I understand the mathematical proof of conservation of momentum. The forces of particles on each other in an isolated group if summed together cancel in pairs. So the vector sum of their momentum stays constant unless an external forces acts on the particles. But, I also have an intuition. A single particle keeps drifting in a direction unless forced to change. And a isolated group of particles too keep drifting as a whole in an average direction unless forced to change. The law for an isolated group of particles is exactly of the same form as that for a single particle.

In the proof for angular momentum, it's the same thing. The torques cancel in pairs and angular momentum is conserved. The proof is quite clear, but the intuition is a lot harder for me. A particle cannot keep moving in a circle on it's own, whereas a group of particles keep rotating as a whole unless forced to change the direction about which they are rotating or the magnitude through an external torque. The law for a single particle leads to a law that is of a different form for the group.

Can somebody give me an intuition as to why this is the case without the proof(though the mathematical proof of itself is quite clear and simple)?

Short version: For a single particle, conservation of angular momentum reducing to conservation of linear momentum. But, for a group, it is a new law. A group rotates as long as there is no external torque. How can I develop an intuition for this?

  • $\begingroup$ For your second paragraph: A particle cannot move in a circle on it's own because the center of mass of any object will move in a straight line in the absence of any external force. Similarly, an object rotating about its own axis will rotate w/ constant omega unless accelerated by an external torque. Note that the individual particles of the body rotating about its axis are acted upon by internal forces. $\endgroup$
    – xasthor
    Sep 14, 2017 at 7:41
  • $\begingroup$ What you are pointing at here is exactly the reason why linear (translational) momentum conservation is not valid in circular motion. There is an external force pulling inwards. But rotational (angular) momentum conservation is valid, since there are no external torques. $\endgroup$
    – Steeven
    Sep 14, 2017 at 8:18

5 Answers 5


Conservation of angular momentum is closely related to the fact that there is no privileged spatial direction. As a consequence, once a direction is defined by an isolated system, it has no reason to change during the system evolution, because there is no distinguished absolute direction "calling it back" so to say. The system has a rotational inertia.

I will get back to this at the end of this discussion. But let's start with your intuition for linear momentum:

A single particle keeps drifting in a direction unless forced to change.

This is because there is no privileged position in space. So, again, an isolated system that defines a position (by being there) has no reason to change it. Observed from another inertial frame, its center of mass moves uniformly in a straight line: this is how "to be not moving" appears in the most general sense. Else we would need to have a reference for absolute rest, which is precisely what we do not have because there is no absolute position.

So the system, having no reason to change the way it "not moves", has inertia: it takes a force to change the class of inertial frames it defines into another one. The force needed is the product of the eventual frame relative speed with respect to the initial one, with the system mass. The mass measures the inertia.

And a isolated group of particles too keep drifting as a whole in an average direction unless forced to change. The law for an isolated group of particles is exactly of the same form as that for a single particle.

Yes, but what law are you talking about? Because conservation of momentum is not the conservation of speed: it is the conservation of the overall motion with inertia of the system (else it would require the same force to stop a fly than to stop a car with the same speed).

And this is why things seem different to you when it comes to angular momentum:

A particle cannot keep moving in a circle on it's own, whereas a group of particles keep rotating as a whole

A group of particles does not keep rotating. See the case described in this question: we have two particles (astronauts there) linked by a taut rope and spinning around their centre of mass. When they let go of the rope, they each follow a straight trajectory: observed from an inertial (non-rotating) frame at rest with respect to their center of mass, their angular speed (with respect to their center of mass) progressively vanishes. Angular momentum is conserved because they move away from their center of mass, not because they keep moving in circle.

The law for a single particle leads to a law that is of a different form for the group.

The single particle is a solid object: its whole mass distribution is maintained by cohesive forces. If you abstractly divide it into different parts, each one of these parts would follow a straight line and move away from the center of rotation, if it was not bound to the other parts. Exactly as the astronauts do: they keep rotating as long as they are bound by the rope.

This has all been well explained in @stafusa's answer.

So the law of angular momentum conservation (that indeed is the same for a single particle and a group) is not the law you are thinking of: it is not a law about keeping moving in circle. It is not about conservation of the angular speed.

What is it then? It is a law about conservation of the overall direction with rotational inertia. Very much like a uniform linear motion is the most general way of "not moving", rotating in such a manner that the total angular momentum does not change is the most general way of "not rotating".

What is not analogous to the conservation of linear momentum is the (false) idea that we can characterize what is being conserved by a class of reference frames, because rotating frames are not inertial. So if we kept rotating along a many-body system, (in the same way as we previously followed a group of particles in inertial motion), we would not so easily observe that the system keeps "not rotating" (as opposed to the previous system center of mass "not moving"). We would have to take into account inertial forces (such as the Coriolis force) that seem to correspond to a change of angular speed, while they actually help preserve both rotation axis and rotational inertia.


What you call an intuition about the conservation of linear momentum is actually little more than being used to it. The ancient Greeks and most small kids wouldn't necessarily agree with you that it's not necessary to have a force acting upon a body to keep it at constant velocity.

That said, there are some observations that might help "getting used" to the conservation of angular momentum:

1) First, in the general situation, developing an intuition from the steps of the proof is actually the best approach, but for acquiring this naturalness you're after, it's probably best to consider a homogeneous, symmetric rigid body, say a sphere, rotating about its axis ("spinning") floating in space, and ask yourself: Why should this ball stop spinning, or start to spin faster, without any force acting on it?

