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

[markie]

So according to you there are two things inside the shell? A proton and, what? The thing which you steadfastly, ignorantly and rather stupidly refuse to acknowledge is that the proton experiences a net zero force from the shell regardless of its location within the shell, as a consequence of the shell theorem, and therefore the arrangement is unstable. Simples!

There doesn't have to be two things inside the shell. The single proton will do. And yes, it is as you say, the proton will experience a net zero force from the shell. The kicker is this: any point on the shell will now experience an inward force from the central proton. Similarly, if there happened to be another particle inside the shell, it too would experience a non uniform field emanating from the proton.

 

[JeanTate]

Heavens no. Photons don't have to remember anything. The configuration of the slits has a real physical effect which in turn alters the statistical probability of where a single photon will go. (Yes we can have statistics with deterministic physics.) This will produce patterns.

The material of the slits are made up of atoms, each atom having an outer orbitsphere. The photon contacts an orbitsphere, is absorbed and reemitted in a direction that is not quite random. Now, some orbitspheres may be less prone to reemit a photon of the same energy ; that would translate into the material being less reflective. But it still would not have a bearing on the direction of the reemitted photon. The banding pattern would still emerge.

(my hilites)

As nearly always with what you post here, markie, a word picture.

Any chance you could flesh that out, with, you know, numbers, equations, mathematics, etc?

 

[jadebox]

There doesn't have to be two things inside the shell. The single proton will do. And yes, it is as you say, the proton will experience a net zero force from the shell. The kicker is this: any point on the shell will now experience an inward force from the central proton.

Every point on the shell will experience some force between the shell and the proton. The net force will, however, still be zero.

 

[JeanTate]

There doesn't have to be two things inside the shell. The single proton will do. And yes, it is as you say, the proton will experience a net zero force from the shell. The kicker is this: any point on the shell will now experience an inward force from the central proton. Similarly, if there happened to be another particle inside the shell, it too would experience a non uniform field emanating from the proton.

Nope.

Not if that particle were a neutrino, or a Z, or ... a photon!

 

[markie]

Wow. So, basically you are arguing that negative zero ia not equal to zero.

It seems you are saying: take any number of charges and their non uniform radial fields, surround the whole thing with a shell of charge, and presto, those interior fields will be nullified and made uniform. Sure. (not)

 

[markie]

Nope.

Not if that particle were a neutrino, or a Z, or ... a photon!

Got me there.

 

[hecd2]

Yes, which underscores my point ; it is not about 'interference' with other photons. Rather it is about the interaction of a single photon with the material slit apparatus. Now, the QM position would be that the single photon (or whatever particle) is interfering with itself, with it own superpositions of quantum potentiality. Either that or it is interfering with real photons from other real parallel universes. I happen to disagree, wholeheartedly.

This is another case in which your incorrigibility signals your ignorance to competent physicists. The fringes of a double slit experiment are nothing like the superposition of the diffraction patterns from two slits of finite width. It is quite clear that the spacing of the fringe pattern is given exactly by the spacing of the slits, the distance from the slits and the wavelength of the light exactly and only as it would be if it were caused by wave interference. Two slits of finite width are mathematically represented by the convolution of two delta functions with a single top hat function of width equal to the slit width. In the far field, the Fourier transform of the convolution of the two functions is given by the product of the Fourier transforms of the functions. The FT of two delta functions is a sinusoid (the fringes). The FT of the top hat function in intensity (amplitude squared) is a sinc squared which modulates the sinusoid. Where the width of the slits is much smaller than the separation of the slits, then the first peak of the sinc squared is much broader than many cycles of the sinusoid, so the influence of diffraction through the finite slit width on the fringes is minimal.

Suffice it to say that both the interference of light passing through the two slits and the diffraction of light passing through a single slit of finite width are fully and uniquely explained by the wave theory of light.

 

[markie]

Every point on the shell will experience some force between the shell and the proton. The net force will, however, still be zero.

In a certain framework, yes. But in the framework of inward and outward, the net inward force is additive to a non zero number.
Just like if the moon were cut in two and the halves put on opposite sides of the earth, the net inward force of the earth on the half moons would be equal to what it was on the single full moon.

 

[jadebox]

It seems you are saying: ...

Actually, I realized that I had misread what you stated, so I edited my response. In the meantime, you replied before I saved the changes.

You are arguing in circles. No one has denied that forces interact between the sphere and what is inside. What you are refusing to understand is that the net effect is zero. So, this idea of proton inside a sphere being stable is nonsensical.

 

[jadebox]

In a certain framework, yes. But in the framework of inward and outward, the net inward force is additive to a non zero number.

The net inward force would be zero just as the net outward force. Are you ignoring that forces are vectors, not scalars on purpose?

