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Wikipedia: angular momentum

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

In physics, the angular momentum of an object rotating about some reference point is the measure of the extent to which the object will continue to rotate about that point unless acted upon by an external torque. In particular, if a point mass rotates about an axis, then the angular momentum with respect to a point on the axis is related to the mass of the object, the velocity and the distance of the mass to the axis. While the motion associated with linear momentum has no absolute frame of reference, the rotation associated with angular momentum is sometimes spoken of as being measured relative to the fixed stars.

Angular momentum is important in physics because it is a conserved quantity: a system's angular momentum stays constant unless an external torque acts on it. Torque is the rate at which angular momentum is transferred in or out of the system. When a rigid body rotates, its resistance to a change in its rotational motion is measured by its moment of inertia. Angular momentum is an important concept in both physics and engineering, with numerous applications. For example, the kinetic energy stored in a massive rotating object such as a flywheel is proportional to the square of the angular momentum. Conservation of angular momentum also explains many phenomena in sports and nature.

Angular momentum in classical mechanics

Relationship between force (F), torque (τ), and momentum vectors (p and L) in a rotating system

Definition

Angular momentum of a particle about some origin is defined as:

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

where:

$\mathbf{L}$ is the angular momentum of the particle,

$\mathbf{r}$ is the position of the particle expressed as a displacement vector from the origin,

$\mathbf{p}$ is the linear momentum of the particle, and

$\times\,$ is the vector cross product.

As seen from the definition, the derived SI units of angular momentum are newton·metre·seconds (N·m·s or kgm²s^-1). Because of the cross product, L is a pseudovector perpendicular to both the radial vector r and the momentum vector p and it is assigned a sign by the right-hand rule.

If a system consists of several particles, the total angular momentum about an origin can be obtained by adding (or integrating) all the angular momenta of the constituent particles. Angular momentum can also be calculated by multiplying the square of the displacement r, the mass of the particle and the angular velocity.

Orbital and spin angular momentum

It is very often convenient to consider the angular momentum of a collection of particles about their center of mass, since this simplifies the mathematics considerably. The angular momentum of a collection of particles is the sum of the angular momenta of each particle:

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

where $R i$ is the distance of particle i from the reference point, $m i$ is its mass, and $V i$ is its velocity. The center of mass is defined by:

$\mathbf{R}=\frac{1}{M}\sum_i m_i \mathbf{R}_i$

where the total mass of all particles is given by

$M=\sum_i m_i\,$

It follows that the velocity of the center of mass is

$\mathbf{V}=\frac{1}{M}\sum_i m_i \mathbf{V}_i\,$

If we define $\mathbf{r}_i$ as the displacement of particle i from the center of mass, and $\mathbf{v}_i$ as the velocity of particle i with respect to the center of mass, then we have

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

and also

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

so that the total angular momentum is

$\mathbf{L}=\sum_i (\mathbf{R}+\mathbf{r}_i)\times m_i (\mathbf{V}+\mathbf{v}_i) = \left(\mathbf{R}\times M\mathbf{V}\right) + \left(\sum_i \mathbf{r}_i\times m_i \mathbf{v}_i\right)$

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 spinning about their center of mass. This second term can be even further simplified if the particles form a rigid body. An analogous result is obtained for a continuous distribution of matter.

Fixed axis of rotation

For many applications where one is only concerned about rotation around one axis, it is sufficient to discard the pseudovector nature of angular momentum, and treat it like a scalar where it is positive when it corresponds to a counter-clockwise rotations, and negative clockwise. To do this, just take the definition of the cross product and discard the unit vector, so that angular momentum becomes:

$L = |\mathbf{r}||\mathbf{p}|\sin \theta_{r,p}$

where θ_r,p is the angle between r and p measured from r to p; an important distinction because without it, the sign of the cross product would be meaningless. From the above, it is possible to reformulate the definition to either of the following:

$L = \pm|\mathbf{p}||\mathbf{r}_{\perp}|$

where r_⊥ is called the lever arm distance to p.

The easiest way to conceptualize this is to consider the lever arm distance to be the distance from the origin to the line that p travels along. With this definition, it is necessary to consider the direction of p (pointed clockwise or counter-clockwise) to figure out the sign of L. Equivalently:

$L = \pm|\mathbf{r}||\mathbf{p}_{\perp}|$

where p_⊥ is the component of p that is perpendicular to r. As above, the sign is decided based on the sense of rotation.

For an object with a fixed mass that is rotating about a fixed symmetry axis, the angular momentum is expressed as the product of the moment of inertia of the object and its angular velocity vector:

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

where

$I\,$ is the moment of inertia of the object (in general, a tensor quantity)

$\mathbf{\omega}$ is the angular velocity.

Conservation of angular momentum

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.

In a closed system angular momentum is constant. This conservation law mathematically follows from continuous directional symmetry of space (no direction in space is any different from any other direction). See Noether's theorem.

