Cont: Brilliant Light Power Going To Market - Free Energy Generator Part 3

[markie]

4 December 2018 markie: "spatial extension as well" and "abstract probability function" lies about QM.

There is no "spatial extension" at all in QM.
There is no "abstract probability function" in QM. The wave function is a mapping of every possible state of a system to complex numbers.

From Ethan's article, my bold:

The secret is that, at a fundamental level, these atomic nuclei don’t behave as particles alone, but rather as waves, too. Each proton is a quantum particle, containing a probability function that describes its location, enabling the two wavefunctions of interacting particles to overlap ever so slightly, even when the repulsive electric force would otherwise keep them entirely apart.

Two wave functions 'overlapping' involve spatial extent. In the abstract. You're just trying to be difficult now.

 

[jadebox]

What demonstrable properties of a hydrino would satisfy you that it was in fact a hydrino?

Existence.

 

[wollery]

Glad someone sees it


Being rigid really has nothing to do with it. Consider: If instead of 2 half moons on opposite sides of the earth it was rather one continuous, rigid ring of matter encircling the earth. As long as the ring was rotating as fast as the orbital velocity of the original moon (or the two half moons), it would remain in orbit. If it wasn't spinning it would come crashing down to the earth of course. That is because the earth is exerting a gravitational force on it whether it is rigid or not.
Again, the shell theorem is about the force of gravity exerted by the shell ; it is not about the gravity of something else exerted on the shell.

Nope.

It also works for the gravitational force exerted on the ring from anything inside it. The same applies to electrostatic forces.

And more importantly it makes a massive difference whether it's a uniform rigid solid or composed of distinct parts.

As a uniform rigid solid any force acting on one part of it is transmitted throughout the whole structure and therefore acts on every other part. The result is that the net force on all the individual elements is the total force on the whole structure. And since the net force on the individual elements is zero, the force on the ring or sphere is zero.

The result is that the net force exerted by the sphere on any object inside it is zero, AND, the net force experienced by the sphere from any object inside it is also zero.



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[markie]

Nope.

It also works for the gravitational force exerted on the ring from anything inside it. The same applies to electrostatic forces.

And more importantly it makes a massive difference whether it's a uniform rigid solid or composed of distinct parts.

As a uniform rigid solid any force acting on one part of it is transmitted throughout the whole structure and therefore acts on every other part. The result is that the net force on all the individual elements is the total force on the whole structure. And since the net force on the individual elements is zero, the force on the ring or sphere is zero.

The result is that the net force exerted by the sphere on any object inside it is zero, AND, the net force experienced by the sphere from any object inside it is also zero.

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You're saying that if you encapsulate an object, like say a neutron star, within a shell, that the neutron star's gravity will somehow be zero at any point on the shell. Dude.

 

[Darat]

You're saying that if you encapsulate an object, like say a neutron star, within a shell, that the neutron star's gravity will somehow be zero at any point on the shell. Dude.

No that is not what he said. Reread his post and pay attention to the "by".

 

[jadebox]

You're saying that if you encapsulate an object, like say a neutron star, within a shell, that the neutron star's gravity will somehow be zero at any point on the shell. Dude.

Dude. Go back and read what you quoted. Look for a phrase like "net effect." If you are confused by that, understand that "net" in this context means something like "final.". If you add a value to minus the value, the net result is zero.


.

 

[UncertainH]

You're conflating a zero 'net force' (arising by symmetry) with the actual gravitational force felt by any point on the shell. Consider: If the moon was broken up into two, each on opposite sides of the earth, the 'net force' would also be zero. But each half moon would still feel the gravitational force of the earth and hence both half moons would be kept in orbit.

Markie your analogy doesn't work. It is true that if Mills model was just a negative electron shell surrounding the nucleus, the positive nucleus would indeed be free to bounce around inside the shell. However, Mills model includes a resonant photon inside the shell which contributes to the central force balance as well as quantization. Clearly any meaningful discussion about the forces inside the orbitsphere must include these resonant photons. All this discussion which does not include the action of these photons is a meaningless diversion which has nothing to do with what Mills is actually proposing.

