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

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From Ethan's article, my bold:
A description aimed at a layman audience need not be accurate physics :eye-poppi!

In actual QM, a quantum particle is described by a wave function which is not a probability function. It is the interpretation of the wave function that gives a probability. This is easily misrepresented - read start of the Wikipedia article and then
In Born's statistical interpretation in non-relativistic quantum mechanics,[8][9][10] the squared modulus of the wave function, |ψ|2, is a real number interpreted as the probability density of measuring a particle's being detected at a given place – or having a given momentum – at a given time, and possibly having definite values for discrete degrees of freedom
4 December 2018 markie: "spatial extension as well" and "abstract probability function" lies about QM remains valid.

Overlapping wave functions are not actual "spatial extension" (as in Mills insanity or even English) in the Copenhagen interpretation because the wave function is snot real! There are no real positions in this interpretation. There is an overlapping probability of measuring positions. An electro is still a point particle. An proton is still treated as a point particle. A quark is still a point particle.
 
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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.
Consider the fact that people have looked at that situation, markie :eye-poppi!
A ring around a central body is unusable just like the sphere. Any perturbation sends it wandering off, usually crashing into the central body. We have known that unattended rings and spheres are impossible because of that instability since Dyson's paper in 1960 and his response the same year.
Dyson sphere: Variants
In fictional accounts, the Dyson-sphere concept is often interpreted as an artificial hollow sphere of matter around a star. This perception is based on a literal interpretation of Dyson's original short paper introducing the concept. In response to letters prompted by some papers, Dyson replied, "A solid shell or ring surrounding a star is mechanically impossible. The form of 'biosphere' which I envisaged consists of a loose collection or swarm of objects traveling on independent orbits around the star."[9]

The classic science fiction Ringworld series by Larry Niven introduced attitude jets to fix that problem.
After the publication of Ringworld, many fans identified numerous engineering problems in the Ringworld as described in the novel. One major one was that the Ringworld, being a rigid structure, was not actually in orbit around the star it encircled and would eventually drift, ultimately colliding with its sun and disintegrating. This led MIT students attending the 1971 Worldcon to chant, "The Ringworld is unstable!" Niven wrote the 1980 sequel The Ringworld Engineers in part to address these engineering issues. In it, the ring is found to have a system of attitude jets atop the rim walls, but the Ringworld has become gravely endangered because most of the jets have been removed by the natives, to power their interstellar ships. (The natives had forgotten the original purpose of the jets.)
 
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A description aimed at a layman audience need not be accurate physics :eye-poppi!

In actual QM, a quantum particle is described by a wave function which is not a probability function. It is the interpretation of the wave function that gives a probability. This is easily misrepresented - read start of the Wikipedia article and then

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

Overlapping wave functions are not actual "spatial extension" (as in Mills insanity or even English) in the Copenhagen interpretation because the wave function is snot real! There are no real positions in this interpretation. There is an overlapping probability of measuring positions. An electro is still a point particle. An proton is still treated as a point particle. A quark is still a point particle.

I think you have described the situation well with your last paragraph. Still, I consider the wave function as giving the point particle an infinite number of potential places to be, so I consider that a kind of space extension, in the abstract.
 
Contemplate real classical physics where the problem is about energy and potential energy, markie. To get over the bar, the high jumper has to expend more energy than the gravitational potential energy of the bar to get their center of gravity above the bar. Classically they will crash into the bar if they have an energy less than the gravitational potential energy (or fall under it!).....
I made a mistake - the jumper has to expend enough energy to exceed their gravitational potential energy for the bar height. In simple terms (for a hypothetical rigid jumper), that gets their center of mass above the bar and they rotate their body around the center of mass to cross over the bar.

Real high jumpers are flexible and a real high jump such as the Fosbury Flop is more complex because their center of mass moves. During the flop, their arched body puts their center of mass below the bar.
 
I think you have described the situation well with your last paragraph. Still, I consider the wave function as giving the point particle an infinite number of potential places to be, so I consider that a kind of space extension, in the abstract.
What you consider wrongly does not matter. The reality is that a point particle even in classic physics is defined as having no extent. QM does not change that fact.

It is also irrelevant to Mills insane theory except maybe as an illustration of how insane it is. We have measured the extent of electrons and electrons are not as big as atoms. Measurements suggest an upper limit to the extent of 10^-22 meters from bound electrons. QED is the most precisely tested theory in physics and has electrons as point particles particles.
 
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I made a mistake - the jumper has to expend enough energy to exceed their gravitational potential energy for the bar height. In simple terms (for a hypothetical rigid jumper), that gets their center of mass above the bar and they rotate their body around the center of mass to cross over the bar.

Real high jumpers are flexible and a real high jump such as the Fosbury Flop is more complex because their center of mass moves. During the flop, their arched body puts their center of mass below the bar.

The trajectory of the centre of mass of a jumper is unalterable after take off. No amount of bodily contortion while in flight can change the parabolic trajectory of his centre of mass. (This ignores minor effects like air resistance.)
 
The trajectory of the centre of mass of a jumper is unalterable after take off.
Almost correct but not what I wrote.
I made a mistake - the jumper has to expend enough energy to exceed their gravitational potential energy for the bar height. In simple terms (for a hypothetical rigid jumper), that gets their center of mass above the bar and they rotate their body around the center of mass to cross over the bar.

Real high jumpers are flexible and a real high jump such as the Fosbury Flop is more complex because their center of mass moves. During the flop, their arched body puts their center of mass below the bar.
A vertical person has a center of mass that is inside their body. High jumpers arch their bodies in the jump so that their center of mass is outside of their body.
The trajectory of their center of mass does not alter after they jump except for probably insignificant wind and wind resistance. It should be a simple parabola.
 