2) It helps also to notice that it's a common misunderstanding$^1$ to link angular momentum to circular (or, at most, elliptical) movement. If you consider the definition

$$\mathbf{L} = \mathbf{r}\times\mathbf{p},$$

and apply it to a free particle, which moves in a straight line with, e.g., velocity $\mathbf{v}=v\,\hat{\imath}$, distant from the $z$ axis by $r_\perp$ (see figure below), and has mass $m$, you can calculate its angular momentum with respect to the $z$ axis: $\mathbf{L} = \mathbf{r}\times\mathbf{p} = m\mathbf{r}\times\mathbf{v} = -mv r_\perp\hat{k}$.

enter image description here

Since $\mathbf{v}$ and $m$ are constant, and so is $r_\perp$, $\mathbf{L}$ is also constant $-$ is conserved in the absence of forces$^2$. So that's a situation where the angular momentum conservation reduces to the linear momentum conservation.

This trivial equivalence is of course not present in general. If that mass was initially rotating clockwise around the origin in the $x$-$y$ plane at constant speed $v$, connected to the origin by a string of length $r_\perp$, then its angular momentum would also be $\mathbf{L} = -mv r_\perp\hat{k}$; and, if this string were cut when $\mathbf{r}=(0,r_\perp,0)$, the mass would start moving over a straight line as depicted in the figure. The removal of the centripetal force changes a periodically evolving $\mathbf{p}$ value to a constant one, but has no effect on $\mathbf{L}$, since it exerts no torque.

$^1$ Reinforced by the textbook tradition of introducing angular momentum and Co. in to context of uniform circular motion, before introducing the more general vector quantities.

$^2$ The example can be taken further and we can consider a force $\mathbf{F}=F\,\hat{\imath}$ exerted on the mass. This increases $v$, has a nonzero torque $-r_\perp F\hat{k}$ and also changes $\mathbf{L}$, yet no proper rotation takes place.

  • $\begingroup$ Why isn't the angular momentum not associated with rotation? Isn't it the case that torques more-or-less produce rotatory movement in a body?? Doesn't change in angular momentum mean either change in axis of rotation or change is magnitude of rotation?? $\endgroup$ Sep 14, 2017 at 10:46
  • 1
    $\begingroup$ @PhyEnthusiast, If, in the straight-line moving mass example, a force $\mathbf{F}$ parallel to its velocity ($\mathbf{F}=F\,\hat{\imath}$) is exerted on the mass, so that $v$ is increased, this force has a nonzero torque $r_\perp F$ and it changes $\mathbf{L}$, yet no proper rotation takes place. That said, of course the concepts are related, like the many usual examples show. What I say is that angular momentum is not associated with circular or elliptical movement exclusively and that the definitions hold in general, independently of the sort of movement. $\endgroup$
    – stafusa
    Sep 14, 2017 at 11:41
  • $\begingroup$ ok I understand now. Thnx for the example $\endgroup$ Sep 15, 2017 at 7:47

Take the earth and the moon system. They are interacting by the force of gravity, rotating about their center of mass point. Once the moon was caught, conservation of angular momentum keeps it in the orbit.

Since you have an intuitive understanding of linear momentum, think of it as linear momenta: the moon moves with momentum p at time t,( where p is a vector) and the earth by -p, the whole system at rest in its center of mass. The force of gravity adds at time t+dt a dp_2 vector to the moon and a -dp_2 to the earth momentum vectors. Linear momentum is still conserved ( sum of earth moon zero) but the vectors have changed direction. Defining angular momentum as rxp (vectors) will give a useful vector which will also be conserved ( add up to zero for the binary system).

Hope this helps.

  • $\begingroup$ I understand all that you point out. But it still doesn't add any intuition in my thinking about angular momentum conservation for a group of particles $\endgroup$ Sep 18, 2017 at 8:29
  • $\begingroup$ a group of particles can always be separated into two sets rotating about the center of mass. take the solar system, where the sun and the group of all planets revolve around the barycenter. $\endgroup$
    – anna v
    Feb 16, 2018 at 6:55

You have successfully developed an intuition for the conservation of linear momentum. As the case of a single particle is a very specialized and ideal case (every particle can be zoomed into and resolved into multiple parts) let us only consider groups of particles:

An isolated group of particles, too, keep drifting as a whole in an average direction unless forced to change.

That is the conservation of linear momentum.

Now, forget about the net (average) direction of the whole and start looking at the parts. (i.e., go to the center of mass frame). And suppose that you have fine threads in your hands, attached to each of the parts so that they don't zoom away. (These are the internal forces). It does not matter how fast you pull at the parts or even if you don't pull at the parts at all.

Now, in general, your threads cannot change the rotations of the parts around you. You can pull them closer, making their orbits tighter, but then they will get faster, as you might have observed if you ever spin when ice-skating or try to do a somersault.

So, in net, something is being conserved. And you can think of that "something" as the area being swept out by the parts per unit time (only works if all motions are on a plane - in the general case, you can take the projection of motions on any plane and the area-rate will remain conserved, since, for a conserved vector all its projections remain conserved).

As long, as there is no external torque - which means nothing external attempts to twist your group of particles in some way, the parts in your system can collide and do whatever and you can pull at them by threads howsoever, that "something" will remain conserved.

Hope this helps.


Why does a particle not keep moving in a circle on its own? The answer is a lack of centripetal force, pulling it to the center of the circle.

With a group of particles, think of the particle on one end as the point of rotation. The rest of the particles maintain their angular momentum/rotation around this particle because of some centripetal force. That centripetal force is the bond holding the particles together.

Another example: Take a beam, with one end nailed to the ground (this is the point of rotation). If you push on the other end of the beam, the beam will obtain an angular momentum by moving in a circle. Assuming no resistance forces, the beam's angular momentum will be conserved. The reason is that there is a centripetal force, which is simply the bonding force between the particles/atoms of the beam.


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