If an electromagnetic wave of 7.5hz travels at c, and has a wavelength of 40,000,000m ... And the earth has a resonant frequency of 7.5hz and a circumference of 40,000,000m ? Would this suggest the speed of light is an angular velocity, not linear? or just a coincidence? or something else?
what the hell are you talking about???
Do bear in mind that many of us are just physicists, and as such we have not yet attained the higher level of consciousness necessary to comprehend your last post.
Do bear in mind that many of us are just physicists, and as such we have not yet attained the higher level of consciousness necessary to comprehend your last post.
OK, having re-read your post, you do make an interesting point. (although the actual schumann resonance of earth is 7.8Hz, so you're fudging it a bit)
The only problem with your hypothesis, you see, is that light has a tendency to travel in straight lines, rather than travelling around planets in big circles.
Edit: Schumann resonances are caused by EM waves propagating around earth's ionosphere (which acts as a waveguide for EM waves), so of course the resonant modes are going to be multiples of the Earth's circumference!
The only problem with your hypothesis, you see, is that light has a tendency to travel in straight lines, rather than travelling around planets in big circles.
Edit: Schumann resonances are caused by EM waves propagating around earth's ionosphere (which acts as a waveguide for EM waves), so of course the resonant modes are going to be multiples of the Earth's circumference!
Thanks for the info. Sorry, I'm a noob, and can't explain things well.
I was thinking all electromagnetic waves have two velocities, the angular velocity an electromagnetic field moves through space, and the linear velocity of the 'electro / magnetic field' junction.
If the field (angular v) moves at c, and the electromagnetic wave is circular wouldn't the linear portion travels slower than c. at 1.90985e8 (c*2/pi)
I was thinking all electromagnetic waves have two velocities, the angular velocity an electromagnetic field moves through space, and the linear velocity of the 'electro / magnetic field' junction.
If the field (angular v) moves at c, and the electromagnetic wave is circular wouldn't the linear portion travels slower than c. at 1.90985e8 (c*2/pi)
Oh, now I understand what you're saying (sorry if I sounded a bit harsh back there, I assumed you were making a troll post or something, and sometimes I like to troll back 
)
When you refer to 'angular velocity' -- I think you actually mean 'angular frequency', which is where each cycle of the wave is divided into 2*pi radians, and you measure the speed of the wave in radians per second (it's just the frequency in Hz * 2pi). This is not c, (it just wouldn't make sense to try to equate it with c
)
The 'linear velocity' you refer to (also known as 'phase velocity') is always equal to c as it propagates through vacuum - it has to be.
(then there's also 'group velocity', which is where you have a combination of more than one wave superposed on top of each other, and the net 'group' of waves appears to move faster or slower than c, but that's all down to wave addition effects rather than anything profound -- I don't think this is what you were getting at, though)
)When you refer to 'angular velocity' -- I think you actually mean 'angular frequency', which is where each cycle of the wave is divided into 2*pi radians, and you measure the speed of the wave in radians per second (it's just the frequency in Hz * 2pi). This is not c, (it just wouldn't make sense to try to equate it with c
)The 'linear velocity' you refer to (also known as 'phase velocity') is always equal to c as it propagates through vacuum - it has to be.
(then there's also 'group velocity', which is where you have a combination of more than one wave superposed on top of each other, and the net 'group' of waves appears to move faster or slower than c, but that's all down to wave addition effects rather than anything profound -- I don't think this is what you were getting at, though)
Thanks for your response, and patience.
I don't think i meant angular frequency (2*pi*f) I'm talking meters per second of the outside edge of an electromagnetic wave ("angular wave length" maybe? not sure what to call it).
I'll try to explain again. If a spherical object's wavelength = its circumference (for example, the earth... wavelength = circumference of 40,000,000m) Now If you convert that 40,000,000m into a full sine wave... The linear distance of the wave or wavelength would be the earth's diameter * 2 ... 12,745,595 * 2 = 25,491,191 meters
Multiple that wave cycle * 7.5 and you get m/s :
7.5 hz * 25,491,191 m = 191,183,936 meters/sec. (linear)
7.5 hz * 40,000,000 m = 300,000,000 meters/sec (angular)
?
