Angular momentum

Velocity of the particle m with respect to the origin O can be resolved into components parallel to (v_∥) and perpendicular to (v_⊥) the radius vector r. The angular momentum of m is proportional to the perpendicular component v_⊥ of the velocity, or equivalently, to the perpendicular distance r_⊥ from the origin.

Angular momentum is a vector quantity (more precisely, a pseudovector) that represents the product of a body's rotational inertia and rotational velocity (in radians/sec) about a particular axis. However, if the particle's trajectory lies in a single plane, it is sufficient to discard the vector nature of angular momentum, and treat it as a scalar (more precisely, a pseudoscalar).[4] Angular momentum can be considered a rotational analog of linear momentum. Thus, where linear momentum p is proportional to mass m and linear speed v,

angular momentum L is proportional to moment of inertia I and angular speed ω measured in radians per second.[5]

Unlike mass, which depends only on amount of matter, moment of inertia is also dependent on the position of the axis of rotation and the shape of the matter. Unlike linear velocity, which does not depend upon the choice of origin, orbital angular velocity is always measured with respect to a fixed origin. Therefore, strictly speaking, L should be referred to as the angular momentum relative to that center.[6]

Because ${\displaystyle I=r^{2}m}$ for a single particle and ${\displaystyle \omega ={\frac {v}{r}}}$ for circular motion, angular momentum can be expanded, ${\displaystyle L=r^{2}m\cdot {\frac {v}{r}},}$ and reduced to,

${\displaystyle L=rmv,}$

the product of the radius of rotation r and the linear momentum of the particle ${\displaystyle p=mv}$, where ${\displaystyle v}$ in this case is the equivalent linear (tangential) speed at the radius (${\displaystyle =r\omega }$).

This simple analysis can also apply to non-circular motion if only the component of the motion which is perpendicular to the radius vector is considered. In that case,

${\displaystyle L=rmv_{\perp },}$

where ${\displaystyle v_{\perp }=v\sin(\theta )}$ is the perpendicular component of the motion. Expanding, ${\displaystyle L=rmv\sin(\theta ),}$ rearranging, ${\displaystyle L=r\sin(\theta )mv,}$ and reducing, angular momentum can also be expressed,

${\displaystyle L=r_{\perp }mv,}$

where ${\displaystyle r_{\perp }=r\sin(\theta )}$ is the length of the moment arm, a line dropped perpendicularly from the origin onto the path of the particle. It is this definition, (length of moment arm)×(linear momentum) to which the term moment of momentum refers.[7]

Another approach is to define angular momentum as the conjugate momentum (also called canonical momentum) of the angular coordinate ${\displaystyle \phi }$ expressed in the Lagrangian of the mechanical system. Consider a mechanical system with a mass ${\displaystyle m}$ constrained to move in a circle of radius ${\displaystyle a}$ in the absence of any external force field. The kinetic energy of the system is

And the potential energy is

${\displaystyle U=0.}$

Then the Lagrangian is

${\displaystyle {\mathcal {L}}\left(\phi ,{\dot {\phi }}\right)=T-U={\frac {1}{2}}ma^{2}{\dot {\phi }}^{2}.}$

The generalized momentum "canonically conjugate to" the coordinate ${\displaystyle \phi }$ is defined by

${\displaystyle p_{\phi }={\frac {\partial {\mathcal {L}}}{\partial {\dot {\phi }}}}=ma^{2}{\dot {\phi }}=I\omega =L.}$

Relationship between force (F), torque (τ), momentum (p), and angular momentum (L) vectors in a rotating system. r is the position vector.

To completely define orbital angular momentum in three dimensions, it is required to know the rate at which the position vector sweeps out angle, the direction perpendicular to the instantaneous plane of angular displacement, and the mass involved, as well as how this mass is distributed in space.[8] By retaining this vector nature of angular momentum, the general nature of the equations is also retained, and can describe any sort of three-dimensional motion about the center of rotation – circular, linear, or otherwise. In vector notation, the orbital angular momentum of a point particle in motion about the origin can be expressed as:

${\displaystyle \mathbf {L} =I{\boldsymbol {\omega }},}$

where

• ${\displaystyle I=r^{2}m}$ is the moment of inertia for a point mass,
• is the orbital angular velocity in radian/second (units 1/second) of the particle about the origin,
• ${\displaystyle \mathbf {r} }$ is the position vector of the particle relative to the origin, ${\displaystyle r=\left\vert \mathbf {r} \right\vert }$,
• ${\displaystyle \mathbf {v} }$ is the linear velocity of the particle relative to the origin, and
• ${\displaystyle m}$ is the mass of the particle.

