| Topic: |
Science > Physics |
| User: |
"rds" |
| Date: |
18 Sep 2006 02:17:19 PM |
| Object: |
Electromagnetic wave and photon spin |
First off I want to say that I am not physics major.
Can someone explain to me the details of how electromagnetic radiation?
Actually occurs. I know the pat answers that the acceleration of
charged particles or change in direction, but that doesn't really
explain the process.
For that matter why does a magnetic field form when current flows at a
constant rate (charge velocity)?
I notice in my car from 0 to 50 MPH , that when I accelerate I feel a
pull in the reverse direction, but when I reach the intended speed the
pull subsides. In some way this is counter intuitive in that I would
assume that the 50 MPH pull would continually cause a drag backwards.
In the same manner is this a parallel to the transmission of EM waves?
In that I get a transmission (radiation) when charges accelerate or
change direction?
Lastly the description of EM wave as abstract photons (all energy no
mass) makes sense to me in that the EM wave propagates as an E field
(sin) + an M field (cos). Is there an actual difference in the nature
of these two fields? Or are they indistinguishable from one another
--more like two E-Fields with a 90 degree phase shift
If the latter is true, and we take any point on the EM wave front an
examine it as it travels toward us, we could easily describe the EM
field as a mass less polarized point in space which rotates like a
plate
spinning with the top of the plate being positive and the bottom of the
plate
being negative. The spinning at the rate of the transmission frequency.
This totally describes the EM field without even referring to a
magnetic
field. As the plate rotates its positive to the 3 and 9 o'clock
positions it
represents the M field peaks, and at the 12 and 6 o'clock position it
represents the E field peaks. I would normally call this point a photon
and the rotation I would call spin. But I read that photon spin is
actually in the direction of travel, and so the spin must be something
different. Any comments clarrifications is welcome.
Thanks.....
.
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| User: "Sorcerer" |
|
| Title: Re: Electromagnetic wave and photon spin |
18 Sep 2006 07:42:41 PM |
|
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"rds" <rdschwarz61@yahoo.com> wrote in message
news:1158607039.490398.66180@e3g2000cwe.googlegroups.com...
| First off I want to say that I am not physics major.
|
| Can someone explain to me the details of how electromagnetic radiation?
| Actually occurs. I know the pat answers that the acceleration of
| charged particles or change in direction, but that doesn't really
| explain the process.
|
| For that matter why does a magnetic field form when current flows at a
| constant rate (charge velocity)?
|
| I notice in my car from 0 to 50 MPH , that when I accelerate I feel a
| pull in the reverse direction, but when I reach the intended speed the
| pull subsides. In some way this is counter intuitive in that I would
| assume that the 50 MPH pull would continually cause a drag backwards.
| In the same manner is this a parallel to the transmission of EM waves?
| In that I get a transmission (radiation) when charges accelerate or
| change direction?
|
| Lastly the description of EM wave as abstract photons (all energy no
| mass) makes sense to me in that the EM wave propagates as an E field
| (sin) + an M field (cos). Is there an actual difference in the nature
| of these two fields? Or are they indistinguishable from one another
| --more like two E-Fields with a 90 degree phase shift
|
| If the latter is true, and we take any point on the EM wave front an
| examine it as it travels toward us, we could easily describe the EM
| field as a mass less polarized point in space which rotates like a
| plate
| spinning with the top of the plate being positive and the bottom of the
| plate
| being negative. The spinning at the rate of the transmission frequency.
| This totally describes the EM field without even referring to a
| magnetic
| field. As the plate rotates its positive to the 3 and 9 o'clock
| positions it
| represents the M field peaks, and at the 12 and 6 o'clock position it
| represents the E field peaks. I would normally call this point a photon
| and the rotation I would call spin. But I read that photon spin is
| actually in the direction of travel, and so the spin must be something
| different. Any comments clarrifications is welcome.
|
| Thanks.....
Sure:
Step 1)
http://www.androcles01.pwp.blueyonder.co.uk/spin.gif
Steps 2) through 9)
are for you to ask questions about until one of us gets bored.
Step 10)
http://www.androcles01.pwp.blueyonder.co.uk/AC/AC.htm
.
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|
| User: "Timo Nieminen" |
|
| Title: Re: Electromagnetic wave and photon spin |
18 Sep 2006 09:18:35 PM |
|
|
On Tue, 18 Sep 2006, rds wrote:
First off I want to say that I am not physics major.
Can someone explain to me the details of how electromagnetic radiation?
Actually occurs. I know the pat answers that the acceleration of
charged particles or change in direction, but that doesn't really
explain the process.
Why do ripples spread out when you drop a rock in a pond? You disturb the
surface of the water, and the disturbance spreads outward. Change the
electromagnetic field in a region, and a similar thing happens. If changes
in the EM field can't travel instantly, then you expect waves, as the
changes travel outwards at the speed of the waves.
[cut]
Lastly the description of EM wave as abstract photons (all energy no
mass) makes sense to me in that the EM wave propagates as an E field
(sin) + an M field (cos). Is there an actual difference in the nature
of these two fields?
Yes. The E field exerts a force on any charge, and the magnetic field
exerts a force on a moving charge; F=q(E+vxB).
Or are they indistinguishable from one another
--more like two E-Fields with a 90 degree phase shift
Not at all. Apart from the difference in effect, they don't have a 90
degree phase shift - in a plane electromagnetic wave, the electric and
magnetic fields are in phase with each other. Their directions are 90
degrees apart (eg E might be straight up, and B horizontal to the right).
[cut]
The spinning at the rate of the transmission frequency.
This totally describes the EM field without even referring to a
magnetic
field. As the plate rotates its positive to the 3 and 9 o'clock
positions it
represents the M field peaks, and at the 12 and 6 o'clock position it
represents the E field peaks.
What you describe is approximately what happens to the two components of
the electric field in a _circularly-polarized_ wave. But these are both
components of the _electric_ field. (The magnetic field of the wave
behaves in the same way, too.) For a plane polarised (or linearly
polarised) plane wave, the average spin of the photons is zero.
I would normally call this point a photon
and the rotation I would call spin. But I read that photon spin is
actually in the direction of travel, and so the spin must be something
different. Any comments clarrifications is welcome.
What is the direction of a rotation? If you have rotation about an axis,
what way does the axis point? There are two directions you could use; the
usual convention is: curl the fingers of your right hand, and stick your
thumb out; if the rotation is in the direction your fingers curl in, the
rotation/spin/angular momentum vector is in the same direction your thumb
is pointing in.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
.
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|
| User: "rds" |
|
| Title: Re: Electromagnetic wave and photon spin |
19 Sep 2006 10:14:40 AM |
|
|
Timo Nieminen wrote:
[cut]
Lastly the description of EM wave as abstract photons (all energy no
mass) makes sense to me in that the EM wave propagates as an E field
(sin) + an M field (cos). Is there an actual difference in the nature
of these two fields?
Yes. The E field exerts a force on any charge, and the magnetic field
exerts a force on a moving charge; F=q(E+vxB).
Or are they indistinguishable from one another
--more like two E-Fields with a 90 degree phase shift
Not at all. Apart from the difference in effect, they don't have a 90
degree phase shift - in a plane electromagnetic wave, the electric and
magnetic fields are in phase with each other. Their directions are 90
degrees apart (eg E might be straight up, and B horizontal to the right).
