Editing Angular momentum (section)

=== Lagrangian formalism ===
In [[Lagrangian mechanics]], angular momentum for rotation around a given axis, is the [[conjugate momentum]] of the [[generalized coordinate]] of the angle around the same axis. For example, <math>L_z</math>, the angular momentum around the z axis, is:
<math display="block">L_z = \frac{\partial \cal{L}}{\partial \dot\theta_z}</math>
where <math>\cal{L}</math> is the Lagrangian and <math>\theta_z</math> is the angle around the z axis.

Note that <math>\dot\theta_z</math>, the time derivative of the angle, is the [[angular velocity]] <math>\omega_z</math>. Ordinarily, the Lagrangian depends on the angular velocity through the kinetic energy: The latter can be written by separating the velocity to its radial and tangential part, with the tangential part at the x-y plane, around the z-axis, being equal to:
<math display="block">\sum_i \tfrac{1}{2}m_i {v_T}_i^2 = \sum_i \tfrac{1}{2} m_i \left(x_i^2 + y_i^2\right) { {\omega_z}_i}^2</math>
where the subscript i stands for the i-th body, and ''m'', ''v''<sub>''T''</sub> and ''ω''<sub>''z''</sub> stand for mass, tangential velocity around the z-axis and angular velocity around that axis, respectively.

For a body that is not point-like, with density ''ρ'', we have instead:
<math display="block">\frac{1}{2}\int \rho(x,y,z) \left(x_i^2 + y_i^2\right) { {\omega_z}_i}^2\,dx\,dy = \frac{1}{2} {I_z}_i  { {\omega_z}_i}^2</math>
where integration runs over the area of the body,<ref>{{cite book |title=Introduction to Classical Mechanics: With Problems and Solutions |author1=David Morin |edition= |publisher=Cambridge University Press |year=2008 |isbn=978-1-139-46837-4 |page=311 |url=https://books.google.com/books?id=Ni6CD7K2X4MC}} [https://books.google.com/books?id=Ni6CD7K2X4MC&pg=PA311 Extract of page 311]</ref> and ''I''<sub>z</sub> is the moment of inertia around the z-axis.

Thus, assuming the potential energy does not depend on  ''ω''<sub>''z''</sub> (this assumption may fail for electromagnetic systems), we have the angular momentum of the ''i''th object:
<math display="block">\begin{align}
  {L_z}_i &= \frac{\partial \cal{L} }{\partial { {\omega_z}_i} } = \frac{\partial E_k}{\partial { {\omega_z}_i} } \\
          &= {I_z}_i \cdot {\omega_z}_i
\end{align}</math>

We have thus far rotated each object by a separate angle; we may also define an overall angle ''θ''<sub>z</sub> by which we rotate the whole system, thus rotating also each object around the z-axis, and have the overall angular momentum:
<math display="block">L_z = \sum_i {I_z}_i \cdot {\omega_z}_i</math>

From [[Euler–Lagrange equation]]s it then follows that:
<math display="block">0 = \frac{\partial \cal{L} }{\partial { {\theta_z}_i} } - \frac{d}{dt}\left(\frac{\partial \cal{L} }{\partial { {\dot\theta_z}_i}}\right) = \frac{\partial \cal{L} }{\partial { {\theta_z}_i} } - \frac{d{L_z}_i}{dt}</math>

Since the lagrangian is dependent upon the angles of the object only through the potential, we have:
<math display="block">\frac{d{L_z}_i}{dt} = \frac{\partial \cal{L}}{\partial { {\theta_z}_i} } = -\frac{\partial V}{\partial { {\theta_z}_i} }</math>
which is the torque on the ''i''th object.

Suppose the system is invariant to rotations, so that the potential is independent of an overall rotation by the angle ''θ''<sub>z</sub> (thus it may depend on the angles of objects only through their differences, in the form <math>V({\theta_z}_i, {\theta_z}_j) = V({\theta_z}_i - {\theta_z}_j)</math>). We therefore get for the total angular momentum:
<math display="block">\frac{d L_z}{dt} = -\frac{\partial V}{\partial {\theta_z} } = 0 </math>
And thus the angular momentum around the z-axis is conserved.

This analysis can be repeated separately for each axis, giving conversation of the angular momentum vector. However, the angles around the three axes cannot be treated simultaneously as generalized coordinates, since they are not independent; in particular, two angles per point suffice to determine its position. While it is true that in the case of a rigid body, fully describing it requires, in addition to three [[Translational symmetry|translational]] degrees of freedom, also specification of three rotational degrees of freedom; however these cannot be defined as rotations around the [[Cartesian coordinate system|Cartesian axes]] (see [[Euler angles]]). This caveat is reflected in quantum mechanics in the non-trivial [[commutation relation]]s of the different components of the [[angular momentum operator]].