This can be expanded, reduced, and by the rules of vector algebra, rearranged:

${\displaystyle {\begin{aligned}\mathbf {L} &=\left(r^{2}m\right)\left({\frac {\mathbf {r} \times \mathbf {v} }{r^{2}}}\right)\\&=m\left(\mathbf {r} \times \mathbf {v} \right)\\&=\mathbf {r} \times m\mathbf {v} \\&=\mathbf {r} \times \mathbf {p} ,\end{aligned}}}$

which is the cross product of the position vector ${\displaystyle \mathbf {r} }$ and the linear momentum ${\displaystyle \mathbf {p} =m\mathbf {v} }$ of the particle. By the definition of the cross product, the ${\displaystyle \mathbf {L} }$ vector is perpendicular to both ${\displaystyle \mathbf {r} }$ and ${\displaystyle \mathbf {p} }$. It is directed perpendicular to the plane of angular displacement, as indicated by the right-hand rule – so that the angular velocity is seen as counter-clockwise from the head of the vector. Conversely, the ${\displaystyle \mathbf {L} }$ vector defines the plane in which ${\displaystyle \mathbf {r} }$ and ${\displaystyle \mathbf {p} }$ lie.

By defining a unit vector ${\displaystyle \mathbf {\hat {u}} }$ perpendicular to the plane of angular displacement, a scalar angular speed ${\displaystyle \omega }$ results, where

${\displaystyle \omega \mathbf {\hat {u}} ={\boldsymbol {\omega }},}$ and

${\displaystyle \omega ={\frac {v_{\perp }}{r}},}$ where ${\displaystyle v_{\perp }}$ is the perpendicular component of the motion, as above.

The two-dimensional scalar equations of the previous section can thus be given direction:

${\displaystyle {\begin{aligned}\mathbf {L} &=I{\boldsymbol {\omega }}\\&=I\omega \mathbf {\hat {u}} \\&=\left(r^{2}m\right)\omega \mathbf {\hat {u}} \\&=rmv_{\perp }\mathbf {\hat {u}} \\&=r_{\perp }mv\mathbf {\hat {u}} ,\end{aligned}}}$

and ${\displaystyle \mathbf {L} =rmv\mathbf {\hat {u}} }$ for circular motion, where all of the motion is perpendicular to the radius ${\displaystyle r}$.

In the spherical coordinate system the angular momentum vector expresses as

${\displaystyle \mathbf {L} =m\mathbf {r} \times \mathbf {v} =mr^{2}\left({\dot {\theta }}\,{\hat {\boldsymbol {\varphi }}}-{\dot {\varphi }}\sin \theta \,\mathbf {\hat {\boldsymbol {\theta }}} \right).}$

Defining angular momentum by using the cross product applies only in three dimensions. Defining it as the bivector L = r ∧ p, where ∧ is the exterior product, is valid in any number of dimensions.

We can also define angular momentum as a rank 2 tensor in any number of dimensions. Namely, if ${\displaystyle x_{i}}$ is a position vector and ${\displaystyle p_{i}}$is the linear momentum vector (classically, ${\displaystyle p_{i}=m{\dot {x}}_{i}}$), then we can define

${\displaystyle M_{ij}=x_{i}p_{j}-p_{i}x_{j}}$

This is a rank 2 antisymmetric tensor with ${\displaystyle n(n-1)/2}$ independent components, where ${\displaystyle n}$ is the number of dimensions. In the usual three-dimensional case it has 3 independent components, which allows us to identify it with a 3 dimensional (pseudo-)vector ${\displaystyle {\vec {M}}=(M_{x},M_{y},M_{z})}$. The components of this vector relate to the components of the rank 2 tensor as follows:

${\displaystyle M_{ij}={\begin{pmatrix}0&M_{z}&-M_{y}\\-M_{z}&0&M_{x}\\M_{y}&-M_{x}&0\end{pmatrix}}}$

Newton's second law of motion can be expressed mathematically,

${\displaystyle \mathbf {F} =m\mathbf {a} ,}$

or force = mass × acceleration. The rotational equivalent for point particles may be derived as follows:

${\displaystyle \mathbf {L} =I{\boldsymbol {\omega }}}$

which means that the torque (i.e. the time derivative of the angular momentum) is

Because the moment of inertia is ${\displaystyle mr^{2}}$, it follows that ${\displaystyle {\frac {dI}{dt}}=2mr{\frac {dr}{dt}}=2rp_{||}}$, and ${\displaystyle {\frac {d\mathbf {L} }{dt}}=I{\frac {d{\boldsymbol {\omega }}}{dt}}+2rp_{||}{\boldsymbol {\omega }},}$ which, reduces to

${\displaystyle {\boldsymbol {\tau }}=I{\boldsymbol {\alpha }}+2rp_{||}{\boldsymbol {\omega }}.}$

This is the rotational analog of Newton's second Law. Note that the torque is not necessarily proportional or parallel to the angular acceleration (as one might expect). The reason for this is that the moment of inertia of a particle can change with time, something that cannot occur for ordinary mass.