 

[markie]

This is another case in which your incorrigibility signals your ignorance to competent physicists. The fringes of a double slit experiment are nothing like the superposition of the diffraction patterns from two slits of finite width. It is quite clear that the spacing of the fringe pattern is given exactly by the spacing of the slits, the distance from the slits and the wavelength of the light exactly and only as it would be if it were caused by wave interference. Two slits of finite width are mathematically represented by the convolution of two delta functions with a single top hat function of width equal to the slit width. In the far field, the Fourier transform of the convolution of the two functions is given by the product of the Fourier transforms of the functions. The FT of two delta functions is a sinusoid (the fringes). The FT of the top hat function in intensity (amplitude squared) is a sinc squared which modulates the sinusoid. Where the width of the slits is much smaller than the separation of the slits, then the first peak of the sinc squared is much broader than many cycles of the sinusoid, so the influence of diffraction through the finite slit width on the fringes is minimal.

Suffice it to say that both the interference of light passing through the two slits and the diffraction of light passing through a single slit of finite width are fully and uniquely explained by the wave theory of light.

Oh yes, the math describing wave diffraction and 'interference' was there well before quantum theory. A classical result. And yes it is as you say: the mere addition of the diffraction pattern from each slit does not produce the interference pattern observed. This is (per my view) because the presence of two slits changes how any one slit will diffract the light, and this is because the photon experiences both slits while ultimately being reemitted through just one slit.

 

[WhatRoughBeast]

Heavens no. Photons don't have to remember anything. The configuration of the slits has a real physical effect which in turn alters the statistical probability of where a single photon will go. (Yes we can have statistics with deterministic physics.) This will produce patterns.

The material of the slits are made up of atoms, each atom having an outer orbitsphere. The photon contacts an orbitsphere, is absorbed and reemitted in a direction that is not quite random. Now, some orbitspheres may be less prone to reemit a photon of the same energy ; that would translate into the material being less reflective. But it still would not have a bearing on the direction of the reemitted photon. The banding pattern would still emerge.

Oh, markie. You've read up on the experiment, but as usual you didn't read enough.

Two questions:

1) If the photon interacts with only one slit, why will covering the other slit produce a different pattern than a single slit? That is, why will it make a difference? And please, before you answer with something foolish, do some more reading to the point that you realize that the sum of two single-slit patterns is not the same as that of a two-slit pattern. "The banding pattern would still emerge." is simply wrong.

2) So riddle me this: since the double slit experiment works for electrons as well as photons, are electrons "absorbed and reemitted" by an orbitsphere? If not, why are the composite band patterns identical with the composite band patterns of single-photon experiments?

 

[Dave Rogers]

Well here you are diverting into a myriad of different topics, each of which deserves it's own thread. But the issue at present is simply ... the applicability of the shell theorem. The shell theorem is saying that a shell will contribute no net force to anything within it ; in other words the shell has no contribution to field differentials inside it. So, with a proton inside the shell, it is the sole contributor to differentials of field potential within the shell. But you seem to be implying that the field is uniform inside the shell no matter what is inside it, in which case any object inside the shell would have no preferred position. If so, you would be mistaken.

The only way that pile of nonsense could possibly work is if it violated Newton's Third Law. If a shell contributes no net force to anything inside it, it cannot exert a restoring force on anything inside it, so anything inside it cannot possibly have a preferred position inside it. Since the shell cannot exert a force on the object inside it, hence the object inside it cannot exert a force on the shell, so there is no preferred position of the shell with respect to the object. Your argument about the uniformity of the field is a red herring, pure and simple; the shell theorem demands that there is no force between the shell and the proton inside it. Since you've already admitted that in your third sentence and then gone on to try and deny it, you are clearly either incompetent, lying or very, very confused.

Dave

 

[JeanTate]

Oh yes, the math describing wave diffraction and 'interference' was there well before quantum theory. A classical result.

As is the one where electrons (buckyballs, ...) are used instead of light?

And where the observed pattern is, over time, the same, even when only one particle is in the apparatus at any time?

And yes it is as you say: the mere addition of the diffraction pattern from each slit does not produce the interference pattern observed. This is (per my view) because the presence of two slits changes how any one slit will diffract the light, and this is because the photon experiences both slits while ultimately being reemitted through just one slit.

Here's where it gets weird: keep the slits, but put a detector next to each. What happens? What if the detectors are switched off? Or only one is on? Or the detectors are switched on after the particle has gone through?

How does a photon (electron, buckyball, etc) know whether a detector is switched on or not?

If Mills can describe these results, quantitatively, I guess he's imported all of QM, not just "quantum" (per you).

 

[Dave Rogers]

And yes, it is as you say, the proton will experience a net zero force from the shell. The kicker is this: any point on the shell will now experience an inward force from the central proton.

Which, according to the shell theorem, is an element of force which, when combined with all other elements of force of the proton on the shell, vector sums to zero. Therefore, either the shell is deformable or there is no force on the proton. If the shell is deformable, then Mills's magical elements of charge are not orbiting in circles, so his entire calculation is nonsense; and if it isn't, then there is no restorative force maintaining the stability of the atom, so his entire model is nonsense.

There isn't actually a third option.

Dave

 

[Dave Rogers]

In a certain framework, yes. But in the framework of inward and outward, the net inward force is additive to a non zero number.

