Merged Theory links Newtonian/Quantum the amazing Frank Znidarsic

[Evilgiraffe]

Here is everything in a pdf
http://www.wbabin.net/ntham/davis.pdf
it explains why the electron doesn't ever crash into the nucleus

I always thought it was because the probability density function, |ψ|², for the electron tends to zero as the radius tends to zero. Effectively the electron spends no time at the nucleus and hence can't crash into it.

If this is wrong, could someone with a better understanding of QM clarify it for me?

 

[JCM]

I always thought it was because the probability density function, |ψ|², for the electron tends to zero as the radius tends to zero. Effectively the electron spends no time at the nucleus and hence can't crash into it.

If this is wrong, could someone with a better understanding of QM clarify it for me?

The paper is stating

The electron has to transition from parent state to daughter state through the transitional (collapsed) subset of the wavefunction. There is no collapsed transitional wavefunction smaller than ground state transitional orbital (rc for the hydrogen atom), because that is the smallest multiple that it can be; only positive integers are allowed for the factor of n. Because there is no transitional state for it to transition to from the ground state, in essence; there is no daughter state for it to transition into during it's transitional (collapsed) wavefunction progression. This essentially locks the electron into the ground state orbital (it can only jump higher, not lower), rendering it into a state of perpetual motion; that if left undisturbed, will stay in motion until the end of the universe.

 

[Dave Rogers]

That's just a regression to the Bohr model of the hydrogen atom, which invokes quantisation without any reasoning. It's less of an explanation than QM.

Really, WTF is so incomprehensible or unbelievable about the concept of a standing wave?

Dave

 

[JCM]

That's just a regression to the Bohr model of the hydrogen atom

The Bohr radius has a real significance to reality, because the point at which the electron cloud's wavefunction predicts the highest probability of finding the electron directly matches the radius predicted by Bohr

In every Bohr orbital the electron carries an integer multiple of Planck's constant. This is because the period of the harmonic motion of the electron equals 2pi / omega

The fine structure constant is the factor between the ground state Bohr radius of hydrogen , the electron's reduced Compton wavelength , and also the classical electron radius

 

[Dave Rogers]

The Bohr model isn't insignificant, it's incomplete. The point is that QM explains both the quantisation of energy levels and the absence of radiation due to the orbital motion of the electron, because the electron orbital is a standing probability wave; in effect, there is no orbital motion. The model you're proselytising discards that explanation and doesn't present an alternative one; it just asserts that electron motion is quantised so the transition can't occur. It doesn't attempt to resolve the contradiction between this assertion and the predictions of classical electromagnetic theory. At least Bohr was honest enough to admit that, rather than pretend to have solved it.

Dave

 

[Dancing David]

Frank's classical radius of the electron shows that the fields of matter are pinned into the structure of matter at that classical radius of the electron like you pin fields in to a superconductor by adding discontinuities. With that model you get the phase velocity equals the velocity the phase velocity equals the velocity of light the group velocity equals the movement of the structure and all of special relativity emerging under that condition
Looking at the Compton frequency, the De broglie wave being the Doppler shifted reflected Compton frequency emerges naturally from this model out of that condition

I see , you are saying he has fudged in classical features, well I guess then he can explain that. And why QM despite all attempts does not match classical models.

The electron probability may have what you could call a 'classical radius', but I am not sure it is really appropriate.

ETA: I see that this also relates to the probability and the radius in Bohr's model, so others have addressed that this is then incomplete. So not so much a fudge as a return to an incomplete model.

 

[Dancing David]

True or not?

The speed of light is invariant, the travel of a photon through a medium is not. There is a difference.

Maybe you should start with Feynman and refreactive indexes.

 

[MattusMaximus]

I was thinking this as well haha. I have gotten ahead of myself and enjoy the jrefs ability to help refine my viewpoint to something more coherent or at least logically presented


I can provide a rough powerpoint Znidarsic made and the papers and the video below which is Frank himself. I obviously cannot offfer a summation for a damn so I will not

Its hard to find the 1.32fm proton radius too.

This is Znidarsic's 25 minute Lecture at John Hopkins only himself speaking. The lecture is old
Part One
Part Two

Dude, I read through Znidarsic's paper, and I found it to be... crap, essentially. Linking to more crappy powerpoints and Youtube videos that contain the same errors and bad arguments isn't going to convince us of anything, except maybe that the crap pile is bigger than we thought.

What you have to do is actually address the criticisms brought up here by those knowledgeable in physics, as opposed to ignoring the criticisms and acting like a broken record

 

[MattusMaximus]

I always thought it was because the probability density function, |ψ|², for the electron tends to zero as the radius tends to zero. Effectively the electron spends no time at the nucleus and hence can't crash into it.

If this is wrong, could someone with a better understanding of QM clarify it for me?

Yup, that's basically it.

 

[MattusMaximus]

The Bohr radius has a real significance to reality, because the point at which the electron cloud's wavefunction predicts the highest probability of finding the electron directly matches the radius predicted by Bohr

In every Bohr orbital the electron carries an integer multiple of Planck's constant. This is because the period of the harmonic motion of the electron equals 2pi / omega

The fine structure constant is the factor between the ground state Bohr radius of hydrogen , the electron's reduced Compton wavelength , and also the classical electron radius

JCM, do you even understand that the Bohr model is, at best, only an approximation for hydrogen atoms - and only hydrogen atoms? As soon as you introduce another electron into the system, Bohr's model goes right out the window. If you want to do anything useful with electrons and more complex atoms, you have to go the route of dealing with electron clouds, probability distributions, and wavefunctions; all of which have much more practical relevance to real life than does the Bohr model.