A rotational analog of Newton's third law of motion might be written, "In a closed system, no torque can be exerted on any matter without the exertion on some other matter of an equal and opposite torque about the same axis."[19] Hence, angular momentum can be exchanged between objects in a closed system, but total angular momentum before and after an exchange remains constant (is conserved).[20]

Seen another way, a rotational analogue of Newton's first law of motion might be written, "A rigid body continues in a state of uniform rotation unless acted by an external influence."[19] Thus with no external influence to act upon it, the original angular momentum of the system remains constant.[21]

The conservation of angular momentum is used in analyzing central force motion. If the net force on some body is directed always toward some point, the center, then there is no torque on the body with respect to the center, as all of the force is directed along the radius vector, and none is perpendicular to the radius. Mathematically, torque ${\displaystyle {\boldsymbol {\tau }}=\mathbf {r} \times \mathbf {F} =\mathbf {0} ,}$ because in this case ${\displaystyle \mathbf {r} }$ and ${\displaystyle \mathbf {F} }$ are parallel vectors. Therefore, the angular momentum of the body about the center is constant. This is the case with gravitational attraction in the orbits of planets and satellites, where the gravitational force is always directed toward the primary body and orbiting bodies conserve angular momentum by exchanging distance and velocity as they move about the primary. Central force motion is also used in the analysis of the Bohr model of the atom.

For a planet, angular momentum is distributed between the spin of the planet and its revolution in its orbit, and these are often exchanged by various mechanisms. The conservation of angular momentum in the Earth–Moon system results in the transfer of angular momentum from Earth to Moon, due to tidal torque the Moon exerts on the Earth. This in turn results in the slowing down of the rotation rate of Earth, at about 65.7 nanoseconds per day,[22] and in gradual increase of the radius of Moon's orbit, at about 3.82 centimeters per year.[23]

The torque caused by the two opposing forces F_g and −F_g causes a change in the angular momentum L in the direction of that torque (since torque is the time derivative of angular momentum). This causes the top to precess.

The conservation of angular momentum explains the angular acceleration of an ice skater as she brings her arms and legs close to the vertical axis of rotation. By bringing part of the mass of her body closer to the axis, she decreases her body's moment of inertia. Because angular momentum is the product of moment of inertia and angular velocity, if the angular momentum remains constant (is conserved), then the angular velocity (rotational speed) of the skater must increase.

The same phenomenon results in extremely fast spin of compact stars (like white dwarfs, neutron stars and black holes) when they are formed out of much larger and slower rotating stars.

Conservation is not always a full explanation for the dynamics of a system but is a key constraint. For example, a spinning top is subject to gravitational torque making it lean over and change the angular momentum about the nutation axis, but neglecting friction at the point of spinning contact, it has a conserved angular momentum about its spinning axis, and another about its precession axis. Also, in any planetary system, the planets, star(s), comets, and asteroids can all move in numerous complicated ways, but only so that the angular momentum of the system is conserved.

Noether's theorem states that every conservation law is associated with a symmetry (invariant) of the underlying physics. The symmetry associated with conservation of angular momentum is rotational invariance. The fact that the physics of a system is unchanged if it is rotated by any angle about an axis implies that angular momentum is conserved.[24]

While angular momentum total conservation can be understood separately from Newton's laws of motion as stemming from Noether's theorem in systems symmetric under rotations, it can also be understood simply as an efficient method of calculation of results that can also be otherwise arrived at directly from Newton's second law, together with laws governing the forces of nature (such as Newton's third law, Maxwell's equations and Lorentz force). Indeed, given initial conditions of position and velocity for every point, and the forces at such a condition, one may use Newton's second law to calculate the second derivative of position, and solving for this gives full information on the development of the physical system with time.[25] Note, however, that this is no longer true in quantum mechanics, due to the existence of particle spin, which is angular momentum that cannot be described by the cumulative effect of point-like motions in space.

As an example, consider decreasing of the moment of inertia, e.g. when a figure skater is pulling in her/his hands, speeding up the circular motion. In terms of angular momentum conservation, we have, for angular momentum L, moment of inertia I and angular velocity ω:

${\displaystyle 0=dL=d(I\cdot \omega )=dI\cdot \omega +I\cdot d\omega }$

Using this, we see that the change requires an energy of:

${\displaystyle dE=d\left({\frac {1}{2}}I\cdot \omega ^{2}\right)={\frac {1}{2}}dI\cdot \omega ^{2}+I\cdot \omega \cdot d\omega =-{\frac {1}{2}}dI\cdot \omega ^{2}}$

so that a decrease in the moment of inertia requires investing energy.

This can be compared to the work done as calculated using Newton's laws. Each point in the rotating body is accelerating, at each point of time, with radial acceleration of:

${\displaystyle -r\cdot \omega ^{2}}$

Let us observe a point of mass m, whose position vector relative to the center of motion is perpendicular to the z-axis at a given point of time, and is at a distance z. The centripetal force on this point, keeping the circular motion, is:

${\displaystyle -m\cdot z\cdot \omega ^{2}}$

Thus the work required for moving this point to a distance dz farther from the center of motion is:

${\displaystyle dW=-m\cdot z\cdot \omega ^{2}\cdot dz=-m\cdot \omega ^{2}\cdot d\left({\frac {1}{2}}z^{2}\right)}$

For a non-pointlike body one must integrate over this, with m replaced by the mass density per unit z. This gives:

${\displaystyle dW=-{\frac {1}{2}}dI\cdot \omega ^{2}}$

which is exactly the energy required for keeping the angular momentum conserved.

