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Angular momentum

From Academic Kids

In physics, angular momentum intuitively measures how much the linear momentum is directed around a certain point called the origin; the moment of momentum. Since angular momentum depends upon the origin of choice, one must be careful when discussing angular momentum to specify the origin and not to combine angular momenta about different origins.


Contents

Angular momentum in classical mechanics

Definition

The traditional mathematical definition of the angular momentum of a particle about some origin is:

<math>\mathbf{L} = \mathbf{r} \times \mathbf{p}<math>

where

L is the angular momentum of the particle,

r is the position of the particle expressed as a displacement vector from the origin

p is the linear momentum of the particle.

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 distance to the point of rotation, the mass of the particle and the angular velocity.

For many applications where one is only concerned about rotation around one axis, it is sufficient to discard the vector 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:

<math>\mathbf{L} = |\mathbf{r}||\mathbf{p}|\sin\theta_{r,p}<math>

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:

<math>\mathbf{L} = \pm|\mathbf{p}||\mathbf{r}_{\mathrm{perpendicular}}|<math>

where rperpendicular 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:

<math>\mathbf{L} = \pm|\mathbf{r}||\mathbf{p}_{\mathrm{perpendicular}}|<math>

where pperpendicular is the component of p that is perpendicular to r. As above, the sign is decided base on the sense of rotation.

For a particle 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 particle and its angular velocity vector:

<math>\mathbf{L}= I \mathbf{\omega} <math>

where

I is the moment of inertia of the particle

ω is the angular velocity.

Conservation of angular momentum

In analogy to Newton's second law for linear momentum, we have the following law about angular momentum:

<math>\frac{d\mathbf{L}}{dt} = \boldsymbol{\tau} <math>

where τ is the net torque about the origin.

This implies that angular momentum is a conserved quantity as long as there is no net torque applied to the particle. What's more, this conservation can be generalized to a system of particles under most conditions so that:

<math>\mathbf{L}_{\mathrm{system}} = \mathrm{constant} \Leftrightarrow \sum \tau_{\mathrm{external}} = 0 <math>

where τexternal is any torque applied to the system of particles.

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

<math>\mathbf{L}_{\mathrm{total}} = \mathbf{L}_{\mathrm{spin}} + \mathbf{L}_{\mathrm{orbit}} <math>

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 amongst 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. In central force motion, two bodies form an isolated system not influenced by outside forces, and the origin is placed somewhere on the line between the two bodies. Since any force the bodies exert on each other must be directed along this line, there can be no net torque, with respect to the aforementioned origin, on either body. Thus, angular momentum is conserved. 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.

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

<math>\sum_i \bold{r}\wedge \bold{p}<math>

(Here, the wedge product is used.).

Angular momentum in quantum mechanics

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

<math>\mathbf{L}=\mathbf{r}\times\mathbf{p}<math>

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

<math>\mathbf{L}=-i\hbar(\mathbf{r}\times\nabla)<math>

where Missing image
Del.gif
Image:del.gif

is the gradient operator. This is a commonly encountered form of the angular momentum operator, though not the most general one. It has the following properties

<math>[L_i, L_j ] = i \hbar \epsilon_{ijk} L_k<math>, <math>\left[L_i, L^2 \right] = 0<math>

and even more importantly commutes with the hamiltonian of such a chargeless and spinless particle

<math>\left[L_i, H \right] = 0<math>.

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

<math>\ L^2 = \frac{1}{\sin\theta}\frac{\partial}{\partial \theta}\left( \sin\theta \frac{\partial}{\partial \theta}\right) + \frac{1}{\sin^2\theta}\frac{\partial^2}{\partial \phi^2} <math>

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

<math> L^2 | l, m \rang = {\hbar}^2 l(l+1) | l, m \rang <math>
<math> L_z | l, m \rang = \hbar m | l, m \rang <math>

where

<math> \lang \theta , \phi | l, m \rang = Y_{l,m}(\theta,\phi)<math>

are the spherical harmonics.

See also

References

  • A.R. Edmonds, Angular Momentum in Quantum Mechanics, (1957) Princeton University Press, ISBN 0-691-07912-9.
  • E. U. Condon and G. H. Shortley, The Theory of Atomic Spectra, (1970) Cambridge at the University Press, ISBN 521-09209-4 See chapter 3.de:Drehimpuls

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