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Comment by alok-g | Hacker News Reader

I have come across the same 'explanation' multiple times but do not find it satisfactory.

>> This "battle of the infinities" cannot be won by either side, so a compromise is reached in which theory tells us that the fall in potential energy is just twice the kinetic energy, and the electron dances at an average distance that corresponds to the Bohr radius.

This "cannot be won by either side" isn't a good explanation of the problem. By saying "a compromise is reached in which theory tells us ...", the article seems going around the answer instead of actually telling the answer to the question.

Assuming relativity is not invoked and given a finite size of the nucleus, what is actually bringing this situation of compromise? What is pushing the electron (or its probability cloud) outwards to counter-balance?

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The opposing effect is the kinetic energy of the electron: as it get squeezed into a smaller area its kinetic energy must go up, assuming energy is constant (a classical system would more or less be the same, think how an elliptical orbit speeds up as it gets closer to the barycenter, but a quantum mechanical system doesn't necessarily have a defined trajectory).

The real sleight of hand in this explanation is it doesn't really explain why the quantum-mechanical electron doesn't radiate energy like the classical electron. That's a bit more subtle, but it's basically due to the fact that the electron is wave-like and confined within the atom (i.e. it does not have enough energy to escape the nucleus). This means that the wave-function has discrete solutions as a series of 'standing waves', which is what causes the quantization in quantum mechanics.

>> The real sleight of hand in this explanation is it doesn't really explain why the quantum-mechanical electron doesn't radiate energy like the classical electron. That's a big more subtle, but it's basically due ...

This subtle part is exactly the piece that I am missing.

So far, we have electron in an orbital. The orbital in itself seems stationary (standing waves), so it may seem nothing is accelerating. I presume that is not it for the answer.

If it is, that still feels like, Oh, I solved Schroginger's equation, and it yields these standing waves for the solution, electron is not really having a certain position and momentum as the wave function hasn't collapsed, and thereby there is no acceleration for the electron to be radiating energy.

Asking a total newbie question. I heard that QM has operators, so momentum also has an operator. Is there one for acceleration? Would it yield a zero?

If so: I see Coulomb's force. Is there another force balancing it? Or is Newton's third law not valid anymore.

> Would it yield a zero?

The electron's position is smeared out over the entire orbital, so in this sense it's being accelerated in all cartesian directions equally — sure, in spherical coordinates, it's always in the radially inwards direction, but it can't go "in" any further because its own wave function is already there, blocking it.

(Operator specifically: IDK if there is one for acceleration, but this is why electrons can't be pushed in once they're at the final energy level).

>> but it can't go "in" any further because its own wave function is already there, blocking it.

This is good food for thought for me. Though I am still not fully clear. If the above were the setup with classical mechanics and electrostatics, with nuclear charge at the center and electronic charge smeared out, I would use Gauss theorem to find electric field and still find that the smeared charge would feel a net force towards the center and want to move that way.

If you squeeze the wavefunction in space, you spread it out in velocity; the classical analogy would be that beyond a certain point, the more it actually moves inwards, the faster it has to be moving outwards.

The relevant piece here is acceleration. I understand the rest. The problem in classical explanation is acceleration resulting in energy loss to radiations. With QM, we have an explanation without that radiation, but I am not clear how and why that is, as posed in my questions above.

The net acceleration in cartesian coordinates is zero.

Am genuinely trying to understand.

It's not zero in other coordinate systems? Or you are saying it does come out zero in all the systems. If the latter, I would love to read more including the maths. :-)

Thanks.

Coordinate grids are arbitrary, for any system you can always put a different grid on the same physical reality.

It's just intuitively clear with a cartesian system that there's no net acceleration in a way it might not be for spherical coordinates. If you imagine a sphere, attach to all points of that sphere a normal unit vector pointing inwards, then all points on the sphere are "pointing inwards"; but as each point on the sphere has a corresponding point opposite it, where "inwards" means the exact opposite in cartesian coordinates, the sum of all those inward vectors is exactly zero. With spherical coordinates, you have to convince yourself that the vectors really do cancel. But they're the same vectors described differently.

Just to be clear though: these are still all euclidian geometry, just with different gridlines drawn on them. As yet, nobody has fully solved QM for curved spaces.