Note, that the above calculation can also be performed per mass, using kinematics only. Thus the phenomena of figure skater accelerating tangential velocity while pulling her/his hands in, can be understood as follows in layman's language: The skater's palms are not moving in a straight line, so they are constantly accelerating inwards, but do not gain additional speed because the accelerating is always done when their motion inwards is zero. However, this is different when pulling the palms closer to the body: The acceleration due to rotation now increases the speed; but because of the rotation, the increase in speed does not translate to a significant speed inwards, but to an increase of the rotation speed.

In Lagrangian mechanics, angular momentum for rotation around a given axis, is the conjugate momentum of the generalized coordinate of the angle around the same axis. For example, ${\displaystyle L_{z}}$, the angular momentum around the z axis, is:

${\displaystyle L_{z}={\frac {\partial {\cal {L}}}{\partial {\dot {\theta }}_{z}}}}$

where ${\displaystyle {\cal {L}}}$ is the Lagrangian and ${\displaystyle \theta _{z}}$ is the angle around the z axis.

Note that ${\displaystyle {\dot {\theta }}_{z}}$, the time derivative of the angle, is the angular velocity ${\displaystyle \omega _{z}}$. Ordinarily, the Lagrangian depends on the angular velocity through the kinetic energy: The latter can be written by separating the velocity to its radial and tangential part, with the tangential part at the x-y plane, around the z-axis, being equal to:

${\displaystyle \sum _{i}{\frac {1}{2}}m_{i}{v_{T}}_{i}^{2}=\sum _{i}{\frac {1}{2}}m_{i}(x_{i}^{2}+y_{i}^{2}){{\omega _{z}}_{i}}^{2}}$

where the subscript i stands for the i-th body, and m, v_T and ω_z stand for mass, tangential velocity around the z-axis and angular velocity around that axis, respectively.

For a body that is not point-like, with density ρ, we have instead:

${\displaystyle {\frac {1}{2}}\int \rho (x,y,z)(x_{i}^{2}+y_{i}^{2}){{\omega _{z}}_{i}}^{2}\,dx\,dy={\frac {1}{2}}{I_{z}}_{i}{{\omega _{z}}_{i}}^{2}}$

where integration runs over the area of the body,[26] and I_z is the moment of inertia around the z-axis.

Thus, assuming the potential energy does not depend on ω_z (this assumption may fail for electromagnetic systems), we have the angular momentum of the ith object:

${\displaystyle {\begin{aligned}{L_{z}}_{i}&={\frac {\partial {\cal {L}}}{\partial {{\omega _{z}}_{i}}}}={\frac {\partial E_{k}}{\partial {{\omega _{z}}_{i}}}}\\&={I_{z}}_{i}\cdot {\omega _{z}}_{i}\end{aligned}}}$

We have thus far rotated each object by a separate angle; we may also define an overall angle θ_z by which we rotate the whole system, thus rotating also each object around the z-axis, and have the overall angular momentum:

${\displaystyle L_{z}=\sum _{i}{I_{z}}_{i}\cdot {\omega _{z}}_{i}}$

From Euler–Lagrange equations it then follows that:

${\displaystyle 0={\frac {\partial {\cal {L}}}{\partial {{\theta _{z}}_{i}}}}-{\frac {d}{dt}}\left({\frac {\partial {\cal {L}}}{\partial {{{\dot {\theta }}_{z}}_{i}}}}\right)={\frac {\partial {\cal {L}}}{\partial {{\theta _{z}}_{i}}}}-{\frac {d{L_{z}}_{i}}{dt}}}$

Since the lagrangian is dependent upon the angles of the object only through the potential, we have:

${\displaystyle {\frac {d{L_{z}}_{i}}{dt}}={\frac {\partial {\cal {L}}}{\partial {{\theta _{z}}_{i}}}}=-{\frac {\partial V}{\partial {{\theta _{z}}_{i}}}}}$

which is the torque on the ith object.

Suppose the system is invariant to rotations, so that the potential is independent of an overall rotation by the angle θ_z (thus it may depend on the angles of objects only through their differences, in the form ${\displaystyle V({\theta _{z}}_{i},{\theta _{z}}_{j})=V({\theta _{z}}_{i}-{\theta _{z}}_{j})}$). We therefore get for the total angular momentum:

${\displaystyle {\frac {dL_{z}}{dt}}=-{\frac {\partial V}{\partial {\theta _{z}}}}=0}$

And thus the angular momentum around the z-axis is conserved.

This analysis can be repeated separately for each axis, giving conversation of the angular momentum vector. However, the angles around the three axes cannot be treated simultaneously as generalized coordinates, since they are not independent; in particular, two angles per point suffice to determine its position. While it is true that in the case of a rigid body, fully describing it requires, in addition to three translational degrees of freedom, also specification of three rotational degrees of freedom; however these cannot be defined as rotations around the Cartesian axes (see Euler angles). This caveat is reflected in quantum mechanics in the non-trivial commutation relations of the different components of the angular momentum operator.

