can
| point out your new one -- which ignores the fact that the vector cross
| product still has a direction in that case. I think your "correction"
| shows that you still don't understand this point.
>Hey, I'm a retired Engineer, you are a Physicist, so we do speak the same
>language, but, perhaps, with a slightly different dialect.
Engineers are supposed to know about the cross product.
ThomasL283
>Wrote in:
>Message-id: <9dvimj$jsa$1...@news.fsu.edu>
>| j...@dirac.csit.fsu.edu (Jim Carr) wrote in <9bobko$4p$1...@news.fsu.edu>:
>| |
>| | "cherring" cherr...@home.com wrote
>| | in article <_PrA6.264754$bb.20...@news1.rdc1.tx.home.com>:
>| | }
>| In article <20010421194838...@ng-cm1.aol.com>
>| thoma...@aol.com (ThomasL283) writes:
>| > It would help with our discussions if you get a copy of the book (Vector
>| >Particles And Nuclear Models).
Jim writes:
>| | <... much obviously snipped by Thomas ...>
>|
>|
>| I don't see how that would keep you from snipping large parts of
>| a discussion you don't want to address.
>
>In article <20010504135601...@ng-fx1.aol.com>
>thoma...@aol.com (ThomasL283) writes:
>>
>>It was useful, because Cherring did get a copy of the book.
> Nonsequitur. Someone getting your book will not help you face
> reality rather than gloss over whatever you do not understand
> or do understand but don't want to deal with because it is in
> conflict with your fantasy theory.
Surely, Jim, you know that the outdated ASCII strait jacket, that these news
groups cling to, is intirely unsatisfactory to present cogent ideas. This
causes intelligent fellows, such as yourself, from really being able to
communicate, or others to properly respond.
>| | } "cherring" <cherr...@home.com> wrote in message
>| | } news:DP2A6.258902$bb.20...@news1.rdc1.tx.home.com...
To me.
>| | } > I agree with your statement that energy must be conserved, that
>| | } > S=0.5[(EcrossH)] implies a zero
>between positive-north and negative-south
>| | } > excursions. I have always had a problem with this also. There is
>nothing
>| | } > with which to pass the energy forward.
>| | In article <20010410002032...@ng-fn1.aol.com>
>| | thoma...@aol.com (ThomasL283) writes:
>| | >Yes the Fig 2.4 page 12 is typical of standard text traveling wave
>diagram.
>| |
Jim said:>| |
>| | Are you pointing out the obvious error made by "cherring" in assuming
>| | that a traveling wave does not travel and that Maxwell's equations
>| | do not lead to a traveling wave when making that observation about
>| | the instantaneous value of S?
>|
I said to Jim:>| >I think the point, we both agreed upon, is that the Poynting
vector has a
>| >problem in that it also describes a static field, and thus does not show
>how
>| >the energy translates, in any case.
>|
Jim said:>| I can see how you might both agree on something that is wrong,
>| but I don't see why you think that this makes it correct.
I said:
>>The Poynting vector has no mechanism for motion, other than by definition,
>>was the point I think we both agreed upon.
Jim said:
> The blind leading the blind, it seems, particularly if you read the
> rest of my article before writing that reply -- as you should have.
>| The Poynting vector does not *describe* a field, it is calculated
>| from it. Big difference. It is a vector, so it is its dot product
>| with an area that gives an energy flux. Details like this matter.
I said:
>>No argument. It merely describes the energy flux density.
Jim said:
> Then why the nonsense about "no mechanism for motion"?
>Snip<
>| That it does not show how a field that is not traveling travels is
>| a rather pointless objection because a static field is not traveling
>| so there is no traveling to show.
>
Tom said:
>>Exactly, the Poynting vector Is the same in both cases. There is nothing
>>to distinguish traveling from static in S=0.5 (E X H)
Jim says
:> You must not have read the next part of my reply. You quote it
> below, but don't reply to it, leaving unclear if you got it.
>| The Poynting vector does show how energy is carried by a traveling
>| electromagnetic wave. That there are zeroes only shows you that
>| what is meaningful for most applications is the time average of S.
>
Tom said:
>>The fine point is that the Poynting vector gives the value and direction,
>>but not a mechanism for travel.
>
Jim says:
> It isn't supposed to. The "fine point" is that Maxwell's equations
> give you the mechanism for travel, as I pointed out in the article
> you are replying to but seem not to have read. You do know that we
> are talking about E+M, right?
Tom said:
>| > (Others have discussed this same fact. See "The Classical
>Electromagnetic
>| >Field" Eyges Addison Wesley 1972, page 195-203)
>|
Jim said:
>| I don't have that book, so I can't tell what you are mistaken about,
>| but perhaps that concerns the fact that it is the divergence of S
>| that tells you about d/dt of the energy density, the observable.
>| This is related to the details I pointed out above.
>
> Did you understand this, Thomas?
>
Tom said:>| |
>| | >Because the vectors are exactly always at 90 degrees the idea of a right
>| | >handed screw and cross product is not applicable.
Jim said:>| |
>| | You have that exactly wrong concerning classical E+M.
>|
Tom said:
>| >The point I was trying to make, Jim, is that when the angle is exactly 90
>| >degrees (as it is in this case) then the vector cross product becomes a
>| >simple product of E times H.
>|
Jim said:
>| Then make that point, rather than make a statement that is nonsense.
>| That would save me the time of pointing out your first error so I can
>| point out your new one -- which ignores the fact that the vector cross
>| product still has a direction in that case. I think your "correction"
>| shows that you still don't understand
>this point.
Tom said:
>>Hey, I'm a retired Engineer, you are a Physicist, so we do speak the same
>>language, but, perhaps, with a slightly different dialect.
>
Jim said:> Engineers are supposed to know about the cross product.
>
Sort of, see my statements to Cherring about the right handed screw, above.
But, hey, it's been 50 years, and I'm not as sharp as you.
Jim Carr
>
>Surely, Jim, you know that the outdated ASCII strait jacket, that these
>news groups cling to, is intirely unsatisfactory to present cogent ideas.
>causes intelligent fellows, such as yourself, from really being able to
>communicate, or others to properly respond.
You have to listen before you respond if what I say is to
be communicated to you. Part of that act of active listening
is to identify what you did not understand and seek clarification.
Simply repeating silly things will lead people like me to ignore
you except to do what I did -- point out to others where you
are going wrong and warn them not to waste their time trying
to make sense out of your nonsense.
<... snip Tom ignoring my advice ...>