# Angular momentum

(Redirected from Conservation of angular momentum)
For a more accessible and less technical introduction to this topic, see Introduction to angular momentum.
This gyroscope remains upright while spinning due to its angular momentum.

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

Angular momentum is related to the rotation or revolution of matter. It is, in effect, a measure of the quantity of rotation of a system of matter, taking into account its mass, speed of rotations, movements and shape. The conservation of angular momentum explains 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 falling cat problem and precession can all be explained in terms of angular momentum conservation. It has numerous applications in physics and engineering, for instance, the gyrocompass, control moment gyroscope, inertial guidance systems and consumer products.

## Angular momentum in classical mechanics

### Definition

#### Scalar

The angular momentum of the particle at P with respect to the origin O is proportional to the perpendicular component $V_\perp$ of the velocity vector $V$.

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. In the simple case of revolution of a particle in a circle about a center of rotation, the particle remaining always in the same plane, it is sufficient to discard the vector nature of angular momentum, and treat it as a scalar.[1] Angular momentum can be considered a rotational analog of linear momentum. Thus, where linear momentum is proportional to mass $m$ and linear speed $v$,

$p = mv,$

angular momentum is proportional to moment of inertia $I$ and angular speed $\omega$, [2]

$L = I\omega.$

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.

Because $I=r^2m$ for a single particle and $\omega=v/r,$ then angular momentum can also be expressed

$L=rmv,$

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

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,

$L=rmv\sin{\theta}$
where
$\theta$ is the angle between the particle's motion and the radius vector - see the graphic
$v_\perp = v\sin\theta$ and $v_\parallel = v\cos\theta$

#### Vector

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

By retaining the 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 particle in motion about the origin of coordinates is defined as

$\mathbf{L} = I\boldsymbol{\omega}$
where $I=r^2m$ is the moment of inertia for a point mass,
$\boldsymbol{\omega}=\frac{\mathbf{r}\times\mathbf{v}}{r^2}$ is the angular velocity of the particle about the origin,
$\mathbf{r}$ is the position vector of the particle relative to the origin, $r=\left\vert\mathbf{r}\right\vert$,
$\mathbf{v}$ is the linear velocity of the particle relative to the origin,
and $m$ is the mass of the particle.

This can be expanded,

$\mathbf{L}=(r^2m)\left(\frac{\mathbf{r}\times\mathbf{v}}{r^2}\right) = m(\mathbf{r}\times\mathbf{v})$,

and by the rules of vector algebra reduced to the form,

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

which is the cross product of the position vector $\mathbf{r}$ and the linear momentum $\mathbf{p}=m\mathbf{v}$ of the particle. By the definition of the cross product, the $\mathbf{L}$ vector is perpendicular to both $\mathbf{r}$ and $\mathbf{p}$. It is directed along the axis of rotation as indicated by the right-hand rule - so that the rotation is seen as counter-clockwise from the head of the vector.

### 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.

Because rotational inertia is a part of angular momentum, it necessarily includes all of the complications of moment of inertia. This is calculated by multiplying elementary bits of the mass by the squares of their distance from the center of rotation.[3] 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 is simply the position of the rotation axis versus the matter of the body. 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, or they may be completely random.

In brief, the more mass and the farther it is from the center of rotation, the greater the moment of inertia, and therefore the greater the angular momentum. In many cases the moment of inertia, and hence the angular momentum, can be simplified by, [4]

$I=k^2m,$
where $k$ is the radius of gyration, the distance from the axis at which the entire mass $m$ may be considered as concentrated.

Similarly, for a point mass $m$ the moment of inertia is defined as,

$I=r^2m$
where $r$ is the radius of the point mass from the center of rotation,

and for any collection of particles $m_i$ as the sum,

$\sum_i I_i = \sum_i r_i^2m_i$

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 kg·m2/s can be reduced to zero rotation (all of the energy can be transferred out of it) by an angular impulse of L kg·m2/s (L torque·seconds).[5]

### Conservation of angular momentum

Conservation follows mathematically from isotropy, or continuous directional symmetry of space, that is, no direction in space is any different from any other direction. See Noether's theorem.[6]

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 compelled by a torque to change its state." Thus angular momentum does not change without an external influence.[7]

Similarly, a rotational analogy 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." The time derivative (the time rate of change) of angular momentum is equal to the torque $(= \mathbf{r} \times \mathbf{F})$:

$\mathbf{\tau} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \frac{\mathrm{d}\mathbf{(r \times p)}}{\mathrm{d}t} = \frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t} \times \mathbf{p} + \mathbf{r} \times \frac{\mathrm{d}\mathbf{p}}{\mathrm{d}t} = 0 + \mathbf{r} \times \mathbf{F} = \mathbf{r} \times \mathbf{F}$
note that $d\mathbf{r}/dt$ is velocity, $d\mathbf{p}/dt$ is force and the cross-product of velocity $\mathrm{d}\mathbf{r}/\mathrm{d}t$ and momentum $\mathbf{p}$ vanishes because the vectors are parallel.

Therefore, a constant (non-changing) angular velocity is equivalent to zero external torque, and requiring the system to be closed is equivalent to requiring that no external influence acts upon it.[8]

Similarly, 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." Hence, angular momentum can be exchanged between objects in a closed system, but total angular momentum before and after an exchange remains the same.[9] It is assumed that internal interaction forces obey Newton's third law of motion in its strong form, that is, that the forces between particles are equal and opposite and act along the line between the particles.

The conservation of angular momentum is used extensively in analyzing what is called central force motion. If the net force on some body is directed always toward some fixed 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. 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. It 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 42 ns/day[citation needed], and in gradual increase of the radius of Moon's orbit, at ~4.5 cm/year[citation needed].

The torque caused by the two opposing forces Fg and −Fg 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. In order for angular momentum to remain constant, the angular velocity (rotational speed) of the skater has to 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.

### 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:

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

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

$\mathbf{L}=\int_V dV \mathbf{r}\times \rho(\mathbf{r}) \mathbf{v}$

### Collection of particles

In a system consisting of multiple particles $m_i$ with position vectors $\mathbf{r}_i$ and linear velocity vectors $\mathbf{v}_i$ the total angular momentum about a point can be obtained by adding the angular momenta of the constituent particles,[10]

$\mathbf{L}=\sum_i \mathbf{r}_i\times m_i \mathbf{v}_i.$

#### Center of mass

The angular momentum of the particles i is the sum of the cross products R x MV + Σri x mivi.

It is convenient to consider the angular momentum of a collection of particles about their center of mass. Given,

$m_i$ is the mass of particle $i$,
$\mathbf{R}_i$ is the position vector of particle $i$ vs the origin,
$\mathbf{V}_i$ is the velocity of particle $i$ vs the origin,
$\mathbf{R}$ is the position vector of the center of mass vs the origin,
$\mathbf{V}$ is the velocity of the center of mass vs the origin,
$\mathbf{r}_i$ is the position vector of particle $i$ vs the center of mass,
$\mathbf{v}_i$ is the velocity of particle $i$ vs the center of mass,

The total mass of the particles is simply their sum,

$M=\sum_i m_i.$

The position vector of the center of mass is defined by,[11]

$M\mathbf{R}=\sum_i m_i \mathbf{R}_i.$

By inspection,

$\mathbf{R}_i = \mathbf{R} + \mathbf{r}_i$ and $\mathbf{V}_i = \mathbf{V} + \mathbf{v}_i.$

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

$\mathbf{L}=\sum_i ( \mathbf{R}_i\times m_i \mathbf{V}_i ).$

Expanding $\mathbf{R}_i$,

\begin{align} \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{align}

Expanding $\mathbf{V}_i$,

\begin{align} \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{align}

It can be shown that (see sidebar),

$\sum_i m_i\mathbf{r}_i = \mathbf{0}$ and $\sum_i m_i\mathbf{v}_i = \mathbf{0},$

therefore the second and third terms vanish,

$\mathbf{L} = \sum_i \mathbf{R} \times m_i\mathbf{V} + \sum_i \mathbf{r}_i \times m_i\mathbf{v}_i .$

The first term can be rearranged,

$\sum_i \mathbf{R} \times m_i\mathbf{V} = \mathbf{R} \times \sum_i m_i\mathbf{V} = \mathbf{R} \times M\mathbf{V},$

and total angular momentum for the collection of particles is finally,[12]

$\mathbf{L} = \mathbf{R} \times M\mathbf{V} + \sum_i \mathbf{r}_i \times m_i\mathbf{v}_i .$

The first term is just the angular momentum of the center of mass. It is the same angular momentum one would obtain if there were just one particle of mass M moving at velocity V located at the center of mass. The second term is the angular momentum that is the result of the particles moving relative to their center of mass. This second term can be even further simplified if the particles form a rigid body, in which case it is the product of moment of inertia and angular velocity of the spinning motion (as above). The same result is true if the discrete point masses discussed above are replaced by a continuous distribution of mass.

