One ball has a certain velocity and is spinning. It hits another ball, and each have a certain friction constant which kind of roughly defines how their surfaces interact.

Now, when I was in high-school and knew less about physics, I could explain the result of this based on the conservation of linear momentum and angular momentum. For linear momentum, this is quite simple and depends on the mass of each ball, for angular momentum there's both the fact that one ball is spinning and the angle at which it hits the other ball (which provides some torque) to take into account as well. The result will be both balls will have a different spin and different velocity but linear momentum and angular momentum are conserved. This can be seen in a way as a consequence of Noether's theorem because of the equivalence of rotated or translated inertial frames.

Now that I know more about physics, this is what happens in my mind and I'm terribly confused why angular momentum is still even a thing:

Both balls are collections of molecules with a certain structure which create intramolecular forces. Regardless of how one ball happened to end up spinning and moving, at the point of contact with the other ball, these molecules will impart linear momentum upon the other ball's molecules, and this linear momentum is caused by the ball spinning and also by the ball's velocity. This linear momentum between molecules needs to be conserved so each will get a share. From that point on, because the balls are squishy, more molecules will become involved and will impart linear momentum, while all the molecules in each ball will still be subject to their intramolecular forces.

Now, the intramolecular bonds in the previously idle ball will start propagating motion through the ball through basically Pauli's principle, and it will basically go at the speed of sound of the material of the ball. This will cause the linear momentum to propagate throughout the ball, but because of the intramolecular forces, they will bend this into a "spin" of the ball as the movement propagates.

As you can see, the final result is explained entirely by linear momentum. There is no such thing as angular momentum involved. Now, I don't want to become the kind of crackpot that says there is no angular momentum. I realise that in another perspective, it could be seen that linear momentum doesn't exist and all of this is just angular momentum being conserved, with a very large radius. However, what I can't see is how both concepts can be fundamental in classical mechanics. I just don't understand anymore how we ever really deal with angular momentum if I can explain everything by molecules imparting linear momentum. Or am I making a mistake?

Note that I have read this post Is Angular Momentum truly fundamental? and it doesn't explain my question.

Thanks for helping me reach deeper understanding,

  • $\begingroup$ It's because initial spinning ball upon hitting second ball - will pass some angular momentum to second ball, which will be forced to spin too. $\endgroup$ Aug 12, 2022 at 10:03
  • $\begingroup$ That… completely sidesteps the complexity of the question. As you can see the motion of the second ball can be explained by linear momenta applied to a bunch of molecules on it (asymmetrically, on it as an isotropic object) $\endgroup$
    – buddhabrot
    Aug 12, 2022 at 10:08
  • $\begingroup$ If ball will rotate around it's COM, then this means that each and every molecule in ball will dance with angular speed too, around COM. It doesn't matter how you started an angular momentum,- by slapping onto ball side with force (. i.e. linear momentum) or otherwise,- as soon as molecules starts to spin around barycenter,- you can't avoid angular momentum concept. $\endgroup$ Aug 12, 2022 at 11:56
  • $\begingroup$ Read also this related answer - What actually is the vector of angular momentum?. $\endgroup$ Aug 12, 2022 at 12:49
  • 1
    $\begingroup$ @buddhabrot to accurately describe the transfer of momentum between two rigid bodies, not only you need the vector of linear momentum (direction and magnitude) but also the location through which this exchange happens (contact point). The location is encoded in angular momentum. It is auxiliary information needed to describe the situation fully. $\endgroup$ Aug 12, 2022 at 13:59

1 Answer 1


The statements below are from the perspective of rigid body dynamics

You are correct that the impulse exchanged between two bodies is pure momentum when described along a specific direction. With no friction, this is the contact normal, and with friction, it is some skew direction starting from the contact point.

But to an outside observer, you need angular momentum to describe the geometry of the problem. Angular momentum (vector) is analogous to torque (vector). Torque describes the geometry of the loading on a rigid body.

  • It describes how far an action occurs. It encodes the minimal distance and direction to the line of action. This is a line in 3D space needed to describe either a force (the line of action) or momentum (axis of percussion).

  • Different observers measure different values because they describe a line in space, which presents itself differently.

  • You recover the point of action of a force $F$ using list item #3 of this answer and the contact point of an impulse $p$ from item #2

    $$\begin{aligned} \vec{r} & = \frac{ \vec{\tau} \times \vec{F} }{ \| \vec{F} \|^2} & & \text{point on line of action of force closest to origin}\\ \vec{r} & = \frac{ \vec{L} \times \vec{p} }{ \| \vec{p} \|^2} & & \text{point on line of action of impulse closest to origin}\\ \end{aligned}$$

  • It is not a real vector, but a pseudo-vector.

Your question is analogous to asking "do we need torque to describe the loading on a body when only a single force is applied?"

The answer is that you need the force vector, and either a point where the force is going through or the equipollent torque $\vec{\tau} = \vec{r} \times \vec{F}$.

Similarly, for any contact where linear momentum is exchanged between two bodies along a line of action, you need to specify either the point of contact, or the equipollent momentum (angular momentum vector) $\vec{L} = \vec{r} \times \vec{p}$

  • $\begingroup$ Thanks for your detailed answer. What I still don’t understand is, isn’t this angular momentum vector just an inaccurate way of describing the in isotropic object, it’s the intramolecular forces that actually cause things to spin! If they go away, things will fly apart! $\endgroup$
    – buddhabrot
    Aug 12, 2022 at 16:00

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