I'm wondering whether angular momentum is just a convenience that I could hypothetically solve any mechanics problems without ever using the concept of angular momentum.

I came up with this question when I saw a problem in my physics textbook today. In the problem, a puck with known velocity hits a lying stick. The puck continues without being deflected, and the stick starts both linear and angular motion. There are three unknowns: velocity of puck and stick after collision, and the angular speed of the stick. So, we need three equations: conservation of linear momentum, kinetic energy, and angular momentum.

So, for instance, is it possible to solve this problem without using angular momentum? Also, how would a physics simulator approach this problem?

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    $\begingroup$ Here is a related question with answers: physics.stackexchange.com/q/678994 $\endgroup$
    – John Darby
    Commented Apr 11, 2022 at 14:33
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    $\begingroup$ The advanced course in mechanics ("theoretical mechanics") introduces generalized coordinates. For example, pendulum offset as measured along its curved trajectory is perfectly legitimate such a coordinate. Another example is pendulum swing angle. From that perspective, angular momentum is just a linear momentum considered along unconventional coordinate. The question then becomes, among numerous coordinate systems, why only the two - Euclidean coordinates, and angles - are distinguished in the Notherian conservation laws. $\endgroup$ Commented Apr 13, 2022 at 15:49

11 Answers 11


I'm wondering whether angular momentum is just a convenience that I could hypothetically solve any mechanics problems without ever using the concept of angular momentum.

If your criterion for something being a convenience is that you could solve problems without it then everything in physics is just a convenience.

There are an infinite number of possible mathematical formulations. So, in principle, it should be possible to convert any mathematical problem into a different formulation that avoids the use of any specific concept that you would like to avoid (or at least hides it so that it is not apparent that you are using the concept).

That said, angular momentum is conserved and it is related (by Noether's theorem) to the fact that the laws of physics are symmetric under spatial rotation. Both conserved quantities and symmetries are very important in modern physics. So even if you classify it as a convenience, it is one of the most important and pervasive conveniences in physics.

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    $\begingroup$ Absolutely agree with this. To supply the asker with a few other things to consider, the standard mechanical idea of angular momentum (as it is related to a cross product of position and linear momentum) is not really as deep as it goes. In Quantum Mechanics, particles can have intrinsic spin, giving them angular momentum. Also, in E/M, angular momentum acts quite differently (See Feynman's Lecture of Field Momentum). So yes, it seems trivial at times in mechanics, but its as deep a concept as any in physics, on par with energy and linear momentum. $\endgroup$ Commented Apr 11, 2022 at 22:28
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    $\begingroup$ More to the point, mathematics spends a great deal of time searching for convenient ways to calculate things. Without this, engineers would be even more expensive than they already are. $\endgroup$
    – EvilSnack
    Commented Apr 12, 2022 at 1:32
  • $\begingroup$ Is spatial rotation different from spatial translation? I can't think of any rotation that can't be replicated by translations $\endgroup$
    – Juan Perez
    Commented Apr 18, 2022 at 0:06
  • $\begingroup$ @JuanPerez yes, they are very different. Even their units are incommensurate. $\endgroup$
    – Dale
    Commented Apr 18, 2022 at 0:48

In classical physics, you can always describe the angular momentum of some extended object in terms of the linear momenta of its constituents.

Take the example of the ice skater whose twirl speeds up as she pulls her arms inward. You could say that her orbital angular momentum $L=I\omega$ is a constant, and because her shape changes to decrease her moment of inertia $I$, her rotational frequency $\omega$ must increase to compensate.

But you could just as well say that the skater’s hands are independent masses, moving in circles because of center-pointing forces which attach the hands to the wrists. When this disarticulated skater pulls her arms in, the additional inward force on her hands changes their motion from circular to spiraling inward. The speeding up is a neat side effect, and I encourage you to compute it without referencing angular momentum.

So in that sense, in classical physics, you can always decompose an angular momentum into an ensemble of linear momenta relative to some reference axis. It’s tedious an inelegant, but possible.

However, in quantum physics, angular momentum turns out to be an irreducible part of the theory. The quantum fields associated with the electron, the proton, the neutron, the photon, etc. all have an intrinsic property which behaves exactly like an angular momentum, even though a “point particle” doesn’t have any components that can “spin.” We call it “spin” anyway. You can’t do quantum mechanics without angular momentum.

