Angular momentum

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This gyroscope remains upright while spinning due to the conservation of its angular momentum.

In physics, angular momentum (rarely, moment of momentum or rotational momentum) is the rotational equivalent of linear momentum. It is an important quantity in physics because it is a conserved quantity – the total angular momentum of a system remains constant unless acted on by an external torque.

In three dimensions, the angular momentum for a point particle is a pseudovector r×p, the cross product of the particle's position vector r (relative to some origin) and its momentum vector p = mv. This definition can be applied to each point in continua like solids or fluids, or physical fields. Unlike momentum, angular momentum does depend on where the origin is chosen, since the particle's position is measured from it. The angular momentum vector of a point particle is parallel and directly proportional to the angular velocity vector ω of the particle (how fast its angular position changes), where the constant of proportionality depends on both the mass of the particle and its distance from origin. For continuous rigid bodies, though, the spin angular velocity ω is proportional but not always parallel to the spin angular momentum of the object, making the constant of proportionality I (called the moment of inertia) a second-rank tensor rather than a scalar.

Angular momentum is additive; the total angular momentum of a system is the (pseudo)vector sum of the angular momenta. For continua or fields one uses integration. The total angular momentum of any rigid body can be split into the sum of two main components: the angular momentum of the centre of mass (with a mass equal to the total mass) about the origin, plus the spin angular momentum of the object about the centre of mass.

Torque can be defined as the rate of change of angular momentum, analogous to force. The conservation of angular momentum helps explain many observed phenomena, for example the increase in rotational speed of a spinning figure skater as the skater's arms are contracted, the high rotational rates of neutron stars, the Coriolis effect, and precession of tops and gyroscopes. Applications include the gyrocompass, control moment gyroscope, inertial guidance systems, reaction wheels, flying discs or Frisbees, and Earth's rotation to name a few. In general, conservation does limit the possible motion of a system, but does not uniquely determine what the exact motion is.

In quantum mechanics, angular momentum is an operator with quantized eigenvalues. Angular momentum is subject to the Heisenberg uncertainty principle, meaning that at any time, only one component can be measured with definite precision; the other two cannot. Also, the "spin" of elementary particles does not correspond to literal spinning motion.^[1]

Angular momentum in classical mechanics

Definition

Scalar — angular momentum in two dimensions

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 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).^[2] Angular momentum can be considered a rotational analog of linear momentum. Thus, where linear momentum ${\displaystyle p}$ is proportional to mass ${\displaystyle m}$ and linear speed ${\displaystyle v}$,

${\displaystyle p=mv,}$

angular momentum ${\displaystyle L}$ is proportional to moment of inertia ${\displaystyle I}$ and angular speed ${\displaystyle \omega }$,^[3]

${\displaystyle L=I\omega .}$

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 speed, which occurs in a straight line, angular speed occurs about a center of rotation. Therefore, strictly speaking, ${\displaystyle L}$ should be referred to as the angular momentum relative to that center.^[4]

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 ${\displaystyle r}$ 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.^[5]

Scalar — angular momentum from Lagrangian mechanics

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

${\displaystyle T={\frac {1}{2}}ma^{2}\omega ^{2}={\frac {1}{2}}ma^{2}{\dot {\phi }}^{2}.}$

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

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

Vector — angular momentum in three dimensions

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

To completely define angular momentum in three dimensions, it is required to know the angle swept out in unit time, the direction perpendicular to the instantaneous plane of angular displacement, and the sense (right- or left-handed) of the angular velocity, as well as the mass involved.^[6] 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 angular momentum of a point particle in motion about the origin is defined as:

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

${\displaystyle I=r^{2}m}$ is the moment of inertia for a point mass,

${\displaystyle {\boldsymbol {\omega }}={\frac {\mathbf {r} \times \mathbf {v} }{r^{2}}}}$ is the angular velocity 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, ${\displaystyle \mathbf {L} =(r^{2}m)\left({\frac {\mathbf {r} \times \mathbf {v} }{r^{2}}}\right),}$ reduced, ${\displaystyle \mathbf {L} =m(\mathbf {r} \times \mathbf {v} ),}$ and by the rules of vector algebra rearranged to the form,

${\displaystyle {\begin{aligned}\mathbf {L} &=\mathbf {r} \times m\mathbf {v} \\&=\mathbf {r} \times \mathbf {p} ,\end{aligned}}}$

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:

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

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

Discussion

Angular momentum can be described as the rotational analog of linear momentum. Like linear momentum it involves elements of mass and displacement. Unlike linear momentum it also involves elements of position and shape.

