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

[W.D.Clinger]

more slapstick from markie, optiongeek, and Mills

With my highlighting of the relevant numbers:

But of course Mills is at least ignorant and probably lying.
Page 314, Table 10.1 Ionization Energies fro some three-electron atoms. He gets every ionization energy wrong ! For example, 5.40390 (experiment value) is not 5.39172 (his value) for Li. Thesis repeated up to twenty electron atoms.

As shown below, Mills made some kind of mistake. I don't know whether his mistake was deliberate, but his mistake was made in a direction that made one of his calculated results appear closer to the experimental value than it actually was.

As I noted in a subsequent post, quoted below, Reality Check inadvertently swapped the experimental value with Mill's calculated value, both from the lithium row of Table 10.1, page 314 of Mills's Big Book of Boo-Boos.

Actually, I get 5.4089 when I do the GUTCP-based calculation. Mills calculated value (Eq 10.25) is 5.4090.

optiongeek said he calculated 5.4089 using Mills's equation. According to Table 10.1, page 314, Mills calculated 5.40390, not 5.4089. I thought optiongeek had simply dropped the digit 3 from both of the numbers highlighted in the above quotation.

When I responded, I noted Reality Check's mistake but corrected optiongeek's apparent mistake without calling attention to it: see the highlighted number below.

RealityCheck inadvertently interchanged the value calculated by Mills with the measured ionization energy. RealityCheck's point stands: the value calculated using standard quantum mechanics is 12 times as close to the experimental value as the value calculated by Mills.


5.39271223 eV.

Comparison:

5.39171495        ; experimental value (per NIST)
5.39271223        ; calculated using quantum mechanics
[HILITE]5.40390[/HILITE]           ; as calculated by Mills in his Big Book

Although I corrected optiongeek's apparent error without rubbing his nose in it, Reality Check noted optiongeek's apparent mistake:

5.4089 is not 5.40390. Either you are missing a '3' (5.40389) or your calculation is wrong since you claim to be using Mills derivation. Or Mills is wrong and you have correctly used Mills derivation.

optiongeek responded by calling our attention to a boo-boo in Mills's Big Book of Boo-Boos:

Think you need to source that "3". I don't see it in the Eq 10.25 that appears my version of GUTCP. If you are accusing me of sloppiness, then fair enough. I've been known to transpose figures from time to time. But when you do, please make sure I'm the one who was the source of sloppiness.

optiongeek is correct! The mistake was made by Mills, not by optiongeek.

According to Mills equation (10.25), page 310, the ionization energy for lithium is 5.3230 + 0.08599, which equals 5.40899, which Mills quite reasonably rounds to 5.4090.

In Table 10.1, however, Mills says the ionization energy he had calculated for lithium is 5.40390, not the 5.4090 claimed by Mills equation (10.25).

According to Table 10.1, the experimental value is 5.39172. By reporting 5.40390 in Table 10.1 instead of the 5.4090 present in Mills equation (10.25), Mills was able to reduce the relative error he reports in Table 10.1 from 0.00319 to 0.00225.

According to both optiongeek and markie (and who could doubt either of them?), Mills equation (10.25) is correct, which means the calculated value reported by Mills in Table 10.1 is wrong, as is the relative error in his calculation.

Is anyone really surprised that Mills made this mistake in a direction that reduced the relative error he reported in Table 10.1?

Optiongeek did not miss a 3. It was Clinger who made the error and put in a 3 when it shouldn't be there.

I apologize for failing to spot Mills's error. I thank you and optiongeek for bringing Mills's error to our attention.

Funny, I got no impression at all that optiongeek was trying to 'impress us', no more than I thought Mills was trying to impress us by retaining the sin(theta)^2 and cos(theta)^2 terms. Optiongeek simply confirmed that the triple integral was performed correctly, because (as I recall) you seemed to have a problem with the magnetic term result and were implying that Mills was just making it up.

Although markie claims to recall a problem I had "with the magnetic term result", I recall no such problem. I am absolutely certain I had not claimed any problem with the particular piece of the energy calculated by equation (10.156). In light of those facts, I remain convinced that optiongeek cited that equation simply because he was impressed that Mills had simplified the triple integral correctly, and thought we would be as impressed as he was by Mills's ability to get the right answer for a very simple exercise in freshman calculus.