  • $\begingroup$ You should include a description of experiments that show them to be the same thing; manipulating the spin of individual electrons in iron to make a dangling iron object spin "out of nowhere". $\endgroup$
    – Yakk
    Commented Apr 12, 2022 at 19:58
  • $\begingroup$ To the editor who suggested gender-neutral pronouns for the skater: I thought about that while writing the post, but it created ambiguity in some sentences about whether “their” referred to the skater or to the skater’s hands. $\endgroup$
    – rob
    Commented Apr 14, 2022 at 17:19

Although I agree with Dale's answer wholeheartedly, the question seems to be free-spirited in the sense of: is the angular-momentum physics that we have today the only physics that could have been possible (or the only one that we would expect if space aliens were to arrive on the scene to teach us their way of doing physics)?

Ultimately the free-spirited answer is: angular momentum is in effect a variant of linear momentum where the linear is the tangent. Atoms of the rotating body are attempting to conserve linear momentum along those tangent lines, but all sorts of forces (e.g., chemical bonds; mechanical interlock; strong nuclear force; weak nuclear force) keep those atoms & their subatomic particles from being free to pursue that tangent path. Hence, those atoms & their subatomic particles deviate from the tangent due to being bonded by whichever force in effect at scale to the rest of the rotating object.

If you were to dispense with the so-called convenience of angular momentum, to instead devise an alternate physics that expressed the angular momentum concept as linear momentum along tangent lines deviated from by internal bonding within the rotating object, then you would need to devise a rather more complex modeling of not only the potential linear momentum along the tangent lines, but also modeling the forces internal to the rotating object that keep it from disintegrating below its redline RPM to borrow the automotive-engine term. To model these internal forces vis a vis the potential linear momentum along tangent lines, you might need such software tools as finite-element analysis (FEA) to model each voxel of the rotating object. Adding such modeling as FEA might arguably prove to be more inconvenient than the back-of-a-napkin modeling that angular momentum (at less than redline RPM) provides.

You might be able to regain some of the back-of-a-napkin convenience that you lost when you dispensed with angular momentum by representing the internal forces (and the tangential potential angular momentum) in equational form via conformal geometric algebra (CGA), but CGA's learning curve might be itself inconveniently steep when compared with the convenience of angular momentum at sub-redline RPM. Conversely, the convenience of CGA's equational non-voxel/non-FEA modeling of both the internal bonding forces within the rotating object and the tangential potential linear momentum might be the most convenient of all if the focus is at & above redline RPM in failure modes instead of during normal operation of the rotating object.

Indeed, angular momentum as potential linear momentum along tangent lines becomes actual linear momentum along tangent lines beyond the redline RPM of the rotating body when the rotating body disintegrates. Indeed, at this point, the physicist (or engineer) must overtly model the conservation of linear momentum more than the conservation of angular momentum to calculating the debris pattern flying off from the rotating object. Example of angular momentum as potential linear momentum along tangent lines becoming actual linear momentum: https://youtu.be/7nSB1SdVHqQ

Let your free-spirit horses roam with alternate schools of thought of physics, but only so far as replacing one convenience (angular momentum) with something else that you find even more useful as either convenience (e.g., CGA equations) or productive drudgery in the minutiæ (e.g., FEA)

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    $\begingroup$ While this is a fun answer, I'm not sure it directly addresses the crux of the question. One must decide whether to deal with the atoms making up some macroscopic body regardless of whether one uses linear or angular momentum. $\endgroup$ Commented Apr 11, 2022 at 15:22
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    $\begingroup$ @AndreasZUERCHER I read it as a synonym for the term used in your answer "free-spirited" $\endgroup$
    – justhalf
    Commented Apr 12, 2022 at 8:43
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    $\begingroup$ No euphemism! Fun =enjoyable to read $\endgroup$ Commented Apr 12, 2022 at 14:38
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    $\begingroup$ This neglects the fact that angular momentum cannot be fully reduced to linear momentum. Things that appear to be point particles have intrinsic spin, to the point where you can create macroscopic spin by manipulating their spin direction. Admittedly it wouldn't occur in the puck situation as far as I can tell. $\endgroup$
    – Yakk
    Commented Apr 12, 2022 at 19:56
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    $\begingroup$ @Yakk, and then we would leave the conformal geometric algebra to instead have one of the Clifford algebras with spinors and/or twistors. $\endgroup$ Commented Apr 12, 2022 at 20:55

Hypothetically, you could model your puck as a collection of $N$ atoms glued together with some kind of force, and solve the equations of motion (Newton's 2nd law) for the collection of atoms when you applied the external force to the puck.

