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Is the electronic interaction the key idea fundamentally underlying the action and reaction law of Newton?

By electronic interaction I mean the Coulomb interaction between charged particles. At a microscopic level, whenever we push on something, we feel a resistance, an equal reaction from the object exerted back onto us. So with this in mind, I can reformulate my first question as follows: had there not been any Coulomb-like interaction between charged particles (equals repulsing, opposites attracting), there would not have been an action-reaction law, right?

So this is saying that a fundamental understanding of "why" there's a reaction force as stated in Newton's law, lies in the nature of matter, namely that ultimately, it is composed of charges that, in most cases completely cancel each other out (statically neutral elements in everyday objects), but when these objects are deformed e.g. by us applying a force on them, the charge distribution is modified and brought out of its overall neutral equilibrium, so the reaction force is in fact the resistance of the object in response to a perturbation of its electronic structure and neutrality.

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  • $\begingroup$ You make very valid observations. I'll add another observation: Newton's third law is valid also for the force of gravity. There's no special connection between the third law and the electrostatic force. $\endgroup$ – garyp Feb 28 '17 at 17:42
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    $\begingroup$ @garyp Please make this an answer. $\endgroup$ – ZeroTheHero Feb 28 '17 at 18:29
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    $\begingroup$ Note that for general electromagnetic interactions, momentum is not conserved unless you factor the momentum carried by the field, and Newton's Third Law does not hold, unless you factor all of this carefully. $\endgroup$ – Jerry Schirmer Mar 1 '17 at 19:26
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    $\begingroup$ @JerrySchirmer thank you for the note. Unfortunately my confusion is deeper and more basic than that: let us step down from the general case, e.g. lets use my daily life example of pushing onto an object: is in this case the action and reaction forces electromagnetic in nature at a microscopic level? And the fact that there's a reaction force (here) is a consequence of Coulomb interaction of charges? in turn preserving momentum conservation. I recall Feynman once explaining all such interactions (exerting a force on an object, car, pillow,...) are of EM nature, but cannot find it anymore :( $\endgroup$ – user929304 Mar 2 '17 at 15:41
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    $\begingroup$ @user929304: fundamentally, yes, you are correct. But thinking about that to understand simple contact forces is going to be more confusing than elucidating. Fundamentally, the idea is that, imagine sitting on your back, on some ice. Now, imagine a 20 lb block. Kick on the block really hard. What happens? You move, and the block moves in the opposite direction. That's fundamentally the idea. And Newton's third law elevates this behaviour to a universal law of nature. $\endgroup$ – Jerry Schirmer Mar 2 '17 at 17:45
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Firstly, although probably obvious, you have forgotten to mention the key feature in the Comlomb force, which I think is key in 'explaining' Newton's third law - that the forces felt by two charges, $q1$ and $q2$, are equal and opposite, since the force is proportional to both $q1$ and $q2$.

You have tried to understand this from a classical viewpoint, which appears to make sense. I will try to address your first question in a classical setting. As far as I can see, the action-reaction law could still exist if there wasn't a Coulombic force. To see this, consider a force that is repulsive on short distances - although seemingly contrived, it would be sufficient to form a microscopic understanding of Newton's third law. This does not require the existence of charges whatsoever.

garyp pointed out that Newton's third law is also applicable in gravity. This is certainly true in Newtonian gravity. In fact, we can also regard Newtonian gravity as a 'Coulomb-like' interaction, with charges replaced by masses (and obviously only attractive). In relativity, however, it is no longer obvious what place Newton's third law would take in the theory.

A full microscopic explanation of Newton's third law, especially more fundamentally "why don't we fall through the ground" or "punch through a wall" (which might seem possible if we are just thinking of microscopic things classically, since much of the atom is empty space!), relies on quantum mechanics, e.g. the Pauli Exclusion Principle. However, your classical understanding is a good one to have, but it must be remembered that classical descriptions tend to break down when you take them too far.

(Edit: I will incorporate the idea of symmetry into this answer after some fruitful discussions with SaudiBombsYemen, which motivated me to think about this topic more deeply.

While Newton's laws hold in any inertial frame, with or without symmetry, there is a close relationship between Newton's laws and symmetries, namely that Newton's laws assert that momentum is conserved in an isolated system, while translational symmetry implies conservation of momentum. This means that, with translational symmetry, Noether's theorem does indeed imply Newton's laws, or at least the momentum aspects of Newton's laws.

However, it is easy to come up with an example that does not have translational symmetry, but Newton's laws, in particular, the third law, still applies. For instance, a particle sitting still in a bowl. Obviously, this system has no translational symmetry, so Noether's theorem does not apply; however, we can apply Newton's third law on the balance between gravity and normal reaction forces (which are a manifestation of the electrostatic repulsion discussed earlier)

Another seemingly contradictory situation that one might think of is: Why do we observe Newton's third law when two particles collide, even if the background is not translationally symmetric? This is because we take the process of collision to be instantaneous, and in such a small (pointlike!) region of space, we would have translational symmetry, and hence conservation of momentum. Note, however, that this conservation of momentum is not the same as conservation of momentum of the whole system.

