The Theory of Invariance

<b>[Bart] "The ether particles move in random direction and collide with each other all the time."</b>

If you spend a little time thinking about the properties of EM waves, you soon realize that a medium such as you describe cannot work. Liquids and gasses comprise particles that "move in random direction and collide with each other all the time". Liquids and gasses can support the propagation of longitudinal waves, but not transverse waves.

(Surface waves are a mixture of longitudinal and transverse waves. As the name implies, they can only occur at the surface of the medium. In the bulk material, only a solid can support transverse waves.)

EM waves are transverse. If the aether exists, it <u>must</u> be a solid. (A very strange solid, for sure.)

Physics is pfun,
LB

Now I go back to Invariance []

SECTION III: BLUE SHIFT, RED SHIFT

Experiment:

A photon with energy E<font size="1">o</font id="size1"> is emitted in a uniform gravitational field g. The photon flies down from A to B then to C. AB = BC = h.

Energy of the photon depend on g and h. g is a constant in this experiment. Now between energy of photon and h, we can write ratio E/E<font size="1">o</font id="size1"> as a function <i>f</i> of h:

E/E<font size="1">o</font id="size1"> = <i>f</i>(h)

or E<font size="1">o</font id="size1">/E = <i>f</i>(-h)

We also have <i>f</i>(2h) = <i>f</i>(h).<i>f</i>(h)

So <i>f</i>(h) is an exponent function,

<i>f</i>(h)= e^(k.h)

E/E<font size="1">o</font id="size1"> = e^(k.h)

Ratio of frequencies also equal to ratio of energies:

f/f<font size="1">o</font id="size1"> = e^(k.h)

Compare this equation to real measurements (or to Newton/Relativity equation f/f<font size="1">o</font id="size1"> <u>~</u> 1 + gh/cc), we have

f/f<font size="1">o</font id="size1"> = e^(gh/cc)


Hence,

In case of a flash of light f<font size="1">o</font id="size1"> is emitted at a distance R<font size="1">o</font id="size1"> from a planet mass M, radius R, then frequency of the flash on surface of the planet is

f = f<font size="1">o</font id="size1">.e^[GM(R<font size="1">o</font id="size1"> - R)/(R<font size="1">o</font id="size1">Rcc)]

In case of a flash of light f<font size="1">o</font id="size1"> is emitted on the surface of the planet, then frequency of the flash in infitity is

f<font size="1">inf</font id="size1"> = f<font size="1">o</font id="size1">.e^(-GM/Rcc)

Consequently, no black hole exists.

For instance,

If a flash of light f<font size="1">o</font id="size1"> is emitted on event horizon of a Schwarzschild "black hole", then frequency of the flash in infinity is

f<font size="1">inf</font id="size1"> = f<font size="1">o</font id="size1">.e^(-0.5)

I agree that under the influence of gravity, the ether particles crowd closer together near masses such as Earth and Sol.
Particles crowding closer together and losing speed goes hand-in-hand (in analogy with the laws of Thermodynamics).
I also agree that is is what dark matter is about.

EM waves are indeed different from longitudinal waves (e.g. show a polarization). I spent quite some time reflecing how transverse waves can propage through the ether and documented my thoughts in the paper "Magnetism-Gravity Working Model" wbabin.net/weuro/leplae.pdf


Started with a model on how magnetism works... from which I deducted how EM waves are formed. One may envision an EM wave as 2 longitudinal waves with opposite phase that start with 1/2 wavelenght separation from each other ...

The model explains why the polarization of EM waves rotates in a magnetic field.

The EM waves have difficulties passing metals because the free electrons absorb the energy of the EM wave.
In the model, I post an hypothesis why exactly at the moment of superconductivity, the EM waves can pass the superconductor.

<b>[Bart] "I spent quite some time reflecing how transverse waves can propage through the ether and documented my thoughts in ..."</b>

Great. Can you provide us with a brief summary here?

<b>[Bart] "One may envision an EM wave as 2 longitudinal waves ..."</b>

Actually, not. The particles involved in the propagation of longitudinal waves vibrate in the wrong direction. The configuration you describe might result in some side to side movement of the particles between the two longitudinal "beams" (due to the out of phase positioning of the pressure zones), but this pseudo-transverse activity would be two dimensional. Actual EM waves are very three dimensional. (With one specific exception.)

