Definition in classical mechanics

Orbital angular momentum in two dimensionsedit

Angular momentum is a vector quantity (more precisely, a pseudovector) that represents the product of a body's rotational inertia and rotational velocity (in radians/sec) 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). Angular momentum can be considered a rotational analog of linear momentum. Thus, where linear momentum p is proportional to mass m and linear speed v,

${\displaystyle p=mv,}$

angular momentum L is proportional to moment of inertia I and angular speed ω measured in radians per second.

${\displaystyle L=I\mathrm{\omega <!--\; \omega \; -->}.}$

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 velocity, which does not depend upon the choice of origin, orbital angular velocity is always measured with respect to a fixed origin. Therefore, strictly speaking, L should be referred to as the angular momentum relative to that center.

Because ${\displaystyle I=r^{2}m}$ for a single particle and ${\displaystyle \mathrm{\omega <!--\; \omega \; -->}=\frac{v}{r}}$ for circular motion, angular momentum can be expanded, ${\displaystyle L=r^{2}m\cdot <!--\; \cdot \; -->\frac{v}{r},}$ and reduced to,

${\displaystyle L=rmv,}$

the product of the radius of rotation r and the linear momentum of the particle ${\displaystyle p=mv}$, where v in this case is the equivalent linear (tangential) speed at the radius (${\displaystyle =r\mathrm{\omega <!--\; \omega \; -->}}$).

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,

${\displaystyle L=rm{v}_{\perp <!--\; \perp \; -->},}$

where ${\displaystyle v_{\perp <!--\; \perp \; -->}=v\sin <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})}$ is the perpendicular component of the motion. Expanding, ${\displaystyle L=rmv\sin <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->}),}$ rearranging, ${\displaystyle L=r\sin <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})mv,}$ and reducing, angular momentum can also be expressed,

${\displaystyle L=r_{\perp <!--\; \perp \; -->}mv,}$

where ${\displaystyle r_{\perp <!--\; \perp \; -->}=r\sin <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})}$ 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.

Scalar—angular momentum from Lagrangian mechanicsedit

Another approach is to define angular momentum as the conjugate momentum (also called canonical momentum) of the angular coordinate ${\displaystyle \mathrm{\varphi <!--\; \varphi \; -->}}$ 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

${\displaystyle T=\frac{1}{2}m{a}^2{\mathrm{\omega <!--\; \omega \; -->}}^2=\frac{1}{2}m{a}^2{\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathrm{\varphi <!--\; \varphi \; -->}}}^2.}$

And the potential energy is

${\displaystyle U=0.}$

Then the Lagrangian is

${\displaystyle \mathcal{L}\left(\mathrm{\varphi <!--\; \varphi \; -->},\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathrm{\varphi <!--\; \varphi \; -->}}\right)=T-<!--\; -\; -->U=\frac{1}{2}m{a}^2{\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathrm{\varphi <!--\; \varphi \; -->}}}^2.}$

The generalized momentum "canonically conjugate to" the coordinate ${\displaystyle \mathrm{\varphi <!--\; \varphi \; -->}}$ is defined by

${\displaystyle p_{\mathrm{\varphi <!--\; \varphi \; -->}}=\frac{\mathrm{\partial <!--\; \partial \; -->}\mathcal{L}}{\mathrm{\partial <!--\; \partial \; -->}\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathrm{\varphi <!--\; \varphi \; -->}}}=m{a}^2\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathrm{\varphi <!--\; \varphi \; -->}}=I\mathrm{\omega <!--\; \omega \; -->}=L.}$

Orbital angular momentum in three dimensionsedit

To completely define orbital angular momentum in three dimensions, it is required to know the rate at which the position vector sweeps out angle, the direction perpendicular to the instantaneous plane of angular displacement, and the mass involved, as well as how this mass is distributed in space. 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 orbital angular momentum of a point particle in motion about the origin can be expressed as:

${\displaystyle \mathbf{L}=I\mathit{\omega <!--\; \omega \; -->},}$

where

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

${\displaystyle \mathit{\omega <!--\; \omega \; -->}=\frac{\mathbf{r}\times <!--\; \times \; -->\mathbf{v}}{r^2}}$ is the orbital angular velocity in radians/sec (units 1/sec) of the particle about the origin,

${\displaystyle \mathbf{r}}$ is the position vector of the particle relative to the origin, ${\displaystyle r=\left|\mathbf{r}\right|}$,

${\displaystyle \mathbf{v}}$ is the linear velocity of the particle relative to the origin, and

${\displaystyle m}$ is the mass of the particle.

This can be expanded, reduced, and by the rules of vector algebra, rearranged:

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 ${\displaystyle \stackrel{\mathbf{^<!--\; ^\; -->}}{\mathbf{u}}}$ perpendicular to the plane of angular displacement, a scalar angular speed ${\displaystyle \mathrm{\omega <!--\; \omega \; -->}}$ results, where

${\displaystyle \mathrm{\omega <!--\; \omega \; -->}\stackrel{\mathbf{^<!--\; ^\; -->}}{\mathbf{u}}=\mathit{\omega <!--\; \omega \; -->},}$ and

${\displaystyle \mathrm{\omega <!--\; \omega \; -->}=\frac{v_{\perp <!--\; \perp \; -->}}{r},}$ where ${\displaystyle v_{\perp <!--\; \perp \; -->}}$ is the perpendicular component of the motion, as above.

The two-dimensional scalar equations of the previous section can thus be given direction:

and ${\displaystyle \mathbf{L}=rmv\stackrel{\mathbf{^<!--\; ^\; -->}}{\mathbf{u}}}$ for circular motion, where all of the motion is perpendicular to the radius ${\displaystyle r}$.

In the spherical coordinate system the angular momentum vector expresses as

${\displaystyle \mathbf{L}=m\mathbf{r}\times <!--\; \times \; -->\mathbf{v}=m{r}^2\left(\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathrm{\theta <!--\; \theta \; -->}}\stackrel{^<!--\; ^\; -->}{\mathit{\phi <!--\; \phi \; -->}}-<!--\; -\; -->\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathrm{\phi <!--\; \phi \; -->}}\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}\stackrel{\mathbf{^<!--\; ^\; -->}}{\mathit{\theta <!--\; \theta \; -->}}\right).}$