The time derivative of angular momentum is called torque:

$\tau = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \mathbf{r} \times \frac{\mathrm{d}\mathbf{p}}{\mathrm{d}t} = \mathbf{r} \times \mathbf{F}$

So requiring the system to be "closed" here is mathematically equivalent to zero external torque acting on the system:

$\mathbf{L}_{\mathrm{system}} = \mathrm{constant} \leftrightarrow \sum \tau_{\mathrm{ext}} = 0$

where $τ e x t$ is any torque applied to the system of particles.

In orbits, the angular momentum is distributed between the spin of the planet itself and the angular momentum of its orbit:

$\mathbf{L}_{\mathrm{total}} = \mathbf{L}_{\mathrm{spin}} + \mathbf{L}_{\mathrm{orbit}}$ ;

If a planet is found to rotate slower than expected, then astronomers suspect that the planet is accompanied by a satellite, because the total angular momentum is shared between the planet and its satellite in order to be conserved.

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, and so the angular momentum of the body about the center is constant. Constant angular momentum is extremely useful when dealing with the orbits of planets and satellites, and also when analyzing the Bohr model of the atom.

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 mass of her body closer to the axis she decreases her body's moment of inertia. Because angular momentum is constant in the absence of external torques, 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 (indeed, decreasing the size of object 10⁴ times results in increase of its angular velocity by the factor 10⁸).

The conservation of angular momentum in Earth-Moon system results in the transfer of angular momentum from Earth to Moon (due to tidal torque Moon exerts of Earth). This in turn results in the slowing down of the rotation rate of Earth (at about 42 nsec/day), and in gradual increase of the radius of Moon's orbit (at ~4.5 cm/year rate).

Angular momentum in relativistic mechanics

In modern (late 20th century) theoretical physics, angular momentum is described using a different formalism. Under this formalism, angular momentum is the 2-form Noether charge associated with rotational invariance (As a result, angular momentum isn't conserved for general curved spacetimes, unless it happens to be asymptotically rotationally invariant). For a system of point particles without any intrinsic angular momentum, it turns out to be

$\sum_i \bold{r}_i\wedge \bold{p}_i$

(Here, the wedge product is used.).

Angular momentum in quantum mechanics

In quantum mechanics, angular momentum is quantized -- that is, it cannot vary continuously, but only in "quantum leaps" between certain allowed values. The angular momentum of a subatomic particle, due to its motion through space, is always a whole-number multiple of $\hbar$ ("h-bar"), defined as Planck's constant divided by 2π. Furthermore, experiments show that most subatomic particles have a permanent, built-in angular momentum, which is not due to their motion through space. This spin angular momentum comes in units of $\hbar/2$ . For example, an electron standing at rest has an angular momentum of $\hbar/2$ .

Basic definition

The classical definition of angular momentum as $\mathbf{L}=\mathbf{r}\times\mathbf{p}$ depends on six numbers: $r x$ , $r y$ , $r z$ , $p x$ , $p y$ , and $p z$ . Translating this into quantum-mechanical terms, the Heisenberg uncertainty principle tells us that it is not possible for all six of these numbers to be measured 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.

Mathematically, angular momentum in quantum mechanics is defined like momentum - not as a quantity but as an operator on the wave function:

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

where r and p are the position and momentum operators respectively. In particular, for a single particle with no electric charge and no spin, the angular momentum operator can be written in the position basis as

$\mathbf{L}=-i\hbar(\mathbf{r}\times\nabla)$

where $\nabla$ is the gradient operator, read as "del," "grad," or "nabla". This orbital angular momentum operator is the most commonly encountered form of the angular momentum operator, though not the only one. It satisfies the following commutation relations

$[L_i, L_j ] = i \hbar \epsilon_{ijk} L_k$ ,

where ε_ijk is the (antisymmetric) Levi-Civita symbol. From this follows

$\left[L_i, L^2 \right] = 0$

Since,

$L_x = -i\hbar (y {\partial\over \partial z} - z {\partial\over \partial y})$

$L_y = -i\hbar (z {\partial\over \partial x} - x {\partial\over \partial z})$

$L_z = -i\hbar (x {\partial\over \partial y} - y {\partial\over \partial x})$

it follows, for example,

Failed to parse (unknown function\begin): \begin{align} \left[L_x,L_y\right] & = -\hbar^2 \left( (y {\partial \over \partial z} - z {\partial\over \partial y})(z {\partial\over \partial x} - x {\partial\over \partial z}) - (z {\partial\over \partial x} - x {\partial\over \partial z})(y {\partial \over \partial z} - z {\partial\over \partial y})\right) \\ & = -\hbar^2 \left( y {\partial\over \partial x} - x {\partial\over \partial y}\right) = i \hbar L_z. \\ \end{align}

Addition of quantized angular momenta

For more details on this topic, see Clebsch-Gordan coefficients.

Given a quantized total angular momentum $\overrightarrow{j}$ which is the sum of two individual quantized angular momenta $\overrightarrow{l_1}$ and $\overrightarrow{l_2}$ ,

$\overrightarrow{j} = \overrightarrow{l_1} + \overrightarrow{l_2}$

the quantum number $j$ associated with its magnitude can range from $| l 1 - l 2 |$ to $l 1 + l 2$ in integer steps where $l 1$ and $l 2$ are quantum numbers corresponding to the magnitudes of the individual angular momenta.

