| Topic: |
Science > Physics |
| User: |
"" |
| Date: |
19 Jun 2005 01:36:17 AM |
| Object: |
EM is an accelerating H field ?? |
Old Man wrote:
curiousjohn4 wrote:
curiousjohn4 is envious, and wants to learn how to do what
Old Man does.
After learning, maybe curiousjohn4 will tell Old Man what the
field strength amplitude (either E or B, since E = c*B) of a
single photon is.
Not the field for 1 photon / m^3, mind you.
You can get that from the equations above. For one photon.
Obviously I can't satisfy your math and QM level. If you want I can
only provide you with an actual number of the E or B field as reported
by NEC2. Although upon thinking of your comment I decided to test one
of my old theories.
Your mention of E = c*B reminded me of my an old theory I had, which
seems to be old hat to Old Man. This theory presumed the far E field
was equal to c * B. That is, I envisioned the magnetic field
traversing at the speed of light. Using the right hand rule I
perceived the electric field equal to B traveling at c in the far
field. So by the right hand rule the electric field would be
perpendicular to both B and the path that the field is traveling. I
wanted to know if the actual real life E field matched calculated E
field by means of c*B. Today I just verified this. This is probably
no big deal in science community, but here's the data via means of NEC2
->
Measured far field at 15000 meters
Frequency = 10MHz
According to NEC, the E field at 15000 meters is 1.6959E-13 V/m and H =
4.5017E-16 A/m
B = H * u0
u0 = 4E-7 * PI Wb * A^-1 * m^-1
B = 4.5017E-16 A/m * 4E-7 * PI Wb * A^-1 * m^-1
B = 5.6570E-022 kg*s^-2*A^-1
E = c * B = 1.6959E-013 kg*m*s^-3*A^-1 = 1.6959E-013 V/m
End result ->
Real life E as reported by NEC = 1.6959E-13 V/m
Calculated E from B*c = 1.6959E-013 V/m
Notice that the two values are an exact match. I believe this means
that the far field at 15000 meters away from the antenna is traveling
at the speed of light. In other words, multiplying the B far field
times c is equal to the electric field. So then I wondered if there
was a transition when the magnetic field accelerated from 0 m/s to c.
To test this theory I ran several measurements at various distances
from the antenna.
Measured far field at 150 meters
Frequency = 10MHz
According to NEC, the E field at 150 meters is 1.6855E-11 V/m and H =
4.4797E-14 A/m
B = H * u0
u0 = 4E-7 * PI Wb * A^-1 * m^-1
B = 4.4797E-14 A/m * 4E-7 * PI Wb * A^-1 * m^-1
B = 5.6294E-020 kg*s^-2*A^-1
E = c * B = 1.6876E-011 kg*m*s^-3*A^-1 = 1.6876E-011 V/m
End result ->
Real life E as reported by NEC = 1.6855E-11 V/m
Calculated E from B*c = 1.6876E-011 V/m
There is a slight difference of 99.88%. This could mean that the
magnetic field at 150 meters away from the antenna is traveling 99.88%
the speed of light.
Measured far field at 15 meters
Frequency = 10MHz
According to NEC, the E field at 15 meters is 1.4610E-10 V/m and H =
4.2932E-13 A/m
B = H * u0
u0 = 4E-7 * PI Wb * A^-1 * m^-1
B = 4.2932E-13 A/m * 4E-7 * PI Wb * A^-1 * m^-1
B = 5.3950E-019 kg*s^-2*A^-1
E = c * B = 1.6174E-010 kg*m*s^-3*A^-1 = 1.6174E-010 V/m
End result ->
Real life E as reported by NEC = 1.4610E-10 V/m
Calculated E from B*c = 1.6174E-010 V/m
At 15 meters there's a 90.33% difference.
Measured far field at 1 meter
Frequency = 10MHz
According to NEC, the E field at 1 meter is 3.6134E-10 V/m and H =
3.9229E-12 A/m
B = H * u0
u0 = 4E-7 * PI Wb * A^-1 * m^-1
B = 3.9229E-12 A/m * 4E-7 * PI Wb * A^-1 * m^-1
B = 4.9297E-018 kg*s^-2*A^-1
E = c * B = 1.4779E-009 kg*m*s^-3*A^-1 = 1.4779E-009 V/m
End result ->
Real life E as reported by NEC = 3.6134E-10 V/m
Calculated E from B*c = 1.4779E-009 V/m
At 1 meter there's a 24.45% difference. I believe this means the
magnetic field is traveling at 24.45% of c at 1 meter away from the
antenna.
With the help of NEC2, I am able to conclude a theory about EM, or at
least one aspect of EM. That is, an accelerating charge transfers
energy to an EM wave by means of accelerating magnetic field. The
magnetic field begins at a zero or near zero velocity and eventually
reaches the speed of light. I suspect that the magnetic field never
reaches the exact speed of light unless you consider an EM wave at
infinite distance from the originating source.