## 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:

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

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:

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

or more compactly in index notation:

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

The angular velocity can also be defined as an antisymmetric second order tensor, with components ωij. The relation between the two antisymmetric tensors is given by the moment of inertia which must now be a fourth order tensor:[13]

$L_{ij} = I_{ijk\ell} \omega_{k\ell} \,.$

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:

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

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

Main article: Spin (physics)
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),[14]

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

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 always have "spin 1/2" (this actually means "spin ħ/2") while photons always 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.

### 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 $\hbar$ is the reduced Planck constant and $\hat{n}$ is any direction vector such as x, y, or z:

 If you measure... The result can be... $L_{\hat{n}}$ $\ldots, -2\hbar, -\hbar, 0, \hbar, 2\hbar, \ldots$ $S_{\hat{n}}$ or $J_{\hat{n}}$ $\ldots, -\frac{3}{2}\hbar, -\hbar, -\frac{1}{2}\hbar, 0, \frac{1}{2}\hbar, \hbar, \frac{3}{2}\hbar, \ldots$ $L^2$ ($= L_x^2+L_y^2+L_z^2$) $(\hbar^2 n(n+1))$, where $n=0,1,2,\ldots$ $S^2$ or $J^2$ $(\hbar^2 n(n+1))$, where $n=0,\frac{1}{2},1,\frac{3}{2}, \ldots$
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.

(There are additional restrictions as well, see angular momentum operator for details.)

The reduced Planck constant $\hbar$ 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 $\mathbf{L}=\mathbf{r}\times\mathbf{p}$, six operators are involved: The position operators $r_x$, $r_y$, $r_z$, and the momentum operators $p_x$, $p_y$, $p_z$. 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 $L_xL_y \neq L_yL_x$. (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: $\mathbf{L}=\mathbf{r}\times\mathbf{p}$, but total angular momentum J is defined in a different, more basic way: J is defined as the "generator of rotations".[15] More specifically, J is defined so that the operator

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

is the rotation operator that takes any system and rotates it by angle $\phi$ about the axis $\hat{\mathbf{n}}$. (The "exp" in the formula refers to operator exponential)

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)

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

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

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

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

## Footnotes

1. ^ Wilson, E. B. (1915). Linear Momentum, Kinetic Energy and Angular Momentum. The American Mathematical Monthly XXII (Ginn and Co., Boston, in cooperation with University of Chicago, et al.)., pg. 190, (at Google books)
2. ^ Worthington, Arthur M. (1906). Dynamics of Rotation. Longmans, Green and Co., London., p. 21. (at Google books)
3. ^ Oberg, Erik et al. (2000). Machinery's Handbook (26th ed.). Industrial Press, Inc., New York. ISBN 0-8311-2625-6. , pg. 143
4. ^ Machinery's Handbook, pg. 146
5. ^ Machinery's Handbook, pg. 161-162
6. ^ Landau, L. D.; Lifshitz, E. M. (1995). The classical theory of fields. Course of Theoretical Physics. Oxford, Butterworth–Heinemann. ISBN 0-7506-2768-9.
7. ^ Worthington, pg. 11
8. ^ Worthington, pg. 12
9. ^ Worthington, chap. VIII, pg. 82
10. ^ Wilson, pg. 190, equation (7)
11. ^ Wilson, pg. 188, equation (3)
12. ^ Wilson, pg. 191, Theorem 8
13. ^ Synge and Schild, Tensor calculus, Dover publications, 1978 edition, p. 161. ISBN 978-0486636122.
14. ^ R.P. Feynman, R.B. Leighton, M. Sands (1964). Feynman's Lectures on Physics (volume 2). Addison–Wesley. pp. 31–7. ISBN 9-780-201-021172.
15. ^ Littlejohn, Robert (2011). "Lecture notes on rotations in quantum mechanics" (PDF). Physics 221B Spring 2011. Retrieved 13 Jan 2012.

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