No, that's completely untrue. The vector sum of the forces is zero whatever so-ordinate system it's calculated in; and the scalar sum of a group of forces exerted in different directions is a meaningless fiction.

Dave

 

[markie]

Oh, markie. You've read up on the experiment, but as usual you didn't read enough.

Two questions:

1) If the photon interacts with only one slit, why will covering the other slit produce a single-slit pattern? That is, why will it make a difference? And please, before you answer with something foolish, do some more reading to the point that you realize that the sum of two single-slit patterns is not the same as that of a two-slit pattern. "The banding pattern would still emerge." is simply wrong.

2) So riddle me this: since the double slit experiment works for electrons as well as photons, are electrons "absorbed and reemitted" by an orbitsphere? If not, why are the composite band patterns identical with the composite band patterns of single-photon experiments?

I think I answered 1) in a very recent post. Regarding 2), the incoming electron itself is not absorbed and reemitted, the effect is more indirect. The approaching electrical field of the free electron impinges on the slit apparatus, there is photon exchange, and the incoming electron deflects appropriately. So ultimately it still depends on photon emission and absorption.
Even with, say, neutrons: Neutrons have a magnetic field that will interact with the slits. Even molecules, like say buckyballs, have (say) paramagnetic or diamagnetic properties and will interact electromagnetically with the slit apparatus. Disclaimer: I don't recall Mills talking about neutrons or atoms or molecules in the context of the double slit experiment, so I'm venturing on my own here.

 

[HappySkeptic99]

Well here you are diverting into a myriad of different topics, each of which deserves it's own thread. But the issue at present is simply ... the applicability of the shell theorem. The shell theorem is saying that a shell will contribute no net force to anything within it ; in other words the shell has no contribution to field differentials inside it. So, with a proton inside the shell, it is the sole contributor to differentials of field potential within the shell.

Yes. However, the proton having its own field is not the issue. The shell can't feel any net force from the proton, nor can the proton feel a net force from the shell.

The thing is, Mills knows this. In fact, he is counting on it. In a Helium atom, there are two spherical S-shells. In a classical Millsian orbitsphere world, the inner electron shell shields exactly 1 unit of proton charge from the outer one, and the other shell contributes nothing (until Mills gets into magnetic effects) to the inner one because it contributes zero field to what is inside. Likewise, the inner shell contributes zero field to what is inside it, meaning the proton can't feel it, but Mills ignores that.

Why is QM fine? Because it doesn't have rigid shells! It has a wave equation that allows motion in the radial direction.

But you seem to be implying that the field is uniform inside the shell no matter what is inside it,

No, the potential due to the spherical shell, within the shell is uniform, and the field owing to the shell is zero. Other particles within the shell can contribute their own fields, but the shell (if it is rigid) cannot feel them, nor can the particles feel the shell.

in which case any object inside the shell would have no preferred position.

Yes, the proton has no preferred position.

If so, you would be mistaken.

No I am not mistaken.

 

[markie]

The only way that pile of nonsense could possibly work is if it violated Newton's Third Law. If a shell contributes no net force to anything inside it, it cannot exert a restoring force on anything inside it, so anything inside it cannot possibly have a preferred position inside it. Since the shell cannot exert a force on the object inside it, hence the object inside it cannot exert a force on the shell, so there is no preferred position of the shell with respect to the object. Your argument about the uniformity of the field is a red herring, pure and simple; the shell theorem demands that there is no force between the shell and the proton inside it. Since you've already admitted that in your third sentence and then gone on to try and deny it, you are clearly either incompetent, lying or very, very confused.

Dave

True, the electron shell itself can exert no restoring force to whatever is in it. That is because the shell itself generates only a uniform field inside it, with no differentials in field strength. The important point however is that as soon as one introduces a charge inside the shell, the field inside is no longer uniform, and so there is indeed a preferred position.

Regarding Newtons third law ; I've only heard of this applied bidirectionally ; so when one is talking about the effect of a charged shell with effects in all inward directions, I wouldn't jump to conclusions. But bidirectionally it would work ; a single point on the shell would experience and equal and opposite reaction to the proton as the (point) proton is experiencing from that point on the shell.

 

[Reality Check]

A "photons remember" delusion when it is the slits that have to remember

Photons don't have to remember anything..

23 January 2019 markie: A "photons remember" delusion when it is the slits and a "probability" delusion.

A source emits a photon at a time into a double slit apparatus. A photon goes thru 1 slit and hits the screen at a position. A photon goes thru the other slit and hits the screen at another position. Repeat. An interference pattern builds up. Each slit has to remember what it did to the previous photons and know what the other slit did to its photons.

markie lies about Mills delusions about the photon double-slit experiment. Mills has deluded "the photon’s electric and magnetic fields give rise to electron or polarization currents at both slits" fantasies. These are deterministic, not markie's "probabilities". A photon going through the apparatus at a lime will be displaced the same amount by his imaginary currents and go to the same point.