If I take a small piece of matter and shake it for a long time, then don't all its electrons and protons emit electromagnetic radiation and lose energy?
To me it seems the rest of the article, i.e. probability density, also means that the electrons undergoe acceleration because their location changes.
The energy they lose by radiating is exactly balanced by the energy they gain from you by accelerating them.
Yes, they do. But it's only a very very tiny amount.
You're shaking both the positive (nucleus) and negative (electrons) parts of the atoms. The electromagnetic fields outside of the atom almost perfectly cancel out. This means that the acceleration only affects the very small residual field and there is very little radiation at far distances, unless you shake it really really fast.
Another way to look at it is that atoms are in fact already constantly shaking by themselves because of thermal movement. And this does in fact result in them emitting radiation: objects at room temperature will radiate in the infrared spectrum because of this.
The trick is that thermal movements of atoms are very fast compared to moving something by hand. Room temperature infrared radiation frequencies are on the order of tens of terahertz. Compare this to manually shaking an object at (say) 10 times a second. With some hand waving math this is like giving it an equivalent temperature of around 10^-10 kelvin. It will radiate extremely long wave radio waves, but such a tiny amount that it will be completely unmeasurable.
If that was a truly workable theory it would have verifiable predictions and compatibility with all prior observation and it would be eagerly explored. Scientists as a whole like to find anomalies and discrepancies; it is what keeps them in business. There is no "conspiracy".
> BLP has announced several times that it was about to deliver commercial products based on Mill's theories but has never delivered any working product.
A skeptic merely rejects claims without evidence.
It's always possible most of them already have done so :0
Seems to me the positive & negative charges would neutralize, and it would no longer be "matter" as we know it.
Not anti-matter, just not actual matter by any stretch of the imagination.
Which is what most of the universe actually consists of, not actual matter.
Matter itself is extremely rare, few, and far between.
Could be all there is left and can be perceived by those composed of it, are the things that can have a physical nature simply because their electrons haven't collapsed into the nucleii.
Yet.
https://www.thegreatcourses.com/courses/einstein-s-relativit...
It is on Kanopy now, for “free” (local library pays). I think there’s a subatomic-particle specific course as well that is more focused.
>> 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?
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.
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.
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).
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.
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.
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.
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.
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.
>> 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
[1] https://courses.physics.illinois.edu/phys580/fa2013/uncertai...
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.
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.
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.
I find it much easier to understand the stability of hydrogen by thinking about ground states. If the electron in H could merge with the proton, they'd make a neutron. But a free neutron is not stable and decays in a few minutes back into a proton, electron and an antineutrino.
With very large, proton-rich nuclei, eventually you get to the opposite situation where the ground state is one less proton, one less electron and one more neutron and the atom decays into that state by, you guessed it, an electron "falling into" the nucleus.
The question of why only makes sense in a framework. Classical EM theory says electrons should fall into the nucleus, but they clearly don't. Asking why they don't is one step towards finding out about quantum mechanics, so it's a really important question to ask.
Yes, at the end of the day, the answer is "because classical EM is wrong and quantum mechanics is right (or at least less wrong)". But there is a lot to learn along the way.
For example you can make a very similar system with a positron and an electron https://en.wikipedia.org/wiki/Positronium . It's stable for a while, and it has very similar properties and levels like an Hydrogen atom. So the "second law" should apply too. But it has a mean lifetime of only 0.12 ns (1.2E-10s).
The second law consider also the properties of the materials. For example there are small bags with sodium acetate that are used as hand heaters https://en.wikipedia.org/wiki/Sodium_acetate#Heating_pad https://www.youtube.com/watch?v=Oj0plwm_NMs You put them in hot water, the content dissolves, and you let them cold to room temperature. When you click the metallic part, they crystallize and release heat.
Another example is connecting a small lamp to a battery. It release light and heat, specially if you can get an old incandescent light bulb.
so, why doesnt an electron fall into a nucleus?
because the nucleus is too small to encompass an electron wave.
senectus1•9mo ago
aabhay•9mo ago