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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.

Johannes Maria Frank
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  • "because obviously the ring does not take any damage when dropped" Are you sure about that? Have you checked with an electron microscope? – Jahan Claes May 21 '19 at 20:11
  • Related if not duplicate: Why doesn't matter pass through other matter if atoms are 99.999% empty space? – Ruslan May 21 '19 at 20:13
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    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. – Solomon Slow May 21 '19 at 20:16
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    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. – Gert May 21 '19 at 20:53
  • @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. – Johannes Maria Frank May 22 '19 at 00:30
  • @Solomon What do you mean by attached? To my knowledge the attachment of nucleus and shell happens due to the EM force alone, there are no other forces governing the outer regions of the atom. – Johannes Maria Frank May 22 '19 at 00:33
  • What do you mean by "destroyed?" Ionized? – hft May 22 '19 at 00:33
  • @hit The nucleus will be not in the center of the atom anymore. Its mass is 4 orders of magnitude heavier than the mass in the shell. The nucleus should simply hit the shell like you would hit the steering wheel if you are in a car crash and you have no seat belt fastened. The law of inertia is still valid. Almost all the kinetic energy is in the nucleus. – Johannes Maria Frank May 22 '19 at 00:39
  • The nucleus should simply hit the shell Except... that it doesn't happen of course. – Gert May 22 '19 at 02:26

1 Answers1

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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.

knzhou
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  • 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. – Johannes Maria Frank May 22 '19 at 01:36
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    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. – knzhou May 22 '19 at 01:58
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    And yes, if you apply a constant force to the Earth and wait a trillion years, you will indeed move the Sun. – knzhou May 22 '19 at 02:00
  • @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? – Gert May 22 '19 at 02:23
  • So you are telling me if I take two attracting particles and I push on one side while they are still apart that will push the other particle in the direction of the push? – Johannes Maria Frank May 22 '19 at 02:29
  • @knzhou Would you be fine if I put this question out on its own here on so. Just to be clear: If I apply a constant force to the earth each time she comes by and wait a trillion years, that will move the sun? Is that the right wording? – Johannes Maria Frank May 22 '19 at 02:36
  • @JohannesMariaFrank You've been a member for 6 years, but only asked 5 questions total, so you don't really know who is who here. Some people are top experts, others don't know much and post wrong answers. If knzhou answers your question, you should feel very lucky, because he is one of a very few here who really knows and also really understands physics. So, if he tells you that by pushing the Earth you'd move the Sun, you should give this a serious thought. Momentum is a product of force and time. Given enough time, even a moderate force can create a huge momentum. – safesphere May 22 '19 at 04:56
  • You don't have to wait a trillion years if your equipment is sensitive enough. If you push the Earth and displace it slightly, the gravitational force on the sun will change, and it will be displaced immediately. Practically speaking, :-) you would probably have to wait a trillion years to be able to detect the movement. – garyp May 22 '19 at 10:57
  • @garyp Interesting, and you really think this will PUSH the sun? The earth can only pull as far as I understood. So the question is outstanding. If you have two attracting particles and you PUSH one of them, you really think this will PUSH the other particle? Is everybody here now inventing new physics? – Johannes Maria Frank May 22 '19 at 19:43
  • @JohannesMariaFrank Here is a classical example. Take a rock on a rope and swing it rapidly in a vertical circle. Gravity provides a constant push on the rock, but the rock isn't falling down. Why? Because the tension force (an effective attraction force between your hand and the rock) does not average out to zero, it averages to the opposite of the weight of the rock. In that way, the force of gravity on the rock is ultimately transmitted to your hand. – knzhou May 22 '19 at 19:50
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    @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. – knzhou May 22 '19 at 19:53
  • @knzhou You are right with your example and yes it is basic physics. When you bind the rock to a rope you limit its degrees of freedom and thus simple vector arithmetic gives you the resulting force acting on the rock. But unfortunately it does not apply as there are no ropes attached to either electrons or the earth. So instead of claiming I do not understand something, explain the question I posed: Two attracting particles, and you push (or accelerate) one of them in addition to the acceleration due to the attraction, you claim this pushes the opposite particle? – Johannes Maria Frank May 23 '19 at 09:08
  • @knzhou I want to add that I appreciate the time you spend on this. It is a difficult question. And you are also right that the opposite particle under the real circumstances has to be pushed, because the only way to push directly is the EM force. But the closer particle will always be affected to a larger effect and thus the question of how to stop the nucleus with a 10**4 heavier mass is still in question. Go away a little from the textbook and start thinking a little more towards falsification. – Johannes Maria Frank May 23 '19 at 09:18