In Newtonian mechanics, is angular momentum an emergent property of linear momentum?

Question

My intuition tells me that angular momentum is merely a consequence of the linear momenta of each constituent particle coupled with the forces pulling each particle towards the center of mass. If this were true, it should be possible to provide a definition of angular momentum in terms of linear momentum and forces. However, I also am aware that, in the modern view, conservation of angular momentum is a consequence of rotational symmetry, while linear momentum is a consequence of translational symmetry, which perhaps implies that they are fundamentally distinct concepts.

Answer 1

However, I also am aware that, in the modern view, conservation of angular momentum is a consequence of rotational symmetry, while linear momentum is a consequence of translational symmetry,

That is correct.

which perhaps implies that they are fundamentally distinct concepts.

That is also correct. Linear and angular momentum are distinct quantities. Neither follows from the other.

(And, in addition, you should re-examine what you understand "emergent property" to mean. The term applies to the collective properties of systems of many particles, which do not exist for any individual component. In your case, both linear and angular momentum are perfectly defined at the single-particle level.)

Answer 2

You may want to look at my recent answer to a similar question. Angular momentum, linear momentum and energy are the seven integrals of motion that arise from

• homogenuity (i.e., translational invariance) of space
• isotropy (i.e., the rotational invariance) of space
• homogenuity of time

In this sense angular momentum is just as fundamental as linear momentum or energy. Note that these symmetries applied well beyond classical physics and mechanics - they can be applied to electromagnetic field, and they are behind the own angular momentum of quantum particles (spin), which cannot be presented as a result of these particles rotating about their axis.

However, the traditional Newtonian mechanics, taught in schools and freshmen physics classes, takes different approach, deducing the properties of angular momentum from those of linear momentum, by decomposing bodies in collections of point particles. The only reason why such approach works is that it is applied to one specific case: rigid bodies, where the two types of momentum are by the constraints on the relative particle positions.

Answer 3

it should be possible to provide a definition of angular momentum in terms of linear momentum

It is possible to map linear and angular momentum. Consider a rotating particle, then :

$$\begin{align} L&=I \cdot \omega \\&= mr^2 \cdot \frac vr \\&=r \cdot mv \\&=rp \end{align} $$

Or if expressed in a vector form, it's a cross-product of position vector and linear momentum : $$\boxed{ \vec L = \vec r \times \vec p } $$

However, it's better to consider linear and angular momentum as different concepts, simply because body can have multiple momenta in a different reference frames at the same time. Consider a soccer ball going to a gate, when ball rotates around it's axis also moving in parabolic trajectory at the same time (kickers can make that happen), so it has additional angular momentum in other reference frame. Then if not enough complexity, try adding Coriolis force. Or analyze ball movement with respect to Sun and Milky Way galaxy center. Or maybe you are kicking a ball in a moving train, in which case ball has linear momentum component with respect to ground too ? Changing reference frame(s) for an object is endless.

Answer 4

This depends upon what you mean - the other answers have explained the sense in which the answer is 'no'. Just to state the obvious, I'll give the sense in which the answer is 'yes'.

In $n-$dimensional point mechanics, we have a phase space with canonical co-ordinates:

$$z = (q_1,\dots,q_n, p_1,\dots,p_n)$$

and all quantities of interest are functions of the coordinates. This includes angular momentum (as well as energy etc). In this sense 'position and momentum are basic, everything else is built from them'. We know this simply because everything is written as a function of the $z$'s and so whatever angular momentum is, it must have some formula $L=L(z)=L(q,p)$. And this statement is special to linear momentum - you could try to invert all the formulas and come up with coordinates on phase space that used the $L$'s but they would fail to be canonical since the components of angular momentum don't commute: $\{L_x,L_y\}=L_z$.

On the other hand, you might ask where the formula $L(z)$ comes from? This is what the other answers get at - you have to define angular momentum as the generator of rotations. This then gives rise to some formula for angular momentum in terms of linear momentum and position, but the specific formula could be different in different cases. So, 'yes' angular momentum does depend on linear momentum in any given scenario, but 'no' that you can't give a general definition based on this that works in all scenarios.

Answer 5

...angular momentum is merely a consequence of the linear momenta of each constituent particle coupled with the forces pulling each particle towards the center of mass.