OK I got you here. I see that you are indeed correct and the E and M
waves are indeed in phase, except for the physical direction. I looked
at the classic picture of an EM wave with 2 sine waves and they do
appear to be in phase with the MAG rotated by 90 degrees is direction.
The sinewave representation is kind of confusing from a picture point
of view. The peak of the sine wave isn't actually a physical movement
of the wave but is more of a representation that the charges are at a
higher (est) speed in the radiator at that point for the E field. The
sinewave itself is just a picture of a constantly
accelerating/decelerating charge in the radiator which in the EM wave
is a voltage which will cause an accelerating/decelrating charge in a
receiving antenna, correct? In fact the sin/cos wave can get EM
radiation to continually occur right? If we used a square wave as a
carrier, I would assume that the only time the signal is being emitted
is during the rise and fall times since that is the only time the
charges in the antenna is actually accelrating or decelrating. However
isnt sinA+cosA esentially a square wave? +1, -1 ? Would the EM wave be
a sin wave at the frequency and a cosine wave at the same frequency?
Also I am still unclear on the E vs M field. Ok they are in phase. But
other than that what does the M field represent? Can it be detected by
changing the polarization on your receiving antenna? What can be done
with the M field that you couldn't do with the E field ? How does the
90 degree direction difference in the Magnetic and Electric fileds
present itself. What is the magnetic filed shaped like. In the antenna,
doesn't it surround the conductor like a donut or ring? How does the 90
degree direction shift get realized?
Are the radiation levels different ie (does E drop off at a ^2 rate and
the M filed drop off at the ^3 rate) ? Is the M field in the EM wave
different from the near magnetic field which I thought was only
surrounding the Tx. antenna?
(of course the antenna can represent any EM radiator , wire, etc)
.
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|
| User: "rds" |
|
| Title: Re: Electromagnetic wave and photon spin |
19 Sep 2006 03:49:42 PM |
|
|
It seems to me that the EM field works like this:
1) Particals are energized with a voltage in lets say a wire.
2) The E field causes charges to move (this is the mystery part)
3) The moving charges causes a magnetic filed (this is the mystery
part)
4) The alternating magnetic field causes a propagating Electric field.
5) Far away from the radiator we get an E and M field whose ratio is
375.
6) This 375 represents the impedance of what used to be called the
ether and accounts for zero point energy lost?
7) The M field directly represents current flow. So like ohms law where
e=I*R we have E=M*375 (in free space).
I can now clearly see now why the change in velocity causes the EM
radiation. But what is still unclear is what or how exactly the
magnetic field is generated when a charge has a velocity (constant
current). What exactly about charge movement causes a magnetic field to
form?
Heres one thought. A charged particle forms force lines. Lets say an
electron (-) forms a force line with another Electron (-) which is in
front of the first electron (-) in the direction of travel that our
field voltage will eventually be applied.
By applying the voltage we cause our electrons to move but not the
force lines This would cause the force lines to overlap or buckle in
the middle of the two particles. it will also cause a gap to be formed
behind both particles.
-----\ gap\--------(-)------\overlap\-----\gap\----(-)-------\overlap\
---->
The overlap means that an excess of force exists in that area causing
the force to eminate 90 degrees away from the initial voltage
direction.The other force vectors going out from the charge at
different angles would get thier force vectors overlapped but because
of the angle the overlaps would not be a s big as the force vector
which was traveling in the direction of the voltage. This can explain
why the magnetic field appears to rotate around the wire in a circular
path as the different magnitudes of the overlaps throw out field
vectors with different values for each part of the circular path.
I'm not sure what the gaps would cause.
Anyway does this make sense?
if the force lines also moved with the particles, then we could
continually add velocity to the force lines. Isn't this simply an
extension of Einstein's C is constant at all velocity references? In
other words, if I was carrying a flashlight and started walking a 3
mph, the speed of light does not change to C + 3 mph does it ? The
light once emitted travels at C in both references. Well if you watched
from the side as I ran the light would never change speed , correct?
You wouldn't see me catching up to the beam, what you'd see is an
overlap or shortening of the entire picture in the left right
direction, and a lengthening in the up down direction. Isn't that why
Einstein says that the M field is the relativisitic result of the E
field? He calls this a time dilation I think. More like a length
dilation I think.
Is this correct or way off base?
.
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|
|
| User: "rds" |
|
| Title: Re: Electromagnetic wave and photon spin |
20 Sep 2006 02:28:46 PM |
|
|
Heres one thought. A charged particle forms force lines. Lets say an
electron (-) forms a force line with another Electron (-) which is in
front of the first electron (-) in the direction of travel that our
field voltage will eventually be applied.
By applying the voltage we cause our electrons to move but not the
force lines This would cause the force lines to overlap or buckle in
the middle of the two particles. it will also cause a gap to be formed
behind both particles.
-----\
gap\------(-)------\overlap\-----\gap\----(-)-------\overlap\---->
The overlap means that an excess of force exists in that area causing
the force to eminate 90 degrees away from the initial voltage
direction.The other force vectors going out from the charge at
different angles would get thier force vectors overlapped but because
of the angle the overlaps would not be as big as the force vector
which was traveling in the direction of the voltage. This can explain
why the magnetic field appears to rotate around the wire in a circular
path as the different magnitudes of the overlaps throw out field
vectors with different values for each part of the circular path.
Further... if the electrons are accelerated the overlap becomes more
and more pronounced and could easily explain the the radiated magnetic
field.
^ ^
|
|
-----\
gap\------(-)------\overlap\-----\gap\----(-)-------\overlap\---->
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|
V V
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| User: "Timo Nieminen" |
|
| Title: Re: Electromagnetic wave and photon spin |
19 Sep 2006 06:33:56 PM |
|
|
On Wed, 19 Sep 2006, rds wrote:
The sinewave representation is kind of confusing from a picture point
of view. The peak of the sine wave isn't actually a physical movement
of the wave but is more of a representation that the charges are at a
higher (est) speed in the radiator at that point for the E field. The
sinewave itself is just a picture of a constantly
accelerating/decelerating charge in the radiator which in the EM wave
is a voltage which will cause an accelerating/decelrating charge in a
receiving antenna, correct?
Pretty much.
In fact the sin/cos wave can get EM
radiation to continually occur right?
It depends on what you mean by continually. Note that the instantaneous
power is given by the Poynting vector, and when E=0 or B=0, this is equal
to zero. You can visualise the transport of energy by a laser beam as a
series of pancakes of energy, 1/2 a wavelength apart, travelling along the
beam at the speed of light.
If we used a square wave as a
carrier, I would assume that the only time the signal is being emitted
is during the rise and fall times since that is the only time the
charges in the antenna is actually accelrating or decelrating. However
isnt sinA+cosA esentially a square wave? +1, -1 ? Would the EM wave be
a sin wave at the frequency and a cosine wave at the same frequency?
Drive the antenna with a square wave, and you'll get a range of
frequencies. Magic words to google for: "fourier series", "fourier
theorem", "fourier series for a square wave".
Also I am still unclear on the E vs M field. Ok they are in phase. But
other than that what does the M field represent? Can it be detected by
changing the polarization on your receiving antenna?
No. A straight wire antenna is essentially an electric field detector.