I don't think i meant angular frequency (2*pi*f) I'm talking meters per second of the outside edge of an electromagnetic wave ("angular wave length" maybe? not sure what to call it).
I'll try to explain again. If a spherical object's wavelength = its circumference (for example, the earth... wavelength = circumference of 40,000,000m) Now If you convert that 40,000,000m into a full sine wave... The linear distance of the wave or wavelength would be the earth's diameter * 2 ... 12,745,595 * 2 = 25,491,191 meters
Multiple that wave cycle * 7.5 and you get m/s :
7.5 hz * 25,491,191 m = 191,183,936 meters/sec. (linear)
7.5 hz * 40,000,000 m = 300,000,000 meters/sec (angular)
?
Any comments on this?
Some thoughts about it...
Radiowaves do follow Earth's curvature if their frequency is under 20 to 30MHz, as they get reflected by the ionosphere and the soil. Called "decametric propagation".
If the frequency is low enough, you have a bit more than half a wavelength in this layer (50km thick), and then waves are easily bent. It becomes a monomode propagation, like in fibre optics. A fibre can be bent reasonably, and light stays in it. Used by old navigation methods, maybe Loran.
The cutoff frequency is around 7.5kHz: under this frequency, no half wavelength fits between the soil and the ionosphere, so no propagation is possible, only fading waves. This explains why international agreements define the use of frequency bands beginning as 7.5kHz; under that, no radiocomm rules.
So: you can't measure anything at 7.5Hz, sorry.
And I guess there is absolutely no coincidence in your frequency, it must be defined just like that.
Radiowaves do follow Earth's curvature if their frequency is under 20 to 30MHz, as they get reflected by the ionosphere and the soil. Called "decametric propagation".
If the frequency is low enough, you have a bit more than half a wavelength in this layer (50km thick), and then waves are easily bent. It becomes a monomode propagation, like in fibre optics. A fibre can be bent reasonably, and light stays in it. Used by old navigation methods, maybe Loran.
The cutoff frequency is around 7.5kHz: under this frequency, no half wavelength fits between the soil and the ionosphere, so no propagation is possible, only fading waves. This explains why international agreements define the use of frequency bands beginning as 7.5kHz; under that, no radiocomm rules.
So: you can't measure anything at 7.5Hz, sorry.
And I guess there is absolutely no coincidence in your frequency, it must be defined just like that.
Thanks for your comment Enthalpy. I was thinking of the earth itself as a resonant cavity, and not so much the ionosphere (waveguide).
A little off topic now but:
Here's the equation for Shumann resonance.
f= c/(2*pi*r) * sqrt(n(n+1))
r=earth's radius= 6.372797e6
n= n-th mode
Looks like a spherical resonant cavity to me.
But, where does the ionosphere come into play here?
If the ionosphere was involved wouldn't the the formula be c/(2*pi*(earth_radius + ionosphere_radius) ?
If that was the case, then the earth's ionosphere (about 300,000m ) + the earth's radius
would make the theoretical Shumann resonant frequency 7.155hz
The theoretical (ideal) Shumann frequency is 7.5hz . Or:
f=c/(2*pi*earth_radius) = 7.4922 hz
But the real Shumann frequency is around 7.8hz ..
Could it be the Shumann frequency has more to do with the earth's upper mantel, than the ionosphere?
If you take the earth radius and subtract the average depth of the upper mantel, you get 250,000m - 6,372,797m = 6,122,797m
Apply this to f=/c/(2*pi* (earth_radius - upper_mantel_depth))
f = 7.8
Any logic to this?
Here's the equation for Shumann resonance.
f= c/(2*pi*r) * sqrt(n(n+1))
r=earth's radius= 6.372797e6
n= n-th mode
Looks like a spherical resonant cavity to me.
But, where does the ionosphere come into play here?
If the ionosphere was involved wouldn't the the formula be c/(2*pi*(earth_radius + ionosphere_radius) ?
If that was the case, then the earth's ionosphere (about 300,000m ) + the earth's radius
would make the theoretical Shumann resonant frequency 7.155hz
The theoretical (ideal) Shumann frequency is 7.5hz . Or:
f=c/(2*pi*earth_radius) = 7.4922 hz
But the real Shumann frequency is around 7.8hz ..
Could it be the Shumann frequency has more to do with the earth's upper mantel, than the ionosphere?