.
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| User: "operator jay" |
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| Title: Re: EM is an accelerating H field ?? |
19 Jun 2005 09:41:07 AM |
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<curiousjohn4@yahoo.com> wrote in message
news:1119162976.992816.171230@g14g2000cwa.googlegroups.com...
Your mention of E = c*B reminded me of my an old theory I had, which
seems to be old hat to Old Man. This theory presumed the far E field
was equal to c * B. That is, I envisioned the magnetic field
traversing at the speed of light. Using the right hand rule I
perceived the electric field equal to B traveling at c in the far
field. So by the right hand rule the electric field would be
perpendicular to both B and the path that the field is traveling. I
wanted to know if the actual real life E field matched calculated E
field by means of c*B. Today I just verified this. This is probably
no big deal in science community, but here's the data via means of NEC2
->
With the help of NEC2, I am able to conclude a theory about EM, or at
least one aspect of EM. That is, an accelerating charge transfers
energy to an EM wave by means of accelerating magnetic field. The
magnetic field begins at a zero or near zero velocity and eventually
reaches the speed of light. I suspect that the magnetic field never
reaches the exact speed of light unless you consider an EM wave at
infinite distance from the originating source.
Just a thought. When you are near to the source of your em waves, is the E
said to have an 'inductive' component in addition to the 'radiation'
component? Could this cause a difference between E and c*B?
j
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| User: "" |
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| Title: Re: EM is an accelerating H field ?? |
19 Jun 2005 11:50:04 AM |
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operator jay wrote:
Just a thought. When you are near to the source of your em waves, is the E
said to have an 'inductive' component in addition to the 'radiation'
component? Could this cause a difference between E and c*B?
You're probably correct. Presently I am nearly certain it is a crazy
thought to think the magnetic field accelerating. Actually that was
just an after thought. Just wanted to confirm to myself that E = c*B
in the far field. I mean, how in the world can we devise an equation
to apply energy to a moving H field. Energy is simply H * m^3 * u.
Where u is the permeability. For free space it's u0 = 4E-7 * PI Wb / H
One last part that still catches my attention is how can the near
inductive field extend past one wavelength? The data shows that E
never reaches c*B in a finite distance. Close but no cigar as they
say.
Another aspect of EM that I can't envision in real macro world is the
particle behavior. In other words, as the wavelength decreases then
the wave & photon behave more like a particle when absorbed. Consider
a gamma ray, which has substantial longitudinal force on when absorbed.
Whereas the effects of radio wave on absorption is mostly a traverse
force. Technically I wonder where what the mechanism is in the EM to
cause this longitudinal force.
Yes, I feel embarrassed that in my excitement I overlooked the obvious.
Like you suggested, the excess B is probably entirely accountable from
the near field-- the inductive component. I'll go hide under my rock
now.
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| User: "Old Man" |
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| Title: Re: EM is an accelerating H field ?? |
19 Jun 2005 03:15:36 PM |
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<curiousjohn4@yahoo.com> wrote in message
news:1119199803.976503.15640@g43g2000cwa.googlegroups.com...
operator jay wrote:
Just a thought. When you are near to the source of your em waves, is the
E
said to have an 'inductive' component in addition to the 'radiation'
component? Could this cause a difference between E and c*B?
You're probably correct. Presently I am nearly certain it is a crazy
thought to think the magnetic field accelerating. Actually that was
just an after thought. Just wanted to confirm to myself that E = c*B
in the far field. I mean, how in the world can we devise an equation
to apply energy to a moving H field. Energy is simply H * m^3 * u.
Where u is the permeability. For free space it's u0 = 4E-7 * PI Wb / H
S = ( c / 4* pi) E x B (Gaussian units)
[Energy Flux] = c * [Energy Density]
[Old Man]
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| User: "Old Man" |
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| Title: Re: EM is an accelerating H field ?? |
19 Jun 2005 03:05:41 PM |
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<curiousjohn4@yahoo.com> wrote in message
news:1119162976.992816.171230@g14g2000cwa.googlegroups.com...
.... Measured far field at 150 meters
Frequency = 10MHz
According to NEC, the E field at 150 meters is 1.6855E-11 V/m and H =
4.4797E-14 A/m
B = H * u0
u0 = 4E-7 * PI Wb * A^-1 * m^-1
B = 4.4797E-14 A/m * 4E-7 * PI Wb * A^-1 * m^-1
B = 5.6294E-020 kg*s^-2*A^-1
E = c * B = 1.6876E-011 kg*m*s^-3*A^-1 = 1.6876E-011 V/m
End result ->
Real life E as reported by NEC = 1.6855E-11 V/m
Calculated E from B*c = 1.6876E-011 V/m
There is a slight difference of 99.88%. This could mean that the
magnetic field at 150 meters away from the antenna is traveling
99.88% the speed of light.