Why? The voltage along an antenna in a uniform electric field is V=EL,
where L is the length of the antenna and E is the electric field in the
direction of the antenna. Where the antenna is smaller than the
wavelength, this is your basic electric dipole antenna. Look on www for
pictures of antennas, and you should find some with loops instead of
straight wires. These are magnetic dipole antennas, and basically work
like an electromagnet in reverse.
[cut]
Are the radiation levels different ie (does E drop off at a ^2 rate and
the M filed drop off at the ^3 rate) ? Is the M field in the EM wave
different from the near magnetic field which I thought was only
surrounding the Tx. antenna?
No, E doesn't drop off at r^2 and B (ie the magnetic field) at r^3.
In the far field (and you might benefit from searching www for "near
field" and "far field"), the _irradiance_ (ie power per unit area) drops
off as r^2. The irradiance is proportional to E^2 (and E times B, and
B^2), so E drops off as r^1. And so does B.
In the near field, you can have a faster drop-off of either E or B,
depending on the details of the antenna. But the near field is not
radiation, in that the energy doesn't travel away from the antenna without
coming back.
In, for example, an electric dipole antenna, the electric field in the
near field drops off as r^3, and in the far field, both E and B drop off
as r^1.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
.
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| User: "rds" |
|
| Title: Re: Electromagnetic wave and photon spin |
20 Sep 2006 09:07:59 AM |
|
|
In the near field, you can have a faster drop-off of either E or B,
depending on the details of the antenna. But the near field is not
radiation, in that the energy doesn't travel away from the antenna without
coming back.
In, for example, an electric dipole antenna, the electric field in the
near field drops off as r^3, and in the far field, both E and B drop off
as r^1.
Is that the defenition of near field? That in the near field the energy
returns to the wire or radiator? This would be the inductive power
transformer field correct? Is this near field radiation still a part of
or the source of the readiated energy or are they two distinct energies?
.
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| User: "Timo A. Nieminen" |
|
| Title: Re: Electromagnetic wave and photon spin |
20 Sep 2006 03:25:32 PM |
|
|
On Thu, 20 Sep 2006, rds wrote:
In the near field, you can have a faster drop-off of either E or B,
depending on the details of the antenna. But the near field is not
radiation, in that the energy doesn't travel away from the antenna without
coming back.
In, for example, an electric dipole antenna, the electric field in the
near field drops off as r^3, and in the far field, both E and B drop off
as r^1.
Is that the defenition of near field?
You could define the far field as where E and B drop off as r^1, for all
practical purposes.
It wouldn't be a good definition of near field, since depending on the
details of the antenna, one can arrange to have the most important
component of the electromagnetic near field being either E or B and
dropping off as r^N, N > 2.
If you have a strong mathematical background, "multipoles" and "multipole
expansion" are the magic words. In particular, compare the usual
electrostatic multipoles (note that they usually written down for the
electrostatic potential, not the fields) and the wave equation multipoles
in the small r limit.
That in the near field the energy
returns to the wire or radiator? This would be the inductive power
transformer field correct?
Yes.
Is this near field radiation still a part of
or the source of the readiated energy or are they two distinct energies?
This is stored-and-recycled energy; it isn't radiated. Thus, one can call
the far-field the "radiation field" and the near field the "inductive
field", although one then risks some confusion, since the radiated energy
is also travelling outwards in the near field. As to how separate they
are, since the far field radiated energy wouldn't be there if it wasn't
for the inductive stored energy in the near field, I think it depends very
much on exactly what you mean by "separate".
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
.
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| User: "srp" |
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| Title: Re: Electromagnetic wave and photon spin |
18 Sep 2006 09:28:59 PM |
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rds a écrit :
First off I want to say that I am not physics major.
Can someone explain to me the details of how electromagnetic radiation?
Actually occurs. I know the pat answers that the acceleration of
charged particles or change in direction, but that doesn't really
explain the process.
For that matter why does a magnetic field form when current flows at a
constant rate (charge velocity)?
I notice in my car from 0 to 50 MPH , that when I accelerate I feel a
pull in the reverse direction, but when I reach the intended speed the
pull subsides. In some way this is counter intuitive in that I would
assume that the 50 MPH pull would continually cause a drag backwards.
In the same manner is this a parallel to the transmission of EM waves?
In that I get a transmission (radiation) when charges accelerate or
change direction?
Lastly the description of EM wave as abstract photons (all energy no
mass) makes sense to me in that the EM wave propagates as an E field
(sin) + an M field (cos).
Since at any point of the wavefront, the sum of M and E is a constant,
it would be more like an E field (sin)^2 + an M field (cos)^2 it seems
to me.
Is there an actual difference in the nature of these two fields?
Apparently, since that minimally they systematically occur normal
to each other, both of them normal to the direction of motion.
Or are they indistinguishable from one another
--more like two E-Fields with a 90 degree phase shift
The related 90 degree phase shift is transverse, not longitudinal.
If the latter is true, and we take any point on the EM wave front an
examine it as it travels toward us, we could easily describe the EM
field as a mass less polarized point in space which rotates like a
plate spinning with the top of the plate being positive and the bottom
of the plate being negative. The spinning at the rate of the transmission
frequency.
This totally describes the EM field without even referring to a
magnetic field.
I don't think so. You would be missing a second plate rotating sidewise
about the 12 - 6 o'clock axis representing the M field to complement
your E field plate rotating about the 3 - 9 o'clock axis.
André Michaud
.
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| User: "Sue..." |
|
| Title: Re: Electromagnetic wave and photon spin |
18 Sep 2006 05:49:23 PM |
|
|
rds wrote:
First off I want to say that I am not physics major.
Can someone explain to me the details of how electromagnetic radiation?
Your spelling is better than mine so you should find this
a breeze:
http://farside.ph.utexas.edu/teaching.html
http://web.mit.edu/8.02t/www/802TEAL3D/teal_tour.htm
Actually occurs. I know the pat answers that the acceleration of
charged particles or change in direction, but that doesn't really
explain the process.
Sun electron shakes, eye electron shakes 8 minutes later.
~~ R.P Feynman
For that matter why does a magnetic field form when current flows at a
constant rate (charge velocity)?
Magnetism is the geometric superposition of Coulomb
forces. The fundamentls are simple but the calculations
grow lots of hair because you can have to do them as
3 D volumes of space.
http://en.wikipedia.org/wiki/Multiple_integral
Where described as a moving charge
The electric field of the moving charges is interacting with other
matter which is ~stationary~ . At the minimum it is matter
in 'all of space' or 'free space'
permeability mu_0
permittivity eps_0
radiation resistance 377 ohms.
I notice in my car from 0 to 50 MPH , that when I accelerate I feel a
pull in the reverse direction, but when I reach the intended speed the
pull subsides. In some way this is counter intuitive in that I would
assume that the 50 MPH pull would continually cause a drag backwards.
In the same manner is this a parallel to the transmission of EM waves?
In that I get a transmission (radiation) when charges accelerate or
change direction?
Not quite. The ~pull~ an electron feels is relatvive to its Coulomb
coupling to nearby matter. A force about 10^32 time larger than
those you feel in your automobile.
Lastly the description of EM wave as abstract photons (all energy no
mass) makes sense to me in that the EM wave propagates as an E field
(sin) + an M field (cos). Is there an actual difference in the nature
of these two fields? Or are they indistinguishable from one another
--more like two E-Fields with a 90 degree phase shift
Yes... You are spot on. The E component diminishes by 1/r^2
and is the farfield or radiative component. The magnetic
component diminishes by 1/r^3 and is effective in the nearfield
only.