If you take the earth radius and subtract the average depth of the upper mantel, you get 250,000m - 6,372,797m = 6,122,797m
Apply this to f=/c/(2*pi* (earth_radius - upper_mantel_depth))
f = 7.8
Any logic to this?
Another idea that may apply to my original thought.
If a electromagnetic wave can travel at ((2*c)/pi)
Then:
((2*c)/pi) / (4*(earth_radius - upper_mantel_average_depth) = 7.8 hz
If there is refraction occuring inside the earth's "iron cavity" + crust, then this could make sense if the 7.8 hz wave can oscillate within the earth (4*r).
This may also explain why the earth is used as an antenna for ELF radio.
Feel free to shoot down my ideas, or tell me what I'm missing, flame, whatever.
Thanks
If a electromagnetic wave can travel at ((2*c)/pi)
Then:
((2*c)/pi) / (4*(earth_radius - upper_mantel_average_depth) = 7.8 hz
If there is refraction occuring inside the earth's "iron cavity" + crust, then this could make sense if the 7.8 hz wave can oscillate within the earth (4*r).
This may also explain why the earth is used as an antenna for ELF radio.
Feel free to shoot down my ideas, or tell me what I'm missing, flame, whatever.
Thanks
I guess you got your information from
http://en.wikipedia.org/wiki/Schumann_resonances
not a very clear entry.
I could indeed imagine a very slow (under 5kHz) resonance involving the soil and the ionosphere. The lowest mode would have a + potential at the soil at one place, a - at the ionosphere above that place, a - at the soil at the antipode and a + at the ionosphere at the antipode. And of course, slow currents flowing in the soil and - in opposite direction - in the ionosphere between the electric poles.
The resonant frequency doesn't depend on the soil-ionosphere distance (which doesn't act as a waveguide in this case, since its cutoff or minimum frequency is around 5kHz) because this distance increases the inductance as it reduces the capacitance equally.
You can compare this fundamental mode to a very short and broad coaxial line with its central conductor (= the soil) open at both ends.
The fundamental resonant frequency would be about half a vacuum wavelength in 20.000km or 7.5Hz. No need for refractive materials.
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Earth used as an antenna in ELF: we have little choice! Antennas must be huge, so they must be horizontal and are unavoidably near to the soil. A perfectly conductive soil would efficiently cancel the signal received by the antenna's wire (put another way: the wave induces the same voltage in the wire and in a conductive soil, so you see no voltage difference at the ends), but the trick is that the soil is lossy. By grounding the wire at its far end, you get a voltage at the near end which corresponds to the losses by the current induced in the soil.
So: no relationship to the iron core nor refraction.
It is an equivalent to the "transfer impedance" of a coax cable. The induced parasitic voltage is due to te losses in the shield.
http://en.wikipedia.org/wiki/Schumann_resonances
not a very clear entry.
I could indeed imagine a very slow (under 5kHz) resonance involving the soil and the ionosphere. The lowest mode would have a + potential at the soil at one place, a - at the ionosphere above that place, a - at the soil at the antipode and a + at the ionosphere at the antipode. And of course, slow currents flowing in the soil and - in opposite direction - in the ionosphere between the electric poles.
The resonant frequency doesn't depend on the soil-ionosphere distance (which doesn't act as a waveguide in this case, since its cutoff or minimum frequency is around 5kHz) because this distance increases the inductance as it reduces the capacitance equally.
You can compare this fundamental mode to a very short and broad coaxial line with its central conductor (= the soil) open at both ends.
The fundamental resonant frequency would be about half a vacuum wavelength in 20.000km or 7.5Hz. No need for refractive materials.
------------------
Earth used as an antenna in ELF: we have little choice! Antennas must be huge, so they must be horizontal and are unavoidably near to the soil. A perfectly conductive soil would efficiently cancel the signal received by the antenna's wire (put another way: the wave induces the same voltage in the wire and in a conductive soil, so you see no voltage difference at the ends), but the trick is that the soil is lossy. By grounding the wire at its far end, you get a voltage at the near end which corresponds to the losses by the current induced in the soil.
So: no relationship to the iron core nor refraction.
It is an equivalent to the "transfer impedance" of a coax cable. The induced parasitic voltage is due to te losses in the shield.