Many problems in physics involve matter in motion about some certain point in space, be it in actual rotation about it, or simply moving past it, where it is desired to know what effect the moving matter has on the point — can it exert energy upon it or perform work about it? Energy, the ability to do work, can be stored in matter by setting it in motion — a combination of its inertia and its displacement. Inertia is measured by its mass, and displacement by its velocity. Their product,

${\displaystyle {\begin{aligned}({\text{amount of inertia}})\times ({\text{amount of displacement}})&={\text{amount of (inertia·displacement)}}\\{\text{mass}}\times {\text{velocity}}&={\text{momentum}}\\m\times v&=p\\\end{aligned}}}$

is the matter's momentum.^[7] Referring this momentum to a central point introduces a complication: the momentum is not applied to the point directly. For instance, a particle of matter at the outer edge of a wheel is, in effect, at the end of a lever of the same length as the wheel's radius, its momentum turning the lever about the center point. This imaginary lever is known as the moment arm. It has the effect of multiplying the momentum's effort in proportion to its length, an effect known as a moment. Hence, the particle's momentum referred to a particular point,


is the angular momentum, sometimes called, as here, the moment of momentum of the particle versus that particular center point. The equation ${\displaystyle L=rmv}$ combines a moment (a mass ${\displaystyle m}$ turning moment arm ${\displaystyle r}$) with a linear (straight-line equivalent) speed ${\displaystyle v}$. Linear speed referred to the central point is simply the product of the distance ${\displaystyle r}$ and the angular speed ${\displaystyle \omega }$ versus the point: ${\displaystyle v=r\omega ,}$ another moment. Hence, angular momentum contains a double moment: ${\displaystyle L=rmr\omega .}$ Simplifying slightly, ${\displaystyle L=r^{2}m\omega ,}$ the quantity ${\displaystyle r^{2}m}$ is the particle's moment of inertia, sometimes called the second moment of mass. It is a measure of rotational inertia.^[8]

Moment of inertia (shown here), and therefore angular momentum, is different for every possible configuration of mass and axis of rotation.

Because rotational inertia is a part of angular momentum, it necessarily includes all of the complications of moment of inertia, which is calculated by multiplying elementary bits of the mass by the squares of their distances from the center of rotation.^[9] Therefore, the total moment of inertia, and the angular momentum, is a complex function of the configuration of the matter about the center of rotation and the orientation of the rotation for the various bits.

For a rigid body, for instance a wheel or an asteroid, the orientation of rotation is simply the position of the rotation axis versus the matter of the body. It may or may not pass through the center of mass, or it may lie completely outside of the body. For the same body, angular momentum may take a different value for every possible axis about which rotation may take place.^[10] It reaches a minimum when the axis passes through the center of mass.^[11]

For a collection of objects revolving about a center, for instance all of the bodies of the Solar System, the orientations may be somewhat organized, as is the Solar System, with most of the bodies' axes lying close to the system's axis. Their orientations may also be completely random.

In brief, the more mass and the farther it is from the center of rotation (the longer the moment arm), the greater the moment of inertia, and therefore the greater the angular momentum for a given angular velocity. In many cases the moment of inertia, and hence the angular momentum, can be simplified by,^[12]

${\displaystyle I=k^{2}m,}$

where ${\displaystyle k}$ is the radius of gyration, the distance from the axis at which the entire mass ${\displaystyle m}$ may be considered as concentrated.

Similarly, for a point mass ${\displaystyle m}$ the moment of inertia is defined as,

${\displaystyle I=r^{2}m}$

where ${\displaystyle r}$ is the radius of the point mass from the center of rotation,

and for any collection of particles ${\displaystyle m_{i}}$ as the sum,

${\displaystyle \sum _{i}I_{i}=\sum _{i}r_{i}^{2}m_{i}}$

Angular momentum's dependence on position and shape is reflected in its units versus linear momentum: kg·m²/s, N·m·s or J·s for angular momentum versus kg·m/s or N·s for linear momentum. Angular momentum's units can be interpreted as torque·seconds, work·seconds, or energy·seconds. An object with angular momentum of L N·m·s can be reduced to zero rotation (all of the energy can be transferred out of it) by an angular impulse of L N·m·s^[13] or equivalently, by torque or work of L N·m for one second, or energy of L J for one second.^[14]

The plane perpendicular to the axis of angular momentum and passing through the center of mass^[15] is sometimes called the invariable plane, because the direction of the axis remains fixed if only the interactions of the bodies within the system, free from outside influences, are considered.^[16] One such plane is the invariable plane of the Solar System.