Equivalently, in Hamiltonian mechanics the Hamiltonian can be described as a function of the angular momentum. As before, the part of the kinetic energy related to rotation around the z-axis for the ith object is:

${\displaystyle {\frac {1}{2}}{I_{z}}_{i}{{\omega _{z}}_{i}}^{2}={\frac {{{L_{z}}_{i}}^{2}}{2{I_{z}}_{i}}}}$

which is analogous to the energy dependence upon momentum along the z-axis, ${\displaystyle {\frac {{{p_{z}}_{i}}^{2}}{{2m}_{i}}}}$.

Hamilton's equations relate the angle around the z-axis to its conjugate momentum, the angular momentum around the same axis:

${\displaystyle {\begin{aligned}{\frac {d{\theta _{z}}_{i}}{dt}}&={\frac {\partial {\cal {H}}}{\partial {L_{z}}_{i}}}={\frac {{L_{z}}_{i}}{{I_{z}}_{i}}}\\{\frac {d{L_{z}}_{i}}{dt}}&=-{\frac {\partial {\cal {H}}}{\partial {\theta _{z}}_{i}}}=-{\frac {\partial V}{\partial {\theta _{z}}_{i}}}\end{aligned}}}$

The first equation gives

${\displaystyle {L_{z}}_{i}={I_{z}}_{i}\cdot {{{\dot {\theta }}_{z}}_{i}}={I_{z}}_{i}\cdot {\omega _{z}}_{i}}$

And so we get the same results as in the Lagrangian formalism.

Note, that for combining all axes together, we write the kinetic energy as:

${\displaystyle E_{k}={\frac {1}{2}}\sum _{i}{\frac {|{\bf {{p}_{i}|^{2}}}}{2m_{i}}}=\sum _{i}\left({\frac {{{p_{r}}_{i}}^{2}}{2m_{i}}}+{\frac {1}{2}}{\bf {{L}_{i}}}^{\textsf {T}}{I_{i}}^{-1}{\bf {{L}_{i}}}\right)}$

where p_r is the momentum in the radial direction, and the moment of inertia is a 3-dimensional matrix; bold letters stand for 3-dimensional vectors.

For point-like bodies we have:

${\displaystyle E_{k}=\sum _{i}\left({\frac {{{p_{r}}_{i}}^{2}}{2m_{i}}}+{\frac {|{\bf {{L}_{i}}}|^{2}}{2m_{i}{r_{i}}^{2}}}\right)}$

This form of the kinetic energy part of the Hamiltonian is useful in analyzing central potential problems, and is easily transformed to a quantum mechanical work frame (e.g. in the hydrogen atom problem).

The angular momentum of the particles i is the sum of the cross products R × MV + Σr_i × m_iv_i.

For a collection of particles in motion about an arbitrary origin, it is informative to develop the equation of angular momentum by resolving their motion into components about their own center of mass and about the origin. Given,

• ${\displaystyle m_{i}}$ is the mass of particle ${\displaystyle i}$,
• ${\displaystyle \mathbf {R} _{i}}$ is the position vector of particle ${\displaystyle i}$ w.r.t. the origin,
• ${\displaystyle \mathbf {V} _{i}}$ is the velocity of particle ${\displaystyle i}$ w.r.t. the origin,
• ${\displaystyle \mathbf {R} }$ is the position vector of the center of mass w.r.t. the origin,
• ${\displaystyle \mathbf {V} }$ is the velocity of the center of mass w.r.t. the origin,
• ${\displaystyle \mathbf {r} _{i}}$ is the position vector of particle ${\displaystyle i}$ w.r.t. the center of mass,
• ${\displaystyle \mathbf {v} _{i}}$ is the velocity of particle ${\displaystyle i}$ w.r.t. the center of mass,

The total mass of the particles is simply their sum,

${\displaystyle M=\sum _{i}m_{i}.}$

The position vector of the center of mass is defined by,[28]

${\displaystyle M\mathbf {R} =\sum _{i}m_{i}\mathbf {R} _{i}.}$

By inspection,

${\displaystyle \mathbf {R} _{i}=\mathbf {R} +\mathbf {r} _{i}}$ and ${\displaystyle \mathbf {V} _{i}=\mathbf {V} +\mathbf {v} _{i}.}$

The total angular momentum of the collection of particles is the sum of the angular momentum of each particle,

Expanding ${\displaystyle \mathbf {R} _{i}}$,

${\displaystyle {\begin{aligned}\mathbf {L} &=\sum _{i}\left[\left(\mathbf {R} +\mathbf {r} _{i}\right)\times m_{i}\mathbf {V} _{i}\right]\\&=\sum _{i}\left[\mathbf {R} \times m_{i}\mathbf {V} _{i}+\mathbf {r} _{i}\times m_{i}\mathbf {V} _{i}\right]\end{aligned}}}$