Just as markie continues to think we are impressed by his repeated denials of interferometry, and by his ignorance of single-photon versions of the double slit experiment, and by his suggestion that we ask a competent physicist whether the shell theorem applies to a situation in which the shell theorem obviously applies.

(Several competent physicists have already replied to that nonsense, for which I thank them.)

 

[markie]

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?

Again, yes. It depends not on if one or many particles are going through at one time, because the interference is not with the other particles but the slits. However with things like electrons and buckyballs there would be interaction and jostling but that would probably only detract from a patterned result.

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).

If the detectors are switched off and not positionally messing with the slits then one will get the normal interference pattern. If one or both are switched 'on' then their EM signal will influence the pattern to at least some degree.

I don't recall Mills saying so, but I would imagine he would say that there would be no pattern generated with neutrinos because they wouldn't react with the slit apparatus to any significant degree. The problem is that they would hardly react with the detector apparatus either hehe.

 

[HappySkeptic99]

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.

Yes, the photon does experience both slits. That's the basis of QM. Your claim that it goes through only one slit is most likely wrong. QM says that "knowing" it goes through one slit destroys the diffraction pattern. There are interesting results involving weak measurements of power flow in electron diffraction, but it is a mistake to say that the particle, when creating a diffraction pattern, goes through one slit.

Even if hidden-variable theory is true, and the particle went through one slit while still diffracting, we cannot know that it did. Knowing would destroy the Heisenberg uncertainty principle, and likely a number of conservation theorems. Taking one path is an observable property (position), which destroys the observable property of momentum (which gives the diffraction).

That is why physicists work in wavefunctions and project results onto observable basis functions. They don't pretend the particle has unobserved properties until observed, because that would be pretending what cannot be tested.

A fun fact -- interference ONLY happens at the level of an individual particle in QM. One particle does not interfere with another (well, an entangled ensemble can interfere as a whole, but then it acts as one unit). This is hard to grasp, but if modeled correctly, it works (even virtual absorption and re-emission is just an entanglement that introduces time delay or phase shift of the wavefunction, not an observation of the original particle wavefunction).

 

[Reality Check]

Truly ignorant gibberish denying the physics of the shell theorem

True, ....

23 January 2019 markie: Massively ignorant gibberish denying the physics of the shell theorem.

The electric field inside a uniformly charged shell is zero, not non-zero as he implies with "uniform" and demanding "no differentials in field strength". The introduction of a proton brings us to the shell theorem with non-iniform (inverse square) electric fields.

 

[HappySkeptic99]

If the detectors are switched off and not positionally messing with the slits then one will get the normal interference pattern. If one or both are switched 'on' then their EM signal will influence the pattern to at least some degree.

Mills isn't wrong. But, show us Mills' equations for electron diffraction. He will either use the same formulas as QM, which prove nothing, or he will have some orbitsphere version, which gives different results which we can test.

What he does instead, is B.S. based on half-truths.

In the decoherence theory modeling of QM, the detector would entangle with the wavefunction, introducing an environmental random phase shift to the portion going through the detector. This phase shift destroys the diffraction pattern. Yes, you can think of that as "messing up the EM wave" if you like, but show me the math. You are either doing the same math as QM or you aren't.

 

[Reality Check]

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.

Another delusion from the classical Millsian orbitsphere world would be the denial of the existence of H- (a hydrogen atom that has gained an election).
Hydrogen anion

How does a Mills orbitsphere become bound to a neutral hydrogen atom when the proton charge is shielded by the first orbitsphere?

 

[optiongeek]

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.

Not commenting on Shell theory, but I do believe the highlighted statement is incorrect. In Mills' model, the two S electrons orbit at the same radius. However, they do occupy different spin orientations.

 

[hecd2]

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.

Where can I find the Millsian physics which predicts the result of a double slit experiment quantitatively and exactly, as exactly as it is predicted by wave interference?

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.

. That’s astonishing. So in Millsian physics, photons which are diffracted by a less reflective material are a different colour (a different energy) from the incident photons? Wow!

But it still would not have a bearing on the direction of the reemitted photon. The banding pattern would still emerge.

So using Millsian physics could you calculate the spacing of the fringes in a double slit experiment at 540nm through slits separated by 4mm at a detection plane 200mm from the slits?

 

[hecd2]

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.

But the existing theory works exactly to predict the outcome. Why should anyone abandon that for some made up nonsense that is incapable of predicting what we observe. That would be insane.

 

[hecd2]

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.

Making it up as you go along. If only you could see yourself as others see you.