Angular momentum and torque

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 is

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

Because angular acceleration is the time derivative of angular velocity, and because the moment of inertia is ${\displaystyle mr^{2}}$ for point particles, the above formula is equivalent to ${\displaystyle {\boldsymbol {\tau }}=I{\frac {d{\boldsymbol {\omega }}}{dt}}+{\frac {d{I}}{dt}}{\boldsymbol {\omega }}.}$ Rearranging into a form suitable for integration, ${\displaystyle {\boldsymbol {\tau }}={\frac {d(I{\boldsymbol {\omega }})}{dt}}}$ and ${\displaystyle {\boldsymbol {\tau }}dt=d(I{\boldsymbol {\omega }}),}$ and integrating with respect to time,

${\displaystyle \int {\boldsymbol {\tau }}dt=I{\boldsymbol {\omega }}+{\text{constant}}.}$

Therefore, a torque acting over time is equivalent to a change in angular momentum, known as angular impulse, by analogy with impulse, which is defined as the change in translational momentum. The constant can be interpreted as the initial angular momentum of the body, before the torque began to act. In particular, if torque ${\displaystyle {\boldsymbol {\tau }}=\mathbf {0} ,}$ then angular momentum ${\displaystyle \mathbf {L} =\mathbf {0} +{\text{constant}}.}$^[17] That is, if no torque acts upon a body, then its angular momentum remains constant. Conversely,

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

or Angular momentum = moment of inertia × angular velocity, and its time derivative is

${\displaystyle {\frac {d\mathbf {L} }{dt}}={\frac {dI}{dt}}{\boldsymbol {\omega }}+I{\frac {d{\boldsymbol {\omega }}}{dt}}.}$

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

${\displaystyle {\frac {d\mathbf {L} }{dt}}=I{\boldsymbol {\alpha }}+2rp_{||}{\boldsymbol {\omega }}.}$

Therefore, the time rate of change of angular momentum about a particular center of rotation is equivalent to applied torque about that center.^[4] If angular momentum is constant, ${\displaystyle {\frac {d\mathbf {L} }{dt}}=0}$ and no torque is applied.

Conservation of angular momentum

A figure skater conserves angular momentum – her rotational speed increases as her moment of inertia decreases by drawing in her arms and legs.

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."^[18] 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).^[19]

Similarly, a rotational analog of Newton's Second law of Motion might be, "A change in angular momentum is proportional to the applied torque and occurs about the same axis as that torque."^[18] Since a torque applied over time is equivalent to a change in angular momentum, then if torque is zero, angular momentum is constant. As above, a system with constant angular momentum is a closed system. Therefore, requiring the system to be closed is equivalent to requiring that no external influence, in the form of a torque, acts upon it.^[20]

A rotational analog of Newton's First Law of Motion might be written, "A body continues in a state of rest or of uniform rotation unless acted by an external torque."^[18] Thus with no external influence to act upon it, the original angular momentum of the system is conserved.^[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. Decrease in the size of an object n times results in increase of its angular velocity by the factor of n².

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]

Angular momentum in orbital mechanics

In astrodynamics and celestial mechanics, a massless (or per unit mass) angular momentum is defined^[25]

${\displaystyle \mathbf {h} =\mathbf {r} \times \mathbf {v} ,}$

called specific angular momentum. Note that ${\displaystyle \mathbf {L} =m\mathbf {h} .}$ Mass is often unimportant in orbital mechanics calculations, because motion is defined by gravity. The primary body of the system is often so much larger than any bodies in motion about it that the smaller bodies have a negligible gravitational effect on it; it is, in effect, stationary. All bodies are apparently attracted by its gravity in the same way, regardless of mass, and therefore all move approximately the same way under the same conditions.

Solid bodies

For a continuous mass distribution with density function ρ(r), a differential volume element dV with position vector r within the mass has a mass element dm = ρ(r)dV. Therefore, the infinitesimal angular momentum of this element is:

${\displaystyle d\mathbf {L} =\mathbf {r} \times dm\mathbf {v} =\mathbf {r} \times \rho (\mathbf {r} )dV\mathbf {v} =dV\mathbf {r} \times \rho (\mathbf {r} )\mathbf {v} }$

and integrating this differential over the volume of the entire mass gives its total angular momentum:

${\displaystyle \mathbf {L} =\int _{V}dV\mathbf {r} \times \rho (\mathbf {r} )\mathbf {v} }$

In the derivation which follows, integrals similar to this can replace the sums for the case of continuous mass.