Expanding ${\displaystyle \mathbf {V} _{i}}$,

${\displaystyle {\begin{aligned}\mathbf {L} &=\sum _{i}\left[\mathbf {R} \times m_{i}\left(\mathbf {V} +\mathbf {v} _{i}\right)+\mathbf {r} _{i}\times m_{i}(\mathbf {V} +\mathbf {v} _{i})\right]\\&=\sum _{i}\left[\mathbf {R} \times m_{i}\mathbf {V} +\mathbf {R} \times m_{i}\mathbf {v} _{i}+\mathbf {r} _{i}\times m_{i}\mathbf {V} +\mathbf {r} _{i}\times m_{i}\mathbf {v} _{i}\right]\\&=\sum _{i}\mathbf {R} \times m_{i}\mathbf {V} +\sum _{i}\mathbf {R} \times m_{i}\mathbf {v} _{i}+\sum _{i}\mathbf {r} _{i}\times m_{i}\mathbf {V} +\sum _{i}\mathbf {r} _{i}\times m_{i}\mathbf {v} _{i}\end{aligned}}}$

It can be shown that (see sidebar),

${\displaystyle \sum _{i}m_{i}\mathbf {r} _{i}=\mathbf {0} }$ and ${\displaystyle \sum _{i}m_{i}\mathbf {v} _{i}=\mathbf {0} ,}$

therefore the second and third terms vanish,

${\displaystyle \mathbf {L} =\sum _{i}\mathbf {R} \times m_{i}\mathbf {V} +\sum _{i}\mathbf {r} _{i}\times m_{i}\mathbf {v} _{i}.}$

The first term can be rearranged,

${\displaystyle \sum _{i}\mathbf {R} \times m_{i}\mathbf {V} =\mathbf {R} \times \sum _{i}m_{i}\mathbf {V} =\mathbf {R} \times M\mathbf {V} ,}$

and total angular momentum for the collection of particles is finally,[29]

The first term is the angular momentum of the center of mass relative to the origin. Similar to § Single particle, below, it is the angular momentum of one particle of mass M at the center of mass moving with velocity V. The second term is the angular momentum of the particles moving relative to the center of mass, similar to § Fixed center of mass, below. The result is general—the motion of the particles is not restricted to rotation or revolution about the origin or center of mass. The particles need not be individual masses, but can be elements of a continuous distribution, such as a solid body.

Rearranging equation (2) by vector identities, multiplying both terms by "one", and grouping appropriately,

${\displaystyle {\begin{aligned}\mathbf {L} &=M(\mathbf {R} \times \mathbf {V} )+\sum _{i}\left[m_{i}\left(\mathbf {r} _{i}\times \mathbf {v} _{i}\right)\right],\\&={\frac {R^{2}}{R^{2}}}M\left(\mathbf {R} \times \mathbf {V} \right)+\sum _{i}\left[{\frac {r_{i}^{2}}{r_{i}^{2}}}m_{i}\left(\mathbf {r} _{i}\times \mathbf {v} _{i}\right)\right],\\&=R^{2}M\left({\frac {\mathbf {R} \times \mathbf {V} }{R^{2}}}\right)+\sum _{i}\left[r_{i}^{2}m_{i}\left({\frac {\mathbf {r} _{i}\times \mathbf {v} _{i}}{r_{i}^{2}}}\right)\right],\\\end{aligned}}}$

gives the total angular momentum of the system of particles in terms of moment of inertia ${\displaystyle I}$ and angular velocity ${\displaystyle {\boldsymbol {\omega }}}$,

In the case of a single particle moving about the arbitrary origin,

${\displaystyle {\begin{aligned}\mathbf {r} _{i}&=\mathbf {v} _{i}=\mathbf {0} ,\\\mathbf {r} &=\mathbf {R} ,\\\mathbf {v} &=\mathbf {V} ,\\m&=M,\end{aligned}}}$

${\displaystyle \sum _{i}\mathbf {r} _{i}\times m_{i}\mathbf {v} _{i}=\mathbf {0} ,}$

${\displaystyle \sum _{i}I_{i}{\boldsymbol {\omega }}_{i}=\mathbf {0} ,}$ and equations (2) and (3) for total angular momentum reduce to,

${\displaystyle \mathbf {L} =\mathbf {R} \times m\mathbf {V} =I_{R}{\boldsymbol {\omega }}_{R}.}$

For the case of the center of mass fixed in space with respect to the origin,

${\displaystyle \mathbf {V} =\mathbf {0} ,}$

${\displaystyle \mathbf {R} \times M\mathbf {V} =\mathbf {0} ,}$

${\displaystyle I_{R}{\boldsymbol {\omega }}_{R}=\mathbf {0} ,}$ and equations (2) and (3) for total angular momentum reduce to,

${\displaystyle \mathbf {L} =\sum _{i}\mathbf {r} _{i}\times m_{i}\mathbf {v} _{i}=\sum _{i}I_{i}{\boldsymbol {\omega }}_{i}.}$

Angular momenta of a classical object.

• Left: "spin" angular momentum S is really orbital angular momentum of the object at every point.
• Right: extrinsic orbital angular momentum L about an axis.
• Top: the moment of inertia tensor I and angular velocity ω (L is not always parallel to ω).[32]
• Bottom: momentum p and its radial position r from the axis. The total angular momentum (spin plus orbital) is J. For a quantum particle the interpretations are different; particle spin does not have the above interpretation.