 

[hecd2]

Again, yes. It depends not on if one or many particles are going through at one time, because the interference is not with the other particles but the slits. However with things like electrons and buckyballs there would be interaction and jostling but that would probably only detract from a patterned result.

Basically Mills’s acolytes descend to the most obvious trumpery in desperate attempts to defend the indefensible. Interaction and jostling that detracts from the patterned result indeed! Do you take us for simpletons?

 

[WhatRoughBeast]

I think I answered 1) in a very recent post.

No, you did not answer it. Instead, you stated, in post 4158, that

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

In other words, you completely ignored the qualitatively different behavior of the system with two slits rather than 1.

So let me try it again. It has been demonstrated experimentally that, if a single electron or a single photon passes through a double-slit apparatus, the resulting target fringes are qualitatively different from the the result of using a single slit. The result is also qualitatively different from the results obtained by illuminating each slit separately and combining the results.

How does a single particle or photon which passes through a slit get affected by the presence or absence of a second slit?

The QM approach at least provides a mechanism for its explanation: a particle or photon has wavelike attributes which allow self-interference. While you don't like the idea, at least it provides a starting point for explaining the observed behavior.

I am giving you a chance to explain how a single Millsian photon, which is restricted to a single location at any time, can possibly be affected by the existence of another slit other than the one it is passing through. So far you have comprehensively refused to address the question.

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.

Ah, you've picked up a new catchphrase:"photon exchange". Can you provide any explanation at all (other than asserting the phrase) how this works? Specifically, what photon energies are involved? And how does this "photon exchange" (again) explain how the process is aware of the existence or nonexistence of a neighboring slit?

So far, as far as I can tell, you are simply ignoring the fact that the fringes produced by a double slit is different from the superposition of the fringes produced by two single slits operating independently. This is certainly convenient for your position ("Effect? I don't see any effect. What are you talking about? I'm certainly not going to discuss any such effect.") but refusing to acknowledge the basic facts of an effect which is well-known and extensively studied establishes you as, to put it as delicately as I can, reality-impaired.

 

[markie]

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.

Every point on the shell will feel and inward force from the proton, and thus the entire shell will feel a net inward force from the proton. The proton will feel an outward force from the electron shell, but it will be the same in all directions and so there will be no net force on the proton.

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.

I may come back to this later.

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

Well, I don't think the QM wave equation is allowing motion at all. There is nothing moving, at least nothing with any defined position and momentum.

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.

I don't know how you can believe that the shell cannot feel the presence of an interior proton as an inward force.

 

[WhatRoughBeast]

I don't know how you can believe that the shell cannot feel the presence of an interior proton as an inward force.

That is simple: he doesn't. However, you are attempting to recast the question.

The question was never about the force on the shell: the shell does feel an inward pull.

The question is, if the shell is spherical (regardless of size) and of uniform charge distribution, does the proton feel a net force in any direction? Does the net force on the proton change with position within the shell?

And the answer to both is a resounding no. This result was established by Isaac Newton, and if you think your understanding of classical physics is superior to Newton's, you have a long uphill battle ahead of you to establish this. Your failure to understand the Shell Theorem, which is a freshman physics level concept, establishes that you are simply not competent at calculus. That you had apparently not even heard of it until you were exposed to it on this thread suggests that your education in the field is appallingly incomplete at a very basic level.

 

[markie]

(snip)

According to both optiongeek and markie (and who could doubt either of them?), Mills equation (10.25) is correct, which means the calculated value reported by Mills in Table 10.1 is wrong, as is the relative error in his calculation.

Is anyone really surprised that Mills made this mistake in a direction that reduced the relative error he reported in Table 10.1?

I apologize for failing to spot Mills's error. I thank you and optiongeek for bringing Mills's error to our attention.

Well now we know Mills made a small transcription error into Table 1. But to imply this was done deliberately by Mills to slightly reduce the relative error is ... just wow.

Although markie claims to recall a problem I had "with the magnetic term result", I recall no such problem. I am absolutely certain I had not claimed any problem with the particular piece of the energy calculated by equation (10.156).

Right, it was not with the Lithium ionization energy that you (and Happyskeptic) implied Mills was just making stuff up to conform with experimental result. It was with the anomalous magnetic moment result.

 

[markie]

Yes, the photon does experience both slits. That's the basis of QM. Your claim that it goes through only one slit is most likely wrong. QM says that "knowing" it goes through one slit destroys the diffraction pattern. There are interesting results involving weak measurements of power flow in electron diffraction, but it is a mistake to say that the particle, when creating a diffraction pattern, goes through one slit.