 

[markie]

No that is not what he said. Reread his post and pay attention to the "by".

That wasn't the part I had in mind. Rather it was this:
"the net force experienced by the sphere from any object inside it is also zero.

 

[markie]

Dude. Go back and read what you quoted. Look for a phrase like "net effect." If you are confused by that, understand that "net" in this context means something like "final.". If you add a value to minus the value, the net result is zero.

.

True, he did say 'net' so he gets a pass. My mistake.

 

[markie]

Markie your analogy doesn't work. It is true that if Mills model was just a negative electron shell surrounding the nucleus, the positive nucleus would indeed be free to bounce around inside the shell. However, Mills model includes a resonant photon inside the shell which contributes to the central force balance as well as quantization. Clearly any meaningful discussion about the forces inside the orbitsphere must include these resonant photons. All this discussion which does not include the action of these photons is a meaningless diversion which has nothing to do with what Mills is actually proposing.

While the trapped, resonant photon plays a role, it is one of degree. So for instance the electric field at a point on the orbitsphere is the sum of the e field of the proton and the e field of the trapped photon (which points radially) which is stuck to the inner side of the orbitsphere. (See equations 2.6 and on)

I don't agree that the nucleus would be free to go anywhere inside the orbitsphere, any more that I believe the earth is free to be anywhere within the moon's orbit.

 

[UncertainH]

While the trapped, resonant photon plays a role, it is one of degree. So for instance the electric field at a point on the orbitsphere is the sum of the e field of the proton and the e field of the trapped photon (which points radially) which is stuck to the inner side of the orbitsphere. (See equations 2.6 and on)

I don't agree that the nucleus would be free to go anywhere inside the orbitsphere, any more that I believe the earth is free to be anywhere within the moon's orbit.

That radial field then is likely what keeps the proton in place because without it, the net field inside the sphere would indeed be exactly zero

 

[wollery]

True, he did say 'net' so he gets a pass. My mistake.

I get a pass?

You're getting the science wrong, you try (unsuccessfully) to to pick a hole in my correction of your error, and you deign to give me a pass?

Holy crapola, those are some mighty large cajones you've got there.

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[wollery]

While the trapped, resonant photon plays a role, it is one of degree. So for instance the electric field at a point on the orbitsphere is the sum of the e field of the proton and the e field of the trapped photon (which points radially) which is stuck to the inner side of the orbitsphere. (See equations 2.6 and on)

I don't agree that the nucleus would be free to go anywhere inside the orbitsphere, any more that I believe the earth is free to be anywhere within the moon's orbit.

Poor analogy.

The Earth isn't free to go anywhere within the Moon's orbit because the Moon isn't a rigid sphere.

Mill's electron orbitsphere is.

Because of that simple fact shell theorem applies and the net force of the orbitsphere anywhere inside it is zero, and the net force of the nucleus on the orbitsphere is zero, regardless of where the nucleus is within the orbitsphere.

This is pretty basic 1st year undergraduate physics Markie.

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[Horatius]

Again, the shell theorem is about the force of gravity exerted by the shell ; it is not about the gravity of something else exerted on the shell.

I guess with the plethora of mistakes being made here, everyone else missed this one.

The math for calculating the force exerted by mass A on shell B is exactly the same as the math for calculating the force exerted by shell B on mass A. They're literally mirror images of each other, there's simply no way one can be zero, and the other non-zero.

Indeed, if you could get a zero in one case and non-zero in the other, we wouldn't need to muck about with this finicky hydrino nonsense, we could get straight-up free energy from any of a number of classical perpetual motion devices that have tried to exploit differences in gravity between one place and another.

 

[markie]

That radial field then is likely what keeps the proton in place because without it, the net field inside the sphere would indeed be exactly zero

We're not talking about the net field over the orbitsphere. We're talking about the force of gravity on any point on the orbitsphere. If two identically massed people are standing on opposite sides of the earth, just because the 'net field' is zero doesn't mean each person experiences zero gravity!
BTW, both the electric field of the nucleus and the electric field of the trapped photon are radial.

