Editing Angular momentum (section)

== Definition in classical mechanics ==
{{redirect|Orbital angular momentum}}
{{redirect|Spin angular momentum}}
Just as for [[angular velocity]], there are two special types of angular [[momentum]] of an object: the '''spin angular momentum''' is the angular momentum about the object's [[center of mass]], while the '''orbital angular momentum''' is the angular momentum about a chosen center of rotation. The [[Earth]] has an orbital angular momentum by nature of revolving around the [[Sun]], and a spin angular momentum by nature of its daily rotation around the polar axis. The total angular momentum is the sum of the spin and orbital angular momenta.  In the case of the Earth the primary conserved quantity is the total angular momentum of the [[Solar System]] because angular momentum is exchanged to a small but important extent among the planets and the Sun. The orbital angular momentum vector of a point particle is always parallel and directly proportional to its orbital angular [[velocity]] vector '''ω''', where the constant of proportionality depends on both the mass of the particle and its distance from origin. The spin angular momentum vector of a rigid body is proportional but not always parallel to the spin angular velocity vector '''Ω''', making the constant of proportionality a second-rank [[tensor]] rather than a scalar.

=== Orbital angular momentum in two dimensions ===
[[File:Ang mom 2d.png|thumb|[[Velocity]] of the [[particle]] ''m'' with respect to the origin ''O'' can be resolved into components parallel to (''v''<sub>∥</sub>) and perpendicular to (''v''<sub>⊥</sub>) the radius vector ''r''. The '''angular momentum''' of ''m'' is proportional to the [[perpendicular component]] ''v''<sub>⊥</sub> of the velocity, or equivalently, to the perpendicular distance ''r''<sub>⊥</sub> from the origin.]]

Angular momentum is a [[Euclidean vector|vector]] quantity (more precisely, a [[pseudovector]]) that represents the product of a body's [[moment of inertia|rotational inertia]] and [[angular velocity|rotational velocity]] (in radians/sec) about a particular axis. However, if the particle's trajectory lies in a single [[plane (geometry)|plane]], it is sufficient to discard the vector nature of angular momentum, and treat it as a [[scalar (mathematics)|scalar]] (more precisely, a [[pseudoscalar]]).<ref name="Wilson">
{{cite journal
 |last1 = Wilson
 |first1 = E. B.
 |title = Linear Momentum, Kinetic Energy and Angular Momentum
 |journal=The American Mathematical Monthly
 |volume=XXII
 |publisher = Ginn and Co., Boston, in cooperation with University of Chicago, et al.
 |date=1915
 |url=https://books.google.com/books?id=nsI0AAAAIAAJ |page= 190
}}</ref> Angular momentum can be considered a rotational analog of linear momentum. Thus, where linear momentum {{mvar|p}} is proportional to [[mass]] {{mvar|m}} and [[speed|linear speed]] {{nowrap|{{mvar|v}},}}
<math display="block" qid=Q41273>p = mv,</math>
angular momentum {{mvar|L}} is proportional to [[moment of inertia]] {{mvar|I}} and [[angular frequency|angular speed]] {{mvar|ω}} measured in radians per second.<ref name="Worthington">{{cite book
 |last1 = Worthington
 |first1 = Arthur M.
 |title = Dynamics of Rotation
 |publisher = Longmans, Green and Co., London
 |date=1906
 |url=https://books.google.com/books?id=eScXAAAAYAAJ|page= 21|via= Google books
}}</ref>
<math display="block" qid=Q161254>L = I\omega.</math>

Unlike mass, which depends only on amount of matter, moment of inertia depends also on the position of the axis of rotation and the distribution 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, {{mvar|L}} should be referred to as the angular momentum ''relative to that center''.<ref name="Taylor90">
{{cite book
 |last1 = Taylor
 |first1 = John R.
 |title = Classical Mechanics
 |url = https://archive.org/details/classicalmechani00jrta
 |url-access = limited
 |publisher = University Science Books, Mill Valley, CA
 |date=2005
 |isbn=978-1-891389-22-1
 |page=[https://archive.org/details/classicalmechani00jrta/page/n104 90]
}}</ref>

In the case of circular motion of a single particle, we can use <math>I = r^2m</math> and <math>\omega = {v}/{r}</math> to expand angular momentum as <math>L = r^2 m \cdot {v}/{r},</math> reducing to:
<math display="block">L = rmv,</math>

the product of the [[radius]] of rotation {{mvar|r}} and the linear momentum of the particle <math>p = mv</math>, where <math>v= r\omega</math> is the [[Speed#Tangential speed|linear (tangential) speed]].

This simple analysis can also apply to non-circular motion if one uses the component of the motion [[perpendicular]] to the [[radius vector]]:
<math display="block">L = rmv_\perp,</math>
where <math>v_\perp = v\sin(\theta)</math> is the perpendicular component of the motion. Expanding, <math>L = rmv\sin(\theta),</math> rearranging, <math>L = r\sin(\theta)mv,</math> and reducing, angular momentum can also be expressed,
<math display="block">L = r_\perp mv,</math>
where <math>r_\perp = r\sin(\theta)</math> is the length of the [[torque#Moment arm formula|''moment arm'']], a line dropped perpendicularly from the origin onto the path of the particle. It is this definition, {{math|(length of moment arm) × (linear momentum)}}, to which the term ''moment of momentum'' refers.<ref name="Dadourian">{{cite book
|last1 = Dadourian
|first1 = H. M.
|title = Analytical Mechanics for Students of Physics and Engineering
|publisher = D. Van Nostrand Company, New York
|date=1913
|url=https://books.google.com/books?id=1b3VAAAAMAAJ|page= 266|via= Google books
}}</ref>