Collection of particles

Center of mass

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 ${\displaystyle i}$ vs the origin,

${\displaystyle \mathbf {V} _{i}}$ is the velocity of particle ${\displaystyle i}$ vs the origin,

${\displaystyle \mathbf {R} }$ is the position vector of the center of mass vs the origin,

${\displaystyle \mathbf {V} }$ is the velocity of the center of mass vs the origin,

${\displaystyle \mathbf {r} _{i}}$ is the position vector of particle ${\displaystyle i}$ vs the center of mass,

${\displaystyle \mathbf {v} _{i}}$ is the velocity of particle ${\displaystyle i}$ vs 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,^[26]

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

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,

${\displaystyle \mathbf {L} =\sum _{i}(\mathbf {R} _{i}\times m_{i}\mathbf {V} _{i})}$     (1)

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

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

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

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

It can be shown that (see sidebar),

Prove that ${\displaystyle \sum _{i}m_{i}\mathbf {r} _{i}=\mathbf {0} }$ ${\displaystyle \mathbf {r} _{i}=\mathbf {R} _{i}-\mathbf {R} }$ ${\displaystyle m_{i}\mathbf {r} _{i}=m_{i}(\mathbf {R} _{i}-\mathbf {R} )}$ ${\displaystyle {\begin{aligned}\sum _{i}m_{i}\mathbf {r} _{i}&=\sum _{i}m_{i}(\mathbf {R} _{i}-\mathbf {R} )\\&=\sum _{i}(m_{i}\mathbf {R} _{i}-m_{i}\mathbf {R} )\\&=\sum _{i}m_{i}\mathbf {R} _{i}-\sum _{i}m_{i}\mathbf {R} \\&=\sum _{i}m_{i}\mathbf {R} _{i}-\left(\sum _{i}m_{i}\right)\mathbf {R} \\&=\sum _{i}m_{i}\mathbf {R} _{i}-M\mathbf {R} \end{aligned}}}$ which, by the definition of the center of mass, is ${\displaystyle =\mathbf {0} ,}$ and similarly for ${\displaystyle \sum _{i}m_{i}\mathbf {v} _{i}.}$

${\displaystyle \sum _{i}m_{i}\mathbf {r} _{i}=\mathbf {0} }$ 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,^[27]

${\displaystyle \mathbf {L} =\mathbf {R} \times M\mathbf {V} +\sum _{i}\mathbf {r} _{i}\times m_{i}\mathbf {v} _{i}}$     (2)

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,

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

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

${\displaystyle \mathbf {L} =I_{R}{\boldsymbol {\omega }}_{R}+\sum _{i}I_{i}{\boldsymbol {\omega }}_{i}.}$     (3)

Simplifications

Single particle

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

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

${\displaystyle \mathbf {r} =\mathbf {R} ,}$

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

${\displaystyle m=M,}$

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

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

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

Fixed center of mass

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 momentum (modern definition)

The 3-angular momentum as a bivector (plane element) and axial vector, of a particle of mass m with instantaneous 3-position x and 3-momentum p.

In modern (20th century) theoretical physics, angular momentum (not including any intrinsic angular momentum – see below) is described using a different formalism, instead of a classical pseudovector. In this formalism, angular momentum is the 2-form Noether charge associated with rotational invariance. As a result, angular momentum is not conserved for general curved spacetimes, unless it happens to be asymptotically rotationally invariant.^{[citation needed]}

In classical mechanics, the angular momentum of a particle can be reinterpreted as a plane element:

${\displaystyle \mathbf {L} =\mathbf {r} \wedge \mathbf {p} \,,}$

in which the exterior product ∧ replaces the cross product × (these products have similar characteristics but are nonequivalent). This has the advantage of a clearer geometric interpretation as a plane element, defined from the x and p vectors, and the expression is true in any number of dimensions (two or higher). In Cartesian coordinates:

${\displaystyle {\begin{array}{rl}\mathbf {L} &=\left(xp_{y}-yp_{x}\right)\mathbf {e} _{x}\wedge \mathbf {e} _{y}+\left(yp_{z}-zp_{y}\right)\mathbf {e} _{y}\wedge \mathbf {e} _{z}+\left(zp_{x}-xp_{z}\right)\mathbf {e} _{z}\wedge \mathbf {e} _{x}\\&=L_{xy}\mathbf {e} _{x}\wedge \mathbf {e} _{y}+L_{yz}\mathbf {e} _{y}\wedge \mathbf {e} _{z}+L_{zx}\mathbf {e} _{z}\wedge \mathbf {e} _{x}\,,\end{array}}}$

or more compactly in index notation:

${\displaystyle L_{ij}=x_{i}p_{j}-x_{j}p_{i}\,.}$

The angular velocity can also be defined as an antisymmetric second order tensor, with components ω_ij. The relation between the two antisymmetric tensors is given by the moment of inertia which must now be a fourth order tensor:^[28]

${\displaystyle L_{ij}=I_{ijk\ell }\omega _{k\ell }\,.}$

Again, this equation in L and ω as tensors is true in any number of dimensions. This equation also appears in the geometric algebra formalism, in which L and ω are bivectors, and the moment of inertia is a mapping between them.

