Is there a mechanistic-type explanation for how forces work? For example, two electrons repel each other. How does that happen? Other than saying that there are force fields that exert forces, how does the electromagnetic force accomplish its effects. What is the interface/link/connection between the force (field) and the objects on which it acts. Or is all we can say is that it just happens: it's a physics primitive?

A similar question was asked here, but I'd like something more intuitive if possible.

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    $\begingroup$ Necessary classic: youtube.com/watch?v=wMFPe-DwULM $\endgroup$
    – Jold
    Apr 19, 2013 at 3:31
  • $\begingroup$ I've tried very hard for some form of an intuitive understanding of force, including gravity. At this point I just accept them but I can't say I have any form of intuitive understanding for them. $\endgroup$ Apr 19, 2013 at 6:28
  • $\begingroup$ For electro-magnetic forces, we can say "photons", but is that really a better explanation than "force fields"? At least "photons" correctly imply quantization, but I don't think that was your objection against the term "force fields". $\endgroup$
    – MSalters
    Apr 19, 2013 at 11:33
  • $\begingroup$ This is also related. These kinds of fundamental terms in physics are human names for things we have observed in nature. $\endgroup$ Apr 19, 2013 at 13:15
  • $\begingroup$ The correct way to describe how to electrons repel is quantum electrodynamics, if you really want to know how it works at a deeper level. The photon couples to the electric charge of the electron. $\endgroup$
    – Dilaton
    Apr 19, 2013 at 13:19

3 Answers 3


Quantum Mechanics says force is not physics primitive. It shows the undelying mechanism for them.

What is a force? It is something that changes the velocity of a particle, with the Newton's second law: $$\vec{F}=m\dfrac{d\vec{v}}{dt}$$ Any other appearances of forces can be reduced to this. For example, when we measure the force with a dynamometer, it is actually two forces aplied to the same particle, and they cancel each other when the particle reaches the offset equilibrium position.

Without any force, the particle would move with the same velocity $\vec{v}=\mathrm{const}$. But the Quantum Mechanics shows a more complicated picture: the particle is distributed in space, and depicted as a wave packet. Its evolution (motion and change) without a force is governed by the Schrödinger wave equation $$i\hbar\dfrac{\partial\Psi}{\partial t}=\dfrac{-\hbar^2}{2m}\nabla^2\Psi\qquad\left[=\dfrac{-\hbar^2}{2m}\dfrac{\partial^2\Psi}{\partial x^2}\quad\text{(in 1D space)}\right]$$ That does say the same thing, $\vec{v}=\mathrm{const}$, but in a sense that all velocity components of the wave packet (that is, its Fourier coefficients) evolve constantly in time, and independently on each other. But that's the mathematical abstraction. Physically, it tells some different story: the wave function oscillates and flows. The rate of oscillations is what we call enegry, and the flow is kept up by the gradient of the phase.

Then, what happens when a force appears on the scene? We should add the potential energy of that force to the Schrödinger equation: $$i\hbar\dfrac{\partial\Psi}{\partial t}=\dfrac{-\hbar^2}{2m}\nabla^2\Psi+U\Psi\qquad\left[=\dfrac{-\hbar^2}{2m}\dfrac{\partial^2\Psi}{\partial x^2}+U\Psi\quad\text{(in 1D space)}\right]$$ What happens to the wave funtion then? It starts to oscillate with higher rate in some points (where $U>0$), while with the same rate in some others (where $U=0$). Because of that, the phase in the first points would outrun the phase the second points. And the gradient of phase tells the wave function to flow away in the direction of the retarding phase. So the particle would run away from the place where the potential energy is high! That's how forces work.

  • $\begingroup$ Isn't a more general rule that a force is something that causes motion to deviate from a geodesic path? $\endgroup$ Jan 10, 2017 at 20:15
  • $\begingroup$ I wonder how do we define $m$ for a particle distributed in space? The rest mass does not exist on a finite point, but distributed without defined bounds (to infinity?!) $\endgroup$ Jan 10, 2017 at 20:21
  • $\begingroup$ @ja72 a force is something that causes motion to deviate from a geodesic path - it is true, but it is a different kind of generalization. General Relativistic, not Quantum Mechanical. And in the General Relativity, force is still physics primitive. That's why I did't talk about that. $\endgroup$
    – firtree
    Jun 29, 2017 at 15:09
  • $\begingroup$ @ja72 $m$ for a distributed particle is just a ratio in its equation of motion, that is, in the Schrödinger wave equation (or Dirac wave equation, or Klein-Gordon wave equation and so on). Roughly speaking, it says, for the given gradient of phase, what the speed the particle has. Thus, the same energy would give heavier particles less speed, and lighter particles greater speed. $\endgroup$
    – firtree
    Jun 29, 2017 at 23:17

The interface between the force(field) and the objects is what we call "charge". Electromagnetism, for example, does not excert any force on an electrically neutral object.

In a very crude way, you could imagine charge as the hooks, on which the springs that mediate the force are hung.

One has to be careful with Newton's graviation, however. It works fine for everyday applications such as falling bodies etc. to picture the graviational charge as "mass". A massless object is weightless from this point of view. This is not true in general relativity! There, mass bends space itself and photons, too, are subject to it, albeit not having any (rest) mass.

The similarities between Newtonian gravitation and electromagnetism go quite far. You could imagine the electromagnetic potential as a mountain at the position of one electron from the point of view of another. The repelling force then acts like a ball rolling up a slope and down again.


A force is the name given to a physical measurement that isn't a physical property of an accelerated mass, but still allows us to predict a value for its acceleration, given the numerical value of a force present.

The only interpretation of force which physically matters is how you assign a number to it.

  • $\begingroup$ This seems like an interesting answer. If I understand the comment correctly it seems to say that force isn't quite the point. What's the point is the fact that acceleration occurs. Since acceleration implies an increase in energy in the accelerated entity, the question then becomes (at least in my mind) how does that energy transfer work/happen? $\endgroup$
    – RussAbbott
    Apr 19, 2013 at 17:43
  • $\begingroup$ @RussAbbott I was trying to say that physical measurements are what matters, and how they enable us to create models of the world. Force is just a physical measurement that allows us to predict what we'll measure for the acceleration of a mass. Wanting to know how force manages to do what it does is open to hand-waving make-believe interpretation. All we can say is that if we make these physical measurements, then we'll measure this other physical measurement. $\endgroup$ Apr 19, 2013 at 22:17

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