At no point in the simulation would the simulator ever use the concept of angular momentum. But (if the simulation was good), if you measured the angular momentum of the solution, you would find it was conserved.

So, no, you don't need to directly use angular momentum to solve a physics problem. But whether or not you use it, it is conserved. And if you use it, you dramatically simplify many problems in rotational motion, to the point where they reduce to "impossible without doing a complicated computer simulation" to "doable with at most a few pages of math."

Additionally, regardless of solving specific problems, knowing angular momentum is conserved gives you a lot of insight. You can use conservation of angular momentum to debug the simulation, for example. Or, you can predict what will happen in a complicated scenario where you can't accurately simulate all of the details, such as predicting how galaxies will rotate as the mass falls in due to gravity. Without insight, we would just be randomly trying to simulate complicated systems, and would not make much progress in understanding Nature.

Finally, in more fundamental areas of physics, like quantum mechanics, angular momentum is absolutely essential for understanding what is going on. Spin, for example, is a form of intrinsic angular momentum. There is no deeper explanation of spin than to say "this is angular momentum that an electron has when it is at rest."

Therefore, while you can solve classical mechanics problems involving rotational motion or rotational stability without ever talking about angular momentum, to do so would be very foolish.


Angular momentum is a fundamental kinematic parameter, the same way that linear momentum is, because rotation is an elementary motion just as translation is.

Because of its locally Euclidean nature, space permits two kinds of distinct (locally) rigid motions - namely, the two just named, and you can't simply reduce one to the other. There is no way to express a rotation purely in terms of translations, and conversely.

Hence, for every translational parameter, one should expect there to be a corresponding rotational parameter. For position, there must be orientation. For velocity, there must be angular velocity, and for momentum, there must be ... Angular Momentum.

Of course, angular momentum is made all the more useful by the fact that the rotational symmetry of the dynamic laws and not just space alone ensure its conservation, but the way I interpret this question is that it is asking why we should not just do everything in terms of translational quantities - and that is because rotation is not translation. In fact, there are spaces such as "taxicab" space where that, while they are not our physical space, rotations, at least as we understand them, simply don't exist.

(Okay, yes, there's 4 individual rotations, as in 4 transformations in total, not 4 ways of transforming, on a taxicab plane. But you get the idea.)


Angular momentum is a fundamentally conserved quantity. One does not need macroscopic, bound objects to see this.

Consider two identical particles moving in opposite directions. In the center-of-momentum frame, their trajectories will be parallel. If they are not on a direct collision course, then their trajectories will be separated by some distance d, and the system will have non-zero angular momentum in the center-of-momentum frame. Given this setup, there is no inertial reference frame with zero angular momentum; it would require a rotating (non-inertial) frame.

The system's angular momentum is preserved at all times and for all interactions, and this property is required to solve final states. For example, if the particles are an electron and positron, they can annihilate to 2 (or more) photons. But the final photons state must have the same angular momentum as the initial state, which forbids some final states that would be allowed following only conservation of energy and linear momentum.


Another small point, implicit in the other answers: it's not exactly that "angular momentum is essential to solve problems", but that it is a convenient concept by which to understand things.

Slightly fancier: neither Lagrangian nor Hamiltonian classical mechanics immediately give "formulas to solve problems", but, rather, amount to discoveries of concepts that apply broadly. In contrast, Newtonian mechanics superficially seems "better" (to a novice?), because it seems to more directly address things. But that turns out not to be the highest virtue. :)


Is angular momentum just a convenience?

Conservation of angular momentum and linear momentum are independent conservation laws. Angular momentum of a system is conserved if the net external torque acting on the system is zero, whereas linear momentum is conserved if the net external force acting on the system is zero.

Angular momentum is not just a convenience, but a necessity for solving some physics problems. You can find some examples here: https://www.concepts-of-physics.com/mechanics/conservation-of-angular-momentum.php

Hope this helps.


Angular momentum is very much like torque.