In relativistic mechanics, the relativistic angular momentum of a particle is expressed as an antisymmetric tensor of second order:

${\displaystyle M_{\alpha \beta }=X_{\alpha }\ P_{\beta }-X_{\beta }P_{\alpha }}$

in the language of four-vectors, namely the four position X and the four momentum P, and absorbs the above L together with the motion of the centre of mass of the particle.

In each of the above cases, for a system of particles, the total angular momentum is just the sum of the individual particle angular momenta, and the centre of mass is for the system.

Angular momentum in quantum mechanics

Angular momentum in quantum mechanics differs in many profound respects from angular momentum in classical mechanics. In relativistic quantum mechanics, it differs even more, in which the above relativistic definition becomes a tensorial operator.

Spin, orbital, and total angular momentum

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).^[29]

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.^[30] (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. Almost all elementary particles have spin. 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, for example electrons have "spin 1/2" (this actually means "spin ħ/2") while photons have "spin 1" (this actually means "spin ħ").

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 fractional values.^[31]

Quantization

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 Bohr model of the atom and was later predicted by Erwin Schrödinger in his Schrödinger equation.

Uncertainty

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.)

Total angular momentum as generator of rotations

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".^[32] 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.

Angular momentum in electrodynamics

When describing the motion of a charged particle in an electromagnetic field, the canonical momentum P (derived from the Lagrangian for this system) is not gauge invariant. As a consequence, the canonical angular momentum L = r × P is not gauge invariant either. Instead, the momentum that is physical, the so-called kinetic momentum (used throughout this article), is (in SI units)

${\displaystyle \mathbf {p} =m\mathbf {v} =\mathbf {P} -e\mathbf {A} }$

where e is the electric charge of the particle and A the magnetic vector potential of the electromagnetic field. The gauge-invariant angular momentum, that is kinetic angular momentum, is given by

${\displaystyle \mathbf {K} =\mathbf {r} \times (\mathbf {P} -e\mathbf {A} )}$

The interplay with quantum mechanics is discussed further in the article on canonical commutation relations.

Angular momentum in optics

In classical Maxwell electrodynamics the Poynting vector is a linear momentum density of electromagnetic field.^[33]

${\displaystyle \mathbf {S} (\mathbf {r} ,t)=\epsilon _{0}c^{2}\mathbf {E} (\mathbf {r} ,t)\times \mathbf {B} (\mathbf {r} ,t).}$

The angular momentum density vector ${\displaystyle \mathbf {L} (\mathbf {r} ,t)}$ is given by a vector product as in classical mechanics:^[34]

${\displaystyle \mathbf {L} (\mathbf {r} ,t)=\mathbf {r} \times \mathbf {S} (\mathbf {r} ,t).}$

The above identities are valid locally , i.e. in each space point ${\displaystyle \mathbf {r} }$ in a given moment ${\displaystyle {t}}$.

History

Newton, in the Principia, hinted at angular momentum in his examples of the First Law of Motion,

A top, whose parts by their cohesion are perpetually drawn aside from rectilinear motions, does not cease its rotation, otherwise than as it is retarded by the air. The greater bodies of the planets and comets, meeting with less resistance in more free spaces, preserve their motions both progressive and circular for a much longer time.^[35]

He did not further investigate angular momentum directly in the Principia,

From such kind of reflexions also sometimes arise the circular motions of bodies about their own centres. But these are cases which I do not consider in what follows; and it would be too tedious to demonstrate every particular that relates to this subject.^[36]

However, his geometric proof of the Law of Areas is an outstanding example of Newton's genius, and indirectly proves angular momentum conservation in the case of a central force.

The Law of Areas

Newton's derivation

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.

Conservation of angular momentum in the Law of Areas

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.

After Newton

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.^[37]

In 1736 Euler, like Newton, touched on some of the equations of angular momentum in his Mechanica without further developing them.^[38]

Bernoulli wrote in a 1744 letter of a "moment of rotational motion", possibly the first conception of angular momentum as we now understand it.^[39]

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,"^[40] 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,^[41] 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.^[42] 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.^[43]