Photons are a mathmatical abstraction not a propagation model.
Your question is to the heart of why Feynman puts wrist watches
on his photons to fix all the mirrors that QM broke. In QED,
the phase information and magnetic components are transported
on the tip of a mathematicians pencil. ;-)
If the latter is true, and we take any point on the EM wave front an
examine it as it travels toward us, we could easily describe the EM
field as a mass less polarized point in space which rotates like a
plate
spinning with the top of the plate being positive and the bottom of the
plate
being negative. The spinning at the rate of the transmission frequency.
This totally describes the EM field without even referring to a
magnetic
field. As the plate rotates its positive to the 3 and 9 o'clock
positions it
represents the M field peaks, and at the 12 and 6 o'clock position it
represents the E field peaks. I would normally call this point a photon
and the rotation I would call spin. But I read that photon spin is
actually in the direction of travel, and so the spin must be something
different. Any comments clarrifications is welcome.
If photons were propgation models you've just made a fairly
good Feynman clock. But they are not. It really helps to have
a good grasp of Maxwell's equations before you assume too
much about photons. Then you can differentiate between what
nature is doing and what a maths operation is doing.
Time-independent Maxwell equations
Time-dependent Maxwell's equations
http://farside.ph.utexas.edu/teaching/em/lectures/lectures.html
Sue...
Thanks.....
.
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| User: "FrediFizzx" |
|
| Title: Re: Electromagnetic wave and photon spin |
19 Sep 2006 02:21:51 AM |
|
|
"Sue..." <suzysewnshow@yahoo.com.au> wrote in message
news:1158619763.325412.221100@d34g2000cwd.googlegroups.com...
rds wrote:
First off I want to say that I am not physics major.
Can someone explain to me the details of how electromagnetic
radiation?
Your spelling is better than mine so you should find this
a breeze:
http://farside.ph.utexas.edu/teaching.html
http://web.mit.edu/8.02t/www/802TEAL3D/teal_tour.htm
Actually occurs. I know the pat answers that the acceleration of
charged particles or change in direction, but that doesn't really
explain the process.
Sun electron shakes, eye electron shakes 8 minutes later.
~~ R.P Feynman
For that matter why does a magnetic field form when current flows at
a
constant rate (charge velocity)?
Magnetism is the geometric superposition of Coulomb
forces. The fundamentls are simple but the calculations
grow lots of hair because you can have to do them as
3 D volumes of space.
http://en.wikipedia.org/wiki/Multiple_integral
Where described as a moving charge
The electric field of the moving charges is interacting with other
matter which is ~stationary~ . At the minimum it is matter
in 'all of space' or 'free space'
permeability mu_0
permittivity eps_0
radiation resistance 377 ohms.
I notice in my car from 0 to 50 MPH , that when I accelerate I feel
a
pull in the reverse direction, but when I reach the intended speed
the
pull subsides. In some way this is counter intuitive in that I would
assume that the 50 MPH pull would continually cause a drag
backwards.
In the same manner is this a parallel to the transmission of EM
waves?
In that I get a transmission (radiation) when charges accelerate
or
change direction?
Not quite. The ~pull~ an electron feels is relatvive to its Coulomb
coupling to nearby matter. A force about 10^32 time larger than
those you feel in your automobile.
Lastly the description of EM wave as abstract photons (all energy no
mass) makes sense to me in that the EM wave propagates as an E field
(sin) + an M field (cos). Is there an actual difference in the
nature
of these two fields? Or are they indistinguishable from one another
--more like two E-Fields with a 90 degree phase shift
In units where c = 1, the E field and B field of a monochromatic EM
radiation plane wave are identical except they are in phase and are
tranverse to the direction of motion. And 90 degrees rotated from each
other. However, that is an ideal case and nature is not usually so
perfect. ;-) But can be used as a good approximation in many cases.
Yes... You are spot on. The E component diminishes by 1/r^2
and is the farfield or radiative component. The magnetic
component diminishes by 1/r^3 and is effective in the nearfield
only.
You need to qualify your 1/r^3. That would only be true for certain
types of source configuration and/or boundary conditions.
Photons are a mathmatical abstraction not a propagation model.
Your question is to the heart of why Feynman puts wrist watches
on his photons to fix all the mirrors that QM broke. In QED,
the phase information and magnetic components are transported
on the tip of a mathematicians pencil. ;-)
Photons are definitely not a math abstraction. The are basically chunks
of propagating EM energy that don't disperse due to the nature of the
relativistic quantum "vacuum" medium. They have only helicity and some
value of momentum as intrinsic properties. All the EM associated with
them comes from the quantum "vacuum".
FrediFizzx
Quantum Vacuum Charge papers;
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.pdf
or postscript
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.ps
http://www.arxiv.org/abs/physics/0601110
http://www.vacuum-physics.com
.
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| User: "Sue..." |
|
| Title: Re: Electromagnetic wave and photon spin |
19 Sep 2006 04:38:58 PM |
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FrediFizzx wrote:
In units where c = 1, the E field and B field of a monochromatic EM
radiation plane wave are identical except they are in phase and are
tranverse to the direction of motion. And 90 degrees rotated from each
other. However, that is an ideal case and nature is not usually so
perfect. ;-) But can be used as a good approximation in many cases.
Nature won't pay any attention to the units you establish
The long range components are electirc as the OP wrote.
Jackson demonstrates with an E plane only antenna.
http://arxiv.org/abs/physics/0506053
Yes... You are spot on. The E component diminishes by 1/r^2
and is the farfield or radiative component. The magnetic
component diminishes by 1/r^3 and is effective in the nearfield
only.
You need to qualify your 1/r^3. That would only be true for certain
types of source configuration and/or boundary conditions.
Having never seen a long range magnet, I don't know how
I would further quality it. The configuration that integrates
the Coulomb forces seem rather inflexible given the finite
speed of light.
http://en.wikipedia.org/wiki/Multiple_integral
Photons are a mathmatical abstraction not a propagation model.
Your question is to the heart of why Feynman puts wrist watches
on his photons to fix all the mirrors that QM broke. In QED,
the phase information and magnetic components are transported
on the tip of a mathematicians pencil. ;-)
Photons are definitely not a math abstraction. The are basically chunks
of propagating EM energy that don't disperse due to the nature of the
relativistic quantum "vacuum" medium. They have only helicity and some
value of momentum as intrinsic properties. All the EM associated with
them comes from the quantum "vacuum".
This may be some conjecture of something more funadamental
than electrons and positrons but if there is any evidence then
someone left the party without their prize.
<<Now, does not the prize to Einstein imply
that the Academy recognised the particle
nature of light? The Nobel Committee says
that Einstein had found that the energy exchange
between matter and ether occurs by atoms emitting
or absorbing a quantum of energy,hv .
As a consequence of the new concept of light quanta
(in modern terminology photons) Einstein proposed the
law that an electron emitted from a substance by
monochromatic light with the frequency has to have
a maximum energy of E=hv-p, where p is the energy needed to
remove the electron from the substance. Robert Andrews
Millikan carried out a series of measurements over a
period of 10 years, finally confirming the validity of this
law in 1916 with great accuracy. Millikan had, however,
found the idea of light quanta to be unfamiliar and strange.