No. The field changes travle at c. As the source distance
increases, the fields asymptotically approach those of a
pure EM wave whereof |E| = c * |B|.
At any distance, the electric field is the sum of two fields. The
first is the "velocity (static) field", which declines as e / R^2 and
points in the radial direction. The second is the "acceleration
(dynamic) field", which declines as e / R and is transverse to the
radial direction.
As source distance, R, increases, the total field asymptotically
approaches that of the acceleration (EM wave) field for which
|E| = c * |B| and for which E and B are transverse to R. The
radiation field is bipolar, sin(theta)^2, about the source. The
diminishing static field is spherically symmetric about the source
There's no need for curiousjohn4's cockeyed "theory". It's all
laid out rather neatly in Jackson's, "Classical Electrodynamics"
Learn some physics. Read it.
[Old Man]
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| User: "" |
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| Title: Re: EM is an accelerating H field ?? |
19 Jun 2005 07:46:22 PM |
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Old Man wrote:
At any distance, the electric field is the sum of two fields. The
first is the "velocity (static) field", which declines as e / R^2 and
points in the radial direction.
Yes, but I believe the electrostatic field term falls off as 1/r^3. I
often refer to it as electric waves. Were you thinking about the
induction term, which I believe is 1/r^2? Also I believe the
electrostatic term is not merely longitudinal but a combination of
traverse and longitudinal.
Not that it makes that much difference, but the E field measurements I
took were only traverse to the wave-- on the same axis of the dipole.
I would think that at far distances, relative to wavelength, the e-term
would become insignificant as if falls of as 1/r^3.
The second is the "acceleration
(dynamic) field", which declines as e / R and is transverse to the
radial direction.
.... the far field.
As source distance, R, increases, the total field asymptotically
approaches that of the acceleration (EM wave) field for which
|E| = c * |B| and for which E and B are transverse to R. The
radiation field is bipolar, sin(theta)^2, about the source. The
diminishing static field is spherically symmetric about the source
I don't know. It's so complex to simulate in the mind. Best left to a
computer, or best yet, I would like to measure the exact speed of an EM
pulse at near distances relative to wavelength. I am fairly certain
that it will be equal to c, which will make you correct as usual Old
Man.
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| User: "Old Man" |
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| Title: Re: EM is an accelerating H field ?? |
20 Jun 2005 07:50:11 PM |
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<curiousjohn4@yahoo.com> wrote in message
news:1119228382.556704.64260@z14g2000cwz.googlegroups.com...
Old Man wrote:
At any distance, the electric field is the sum of two fields. The
first is the "velocity (static) field", which declines as e / R^2 and
points in the radial direction.
Yes, but I believe the electrostatic field term falls off as 1/r^3.
If the source is a vibrating electric dipole (zero monopole),
dipole length = d + A * sin(w * t)
where, A < d, then the static electric field is dipolar, and
declines as d / r^3.
[Old Man]
I often refer to it as electric waves. Were you thinking about the
induction term, which I believe is 1/r^2? Also I believe the
electrostatic term is not merely longitudinal but a combination of
traverse and longitudinal.
Not that it makes that much difference, but the E field measurements I
took were only traverse to the wave-- on the same axis of the dipole.
I would think that at far distances, relative to wavelength, the e-term
would become insignificant as if falls of as 1/r^3.
The second is the "acceleration
(dynamic) field", which declines as e / R and is transverse to the
radial direction.
... the far field.
As source distance, R, increases, the total field asymptotically
approaches that of the acceleration (EM wave) field for which
|E| = c * |B| and for which E and B are transverse to R. The
radiation field is bipolar, sin(theta)^2, about the source. The
diminishing static field is spherically symmetric about the source
I don't know. It's so complex to simulate in the mind. Best left to a
computer, or best yet, I would like to measure the exact speed of an EM
pulse at near distances relative to wavelength. I am fairly certain
that it will be equal to c, which will make you correct as usual Old
Man.
.
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| User: "" |
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| Title: Re: EM is an accelerating H field ?? |
19 Jun 2005 01:51:45 AM |
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wrote:
There is a slight difference of 99.88%.
At 15 meters there's a 90.33% difference.
At 1 meter there's a 24.45% difference.
Correction. I didn't mean to write "difference" but you get the
picture; e.g., 99.88% should be 0.12% or 99.88% c.
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| User: "" |
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| Title: Re: EM is an accelerating H field ?? |
19 Jun 2005 02:01:05 AM |
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Another side note according to NEC2. The values are relative to
frequency. In other words, take the measured values at 150 meters at
10MHz. At 150 meters the magnetic field is traveling at 99.88% c, or
so the above-mentioned theory suggests. Considering 500THz, then the
magnetic field would be traveling at 99.88% c at 3.00E-006 meters.
150 m * 10MHz / 500THz = 3.00E-006 meters.
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