The angular momentum as a conserved quantity is more than a rotating related stuff, as expressed by: linear momentum + centripetal forces. It is true that for an orbit, there is centripetal forces for an eventual satellite spin, and also a centripetal force (gravity) for the orbital angular momentum.

But even a free moving non rotating body, and consequentely without centripetal forces, has angular momentum, depending on the origin: $\mathbf L = \mathbf r \times \mathbf p$

It is important because if 2 non spinning objects collide, both can spin in the same way (both clockwise for example) afterwards. The total angular momentum doesn't change however, if measured from the same point of an inertial frame.

There is no rotation involved in part of the angular momentum that is conserved in cases like that.

Answer 6

The direct answer is no; but it was not an issue that was clarified until around the 1950's and afterwards when it was finally understood that the correct symmetry group describing kinematics for non-relativistic theory is not the Galilei group, but the Bargmann group; and when it was understood that a generalization of the Wigner classification of Relativistic Quantum Theory also existed for both classical systems and for non-relativistic theory (both classical and quantum).

In both classical and quantum theory, the types of fundamental systems that accord with the symmetries given in a kinematic transformation group (which include: spatial rotations, spatial translations, time translations and "boosts", i.e. transforms from one frame to another moving uniformly with respect to it) are called its "representations" and are modelled using symplectic geometry. Even for quantum theory, this method of classification largely subsumes the method based on Hilbert space and Hilbert space representations, so it provides a universal framework that transcends the "classical" versus "quantum" paradigm divide ... as well as the "relativistic" versus "non-relativistic" paradigm divide.

What emerged from the newer understanding is that even in classical physics, there is a notion of "spin"; and that, moreover, this notion also applies to systems in non-relativistic theory, not just relativity.

The kinematic groups associated respectively with Relativity and non-relativistic theory are the Poincaré and Bargmann groups. Each of them is a Lie group, and like all Lie groups is associated with a Poisson manifold. Poisson manifolds, in turn, decompose into a layering of symplectic leaves, each one of which provides a description of a type of elementary system. For the Poincaré group, the elementary systems coincide, one-to-one, with the "irreducible representations" given by Hilbert spaces and its representation theory.

The classification, for both Relativistic and non-Relativistic theory, includes elementary systems that are homogeneous in space and time. Of these, the one which is isotropic (i.e. invariant under spatial rotations) and boost-invariant is called the "vacuum". There are also non-isotropic and boost non-invariant homogeneous systems. None of them have any kind of well-defined localization in space or time, but do have a well-defined angular momentum and moment.

For all homogeneous systems, the mass and momentum are 0 and they have a well-defined, and invariant, energy - the vacuum energy. For Relativity, this is 0. But Relativity can be expanded by expanding Poincaré into Poincaré × E(1). This probably should be done, because only the latter has the Bargmann group as its correspondence limit, while the former does not. In this expanded version of Relativity, there is also a notion of vacuum energy (as well as an additional "rest energy" for the non-homogeneous systems, which I'm about to describe next.)

Of the non-homogeneous systems, are those that have rest frames - frame in which the momentum is 0; one of which is also the center of mass frame - in which the moment is 0. In this frame, in general, the value of the angular momentum is the system's spin, and it need not be 0.

In frames other than the center of mass / rest frame, the system's angular momentum decomposes into its orbital part and its spin, with the same formula applying across the board to both relativity and non-relativistic theory, to quantum and classical systems. The orbital part of the angular momentum depends on the momentum (and position), while the spin part does not. Similarly, the moment depends on momentum, position, mass and also has an explicit dependence on time. For Relativity, the moment also has a dependence on spin - a feature that distinguishes relativistic from non-relativistic systems.

One of the ways in which the independence of angular momentum (and spin) from momentum shows up in non-relativistic theory is as a large gap in Newton's Laws. It is not possible to derive angular momentum conservation from Newton's laws, because there is no "third law" for helical angular momentum! That's the component of angular momentum aligned with the axis separating two bodies.

A third law for helical angular momentum would read like this: "for every two bodies, to each torque impressed upon one, by the other, along an axis aligned with the separation of the two bodies, is an equal and opposite torque impressed upon the other."

It cannot be derived from Newton's Laws; particularly, not for spin, especially not if the system in question is elementary.