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.

What I wrote is correct. What you wrote deserves to be explored.

What you say here is correct: 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.

and in accordance to Newton's law of equal and opposite forces.

But let's get into the details. The proton exerts a certain force on a point on the orbitsphere. By Newton, that point on the orbitsphere exerts an equal and opposite force on the proton. But it is not symmetric as it may seem at first. The difference is this: The net force of the entire proton on that point on the orbitsphere is not zero, while the net force of the entire orbitsphere on the proton (or indeed any place inside the orbitsphere) is zero.
 
Almost correct but not what I wrote.

A vertical person has a center of mass that is inside their body. High jumpers arch their bodies in the jump so that their center of mass is outside of their body.
The trajectory of their center of mass does not alter after they jump except for probably insignificant wind and wind resistance. It should be a simple parabola.

Agree.
 
....The difference is this: The net force of the entire proton on that point on the orbitsphere is not zero, while the net force of the entire orbitsphere on the proton (or indeed any place inside the orbitsphere) is zero.
That is just non-science assertions markie.
Start by ignoring Mills insanity. Now read the shell theorem
Isaac Newton proved the shell theorem[1] and stated that:
1.A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its centre.
2.If the body is a spherically symmetric shell (i.e., a hollow ball), no net gravitational force is exerted by the shell on any object inside, regardless of the object's location within the shell.
By the shell theorem:
1. We have a proton and an external charged body (the shell). We treat the proton as a charged point.
2. No net electrostatic force is exerted by the shell on any proton inside, regardless of the proton's location within the shell.
Newton's third law gives us that there is no net electrostatic force exerted by the proton on the shell.

Mills insanity may include that the proton is not a uniformly charged body. If so it is up to him or you to use physics to show that a realistic proton cannot crash into the shell regardless of whatever impacts the hydrogen atom.
 
That is just non-science assertions markie.
Start by ignoring Mills insanity. Now read the shell theorem

By the shell theorem:
1. We have a proton and an external charged body (the shell). We treat the proton as a charged point.
2. No net electrostatic force is exerted by the shell on any proton inside, regardless of the proton's location within the shell.
Newton's third law gives us that there is no net electrostatic force exerted by the proton on the shell.

Mills insanity may include that the proton is not a uniformly charged body. If so it is up to him or you to use physics to show that a realistic proton cannot crash into the shell regardless of whatever impacts the hydrogen atom.

I did show it, and my reasoning doesn't require the proton to be anything other than a point. You just weren't paying attention that's all.

Hint: You reasoning is incomplete when you say:
"Newton's third law gives us that there is no net electrostatic force exerted by the proton on the shell."

Again, it's not about the net electrostatic force exerted by the proton on the entire orbitsphere; it's about the force exerted by the proton on any given point on the orbitsphere.
 
Just popping in to make an observation. All the QM discussion is fascinating, but it's rather like asking if unicorns can fart rainbows outside of showing that unicorns actually exist.

To date, only Mills has claimed to have produced hydrinos, but no other lab in the world has. There are no conferences exploring the properties of hydrinos, no papers in journals proving the hydrino's mass, no reports on the benefits or toxicity of hydrinos. Just some spectral lines on a machine that wasn't even capable of showing information in range where the hydrino is speculated to exist.

Mills might as well be studying N-rays.
 
...it's about the force exerted by the proton on any given point on the orbitsphere.
Which when added to all the other forces on all the other given points on the orbitsphere sum to zero.

And being a single solid object, since the sum of forces on it equals zero it experiences no force as a whole.

Okay, we've now strayed into high school physics that you're getting wrong!



Sent from my SM-J320FN using Tapatalk
 
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.

And in that case the satelite will have a different orbit from the one it had before. Before it had a perfect circular orbit (according to he base situation you described), now it will have a more eliptical orbit with a different speed.
 
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.

Now try generalising your thought experiment into launching a spherical shell composed entirely of satellites, set up so that in every orbit there are as many satellites moving one way as the other, and watch them crash into each other and fall back to Earth. What have we learned here, children? That Mills's fantasy is impossible to represent in terms of individual isolated objects in circular orbits.

Dave
 
...
To date, only Mills has claimed to have produced hydrinos, but no other lab in the world has. There are no conferences exploring the properties of hydrinos, no papers in journals proving the hydrino's mass, no reports on the benefits or toxicity of hydrinos. Just some spectral lines on a machine that wasn't even capable of showing information in range where the hydrino is speculated to exist.

Mills might as well be studying N-rays.

This.
 
Now try generalising your thought experiment into launching a spherical shell composed entirely of satellites, set up so that in every orbit there are as many satellites moving one way as the other, and watch them crash into each other and fall back to Earth. What have we learned here, children? That Mills's fantasy is impossible to represent in terms of individual isolated objects in circular orbits.

Dave
You'd almost think “classical“ physics can't describe the reality of the atom!
 
Which when added to all the other forces on all the other given points on the orbitsphere sum to zero.

And being a single solid object, since the sum of forces on it equals zero it experiences no force as a whole.

Okay, we've now strayed into high school physics that you're getting wrong!

Sent from my SM-J320FN using Tapatalk

You say that, but you can't refute my satellite swarm analogy and the laws or orbital mechanics.
 
And in that case the satelite will have a different orbit from the one it had before. Before it had a perfect circular orbit (according to he base situation you described), now it will have a more eliptical orbit with a different speed.

This is a good point, I think you're correct in that particular scenario.
 
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