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When a particle moves on a gravitational potential subject to no contact and tidal forces, the particule clearly is in inertial movement.

But what about the "motion" of an electron bound into an atom around its nucleus?

I know that in quantum mechanics we cannot strictly speak of a classical trajectory, of the electron around its nucleus. But there is motion, as many electronic atomic states have non-zero expected values for $\hat{\mathbf{p}}$ and even for $\hat{\mathbf{L}}$. So there is some type of motion.

My question is: in an isolated atom, say H, under no external forces, is the electron in inertial motion?

EDIT: To be more specific, if I could install an accelerometer on the atom's electron, would this accelerometer read zero (inertial frame) or non-zero (non-inertial frame).

Does the electron free fall in its "orbit" around the nucleus?

Arc
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2 Answers2

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An planet moving around a sun is considered to be in an inertial frame because of the equivalence principle, which implies that cancellation of inertial mass and gravitational charge in the Newton's law* is exact. So, regardless of its mass, an object in a gravitational field experiences the same acceleration. This precludes detection of such acceleration by an accelerometer.

On the other hand, an electric charge moving in an electric field will have different acceleration depending on its charge-to-mass ratio. This lets one detect acceleration by some kind of accelerometer, which could contain particles of different charge-to-mass ratios and measure their relative motion to infer acceleration. This fact doesn't depend on whether we talk about a classical or quantum particle.


* Newton's law is classical, but it's a valid small-field approximation to GR.

Ruslan
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  • Thus, due to the presence of the electric field, electron motion is non-inertial, check? – Arc Feb 15 '22 at 16:49
  • @Arc exactly so – Ruslan Feb 15 '22 at 19:56
  • Ah! Now I get another point of your argument (2nd paragraph): for gravity alone we have $m a = m g$, so the mass cancel out to give $a = g$, and if the mass is gone then accelerometers measure zero. But for an electric field $m a = q E$ and thus $a = (q/m) E$, this is the charge-to-mass ratio you mention, so the mass is not gone and accelerometers measure that ratio, and thus the frame is non-inertial. Check? – Arc Feb 15 '22 at 20:33
  • I guess this is the reason why synchrotrons produce radiation: the accelerated charged particules are in non-inertial motion, whereas a charged particule in circular orbit with gravitation does not emit radiation because it's in inertial motion, albeit it's angular motion. – Arc Feb 15 '22 at 20:34
  • @Arc this is right, although I don't know of any experiment that was actually performed to check this scenario with a charged particle in a gravitational orbit. – Ruslan Feb 15 '22 at 20:46
  • I know there's a lot of controversy about this issue, it seems to be a paradox, solved by Fritz Rohrlich: "... shows that a charged particle and a neutral particle fall equally fast in a gravitational field. Likewise, a charged particle at rest in a gravitational field does not radiate in its rest frame, but it does so in the frame of a free falling observer." – Arc Feb 16 '22 at 01:57
  • And Boulware later about radiation emited: "that the radiation goes into a region of spacetime inaccessible to the co-accelerating, supported observer. In effect, a uniformly accelerated observer has an event horizon, and there are regions of spacetime inaccessible to this observer." – Arc Feb 16 '22 at 01:59
  • Intuitively, the electron on a gravitational field doesn't know it's being accelerated, it feels weightless, no extrenal compression or dilation promoted, and can't distinguish between gravity and empty space. So, I think it doesn't radiate in gravitational orbit. – Arc Feb 16 '22 at 02:04
  • There have been many discussions here about this very interesting problem (for example, see here, here, and here). But anyways, your answer has been very helpful! Thanks. – Arc Feb 16 '22 at 02:04
  • @Arc if this answer resolved your question, you can accept it by clicking the tickmark to its left. – Ruslan Feb 16 '22 at 07:55
  • Yes! You did answer the question, but I usually sleep over answers a night or two before accepting them (a practice I learnt from StackOverflow). Regarding this matter - and within the context of your answer and our discussion here on the commentaries - I have posted a related question. – Arc Feb 16 '22 at 16:48
  • Ruslan, could you please clarify how could you define the quantum acceleration operator for an electron's bound state in the hydrogen atom, i.e. in the context of the OP question? I admit I'm drawing a blank here. – kkm -still wary of SE promises Feb 17 '22 at 00:39
  • @kkm, see here. – Arc Feb 17 '22 at 04:02
  • @Arc, thanks, makes some sense. So you solve the t-dependent SE. That's interesting, wondering what types of (meta)stable solutions may exist. E.g., the s-orbital ringing with surface waves like a struck drum? Something interesting happening with Rydberg states, like the orbital's precession? Not all initial conditions are physically attainable, but math is interesting. But I'm even more curious what this "acceleration" means if you apply it to a state. It's complex-valued and spread across space. It's not a physical acceleration for sure, since the bound electron doesn't radiate. Weird. :) – kkm -still wary of SE promises Feb 19 '22 at 04:02
  • @kkm, agree, it can become beautifully weird! We are too used to textbook examples, If you put t-dep into play all sorts of weird phenomena will happen. Nice picture of an "s-orbital ringing with surface waves like a struck drum". But if you consider that t-dep SE solutions are a t-dep linear combination of t-indep SE solutions, you may even have an atom ringing between s-, p-, d-, etc. orbitals, in a cyclic shape-shifiting mix of orbitals, for example, after a high energy collision with a wall. I can even picture delta spikes peeking at random far away from nucleus from time to time. – Arc Feb 19 '22 at 16:49
  • You can have orbital precession in a Rydberg H, which causes nonhydrogenic properties (see this paper and this visualization... figure 6 is the orbital ringing you wanted :) As to the acceleration, well, formally I'd say it's $d_t