With QM it is presumably not the actual photon going through both slits. It is the wave function, as abstract as it is. Otherwise one would have two photons instead of one, violating conservation of energy among other things.

Even if hidden-variable theory is true, and the particle went through one slit while still diffracting, we cannot know that it did. Knowing would destroy the Heisenberg uncertainty principle, and likely a number of conservation theorems. Taking one path is an observable property (position), which destroys the observable property of momentum (which gives the diffraction).

Mills theory does not involve hidden variables btw. And what you are describing, as you know, is standard fare QM.

That is why physicists work in wavefunctions and project results onto observable basis functions. They don't pretend the particle has unobserved properties until observed, because that would be pretending what cannot be tested.

In other words we can't presume to know what nature is doing when we're not looking. But QMers are presuming. They're presuming that nature actually does not afford (say) an electron travelling though space a definite trajectory, because that would imply a simultaneously real position and momentum. To which I say ... fake news.

A fun fact -- interference ONLY happens at the level of an individual particle in QM. One particle does not interfere with another (well, an entangled ensemble can interfere as a whole, but then it acts as one unit). This is hard to grasp, but if modeled correctly, it works (even virtual absorption and re-emission is just an entanglement that introduces time delay or phase shift of the wavefunction, not an observation of the original particle wavefunction).

Don't get me started on virtual photons.

 

[markie]

Basically Mills’s acolytes descend to the most obvious trumpery in desperate attempts to defend the indefensible. Interaction and jostling that detracts from the patterned result indeed! Do you take us for simpletons?

The more coherent and less noisy, the better for patterns.

 

[markie]

Where can I find the Millsian physics which predicts the result of a double slit experiment quantitatively and exactly, as exactly as it is predicted by wave interference?

In the book. But beware, it's old fashioned classical mechanics.

. That’s astonishing. So in Millsian physics, photons which are diffracted by a less reflective material are a different colour (a different energy) from the incident photons? Wow!

No, but less reflective materials would tend to absorb more photons and heat up, perhaps releasing in the infrared.

So using Millsian physics could you calculate the spacing of the fringes in a double slit experiment at 540nm through slits separated by 4mm at a detection plane 200mm from the slits?

Afaik Mills uses the same formulae and methods that produce the classical wave equation results, so yeah.

 

[markie]

(snip) the shell does feel an inward pull.

The question is, if the shell is spherical (regardless of size) and of uniform charge distribution, does the proton feel a net force in any direction? Does the net force on the proton change with position within the shell?

And the answer to both is a resounding no.

I agree with all of the above, but it doesn't appear HappySkeptic will agree with the very first part.

 

[Red Baron Farms]

Well now we know Mills made a small transcription error into Table 1. But to imply this was done deliberately by Mills to slightly reduce the relative error is ... just wow.

You seem surprised, yet you know he made up the whole cryodistilled dihydrino liquid thing as evidenced by your silence once it was pointed out.

You also know he made up the whole "we have working prototypes" thing, as evidenced by the fact that was over 10 years ago and yet still no one has ever seen it operating.

You also know last year he proposed as a goal to do things he previously claimed he already did long ago. So not only were the initial claims false, the goals for last year were also not met...yet again.

With this large amount of fraud and wishful thinking in evidence, why in the world would you be surprised to find out he could have also purposely done it on his charts as well. I don't claim it was purposeful, but it certainly could have been. It absolutely would not surprise me in the least.

I am absolutely certain that Mills is indeed a fraud. What I don't know is if he has deluded himself in a sort of "beautiful mind" way or if it is deliberately done with malice. However, it is so obvious that it actually does require him to have a pathologically delusional mind in order for it to not be purposeful.

As for his followers like you and others... I'll simply quote it yet again since he said it best:

“It's easier to fool people than to convince them that they have been fooled.”
― Mark Twain

Oh and just to be clear. I am a skeptic, so I have already researched this Mark Twain quote. He never wrote it down in that exact wording but many times he wrote the same thought. So this could have been either a verbal quote or a paraphrased "quote".

For certain he did publish this though:

"How easy it is to make people believe a lie, and hard it is to undo that work again!"- Samuel Clemens – Autobiographical dictation, 2 December 1906.

Which actually just might be more to the point regarding hydrinos.