Angular momentum

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 Classical mechanics $\vec{F} = \frac{\mathrm{d}}{\mathrm{d}t}(m \vec{v})$ Newton's Second Law History of ...
 This gyroscope remains upright while spinning due to its angular momentum.

The angular momentum of a system of particles is the sum of those of the particles within it.

In 2 dimensions the angular momentum of a particle of mass m with respect to a chosen origin is given by:

$L = mvr \sin \theta \;$

where m is the mass, v is the speed, r is the distance from the origin and θ is the angle between the velocity and the radius vector.

In 3 dimensions the angular momentum of a particle about an origin is a vector quantity related to rotation, equal to the mass of the particle multiplied by the cross product of the position vector of the particle with its velocity vector.

Angular momentum is important in physics because it is a conserved quantity: the angular momentum of an isolated system stays constant unless an external torque acts on it. Rotational symmetry of space is related to the conservation of angular momentum by Noether's theorem. The conservation of angular momentum explains many phenomena found in nature and angular momentum has numerous applications in physics and engineering.

## Angular momentum in classical mechanics

### Definition

Angular momentum (also known as moment of momentum) of a particle about a given 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 vector of the particle relative to 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 kg·ms) or joule seconds. 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.

### Angular momentum of a collection of particles

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.

### Angular momentum in the centre of mass frame

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

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

where Ri is the distance of particle i from the reference point, mi is its mass, and Vi 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 centre 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 centre of mass, and $\mathbf{v}_i$ as the velocity of particle i with respect to the centre 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 centre 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 centre 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 a spin appears. 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 θr,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 $\mathbf{r}_{\perp}$ 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 $\mathbf{p}_{\perp}$ 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.

As the kinetic energy K of a massive rotating body is given by

$\mathbf{K}= I \mathbf{\omega^2}/2$

it is proportional to the square of the angular velocity.

### Conservation of angular momentum

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 | l1l2 | to l1 + l2 in integer steps where l1 and l2 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 Lz are therefore $e^{i m_l \phi}$, and since φ has a period of , ml 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 Ji = Li + Si Being an angular momentum, J satisfies the same commutation relations as L, as will be explained below. namely

$[J_\ell, J_m ] = i \hbar \varepsilon_{lmn} J_n$

from which follows

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

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 dφ, the rotated wavefunction is ψ + idφJzψ. 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

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

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 mj. Indeed for fermions the spin S and total angular momentum J are half-integer. In fact this is the most general case: j and mj are either integers or half-integers.

Technically, this is because the universal cover of SO(3) is isomorphic to SU(2), and the representations of the latter are fully known. Ji span the Lie algebra and J is the Casimir invariant, and it can be shown that if the eigenvalues of Jz and J are mj and j(j+1) then mj and j are both integer multiples of one-half. j is non-negative and mj 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.

## Footnotes

Published - July 2009