The classical definition of angular momentum as ${\displaystyle \mathbf {L} =\mathbf {r} \times \mathbf {p} }$ can be carried over to quantum mechanics, by reinterpreting r as the quantum position operator and p as the quantum momentum operator. L is then an operator, specifically called the orbital angular momentum operator. The components of the angular momentum operator satisfy the commutation relations of the Lie algebra so(3). Indeed, these operators are precisely the infinitesimal action of the rotation group on the quantum Hilbert space.[33] (See also the discussion below of the angular momentum operators as the generators of rotations.)

However, in quantum physics, there is another type of angular momentum, called spin angular momentum, represented by the spin operator S. Spin is often depicted as a particle literally spinning around an axis, but this is a misleading and inaccurate picture: spin is an intrinsic property of a particle, unrelated to any sort of motion in space and fundamentally different from orbital angular momentum. All elementary particles have a characteristic spin (possibly zero),[34] and almost all elementary particles have nonzero spin.[35] For example electrons have "spin 1/2" (this actually means "spin ħ/2"), photons have "spin 1" (this actually means "spin ħ"), and pi-mesons have spin 0.[36]

Finally, there is total angular momentum J, which combines both the spin and orbital angular momentum of all particles and fields. (For one particle, J = L + S.) Conservation of angular momentum applies to J, but not to L or S; for example, the spin–orbit interaction allows angular momentum to transfer back and forth between L and S, with the total remaining constant. Electrons and photons need not have integer-based values for total angular momentum, but can also have half-integer values.[37]

In molecules the total angular momentum F is the sum of the rovibronic (orbital) angular momentum N, the electron spin angular momentum S, and the nuclear spin angular momentum I. For electronic singlet states the rovibronic angular momentum is denoted J rather than N. As explained by Van Vleck,[38] the components of the molecular rovibronic angular momentum referred to molecule-fixed axes have different commutation relations from those for the components about space-fixed axes.

In quantum mechanics, angular momentum is quantized – that is, it cannot vary continuously, but only in "quantum leaps" between certain allowed values. For any system, the following restrictions on measurement results apply, where ${\displaystyle \hbar }$ is the reduced Planck constant and ${\displaystyle {\hat {n}}}$ is any Euclidean vector such as x, y, or z:

If you measure...	The result can be...
${\displaystyle L_{\hat {n}}}$	${\displaystyle \ldots ,-2\hbar ,-\hbar ,0,\hbar ,2\hbar ,\ldots }$
${\displaystyle S_{\hat {n}}}$ or ${\displaystyle J_{\hat {n}}}$	${\displaystyle \ldots ,-{\frac {3}{2}}\hbar ,-\hbar ,-{\frac {1}{2}}\hbar ,0,{\frac {1}{2}}\hbar ,\hbar ,{\frac {3}{2}}\hbar ,\ldots }$
${\displaystyle {\begin{aligned}&L^{2}\\={}&L_{x}^{2}+L_{y}^{2}+L_{z}^{2}\end{aligned}}}$	${\displaystyle \left[\hbar ^{2}n(n+1)\right]}$, where ${\displaystyle n=0,1,2,\ldots }$
${\displaystyle S^{2}}$ or ${\displaystyle J^{2}}$	${\displaystyle \left[\hbar ^{2}n(n+1)\right]}$, where ${\displaystyle n=0,{\frac {1}{2}},1,{\frac {3}{2}},\ldots }$

In this standing wave on a circular string, the circle is broken into exactly 8 wavelengths. A standing wave like this can have 0,1,2, or any integer number of wavelengths around the circle, but it cannot have a non-integer number of wavelengths like 8.3. In quantum mechanics, angular momentum is quantized for a similar reason.

The reduced Planck constant ${\displaystyle \hbar }$ is tiny by everyday standards, about 10⁻³⁴ J s, and therefore this quantization does not noticeably affect the angular momentum of macroscopic objects. However, it is very important in the microscopic world. For example, the structure of electron shells and subshells in chemistry is significantly affected by the quantization of angular momentum.

Quantization of angular momentum was first postulated by Niels Bohr in his model of the atom and was later predicted by Erwin Schrödinger in his Schrödinger equation.

In the definition ${\displaystyle \mathbf {L} =\mathbf {r} \times \mathbf {p} }$, six operators are involved: The position operators ${\displaystyle r_{x}}$, ${\displaystyle r_{y}}$, ${\displaystyle r_{z}}$, and the momentum operators ${\displaystyle p_{x}}$, ${\displaystyle p_{y}}$, ${\displaystyle p_{z}}$. However, the Heisenberg uncertainty principle tells us that it is not possible for all six of these quantities to be known simultaneously with arbitrary precision. Therefore, there are limits to what can be known or measured about a particle's angular momentum. It turns out that the best that one can do is to simultaneously measure both the angular momentum vector's magnitude and its component along one axis.

The uncertainty is closely related to the fact that different components of an angular momentum operator do not commute, for example ${\displaystyle L_{x}L_{y}\neq L_{y}L_{x}}$. (For the precise commutation relations, see angular momentum operator.)

