I made the following thought experiment: Dropping a gold ring on a wooden table. It drops, hits the table, bounces off, hits again with less velocity and so on until it finally rests.

Now consider an gold atom inside the ring. It will of course be accelerated and there is no problem with the nucleus and shell having a huge mass difference as the gravitational acceleration is independent of the mass.

Assume it is a carbon atom that is hit when the ring hits the table. This is reasonable as there are a lot of carbon atoms in wood.

Now the only force that can stop our ring is the electromagnetic force, since we only have four forces, there is no anti-gravity and the weak and the strong force do not extend to the outer shell. From the geometry of the two atoms the one shell interacts therefore with the other shell first.

The gold atom is a lot heavier than the carbon atom so the carbon atom will start moving and will in turn move other atoms which distributes the force so the counter force starts to increase and in the end will balance forces which brings the gold atom to a halt and then even pushes back so the ring bounces back. The rules are governed by Hooke’s law, the table acts like a spring.

But the atom is not a solid sphere, it is like a solar system with all the mass centered in the center. And here I am not understanding how this can actually work.

If the electromagnetic force is stopping the atom it can only act on the shell first (because of the speed of light being finite) and therefore the nucleus is simply continuing to follow his trajectory because of the law of inertia. It is thus suddenly pushed out of the center of the atom and even if I ignore that now one side of the shell is pulling harder on the nucleus than the other, the shear difference in mass must just lead to the nucleus crushing through the shells of several atoms.

It is like trying to stop a Mercedes by pushing against the star mounted on the bonnet.

So what is preventing the atom from being destroyed? How is the force that stops the shell actually put on the nucleus, because obviously the ring does not take any damage when dropped.

  • $\begingroup$ "because obviously the ring does not take any damage when dropped" Are you sure about that? Have you checked with an electron microscope? $\endgroup$ May 21, 2019 at 20:11
  • $\begingroup$ Related if not duplicate: Why doesn't matter pass through other matter if atoms are 99.999% empty space? $\endgroup$
    – Ruslan
    May 21, 2019 at 20:13
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    $\begingroup$ Re, "It is like trying to stop a Mercedes by pushing against the star mounted on the bonnet." Perhaps you underestimate how strongly the "star" is attached. A gold atom is not a teeny-tiny Mercedes Benz, and the forces that hold its component parts together operate on a much different scale from the forces with which you are familiar on a human scale. $\endgroup$ May 21, 2019 at 20:16
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    $\begingroup$ If the electromagnetic force is stopping the atom it can only act on the shell, and therefore the nucleus is simply continuing to follow his trajectory because of the law of inertia. No. The Coulombic forces act on the electrons AND the nucleus. It's a reciprocal force, so the nucleus is not free to fall AT ALL. $\endgroup$
    – Gert
    May 21, 2019 at 20:53
  • $\begingroup$ @Gert Thanks for pointing this wording imprecision out. Of course the nucleus will at one time be exposed to the EM force but when he does significantly the shell is already under a very strong force due to the nature of the 1/r**2 of the EM force. $\endgroup$ May 22, 2019 at 0:30

1 Answer 1


The appropriate intuition here is that small objects operate on faster timescales. You might not be able to stop a Mercedes by pushing on the star instantly, but you certainly can if you gradually push on it for a couple of centuries.

In the case of an atom, the appropriate timescales are given by the de Broglie relation $E = \hbar \omega$, so $$t \sim \frac{1}{\omega} \sim \frac{\hbar}{E} \sim \frac{10^{-34} \, \text{J s}}{1 \, \text{eV}} \sim 10^{-15} \, \text{s}.$$ If an impact takes a few milliseconds, then in a classical picture, during the collision the electron can go around the nucleus a trillion times. In our solar system, the equivalent timescale for the Sun and the Earth would be a trillion years.

The same intuition holds in the quantum case. The collision is not sudden at all, and there's no reason that impulse can't be gradually transferred from the electron to the nucleus. In fact, for both the classical and quantum cases, this intuition can be formalized by the adiabatic theorem, whose conditions are satisfied extremely well here.

  • $\begingroup$ Are you kidding me? What kind of impulse transfer? Between a positive and a negative particle? And according to your answer pushing against the earth will stop the sun? May I remind you that the absolute difference in mass between an electron and a gold nucleus is 5 orders of magnitude. $\endgroup$ May 22, 2019 at 1:36
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    $\begingroup$ Well, you didn’t have to remind me, I was perfectly aware of that. The point is that it still works out, just because of the huge ratio of timescales involved. If you don’t believe it’s intuitive, you can look up more formal treatments. For example, the idea that electrons can push nuclei around is the foundation of the Born-Oppenheimer approximation. $\endgroup$
    – knzhou
    May 22, 2019 at 1:58
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    $\begingroup$ And yes, if you apply a constant force to the Earth and wait a trillion years, you will indeed move the Sun. $\endgroup$
    – knzhou
    May 22, 2019 at 2:00
  • $\begingroup$ @JohannesMariaFrank: with all due respect Sir but you are making a bit of a fool of yourself. Just because you don't seem to understand what the answer is about doesn't make it wrong, you know? $\endgroup$
    – Gert
    May 22, 2019 at 2:23
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    $\begingroup$ @JohannesMariaFrank By the way, this is all at the level of a first physics course. You need a little more humility. I get that you can code and so you understand basic logic, but that doesn't automatically mean you know everything. If you don't understand something from first year physics, it doesn't mean that all of physics is wrong, it means you don't know physics. $\endgroup$
    – knzhou
    May 22, 2019 at 19:53

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