The Nobel Committee avoids committing itself to the
particle concept. Light-quanta or with modern terminology,
photons, were explicitly mentioned in the reports on
which the prize decision rested only in connection with
emission and absorption processes. The Committee says
that the most important application of Einstein's photoelectric
law and also its most convincing confirmation has come from
the use Bohr made of it in his theory of atoms, which explains
a vast amount of spectroscopic data. >>
http://nobelprize.org/physics/articles/ekspong/index.html
.... and your no-dispersing ~photons~ can't take any
cell-phones into the mirror spaces to conspire about
their phase.
http://www.eso.org/projects/vlti/ ;-)
Sue...
FrediFizzx
Quantum Vacuum Charge papers;
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.pdf
or postscript
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.ps
http://www.arxiv.org/abs/physics/0601110
http://www.vacuum-physics.com
.
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| User: "FrediFizzx" |
|
| Title: Re: Electromagnetic wave and photon spin |
20 Sep 2006 01:27:35 AM |
|
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"Sue..." <suzysewnshow@yahoo.com.au> wrote in message
news:1158701938.537480.18070@k70g2000cwa.googlegroups.com...
FrediFizzx wrote:
In units where c = 1, the E field and B field of a monochromatic EM
radiation plane wave are identical except they are in phase and are
tranverse to the direction of motion. And 90 degrees rotated from
each
other. However, that is an ideal case and nature is not usually so
perfect. ;-) But can be used as a good approximation in many cases.
Nature won't pay any attention to the units you establish
Well, maybe nature doesn't be we sure can "pay attention". ;-)
The long range components are electirc as the OP wrote.
I don't know what that means. You mean *only* electric? Sorry, that is
flat out wrong if so.
Jackson demonstrates with an E plane only antenna.
http://arxiv.org/abs/physics/0506053
Yes... You are spot on. The E component diminishes by 1/r^2
and is the farfield or radiative component. The magnetic
component diminishes by 1/r^3 and is effective in the nearfield
only.
You need to qualify your 1/r^3. That would only be true for certain
types of source configuration and/or boundary conditions.
Having never seen a long range magnet, I don't know how
I would further quality it. The configuration that integrates
the Coulomb forces seem rather inflexible given the finite
speed of light.
Read Timo's reply to rds in this thread. He has it exactly right.
Magnets are dipoles. The B field of EM radiation doesn't necessarilly
come from a magnet.
http://en.wikipedia.org/wiki/Multiple_integral
Photons are a mathmatical abstraction not a propagation model.
Your question is to the heart of why Feynman puts wrist watches
on his photons to fix all the mirrors that QM broke. In QED,
the phase information and magnetic components are transported
on the tip of a mathematicians pencil. ;-)
Photons are definitely not a math abstraction. The are basically
chunks
of propagating EM energy that don't disperse due to the nature of
the
relativistic quantum "vacuum" medium. They have only helicity and
some
value of momentum as intrinsic properties. All the EM associated
with
them comes from the quantum "vacuum".
This may be some conjecture of something more funadamental
than electrons and positrons but if there is any evidence then
someone left the party without their prize.
Seems to be just plain common sense to me. The evidence is as plain as
just opening your eyes. ;-)
<<Now, does not the prize to Einstein imply
that the Academy recognised the particle
nature of light? The Nobel Committee says
that Einstein had found that the energy exchange
between matter and ether occurs by atoms emitting
or absorbing a quantum of energy,hv .
As a consequence of the new concept of light quanta
(in modern terminology photons) Einstein proposed the
law that an electron emitted from a substance by
monochromatic light with the frequency has to have
a maximum energy of E=hv-p, where p is the energy needed to
remove the electron from the substance. Robert Andrews
Millikan carried out a series of measurements over a
period of 10 years, finally confirming the validity of this
law in 1916 with great accuracy. Millikan had, however,
found the idea of light quanta to be unfamiliar and strange.
The Nobel Committee avoids committing itself to the
particle concept. Light-quanta or with modern terminology,
photons, were explicitly mentioned in the reports on
which the prize decision rested only in connection with
emission and absorption processes. The Committee says
that the most important application of Einstein's photoelectric
law and also its most convincing confirmation has come from
the use Bohr made of it in his theory of atoms, which explains
a vast amount of spectroscopic data. >>
http://nobelprize.org/physics/articles/ekspong/index.html
... and your no-dispersing ~photons~ can't take any
cell-phones into the mirror spaces to conspire about
their phase.
This is supposed to mean something?
http://www.eso.org/projects/vlti/ ;-)
Wrong direction to the mirror spacetimes.
"Higgs-field Portal into Hidden Sectors"
http://www.arxiv.org/abs/hep-ph/0605188
FrediFizzx
Quantum Vacuum Charge papers;
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.pdf
or postscript
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.ps
http://www.arxiv.org/abs/physics/0601110
http://www.vacuum-physics.com
.
|
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| User: "Sue..." |
|
| Title: Re: Electromagnetic wave and photon spin |
20 Sep 2006 02:20:26 PM |
|
|
FrediFizzx wrote:
"Sue..." <suzysewnshow@yahoo.com.au> wrote in message
news:1158701938.537480.18070@k70g2000cwa.googlegroups.com...
FrediFizzx wrote:
In units where c = 1, the E field and B field of a monochromatic EM
radiation plane wave are identical except they are in phase and are
tranverse to the direction of motion. And 90 degrees rotated from
each
other. However, that is an ideal case and nature is not usually so
perfect. ;-) But can be used as a good approximation in many cases.
Nature won't pay any attention to the units you establish
Well, maybe nature doesn't be we sure can "pay attention". ;-)
The long range components are electirc as the OP wrote.
I don't know what that means. You mean *only* electric? Sorry, that is
flat out wrong if so.
Correct. *Only* electric. At visible light frequencies gas
molecules are effective coupling structures so it is their
nearfield that interacts to give farfield light some magnetic
properties like Faraday rotatation. Perhaps that has you
confused. Emitters and detectors of visible light are usually
atoms, so that adds to the illusion of a particle being
propagated.
Jackson demonstrates with an E plane only antenna.
http://arxiv.org/abs/physics/0506053
Yes... You are spot on. The E component diminishes by 1/r^2
and is the farfield or radiative component. The magnetic
component diminishes by 1/r^3 and is effective in the nearfield
only.
You need to qualify your 1/r^3. That would only be true for certain
types of source configuration and/or boundary conditions.
Having never seen a long range magnet, I don't know how
I would further quality it. The configuration that integrates
the Coulomb forces seem rather inflexible given the finite
speed of light.
Read Timo's reply to rds in this thread. He has it exactly right.
Magnets are dipoles. The B field of EM radiation doesn't necessarilly
come from a magnet.
I did read his reply and didn't find anything wrong with it.
It is doesn't matter the what the magnetism comes from;
Ancent aodestones, MRI magnets or fiber optic couplers.
The geometry to superposition one charge from two
positions establishes the 1/r^3 attenuation, not the
material used in the structures.
http://en.wikipedia.org/wiki/Multiple_integral
Photons are a mathmatical abstraction not a propagation model.