Angular momentum as a generator of rotations

If $φ$ is the angle around a specific axis, for example the azimuthal angle around the z axis, then the angular momentum along this axis is the generator of rotations around this axis:

$L_z = -i\hbar {\partial\over \partial \phi}.$

The eigenfunctions of L_z are therefore $e^{i m_l \phi}$ , and since $φ$ has a period of $2π$ , m_l must be an integer.

For a particle with a spin S, this takes into account only the angular dependence of the location of the particle, for example its orbit in an atom. It is therefore known as orbital angular momentum. However, when one rotates the system, one also changes the spin. Therefore the total angular momentum, which is the full generator of rotations, is $J i = L i + S i$ Being an angular momentum, J satisfies the same commutation relations as L, as will explained below. namely

$[J_i, J_j ] = i \hbar \epsilon_{ijk} J_k$

from which follows

$\left[J_i, J^2 \right] = 0.$

Acting with J on the wavefunction $ψ$ of a particle generates a rotation: $e^{i \phi J_z} \psi$ is the wavefunction $ψ$ rotated around the z axis by an angle $φ$ . For an infinitesmal rotation by an angle $d φ$ , the rotated wavefunction is $ψ + i d φ J z ψ$ . This is similarly true for rotations around any axis.

In a charged particle the momentum gets a contribution from the electromagnetic field, and the angular momenta L and J change accordingly.

If the Hamiltonian is invariant under rotations, as in spherically symmetric problems, then according to Noether's theorem, it commutes with the total angular momentum. So the total angular momentum is a conserved quantity

$\left[J_i, H \right] = 0$

Since angular momentum is the generator of rotations, its commutation relations follow the commutation relations of the generators of the three-dimensional rotation group SO(3). This is why J always satisfies these commutation relations. In d dimensions, the angular momentum will satisfy the same commutation relations as the generators of the d-dimensional rotation group SO(d).

SO(3) has the same Lie algebra (i.e. the same commutation relations) as SU(2). Generators of SU(2) can have half-integer eigenvalues, and so can m $j$ . Indeed for fermions the spin S and total angular momentum J are half-integer. In fact this is the most general case: j and m $j$ are either integers or half-integers.

Technically, this is because the universal cover of SO(3) is isomorphic SU(2), and the representations of the latter are fully known. J_i span the Lie algebra and J² is the Casimir invariant, and it can be shown that if the eigenvalues of J_z and J² are m_j and j(j+1) then m_j and j are both integer multiples of one-half. j is non-negative and m_j takes values between -j and j.

Relation to spherical harmonics

Angular momentum operators usually occur when solving a problem with spherical symmetry in spherical coordinates. Then, the angular momentum in space representation is:

$L^2 = -\frac{\hbar^2}{\sin\theta}\frac{\partial}{\partial \theta}\left( \sin\theta \frac{\partial}{\partial \theta}\right) - \frac{\hbar^2}{\sin^2\theta}\frac{\partial^2}{\partial \phi^2}$

When solving to find eigenstates of this operator, we obtain the following

$L^2 | l, m \rang = {\hbar}^2 l(l+1) | l, m \rang$

$L_z | l, m \rang = \hbar m | l, m \rang$

where

$\lang \theta , \phi | l, m \rang = Y_{l,m}(\theta,\phi)$

are the spherical harmonics.

Angular momentum in electrodynamics

When describing the motion of a charged particle in the presence of an electromagnetic field, the "kinetic momentum" p is not gauge invariant. As a consequence, the canonical angular momentum $\mathbf{L} = \mathbf{r} \times \mathbf{p}$ is not gauge invariant either. Instead, the momentum that is physical, the so-called canonical momentum, is

$\mathbf{p} -\frac {e \mathbf{A} }{c}$

where $e$ is the electric charge, c the speed of light and A the vector potential. Thus, for example, the Hamiltonian of a charged particle of mass m in an electromagnetic field is then

$H =\frac{1}{2m} \left( \mathbf{p} -\frac {e \mathbf{A} }{c}\right)^2 + e\phi$

where $φ$ is the scalar potential. This is the Hamiltonian that gives the Lorentz force law. The gauge-invariant angular momentum, or "kinetic angular momentum" is given by

$K= \mathbf{r} \times \left( \mathbf{p} -\frac {e \mathbf{A} }{c}\right)$

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

External links

Conservation of Angular Momentum - a chapter from an online textbook

Angular Momentum in a Collision Process - derivation of the three dimensional case

References

E. U. Condon and G. H. Shortley, The Theory of Atomic Spectra, (1935) Cambridge at the University Press, ISBN 0-521-09209-4 See chapter 3.
Edmonds, A.R., Angular Momentum in Quantum Mechanics, (1957) Princeton University Press, ISBN 0-691-07912-9.
Serway, Raymond A.; Jewett, John W. (2004). Physics for Scientists and Engineers (6th ed.). Brooks/Cole. ISBN 0-534-40842-7.
Tipler, Paul (2004). Physics for Scientists and Engineers: Mechanics, Oscillatiions and Waves, Thermodynamics (5th ed.). W. H. Freeman. ISBN 0-7167-0809-4.

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angular momentum