And, the energy contained in the two longitudinal beams would be very evident if they existed physically. You probably need to spend less time with the math and more time with the physics. It will help you avoid simple mistakes like this.

If the aether exists, it must behave like a solid. At least as far as EM wave propagation is concerned. Other observational constraints on its properties suggest that it <u>also</u> needs to behave like a vacuum. At least as far as detectable levels of drag on planets and stuff is concerned.

Hmmm. How can this be?

LB

[Larry] "The configuration you describe might result in some side to side movement of the particles"

The actual movement is a circular path:
- top : a 'high' pressure moving forward with the speed of light
(a push // forward pointing force)
- bottom : a 'low' pressure moving forward with the speed of light
(a pull // backward pointing force)
- middle: side-to-side movement (due to the out of phase positioning of the pressure zones)

This circular movement 'self-replicates' itself: above and below the intial circular path, new circular paths are created which turn in opposite direction relative to the initial circular path. This 'self-replication' is caused by entrainment and explains why the EM wave spreads.

The above is true for EM waves initiated through an antenna that sends continuous sinusoidal waves. For single photons, the 'self-replication' creates a wave pattern of which the circles gradually change in dimension. This change in dimension is causing the self-replication to stop as a result of which the EM wave associated with a single photon not spread.

A wave (pattern) looks like a gearbox of which the individual gears turn in opposite direction relative to each other. The gearbox travels with the spead of light.

And you are right: the aether behaves like a solid for EM waves.
At the moment the wave pattern of a single photon somewhere hits a detector, the momentum felt by the detector is the combined momentum of 'all of the gears in the gearbox'. This is what explains the photoelectric effect which is an illustration of the wave-particle duality.

Magnetism is explained as follows:
- the aether is dragged by moving electrons
- If we have 2 parallel wires with electrons travelling in the same direction, than the aether between the 2 wires will move faster than at the opposite sides of the wires. This causes the wires to be attracted to each other as a result of the Bernouilli Effect.
- If we have 2 coils with electrons orbiting in the same direction around the coils, than the aether between the coils will rotate faster than the eather at the opposite sides. This will cause the wires to be attracted to each other as a result of the Bernouilli Effect (compare to the lift a frisbee)

In summary: aether that circulates = a magnetic field.

An object rotating in the aether induces a magnetic field as well: the Earth magnetism is one example; the 'London moment' is another.

The above also explains why gravitomagnetism is similar to magnetism. The only difference being that for gravitomagnetism the aether is entrained by a mass instead of by electrons.

So, why are EM waves pictured as sinusoidal waves?

Looking form the perspective of a 'static observer': the cirular eather movements pass by with the speed of light. As each every of the circles is moving in opposite direction, the associated magnetic field is reversed with every circle passing by. Depicting the strenght and direction of the magnetic field results into a perfect sinusoidal representation.

Bart

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Cindy</i>

Compare this equation to real measurements (or to Newton/Relativity equation f/f<font size="1">o</font id="size1"> <u>~</u> 1 + gh/cc), we have

f/f<font size="1">o</font id="size1"> = e^(gh/cc)


Hence,

In case of a flash of light f<font size="1">o</font id="size1"> is emitted at a distance R<font size="1">o</font id="size1"> from a planet mass M, radius R, then frequency of the flash on surface of the planet is

f = f<font size="1">o</font id="size1">.e^[GM(R<font size="1">o</font id="size1"> - R)/(R<font size="1">o</font id="size1">Rcc)]

In case of a flash of light f<font size="1">o</font id="size1"> is emitted on the surface of the planet, then frequency of the flash in infitity is

f<font size="1">inf</font id="size1"> = f<font size="1">o</font id="size1">.e^(-GM/Rcc)

Consequently, no black hole exists.

For instance,

If a flash of light f<font size="1">o</font id="size1"> is emitted on event horizon of a Schwarzschild "black hole", then frequency of the flash in infinity is

f<font size="1">inf</font id="size1"> = f<font size="1">o</font id="size1">.e^(-0.5)


<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

Hi everyone,

Are there member who disagree the red shift/blue shift equations above?