As mentioned above, orbital angular momentum L is defined as in classical mechanics: ${\displaystyle \mathbf {L} =\mathbf {r} \times \mathbf {p} }$, but total angular momentum J is defined in a different, more basic way: J is defined as the "generator of rotations".[39] More specifically, J is defined so that the operator

${\displaystyle R({\hat {n}},\phi )\equiv \exp \left(-{\frac {i}{\hbar }}\phi \,\mathbf {J} \cdot {\hat {\mathbf {n} }}\right)}$

is the rotation operator that takes any system and rotates it by angle ${\displaystyle \phi }$ about the axis ${\displaystyle {\hat {\mathbf {n} }}}$. (The "exp" in the formula refers to operator exponential) To put this the other way around, whatever our quantum Hilbert space is, we expect that the rotation group SO(3) will act on it. There is then an associated action of the Lie algebra so(3) of SO(3); the operators describing the action of so(3) on our Hilbert space are the (total) angular momentum operators.

The relationship between the angular momentum operator and the rotation operators is the same as the relationship between Lie algebras and Lie groups in mathematics. The close relationship between angular momentum and rotations is reflected in Noether's theorem that proves that angular momentum is conserved whenever the laws of physics are rotationally invariant.

Newton's derivation of the area law using geometric means

As a planet orbits the Sun, the line between the Sun and the planet sweeps out equal areas in equal intervals of time. This had been known since Kepler expounded his second law of planetary motion. Newton derived a unique geometric proof, and went on to show that the attractive force of the Sun's gravity was the cause of all of Kepler's laws.

During the first interval of time, an object is in motion from point A to point B. Undisturbed, it would continue to point c during the second interval. When the object arrives at B, it receives an impulse directed toward point S. The impulse gives it a small added velocity toward S, such that if this were its only velocity, it would move from B to V during the second interval. By the rules of velocity composition, these two velocities add, and point C is found by construction of parallelogram BcCV. Thus the object's path is deflected by the impulse so that it arrives at point C at the end of the second interval. Because the triangles SBc and SBC have the same base SB and the same height Bc or VC, they have the same area. By symmetry, triangle SBc also has the same area as triangle SAB, therefore the object has swept out equal areas SAB and SBC in equal times.

At point C, the object receives another impulse toward S, again deflecting its path during the third interval from d to D. Thus it continues to E and beyond, the triangles SAB, SBc, SBC, SCd, SCD, SDe, SDE all having the same area. Allowing the time intervals to become ever smaller, the path ABCDE approaches indefinitely close to a continuous curve.

Note that because this derivation is geometric, and no specific force is applied, it proves a more general law than Kepler's second law of planetary motion. It shows that the Law of Areas applies to any central force, attractive or repulsive, continuous or non-continuous, or zero.

The proportionality of angular momentum to the area swept out by a moving object can be understood by realizing that the bases of the triangles, that is, the lines from S to the object, are equivalent to the radius r, and that the heights of the triangles are proportional to the perpendicular component of velocity v_⊥. Hence, if the area swept per unit time is constant, then by the triangular area formula 1/2(base)(height), the product (base)(height) and therefore the product rv_⊥ are constant: if r and the base length are decreased, v_⊥ and height must increase proportionally. Mass is constant, therefore angular momentum rmv_⊥ is conserved by this exchange of distance and velocity.

In the case of triangle SBC, area is equal to 1/2(SB)(VC). Wherever C is eventually located due to the impulse applied at B, the product (SB)(VC), and therefore rmv_⊥ remain constant. Similarly so for each of the triangles.

Leonhard Euler, Daniel Bernoulli, and Patrick d'Arcy all understood angular momentum in terms of conservation of areal velocity, a result of their analysis of Kepler's second law of planetary motion. It is unlikely that they realized the implications for ordinary rotating matter.[45]

In 1736 Euler, like Newton, touched on some of the equations of angular momentum in his Mechanica without further developing them.[46]

Bernoulli wrote in a 1744 letter of a "moment of rotational motion", possibly the first conception of angular momentum as we now understand it.[47]

In 1799, Pierre-Simon Laplace first realized that a fixed plane was associated with rotation—his invariable plane.

Louis Poinsot in 1803 began representing rotations as a line segment perpendicular to the rotation, and elaborated on the "conservation of moments".

In 1852 Léon Foucault used a gyroscope in an experiment to display the Earth's rotation.

William J. M. Rankine's 1858 Manual of Applied Mechanics defined angular momentum in the modern sense for the first time:

...a line whose length is proportional to the magnitude of the angular momentum, and whose direction is perpendicular to the plane of motion of the body and of the fixed point, and such, that when the motion of the body is viewed from the extremity of the line, the radius-vector of the body seems to have right-handed rotation.

In an 1872 edition of the same book, Rankine stated that "The term angular momentum was introduced by Mr. Hayward,"[48] probably referring to R.B. Hayward's article On a Direct Method of estimating Velocities, Accelerations, and all similar Quantities with respect to Axes moveable in any manner in Space with Applications,[49] which was introduced in 1856, and published in 1864. Rankine was mistaken, as numerous publications feature the term starting in the late 18th to early 19th centuries.[50] However, Hayward's article apparently was the first use of the term and the concept seen by much of the English-speaking world. Before this, angular momentum was typically referred to as "momentum of rotation" in English.[51]