I've been told when considering the net force on a system of particles, only the external forces on all the particles matter. Why don't the internal forces on each particle matter? I've been told it is because of Newton's third law, that:

$$F_{ij}^{int} = -F_{ji}^{int}$$

However it's not clicking with me, mainly because I can't seem to piece together why the above is true. $j$ and $i$ refers to particles, so $F_{ij}$ refers to the force on the $i$th particle due to the $j$th particle. From what I gather, it's saying when some particle $i$ pushes on another particle $j$, that pushed particle $j$ exerts an equal and opposite directed force on the $i$th particle. So the $i$th particle stops accelerating, since it feels an equal and opposite force on impact, and the $j$th particle starts to accelerate. How does this explain why internal forces do not be considered? I feel like I have a fundamental misunderstanding of Newton's Third Law here, or understanding of what's going on here with the particles themselves.

However my logic behind this is that, if all the particles are uniform, and the forces on it act on all particles, and so no particles will accelerate relative to any others. For example, if a clump of particles were dropped from a building (neglecting air resistance) no particles would fall faster than the others, so only external forces can be considered.


I'm going to try to illustrate a situation describing forces on individual particles that amount to an external force:

Suppose a particle in a system of many particles in moving inside the multi-particle mass at some velocity $v$, and suppose it has such a velocity from experiencing an external force that affected it and some particles near the point of impact. Suppose it collides with another particle elastically. The original particle now moves in velocity $-.5v$ while the particle hit goes at $.5v$. Both move until colliding at the opposite ends of the connected mass. This causes a net force of $0$ on the mass as a whole. But the mass as a whole was struck by an external force? It should be experiencing a net force in the direction of the external force? And suppose this scenario happened to all the initial particles near the external force when it made impact, so no bulk motion occurred. Where am I going wrong with this? Sure, some collisions could be inelastic but some wouldn't, surely? And if so this would happen which seems bizarre to me.

  • $\begingroup$ I can't understand what is going on in your edit. I think it would be more clear to me if you explained in more detail what is going on and why. $\endgroup$ – Brian Moths Nov 6 '17 at 13:03
  • $\begingroup$ @NowIGetToLearnWhatAHeadIs My edit is my illustration of a situation describing forces on individual particles that amount to an external force. It's fallacious, of course, as my thinking is, but needs to rectified. $\endgroup$ – sangstar Nov 6 '17 at 18:33

I am not sure why you are confused. You need to understand two things. There first is that the total momentum of the system is the sum of each particle's momentum. The second thing is that the rate of change in each particle's momentum is the force that acts on it.

From these two facts it follows that the rate of change of total momentum is the sum of all the forces on each particle. This sum can be decomposed into two pieces. The first piece is the sum of external forces on each particle and the second is the sum of internal forces. The second sum, the one of internal forces, can be grouped as action reaction pairs. Each pair must sum to zero and therefore the total sum just be zero. Since the sum over internal forces us zero, the total rate of momentum change must be the sum of external forces.

As an example, consider two billiards balls (mass $m$), one above the other falling under gravity. Then the external force is $2mg$ pointing down. Now suppose we add a compressed spring between the two balls. This spring exerts a force $F$ on the top ball pointing up and it must exert an opposite force on the bottom ball, namely a force $F$ pointing down. Therefore the total force is $mg -F$ pointing down from the top ball plus $mg+F$ pointing down from the bottom ball. In the sum, the $F$s cancel out so the total force is again $2mg$. Here you can see the internal forces for not need to be considered because the cancel out when doing the sum.

  • $\begingroup$ "Each pair must sum to zero and therefore the total sum just be zero." But why? $\endgroup$ – sangstar Nov 6 '17 at 15:08
  • $\begingroup$ @sangstar By newton's third law if one particle exerts a force $\mathbf{F}$ on a second object, then the second object must exert a force $-\mathbf{F}$ on the first. Since these two forces are opposite, they have to sum to zero. $\endgroup$ – Brian Moths Nov 6 '17 at 17:30
  • $\begingroup$ What if internal non conservative force acts on a system, will the center of mass remain at the same place. $\endgroup$ – user203191 Sep 8 '18 at 1:01
  • $\begingroup$ Yes, with the additional assumption that no external forces are acting. If there are no external forces, then no matter whether the internal forces are conservative or non-conservative, the center of mass cannot change. Conservation of momentum is totally guaranteed unlike conservation of kinetic + potential energy in a mechanical system, which can be violated, for example, when friction is present. $\endgroup$ – Brian Moths Sep 8 '18 at 21:26

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