=== Scalar angular momentum from Lagrangian mechanics ===
Another approach is to define angular momentum as the conjugate momentum (also called '''canonical momentum''') of the angular coordinate <math>\phi</math> expressed in the [[Lagrangian mechanics|Lagrangian]] of the mechanical system. Consider a mechanical system with a mass <math>m</math> constrained to move in a circle of radius <math>r</math> in the absence of any external force field. The kinetic energy of the system is
<math display="block" qid="Q46276">T = \tfrac{1}{2}mr^2 \omega^2 = \tfrac{1}{2}mr^2 \dot{\phi}^2.</math>

And the potential energy is
<math display="block">U = 0.</math>

Then the Lagrangian is
<math display="block">\mathcal{L}\left(\phi, \dot{\phi}\right) = T - U = \tfrac{1}{2}mr^2 \dot{\phi}^2.</math>

The ''generalized momentum'' "canonically conjugate to" the coordinate <math>\phi</math> is defined by
<math display="block">p_\phi = \frac{\partial \mathcal{L}}{\partial \dot{\phi}} = mr^2 \dot{\phi} = I\omega = L.</math>

=== Orbital angular momentum in three dimensions ===
[[File:Torque animation.gif|frame|right|Relationship between [[force]] ('''F'''), [[torque]] ('''τ'''), [[momentum]] ('''p'''), and angular momentum ('''L''') vectors in a rotating system. '''r''' is the [[Position (vector)|position vector]].]]

To completely define orbital angular momentum in [[Three-dimensional space|three dimensions]], it is required to know the rate at which the position vector sweeps out an 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.<ref>
{{cite book
 |last1 = Watson
 |first1 = W.
 |title = General Physics
 |publisher = Longmans, Green and Co., New York
 |date=1912
 |url=https://books.google.com/books?id=Yb8zAQAAMAAJ
 |page=33
 |via=Google books
}}</ref> By retaining this [[Euclidean vector|vector]] nature of angular momentum, the general nature of the equations is also retained, and can describe any sort of three-dimensional [[Motion (physics)|motion]] about the center of rotation – [[Circular motion|circular]], [[Linear motion|linear]], or otherwise. In [[vector notation]], the orbital angular momentum of a [[point particle]] in motion about the origin can be expressed as:
<math display="block">\mathbf{L} = I\boldsymbol{\omega},</math>
where
* <math>I = r^2m</math> is the [[moment of inertia]] for a [[Point particle|point mass]],
* <math qid=Q161635>\boldsymbol{\omega}=\frac{\mathbf{r}\times\mathbf{v}}{r^2}</math> is the orbital [[angular velocity]] of the particle about the origin,
* <math>\mathbf{r}</math> is the position vector of the particle relative to the origin, and <math>r=\left\vert\mathbf{r}\right\vert</math>,
* <math>\mathbf{v}</math> is the linear velocity of the particle relative to the origin, and
* <math>m</math> is the mass of the particle.

This can be expanded, reduced, and by the rules of [[vector calculus|vector algebra]], rearranged:
<math display="block">\begin{align}
  \mathbf{L} &= \left(r^2m\right)\left(\frac{\mathbf{r}\times\mathbf{v}}{r^2}\right) \\
             &= m\left(\mathbf{r}\times\mathbf{v}\right) \\
             &= \mathbf{r}\times m\mathbf{v} \\
             &= \mathbf{r}\times\mathbf{p},
\end{align}</math>
which is the [[cross product]] of the position vector <math>\mathbf{r}</math> and the linear momentum <math>\mathbf{p} = m\mathbf{v}</math> of the particle. By the definition of the cross product, the <math>\mathbf{L}</math> vector is [[perpendicular]] to both <math>\mathbf{r}</math> and <math>\mathbf{p}</math>. 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 [[clockwise|counter-clockwise]] from the head of the vector. Conversely, the <math>\mathbf{L}</math> vector defines the [[Plane (geometry)|plane]] in which <math>\mathbf{r}</math> and <math>\mathbf{p}</math> lie.

By defining a [[unit vector]] <math>\mathbf{\hat{u}}</math> perpendicular to the plane of angular displacement, a [[angular frequency|scalar angular speed]] <math>\omega</math> results, where
<math display="block">\omega\mathbf{\hat{u}} = \boldsymbol{\omega},</math> and
<math display="block">\omega = \frac{v_\perp}{r},</math> where <math>v_\perp</math> is the perpendicular component of the motion, as above.

The two-dimensional scalar equations of the previous section can thus be given direction:
<math display="block">\begin{align}
  \mathbf{L} &= I\boldsymbol{\omega}\\
             &= I\omega\mathbf{\hat{u}}\\
             &= \left(r^2m\right)\omega\mathbf{\hat{u}}\\
             &= rmv_\perp \mathbf{\hat{u}}\\
             &= r_\perp mv\mathbf{\hat{u}},
\end{align}</math>
and <math>\mathbf{L} = rmv\mathbf{\hat{u}}</math> for circular motion, where all of the motion is perpendicular to the radius <math>r</math>.

In the [[spherical coordinate system]]  the angular momentum vector expresses as
: <math>\mathbf{L} =
  m \mathbf{r} \times \mathbf{v} = m r^2 \left(\dot\theta\,\hat{\boldsymbol\varphi} - \dot\varphi \sin\theta\,\mathbf{\hat{\boldsymbol\theta}}\right).                            
</math>