Credits to SaudiBombsYemen for much of this content. )

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  • $\begingroup$ Very helpful, thanks. May I ask how momentum conservation plays into this picture? (As it seems to be suggested by the other answers). So with my description, and your refined picture here, are we in fact explaining "how" momentum is conserved in such circumstances? $\endgroup$ – user929304 Mar 2 '17 at 15:48
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You made some good observations, but as garyp said Newton's third law is far more general, in fact what lies behind Newtons third law is a deep fact about space, it applies everywhere in the universe, to all forces. In fact, the third law is just the conservation of momentum. Because the total momentum of the universe is conserved with time, its change is zero, and thus the total force (momentum's derivative) is cero. In other words, whenever there is a force in one direction, there must be another in the opposite direction so that they cancel out summing 0.

Now the great question: why is momentum conserved? It is a consequence of the properties of space: the laws of physics are the same everywhere in the universe. This statement is incredibly important, it means that we can translate all the particles of the universe a certain distance in any direction and nothing would change, the particles would behave the same way because the laws that govern them are the same and because their velocities and positions relative to each other didn't change (and thus potential energy and forces between them). This, is called a symmetry. This symmetry implies that changing the system with respect to these variables will leave it unchanged.

You might guess that this is related to some sort of conservation, a derivative that vanishes, and you would be right, this results in the derivative of a quantity being zero. Mathematically what you would do is compute a small change in the lagrangian (which is a function that conveniently packages all the information of a system, from which one can retrieve, using the Euler-Lagrange equations, the equations of motion):

$$\delta\,L=\sum \left(\frac{\partial L}{\partial \dot x_i}\delta \dot x_i + \frac{\partial L}{\partial x_i}\delta x_i \right)$$

where $\dot x$ and $x$ are the $ith$ particle's velocity and position,
and manipulate it, using the fact that $\delta L=0$ due to the symmetry, to end up with a statement of the this kind $\frac{dQ}{dt}=0$. Where Q is the conserved quantity. Behold Noether's theorem. Now, it depends what kind of changes you are making, if it is a translation in space, rotation or translation in time. Translation in space results in the conservation of linear momentum, rotation results in conservation of angular momentum and translation in time in energy conservation (this last one at least in classical mechanics).

I hope I managed to clear some of your doubts, in the process of trying to explain it intuitively I might have messed a bit with the underlying mathematics and there may be things which require some clarification. If you wish to learn more on the subject I recommend the classic in the subject: Goldstein - Classical Mechanics. But if you don't have time for a rigorous book, I recommend Leonard Susskind book What you need to know to start doing physics, based on the first set of lectures he gave at Stanford.

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  • $\begingroup$ @John If the laws of physics dictated different forces for different positions, the force they would exert on each other would be different and thus action would necessary equal reaction. $\endgroup$ – SaudiBombsYemen Mar 2 '17 at 13:24
  • $\begingroup$ @John If there is not translational symmetry you can end up with cases as the one I stated above. I forgot to mention it was force between 2 particles for example. If forces depended on position rather that relative distance newton's third law wouldn't be true $\endgroup$ – SaudiBombsYemen Mar 2 '17 at 14:56
  • $\begingroup$ @John The particle doesn't rebound with the same speed! Since the box is usually much more massive than the particle, its final speed is approximated as the original one in the opposite direction, but in reality is not. Make the calculations. Momentum is indeed conserved! In a mass spring system momentum varies, sure, there is a force, a change in momentum, but if you take into account the whole system, even our entire planet momentum will be conserved. $\endgroup$ – SaudiBombsYemen Mar 2 '17 at 20:53
  • $\begingroup$ @John (continuation) If you only consider half of the system Newton's third law isn't true either. I think you don't realize Newton's third law is just the statement that total momentum doesn't change over time, they are equivalent, one implies the other. I know they taught us it as an axiom, but since brilliant's Emmy Noether's theorem we know it can be derived from deeper principles. $\endgroup$ – SaudiBombsYemen Mar 2 '17 at 20:57
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No. Newton's 3rd law applies to all forces and Coulomb's law is just one example, not the underlying cause.

Here's an example I use with my students. I hold out my pointer finger and challenge them to touch my finger, with their pointer finger, without letting me touch them. They quickly find that it's impossible. For instance, they might try wearing gloves, but then if they claim I didn't touch them, it likewise means they didn't touch me. The conclusion is that "contact" is a mutual interaction.

The thing Newton's 3rd law adds to that concept is that "force" is a mutual interaction of equal magnitude.

To the best of my understanding, the reason this must be true is more about logic and symmetry rather than physics, by which I mean, if the two forces were unequal, it would result in contradictions.

Or perhaps this is a more satisfying answer: A force can only happen when two objects interact. (Coulomb's law has two charges, the law of universal gravitation has two masses). So we should think of a force as a single interaction between two objects. Is the same size on both because there's only one force to begin with and it pushes objects apart or pulls them together. When incorporating that force into a free body diagram and/or Newton's 2nd law, the rules of FBD's require us to isolate the object and the rules of the 2nd law require that vectors have a single specific direction, so we take what is physically a single two way interaction and break it into two one way interactions.

The issues you raised would be a good explanation of why, in my example that all forces require the interaction of two objects, I only had to mention Coulomb's law and gravitation. All other forces (outside of the strong and weak interactions) are manifestations of those two, and in the case of things like friction, tension, normal force, etc, they are all manifestations of electric forces.

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