Torque describes the moment of a force, and angular momentum describes the moment of momentum. Here "moment of ..." has the meaning of the perpendicular distance to a line. You see this from the equations used to derived the moment-of

$$ \begin{aligned} \vec{L} & = \vec{r} \times \vec{p} & & \text{ang. momentum of momentum }\vec{p}\\ \vec{\tau} & = \vec{r} \times \vec{F} & & \text{torque of force }\vec{F}\\ \end{aligned}$$

Any system of forces/torques applied to a rigid body can be resolved to a single force and a single torque about an arbitrary point.

Or a system of forces can be used in place of the torque. For example, the two loading systems below are equivalent if

$$ \begin{aligned} Fx &= Fx_1 + Fx_2 \\ Fy &= Fy_1 + Fy_2 \\ Mz &= r\,(Fx_2 - Fx_1) + r\,(Fy_1-Fy_2) \end{aligned} $$


And similarly for all three planes for the 3D case. You need 6 forces placed in a star shape to produce the equipollent force and torque at the center.

So the same goes for angular momentum. You can replace the single angular momentum of a rigid body, with 6 moving point particles that are connected together.

Developing such a system and the equations of motion is rather straightforward, although a bit futile, knowing how the concept of angular momentum (just as torque) yields a lot of simplifications.


Fundamental particles (electrons, quarks etc.) have intrinsic angular momentum that constrain the interactions they participate in, so this characteristic of the world is not merely a feature of the mathematical formulation of mechanics.


Conservation of Linear Momentum is a special case of, and subservient to, conservation of Angular Momentum .

For Particles A and B, conservation of Linear Momentum can be described as ( force of A on B ) = - ( force of B on A ), resulting in ( impulse of A on B ) = - ( impulse of B on A ), resulting in no net delta Linear momentum .

Conservation of Angular Momentum is the same, but ADDs the proviso that the force be parallel along the path between A and B .

If we ignore the proviso, it is simple to construct an imaginary example where Linear momentum is conserved, but not angular momentum. If particle A is unmoving at x = 1, y= 0 and equal mass particle B is unmoving at x = -1, y = 0 , their center of mass is at (0,0) . Then particle A interacts with B , imparting a negative Y velocity on B, and because Linear momentum is conserved, an equal positive Y velocity on itself . The center of mass remains unmoving at (0,0) . But now they have equal Angular Momentum, not opposite. Net Angular Momentum appeared from nowhere, because the magic imaginary force was not in the X direction . Particles in our universe cannot do this .

Conservation of angular momentum implies conservation of linear momentum . And if you only look at examples where forces act along the lines separating bodies, it would be easy to convince yourself that conservation of linear momentum was adequate to explain the behavior. But it is conservation of Angular Momentum that prevents forces from providing a net torque .

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    $\begingroup$ "Conservation of angular momentum implies conservation of linear momentum" no it doesn't. A body moving in a 1/r potential has a conserved angular momentum but not a conserved linear momentum. The conservation laws originate from two different symmetries of the system. Angular momentum comes from spherical symmetry and linear momentum from translational symmetry. $\endgroup$ Commented Apr 14, 2022 at 8:45
  • $\begingroup$ @JohnRennie A body moving in a 1/r potential is interacting with a unnamed other body, so the thought experiment is therefore incomplete, and does not answer whether linear momentum is conserved. Further, that example does not conserve Angular momentum from all viewpoints, only from your special chosen viewpoint, so it does not represent 'Conservation of angular momentum'. $\endgroup$ Commented Apr 14, 2022 at 11:33
  • $\begingroup$ @Andrew if the simulation only allows forces along the path between particles, then it explicitly enforces conservation of angular momentum, so it is not a coincidence that angular momentum is conserved at the end. $\endgroup$ Commented Apr 14, 2022 at 11:37
  • $\begingroup$ @JohnRennie I ask that you remove the downvote until you provide a real counter argument. Your comment defeated a straw man, having little to do with my post about the interaction of particles. My best interpretation of your argument is, "If a pebble orbits a star, and we ignore the influence the pebble has on the star, we can pretend that linear momentum is violated". I do not have stackExchange reputation to defend my post, but I provided valid math. I hope that you will actually read it, and provide a comment that applies to what I wrote. $\endgroup$ Commented Apr 15, 2022 at 12:12
  • $\begingroup$ I did not downvote your post. $\endgroup$ Commented Apr 15, 2022 at 14:05

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