Your question is to the heart of why Feynman puts wrist watches
on his photons to fix all the mirrors that QM broke. In QED,
the phase information and magnetic components are transported
on the tip of a mathematicians pencil. ;-)
Photons are definitely not a math abstraction. The are basically
chunks
of propagating EM energy that don't disperse due to the nature of
the
relativistic quantum "vacuum" medium. They have only helicity and
some
value of momentum as intrinsic properties. All the EM associated
with
them comes from the quantum "vacuum".
This may be some conjecture of something more funadamental
than electrons and positrons but if there is any evidence then
someone left the party without their prize.
Seems to be just plain common sense to me.
The evidence is as plain as
just opening your eyes. ;-)
My eyes tend to fall on URL's that have an ec edu org or gov in
in them. You wouldn't have a dog in this fight would you? ;-)
<<Now, does not the prize to Einstein imply
that the Academy recognised the particle
nature of light? The Nobel Committee says
that Einstein had found that the energy exchange
between matter and ether occurs by atoms emitting
or absorbing a quantum of energy,hv .
As a consequence of the new concept of light quanta
(in modern terminology photons) Einstein proposed the
law that an electron emitted from a substance by
monochromatic light with the frequency has to have
a maximum energy of E=hv-p, where p is the energy needed to
remove the electron from the substance. Robert Andrews
Millikan carried out a series of measurements over a
period of 10 years, finally confirming the validity of this
law in 1916 with great accuracy. Millikan had, however,
found the idea of light quanta to be unfamiliar and strange.
The Nobel Committee avoids committing itself to the
particle concept. Light-quanta or with modern terminology,
photons, were explicitly mentioned in the reports on
which the prize decision rested only in connection with
emission and absorption processes. The Committee says
that the most important application of Einstein's photoelectric
law and also its most convincing confirmation has come from
the use Bohr made of it in his theory of atoms, which explains
a vast amount of spectroscopic data. >>
http://nobelprize.org/physics/articles/ekspong/index.html
... and your no-dispersing ~photons~ can't take any
cell-phones into the mirror spaces to conspire about
their phase.
This is supposed to mean something?
They rock that array on tracks and watch the
emission of a *single* atom interfer with itself
after splitting four ways. You said photons are
indivisible so they must be catching something
else.... probably pigeon poop. ;-)
http://www.eso.org/projects/vlti/ ;-)
<< Wrong direction to the mirror spacetimes.>>
Email them tell 'em to read the instructions
the next time they try to put something like that
together. :o)
"Higgs-field Portal into Hidden Sectors"
http://www.arxiv.org/abs/hep-ph/0605188
I'll enjoy reading that but 10 layers of abstraction
about conjectural particles isn't going to disprove what
we can demonstrate on any kitchen table.
Sue...
FrediFizzx
Quantum Vacuum Charge papers;
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.pdf
or postscript
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.ps
http://www.arxiv.org/abs/physics/0601110
http://www.vacuum-physics.com
.
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| User: "Autymn D. C." |
|
| Title: Re: Electromagnetic wave and photon spin |
20 Sep 2006 04:52:28 PM |
|
|
Sue... wrote:
Where described as a moving charge
The electric field of the moving charges is interacting with other
matter which is ~stationary~ . At the minimum it is matter
in 'all of space' or 'free space'
permeability mu_0
permittivity eps_0
radiation resistance 377 ohms.
|377O = V/I = kqt/qs = k/c| = 4pc)7 = yup
However in deep space I expect that c_(+) < c_( ) < c_(H). Dammit,
there used to be something about energy density and a c as a few m/s
greater on <http://scienceworld.wolfram.com/physics/SpeedofLight.html>
but it disappeard. The site shuts out the Wayback Machine too.
-Aut
.
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| User: "Sue..." |
|
| Title: Re: Electromagnetic wave and photon spin |
20 Sep 2006 07:19:36 PM |
|
|
Autymn D. C. wrote:
Sue... wrote:
Where described as a moving charge
The electric field of the moving charges is interacting with other
matter which is ~stationary~ . At the minimum it is matter
in 'all of space' or 'free space'
permeability mu_0
permittivity eps_0
radiation resistance 377 ohms.
|377O = V/I = kqt/qs = k/c| = 4pc)7 = yup
However in deep space I expect that c_(+) < c_( ) < c_(H). Dammit,
there used to be something about energy density and a c as a few m/s
greater on <http://scienceworld.wolfram.com/physics/SpeedofLight.html>
but it disappeard. The site shuts out the Wayback Machine too.
Use the Tardis when that happens.
http://scienceworld.wolfram.com/physics/D.html
http://scienceworld.wolfram.com/physics/H.html
cgs -->> MKS
http://scienceworld.wolfram.com/physics/ElectricSusceptibility.html
<< The interstellar gas consists partly of neutral atoms
and molecules, as well as charged particles, such as
ions and electrons. This gas is extremely dilute, with an
average density of about 1 atom per cubic centimeter.
(For comparison, the air we breathe has a density of
approximately 30,000,000,000,000,000,000 molecules
per cubic centimeter.) >>
http://www-ssg.sr.unh.edu/ism/what1.html
Yer falling down on the job. :-(
I've miss-spelt at least 30 words this weak sew far
and I can't tell you've corrected even a forth of them.
Sigh... If you want something done write, it seams
ewe awl ways have too due it your self.
Sue...
-Aut
.
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| User: "RP" |
|
| Title: Re: Electromagnetic wave and photon spin |
19 Sep 2006 06:15:23 PM |
|
|
rds wrote:
First off I want to say that I am not physics major.
Can someone explain to me the details of how electromagnetic radiation?
Actually occurs. I know the pat answers that the acceleration of
charged particles or change in direction, but that doesn't really
explain the process.
For that matter why does a magnetic field form when current flows at a
constant rate (charge velocity)?
I notice in my car from 0 to 50 MPH , that when I accelerate I feel a
pull in the reverse direction, but when I reach the intended speed the
pull subsides. In some way this is counter intuitive in that I would
assume that the 50 MPH pull would continually cause a drag backwards.
In the same manner is this a parallel to the transmission of EM waves?
In that I get a transmission (radiation) when charges accelerate or
change direction?
Lastly the description of EM wave as abstract photons (all energy no
mass) makes sense to me in that the EM wave propagates as an E field
(sin) + an M field (cos). Is there an actual difference in the nature
of these two fields? Or are they indistinguishable from one another
--more like two E-Fields with a 90 degree phase shift
If the latter is true, and we take any point on the EM wave front an
examine it as it travels toward us, we could easily describe the EM
field as a mass less polarized point in space which rotates like a
plate
spinning with the top of the plate being positive and the bottom of the
plate
being negative. The spinning at the rate of the transmission frequency.
This totally describes the EM field without even referring to a
magnetic
field. As the plate rotates its positive to the 3 and 9 o'clock
positions it
represents the M field peaks, and at the 12 and 6 o'clock position it
represents the E field peaks. I would normally call this point a photon
and the rotation I would call spin. But I read that photon spin is
actually in the direction of travel, and so the spin must be something
different. Any comments clarrifications is welcome.
Thanks.....
There is nothing spinning in the frame of the wave. In that frame you
would experience a constant B field and little to no E field at all,
unless you are located within the near field, in which case you would
experience strong inductive effects. This is why an rf detector
shouldn't be used in the near vicinity of an antenna. You will get a
false reading of the output energy there.