    = <F(x,t)>$, so for highly localized momentum it approximately follows the classical acceleration, but with caveats. For a visualization see these wavy plots.

    – Arc Feb 19 '22 at 16:58
  • @kkm, by the way, concerning our discussion on frames of reference on quantum mechanics, it seems we were both wrong... the answer is neither yes or no, it is far more complicated than that: Quantum mechanics and the covariance of physical laws in quantum reference frames. There can be inertial and non-inertial frames of reference in QM but they can be in a coherent superposition of frames, which render effects like entanglement and superposition frame-dependent! :O – Arc Feb 19 '22 at 17:04
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In classical physics and special relativity, an inertial frame of reference is a frame of reference that is not undergoing acceleration. In an inertial frame of reference, a physical object with zero net force acting on it moves with a constant velocity (which might be zero)—or, equivalently, it is a frame of reference in which Newton's first law of motion holds

Italics mine.

The moon around the earth is not in an inertial frame, as it is undergoing acceleration, $d\vec p/dt$, $\vec p$ the vector of momentum, going around the earth. Thus even in the Bohr model which has the electron rotating about the nucleus, the electron is not in an inertial frame.

Edit after long discussions in comments:

Arc in a comment:

theories overlap, so we should make them at least partially consistent, that's why I believe we should be able to speak of inertial frames in quantum mechanics, at least in some sense that matches classical mechanics on these boundaries

Quantum mechanics is a probabilistic theory. It can only calculate the probability of a specific event happening at $(t,x,y,z)$ , with the four vector at $(E,p_x,p_y,p_z)$ for energy momentum through the calculations of the wavefunction of the system.

The classical and quantum frames overlap for free electrons, whose "track" can be fitted with classical formulas in the detection chambers. An electron with no fields in the chamber goes in a straight line, and an inertial frame can be calculated. ( putting accelerometers is out of the question because quantum mechanics enters and the state of the electron changes by the existence of a macroscopic machine with billions of molecules.)

The theories should overlap and give the same predictions where they hold , within the limits of the variables where they are valid. Quantum mechanics theories became necessary because Newtonian mechanics was unable to describe the measurements in small dimensions, at the level of single electrons , molecules etc.

Quantum mechanics,was invented as a theory eventually , starting with the inability of classical mechanics and electrodynamics to model black body radiation, the photoelectric effect and atomic spectra.

A bound electron, if measured, at time t will have a probability of being found at rest with the laboratory, at a given (t,x,y,z) that is all. Probability. where it will be at t+dt is also a probability. That is why there are no orbits in the atom, but orbitals.

orbitalshydr

anna v
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