 

[markie]

For those of you who think that the proton is free to migrate willy nilly within the orbitsphere, here's a thought experiment.

We want to launch a satellite into earth's orbit. We want the satellite's orbital speed to be a constant, exact value. Simple orbital mechanics will then dictate the the satellite *must* orbit at a certain fixed distance from the earth. (No matter what its mass).

Now say we want to launch a swarm of such satellites. Different orbital trajectories but same constant speed as the original satellite. Again, simple orbital mechanics demands that each and every satellite must orbit at the same fixed distance from earth. So we get a swarming shell of satellites going around the earth at the same fixed distance. The earth is not free to move within this shell, or else that would violate orbital mechanics. If a satellite is perturbed, say bumped towards the earth, but retains its original tangental speed, it will quickly oscillate its way back to its original orbit distance.

Similarly, the proton is not free to move within the volume defined by the orbitsphere. The atom is stable. Incredibly stable.

 

[erwinl]

For those of you who think that the proton is free to migrate willy nilly within the orbitsphere, here's a thought experiment.

We want to launch a satellite into earth's orbit. We want the satellite's orbital speed to be a constant, exact value. Simple orbital mechanics will then dictate the the satellite *must* orbit at a certain fixed distance from the earth. (No matter what its mass).

Now say we want to launch a swarm of such satellites. Different orbital trajectories but same constant speed as the original satellite. Again, simple orbital mechanics demands that each and every satellite must orbit at the same fixed distance from earth. So we get a swarming shell of satellites going around the earth at the same fixed distance. The earth is not free to move within this shell, or else that would violate orbital mechanics. If a satellite is perturbed, say bumped towards the earth, but retains its original tangental speed, it will quickly oscillate its way back to its original orbit distance.

Similarly, the proton is not free to move within the volume defined by the orbitsphere. The atom is stable. Incredibly stable.

Could you please show with some arrows (before and after) how you can give a speeding satelite of a given speed a bump towards the earth while retaining its original orbital speed?

 

[Myriad]

So we get a swarming shell of satellites going around the earth at the same fixed distance.

Which then crash into each other, because you set them orbiting at the same fixed distance but on different trajectories.

If you want them to form an equidistant spherical shell without them crashing into each other, you can connect them to one another into a rigid sphere. But, once you've done that, most of them (excepting only the ones on the shell's equator, if you've chosen the right rotational period for the radius of the shell) are no longer in free-fall orbits. So the system is not stable any more. Incredibly not stable.

 

[markie]

Could you please show with some arrows (before and after) how you can give a speeding satelite of a given speed a bump towards the earth while retaining its original orbital speed?

Rather tangental, but:
Consider a pebble sized meteorite coming in towards the earth with a certain velocity. It is just beginning to impact the satellite. At that moment, if the projection of the meteorite's velocity vector onto the tangent line of the satellite's orbit just so happens to be the same magnitude as the satellite's speed, it won't alter the satellite's tangental velocity, but it will still give the satellite an abrupt push towards the earth. But hey I'm no rocket scientist.

 

[markie]

Which then crash into each other, because you set them orbiting at the same fixed distance but on different trajectories.

If you want them to form an equidistant spherical shell without them crashing into each other, you can connect them to one another into a rigid sphere. But, once you've done that, most of them (excepting only the ones on the shell's equator, if you've chosen the right rotational period for the radius of the shell) are no longer in free-fall orbits. So the system is not stable any more. Incredibly not stable.

Of course these satellites cannot be connected. But that is rather beside the point of the earth having a fixed position at the centre of the satellite swarm.

Now with the electron orbitsphere, orbital 1D rings of moving change can cross other rings with no problem, just like any number of 1D electric field lines from different sources can superimpose at any point without a problem. No crashie crash.
Edit addition: the orbitsphere is 'rigid' only in the sense that its shape is fixed. It is a surface full of motion.