But if at rest wrt the source and located in the far field, then the B
field is then changing in your frame, and an E field is generated
within you by the changing B field. This is a qvB field, which is
dimensionally equivalent to an E field. Thus the E component doesn't
exist in the far field until a real charge is affected by the passing B
wave. The generated E field is rotated 90deg wrt B, which has nothing
to do with waves per se, it's simply the relationship of E and B
anytime that a B field generates a current. Could be a conductor moving
near a magnet, same result.
This is the way I see it, but then on a deeper level there are neither
B nor E fields. I won't pain you with the lengthy explanation in terms
of that more fundamental field. Not this time. Everyone, say "whewww!"
:)
Richard Perry
.
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| User: "rds" |
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| Title: Re: Electromagnetic wave and photon spin |
20 Sep 2006 08:27:45 AM |
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|
Mr. Perry,
I am interested to hearing your theory/ideas. I am really trying to get
a firmer grip on EM and particle and fields in general. Your latest
text was thought out. Please feel free to expound.
Rick
RP wrote:
rds wrote:
First off I want to say that I am not physics major.
Can someone explain to me the details of how electromagnetic radiation?
Actually occurs. I know the pat answers that the acceleration of
charged particles or change in direction, but that doesn't really
explain the process.
For that matter why does a magnetic field form when current flows at a
constant rate (charge velocity)?
I notice in my car from 0 to 50 MPH , that when I accelerate I feel a
pull in the reverse direction, but when I reach the intended speed the
pull subsides. In some way this is counter intuitive in that I would
assume that the 50 MPH pull would continually cause a drag backwards.
In the same manner is this a parallel to the transmission of EM waves?
In that I get a transmission (radiation) when charges accelerate or
change direction?
Lastly the description of EM wave as abstract photons (all energy no
mass) makes sense to me in that the EM wave propagates as an E field
(sin) + an M field (cos). Is there an actual difference in the nature
of these two fields? Or are they indistinguishable from one another
--more like two E-Fields with a 90 degree phase shift
If the latter is true, and we take any point on the EM wave front an
examine it as it travels toward us, we could easily describe the EM
field as a mass less polarized point in space which rotates like a
plate
spinning with the top of the plate being positive and the bottom of the
plate
being negative. The spinning at the rate of the transmission frequency.
This totally describes the EM field without even referring to a
magnetic
field. As the plate rotates its positive to the 3 and 9 o'clock
positions it
represents the M field peaks, and at the 12 and 6 o'clock position it
represents the E field peaks. I would normally call this point a photon
and the rotation I would call spin. But I read that photon spin is
actually in the direction of travel, and so the spin must be something
different. Any comments clarrifications is welcome.
Thanks.....
There is nothing spinning in the frame of the wave. In that frame you
would experience a constant B field and little to no E field at all,
unless you are located within the near field, in which case you would
experience strong inductive effects. This is why an rf detector
shouldn't be used in the near vicinity of an antenna. You will get a
false reading of the output energy there.
But if at rest wrt the source and located in the far field, then the B
field is then changing in your frame, and an E field is generated
within you by the changing B field. This is a qvB field, which is
dimensionally equivalent to an E field. Thus the E component doesn't
exist in the far field until a real charge is affected by the passing B
wave. The generated E field is rotated 90deg wrt B, which has nothing
to do with waves per se, it's simply the relationship of E and B
anytime that a B field generates a current. Could be a conductor moving
near a magnet, same result.
This is the way I see it, but then on a deeper level there are neither
B nor E fields. I won't pain you with the lengthy explanation in terms
of that more fundamental field. Not this time. Everyone, say "whewww!"
:)
Richard Perry
.
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|
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| User: "RP" |
|
| Title: Re: Electromagnetic wave and photon spin |
20 Sep 2006 12:06:35 PM |
|
|
rds wrote:
Mr. Perry,
I am interested to hearing your theory/ideas. I am really trying to get
a firmer grip on EM and particle and fields in general. Your latest
text was thought out. Please feel free to expound.
Rick
I didn't mean to mislead you in the present argument, I was only making
light of the previous replies that stated that the far field was purely
an E field. In reality, it can be interpreted to be either, or even
both, it depends entirly upon your frame of reference and upon your
philosphical interpretation. Does a changing B field cause an E field,
or is it the emerging E field that destroys the B field? Or is it
rather that these changes aren't directly linked to each other in the
sense that one causes the other, but rather that they simply occur
simultaneously? There are many heavy questions involved, and if you
immerse yourself in the abstract mathematical formalism you will get
the correct forces, but you won't necessarily be able to formulate an
absolute physical model corresponding to that math. When the changes
in two values occur simultaneously, then it is difficult to imagine
that the one caused the other. It would seem rather that both are
caused by yet something else that preceeded them both.
Faradays force equation only relates a changing B field to a changing E
field, it doesn't attempt to qualify the events, only to quantify them.
If we break the E field down, it is simply an expression of the force
that will be exerted on some charge, which occurs for apparently no
reason at all other than that the charge exists within that field.
Thus, we could interpret the E field to exist anytime a charge
experiences accelleration. However, if we are allowed to do this, then
why would we still need the B field? Couldn't we just nterpret the B
field to be an E field that only acts perpendicular to the line of
motion of the charge? Sure we could. And in fact this is how Purcell
quantifies the B field, i.e. in terms of relativistically distorted E
fields. Thus from this perspective one would have to regard Timo's
statement to be correct, the em wave consists only of an E field.
But conversely, the E field of Purcell's can be interpreted as a
relativistic distortion of B fields. E field can be found to have a B
component when transforming to a different inertial frame. This is the
classical means to unification of these two fields, namely that they
are just different perspectives of the singular em field. The
magnitudes of E and B for any given em field are entirely frame
dependent. It is for this reason that conclude that these fields are
not fundamental, but are instead macroscopic effects generated by
something more fundamental. Nothing that is frame dependent can be
anything more than a perspective of something that isn't frame
dependent, because frames of reference have no physical influence on
nature.
You could always go with QED, but it too suffers from many
philosophical flaws, but it is still a hell of a lot better description
of these events than the E and B field interpretation.
I don't have time at the moment to get into a deep discussion about
bare electrons and thier fields. Maybe later. Thanks for your interest
though.
Richard Perry
.
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| User: "PD" |
|
| Title: Re: Electromagnetic wave and photon spin |
20 Sep 2006 11:41:20 AM |
|
|
rds wrote:
First off I want to say that I am not physics major.
Can someone explain to me the details of how electromagnetic radiation?
Actually occurs. I know the pat answers that the acceleration of
charged particles or change in direction, but that doesn't really
explain the process.
Details? Without invoking a lot of math? That gets tricky. Plain words
only carry you so far before the details get muddy and can be
confusing.
For that matter why does a magnetic field form when current flows at a
constant rate (charge velocity)?
Well, the way this is usually answered is the other way around. Two
currents in proximity to each other exert a force on each other, and
the question is why? A current causes a deflection in a nearby compass
needle, and the question is why? The response to this is that it is
caused by the presence of a field, but not one that behaves like the
electric field. And so we presume that there is another field that has
current, not just charge, as the source.
I notice in my car from 0 to 50 MPH , that when I accelerate I feel a
pull in the reverse direction, but when I reach the intended speed the
pull subsides. In some way this is counter intuitive in that I would
assume that the 50 MPH pull would continually cause a drag backwards.
