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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

by alok-g 1y ago

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?

rcxdude 1y ago
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.
- alok-g 1y ago
  >> 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.
  - ben_w 1y ago
    > 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).
    - alok-g 1y ago
      >> 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.
      - ben_w 1y ago
        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.
        alok-g 1y ago
        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.
        ben_w 1y ago
        The net acceleration in cartesian coordinates is zero.
Ygg2 1y ago
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.
- alok-g 1y ago
  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 https://news.ycombinator.com/item?id=43819506
mr_mitm 1y ago
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.
- alok-g 1y ago
  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.
  - mr_mitm 1y ago
    Oh, I thought you were asking something else. In a spiral motion, the infinities never come into play. Nevermind.

by alok-g 1y ago

rcxdude 1y ago

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.

alok-g 1y ago
>> 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.
- ben_w 1y ago
  > 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).
  - alok-g 1y ago
    >> 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.
    - ben_w 1y ago
      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.
      - alok-g 1y ago
        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.
        ben_w 1y ago
        The net acceleration in cartesian coordinates is zero.

Ygg2 1y ago

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.

alok-g 1y ago
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 https://news.ycombinator.com/item?id=43819506

mr_mitm 1y ago

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.

alok-g 1y ago
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.
- mr_mitm 1y ago
  Oh, I thought you were asking something else. In a spiral motion, the infinities never come into play. Nevermind.