Apparently, Znidarsic doesn't understand this fact either, yet he wishes to take down QM; it's as bad as creationists wanting to tear down evolution, yet they have no clue about the very theory they are against

 

[Skwinty]

Pedantic question here wrt why the electron doesn't crash into the nucleus.

If the Copenhagen interpretation is being used to explain why the electron stays away from the nucleus, then the Copenhagen interpretations claim to only correlate statistical patterns with experience has been stretched, because that interpretation is only describing the statistical correlation to the observation.

Would it not be more accurate to say that the Heisenberg uncertainty principle and the Pauli exclusion principle is the reason.

End of pedantic question.

 

[Evilgiraffe]

Yup, that's basically it.

Thanks Mattus. It's nice to know that at least some of the QM lectures lodged in my brain.

 

[RussDill]

Its pretty clear whoever wrote the paper hasn't studied much physics at all:

One of the greatest mysteries in quantum mechanics is why the energy contained in photon is proportional to its frequency.

It is no such mystery. It is actually really, really clear why it must be so. The rest of the paper contains similar straw men.

And of course more misunderstandings:

A photon in free space is known to have a wavelength, but standard
science does not consider its wavewidth to be relevant. The width
of the wave of the uncollapsed photon is known to exist though, as
the two-slit experiment proves.

[...]

A photon is an oscillating positive and negative charge. In the above
gure, on the y-axis (vertical) above the zero line of the photon rep-
resents a positive charge. When the photon dips below the y-axis,
its charge becomes negative.

Really JCM? Really? You read this and went "WOW!"

 

[Vorticity]

I always thought it was because the probability density function, |ψ|², for the electron tends to zero as the radius tends to zero. Effectively the electron spends no time at the nucleus and hence can't crash into it.

If this is wrong, could someone with a better understanding of QM clarify it for me?

Yup, that's basically it.

Actually, I think that's only true for states with non-zero angular momentum. All other states have a maximum of density at the nucleus.

ETA:

More info..

 

[Evilgiraffe]

It's true that states with non-zero angular momentum have a node at the nucleus and so the probability is identically zero there.

Zero angular momentum 's' states also have (essentially) zero density at the nucleus because to find the probability, one must multiply
|ψ|² by a spherical volume element 4πr²dr. This obviously tends to zero as the radius decreases, reducing the probability of finding the electron as one approaches the nucleus.

 

[Vorticity]

It's true that states with non-zero angular momentum have a node at the nucleus and so the probability is identically zero there.

Agree.

Zero angular momentum 's' states also have (essentially) zero density at the nucleus

Disagree.

You're confusing the total probability density for all three polar coordinates with the marginal probability density you get for r alone. These two things are distinct. The total density |ψ|² for an s orbital is unambiguously non-zero at the origin.

because to find the probability, one must multiply
|ψ|² by a spherical volume element 4πr²dr. This obviously tends to zero as the radius decreases, reducing the probability of finding the electron as one approaches the nucleus.

Agree and disagree. It's a funny question, actually.

If I'm dealing with the n = 1 wave function in spherical coordinates, I can ask "What is the most probable value for the radius r?" I can then calculate the probability density function (pdf) for the radius r of the electron. It has 0 pdf at r = 0, and a maximum at the Bohr radius.

Now, if I'm dealing with that same wave function in Cartesian coordinates (which is a perfectly valid - if inconvenient - thing to do), I can ask "What is the most probable value for one of the Cartesian coordinates - say x?" I can then calculate the pdf of x, and find that it's most probable position is at x = 0! (ETA: In fact, I just did a little figuring, and it looks like the pdf of x diverges logarithmically at the origin!)

Is there a contradiction between these two things? No. It's merely an artifact of the spherical coordinate system: There's "more space" at higher r than at lower r, hence the 4πr²dr factor you mention above. When we switch to Cartesian coordinates, for which the underlying "space" is homogeneous, this effect disappears.

 

[Evilgiraffe]

Ah, thanks Vorticity. So the question now is, if it's not the lack of electron density at the nucleus that stops s electrons crashing into the nucleus, what does?

 

[Vorticity]

Ah, thanks Vorticity. So the question now is, if it's not the lack of electron density at the nucleus that stops s electrons crashing into the nucleus, what does?

Lack of a discrete electron.

If your electron is a nebulous cloud of probability rather than a discrete classical particle, what would it even mean for the electron to crash into the nucleus?

 

[Tubbythin]

Ah, thanks Vorticity. So the question now is, if it's not the lack of electron density at the nucleus that stops s electrons crashing into the nucleus, what does?

Nothing. Or at least the wave-function penetrates the nuclear interior. "You" can actually use this to measure some nuclear properties. Roughly, the wavefunction and corresponding atomic energy level will differ between a point nucleus approximation and one with a finite size that you measure (think Gauss' law). You can then use that difference to work out (with a lot of maths) the charge radius of the nucleus in question.

 

[MattusMaximus]

Thanks for the correction & clarification Vorticity!

Lack of a discrete electron.

If your electron is a nebulous cloud of probability rather than a discrete classical particle, what would it even mean for the electron to crash into the nucleus?

Not to mention, you have to recall that in this view we're looking at the electron in terms of a wavefunction, and it is a fundamental property of waves that when they interact they pass through each other and they are non-localized. Perhaps not the best way to think about it, but certainly better than thinking of the electron as a billiard ball (at least in this case).

ETA: What Tubbythin said