This observation is central to Newtonian mechanics. What you feel as a
pull backwards is an illusion. What's really going on is that the seat
back is pushing harder *forward* than what you're used to. Also, some
of the blood in your head gets left behind because your solid tissue
gets accelerated by the seat back, while the blood isn't accelerated as
readily, and you mistake the *lack* of an effective push forward on
your blood to be a *pull* backwards. The key thing is that when you get
to a steady 50 mph, all that stops, there is no more acceleration, and
aside from a little vibration you can detect no difference between
moving at 50 mph and being stationary at 0 mph. This is why you can
fall asleep on a plane but wake up the moment the pilot throttles down
for descent. In fact, there is no physical distinction between 50 mph
and 0 mph -- to the point where you actually have to be careful what
you mean when you say 0 mph. 0 mph with respect to WHAT? For example,
are you aware of the 66,000 mph that you are traveling around the sun?
It's fortuitous that you brought this up in the context of EM waves,
because this utter equivalence between the physics at 50 mph and at 0
mph, is precisely what led to the direct jump from the laws of
electromagnetism to special relativity.
In the same manner is this a parallel to the transmission of EM waves?
In that I get a transmission (radiation) when charges accelerate or
change direction?
Lastly the description of EM wave as abstract photons (all energy no
mass) makes sense to me in that the EM wave propagates as an E field
(sin) + an M field (cos).
Both sin or both cos, actually. The energy is stored in the electric
and magnetic fields, and the point is that the fields won't propagate
unless there is both. The surprise is that the energy stored in the
field is not distributed smoothly and continuously through space. It is
stored in *chunks*, and these chunks that deliver their energy and
momentum all at once when they interact with something are called
photons.
The fact that both E and M fields are required for propagation is what
comes from understanding what the mathematics of Maxwell's equations
mean. At this point, words would become sloppy.
Is there an actual difference in the nature
of these two fields? Or are they indistinguishable from one another
--more like two E-Fields with a 90 degree phase shift
If the latter is true, and we take any point on the EM wave front an
examine it as it travels toward us, we could easily describe the EM
field as a mass less polarized point in space which rotates like a
plate
spinning with the top of the plate being positive and the bottom of the
plate
being negative. The spinning at the rate of the transmission frequency.
This totally describes the EM field without even referring to a
magnetic
field. As the plate rotates its positive to the 3 and 9 o'clock
positions it
represents the M field peaks, and at the 12 and 6 o'clock position it
represents the E field peaks. I would normally call this point a photon
and the rotation I would call spin. But I read that photon spin is
actually in the direction of travel, and so the spin must be something
different. Any comments clarrifications is welcome.
Thanks.....
.
|
|
|
| User: "Sorcerer" |
|
| Title: Re: Electromagnetic wave and photon spin |
20 Sep 2006 12:01:42 PM |
|
|
"PD" <TheDraperFamily@gmail.com> wrote in message
news:1158770480.521993.261410@i42g2000cwa.googlegroups.com...
|
| rds wrote:
| > First off I want to say that I am not physics major.
| >
| > Can someone explain to me the details of how electromagnetic radiation?
| > Actually occurs. I know the pat answers that the acceleration of
| > charged particles or change in direction, but that doesn't really
| > explain the process.
|
| Details? Without invoking a lot of math? That gets tricky. Plain words
| only carry you so far before the details get muddy and can be
| confusing.
|
| >
| > For that matter why does a magnetic field form when current flows at a
| > constant rate (charge velocity)?
|
| Well, the way this is usually answered is the other way around. Two
| currents in proximity to each other exert a force on each other, and
| the question is why? A current causes a deflection in a nearby compass
| needle, and the question is why?
HAHAHA!
http://www.androcles01.pwp.blueyonder.co.uk/spin.gif
A power station causes a deflection in a DISTANT compass
needle, and the answer is copper wire. It makes distant tunsten wire
connected to the copper wire glow, too.
Androcles
The response to this is that it is
| caused by the presence of a field, but not one that behaves like the
| electric field. And so we presume that there is another field that has
| current, not just charge, as the source.
|
| >
| > I notice in my car from 0 to 50 MPH , that when I accelerate I feel a
| > pull in the reverse direction, but when I reach the intended speed the
| > pull subsides. In some way this is counter intuitive in that I would
| > assume that the 50 MPH pull would continually cause a drag backwards.
|
| This observation is central to Newtonian mechanics. What you feel as a
| pull backwards is an illusion. What's really going on is that the seat
| back is pushing harder *forward* than what you're used to. Also, some
| of the blood in your head gets left behind because your solid tissue
| gets accelerated by the seat back, while the blood isn't accelerated as
| readily, and you mistake the *lack* of an effective push forward on
| your blood to be a *pull* backwards. The key thing is that when you get
| to a steady 50 mph, all that stops, there is no more acceleration, and
| aside from a little vibration you can detect no difference between
| moving at 50 mph and being stationary at 0 mph. This is why you can
| fall asleep on a plane but wake up the moment the pilot throttles down
| for descent. In fact, there is no physical distinction between 50 mph
| and 0 mph -- to the point where you actually have to be careful what
| you mean when you say 0 mph. 0 mph with respect to WHAT? For example,
| are you aware of the 66,000 mph that you are traveling around the sun?
|
| It's fortuitous that you brought this up in the context of EM waves,
| because this utter equivalence between the physics at 50 mph and at 0
| mph, is precisely what led to the direct jump from the laws of
| electromagnetism to special relativity.
|
| > In the same manner is this a parallel to the transmission of EM waves?
| > In that I get a transmission (radiation) when charges accelerate or
| > change direction?
| >
| > Lastly the description of EM wave as abstract photons (all energy no
| > mass) makes sense to me in that the EM wave propagates as an E field
| > (sin) + an M field (cos).
|
| Both sin or both cos, actually. The energy is stored in the electric
| and magnetic fields, and the point is that the fields won't propagate
| unless there is both. The surprise is that the energy stored in the
| field is not distributed smoothly and continuously through space. It is
| stored in *chunks*, and these chunks that deliver their energy and
| momentum all at once when they interact with something are called
| photons.
|
| The fact that both E and M fields are required for propagation is what
| comes from understanding what the mathematics of Maxwell's equations
| mean. At this point, words would become sloppy.
|
| > Is there an actual difference in the nature
| > of these two fields? Or are they indistinguishable from one another
| > --more like two E-Fields with a 90 degree phase shift
| >
| > If the latter is true, and we take any point on the EM wave front an
| > examine it as it travels toward us, we could easily describe the EM
| > field as a mass less polarized point in space which rotates like a
| > plate
| > spinning with the top of the plate being positive and the bottom of the
| > plate
| > being negative. The spinning at the rate of the transmission frequency.
| > This totally describes the EM field without even referring to a
| > magnetic
| > field. As the plate rotates its positive to the 3 and 9 o'clock
| > positions it
| > represents the M field peaks, and at the 12 and 6 o'clock position it
| > represents the E field peaks. I would normally call this point a photon
| > and the rotation I would call spin. But I read that photon spin is
| > actually in the direction of travel, and so the spin must be something
| > different. Any comments clarrifications is welcome.
| >
| > Thanks.....
|
.
|
|
|
|
|
|
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|
|