I understand. Thanks!

Will try to find the math that shows acceleration in the cartesian coordinates (and with Euclidian geometry) comes out to be zero. That then hopefully will help me understand how is QM solving the original problem.

Here is a simple explanation: They can't fall in due to Heisenberg's uncertainty. For that, I'd need you to accept as an axiom[1] (unprovable truth) that the following principle is true: ΔpΔx (uncertainty of momentum times uncertainty of position) is roughly equal (or greater) to some constant, we'll call it H.

Due to this, if one uncertainty grows smaller, the other grows larger.

If the electron is located in the nucleus, its position (Δx) would be much more narrow than if it was around the nucleus. Since the uncertainty of Δx goes down, Δp must go up co compensate. Turns out, this Δp is enough to give it an enough momentum to overcome attractive force.

But then comes a smart observer, and says, but what if an electron managed to lodge itself exactly into the center of a proton. Since Coulomb's law says F= q1q2/r^2, and r is 0[2], that's Infinity! You can't escape infinite charge attractiveness!

To that, you can notice that as r and Δx approaches 0, Δp also approaches infinity, so it will have more chance to escape before that happens. But in some rare cases, it will interact and form a neutron, with some energy being emitted as a neutrino.

[1] It's not an axiom, it's an observation derived from experiments. However, why is the matter behaving like that out of physics wheelhouse. It can tell you a lot about laws, but very little WHY are laws like that. So it might as well be an axiom.

[2] This is, of course, assuming that space is infinitely divisible, which is yet to be confirmed or denied.

I have heard this too and still feel that there's more to this that is critically missed.

>> For that, I'd need you to accept as an axiom (unprovable truth) that the following principle is true: ΔpΔx is roughly equal (or greater) to some constant, we'll call it H.

As far as I know, Heisenberg's principle isn't exactly an axiom. It can largely be derived!

A waveform in time and its Fourier transform have such a relationship, which is entirely mathematical. More confined a time waveform is, more spread out the frequency spectrum is, and vice versa.

Now de Broglie principle linked momentum of a particle to its frequency! That makes the trick. So far, in Newtonian mechanics, momentum just related to position via derivative of position, i.e., velocity. Both were time waveforms.

But with de Broglie principal, momentum (also?) links to frequency domain, while the position remains in time domain. Now the Schrodinger's equation yields the said relation between position and momentum.

What I'll write next fits better in another comment box, so will do that.

(Edit:) Wrote here: https://news.ycombinator.com/item?id=43819506

How can something be an axiom and at the same time has a mathematical derivation well understood from Quantum Mechanics [1] and are supported by experimental observations ?

[1] https://courses.physics.illinois.edu/phys580/fa2013/uncertai...

> How can something be an axiom and at the same time has a mathematical derivation well understood from Quantum Mechanics

See my notes, it's not an axiom, just take it for granted for the explanation to work. You can derive it from experiments like squeezing light, or you can derive it from other things as you mentioned.

The idea of axioms in mathematics is not quite the same as it is with physics.

Physicists don't need the type of mathematical rigor such as the set theory axioms to build up all of mathematics. In fact, by keeping in mind the multitudes of paths from which one part of physics could be related or derived from another part, you are exploring the possibly undiscovered laws in nature.

It will only be relevant to axiomatically define physics (from a set of basic axioms to reach all laws) after we confirmed to have discovered all of physics. That day hasn't come yet - there's no grand unified theory so far, and there's still stuff that we dont know surely.

The way I see it is that it could only actually collapse if the electron was headed precisely at the nucleus (the nucleus has diameter zero in this simplified model). Even in classical physics, the electron would move on an ellipsis around the nucleus in almost all cases, so the average distance would be non-zero.

Being pointed exactly at the nucleus is not possible in quantum mechanics, because that would require infinite precision, which is precisely what we give up in this model, so the average distance is yet again non-zero.

This explanation is not correct. Here, the electron is again accelerating, will lose energy, and ultimately spiral into the nucleus. This does not require it to be pointing exactly into the nucleus.

Oh, I thought you were asking something else. In a spiral motion, the infinities never come into play. Nevermind.

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