Editing Angular momentum (section)

== Conservation of angular momentum <span class="anchor" id="conservation_of_angr_mntm_anchor"></span> ==
<!-- This section is linked from [[Conservation law]] and [[Conservation of angular momentum]] -->
[[File:Cup of Russia 2010 - Yuko Kawaguti (2).jpg|thumb|upright|A [[Figure skating|figure skater]] in a spin uses conservation of angular momentum – decreasing her [[moment of inertia]] by drawing in her arms and legs increases her [[angular velocity|rotational speed]].]]

=== General considerations ===
A rotational analog of [[Newton's laws of motion#Newton's third law|Newton's third law of motion]] might be written, "In a [[closed system]], no torque can be exerted on any matter without the exertion on some other matter of an equal and opposite torque about the same axis."<ref name="Crew" /> Hence, ''angular momentum can be exchanged between objects in a closed system, but total angular momentum before and after an exchange remains constant (is conserved)''.<ref>
{{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= 82|via= Google books
}}</ref>

Seen another way, a rotational analogue of [[Newton's laws of motion#Newton's first law|Newton's first law of motion]] might be written, "A rigid body continues in a state of uniform rotation unless acted upon by an external influence."<ref name="Crew">
{{cite book
| url =https://books.google.com/books?id=sv6fAAAAMAAJ
| title =The Principles of Mechanics: For Students of Physics and Engineering
| publisher = Longmans, Green, and Company, New York
| last1 =Crew
| first1 =Henry
| date =1908
|page=88
|via= Google books}}</ref> Thus ''with no external influence to act upon it, the original angular momentum of the system remains constant''.<ref>{{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= 11|via= Google books}}</ref>

The conservation of angular momentum is used in analyzing [[Classical central-force problem|''central force motion'']]. If the net force on some body is directed always toward some point, the ''center'', then there is no torque on the body with respect to the center, as all of the force is directed along the [[Position (vector)|radius vector]], and none is [[perpendicular]] to the radius. Mathematically, torque <math>\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F} = \mathbf{0}, </math> because in this case <math>\mathbf{r}</math> and <math>\mathbf{F}</math> are parallel vectors. Therefore, the angular momentum of the body about the center is constant. This is the case with [[gravity|gravitational attraction]] in the [[orbit]]s of [[planet]]s and [[satellite]]s, where the gravitational force is always directed toward the primary body and orbiting bodies conserve angular momentum by exchanging distance and velocity as they move about the primary. Central force motion is also used in the analysis of the [[Bohr model]] of the [[atom]].

For a planet, angular momentum is distributed between the spin of the planet and its revolution in its orbit, and these are often exchanged by various mechanisms. The conservation of angular momentum in the [[Lunar theory|Earth–Moon system]] results in the transfer of angular momentum from Earth to Moon, due to [[Tidal acceleration|tidal torque]] the Moon exerts on the Earth. This in turn results in the slowing down of the rotation rate of Earth, at about 65.7 nanoseconds per day,<ref>
{{cite journal
| title =Long-term changes in the rotation of the earth – 700 B.C. to A.D. 1980
|journal = Philosophical Transactions of the Royal Society
|volume = 313
| issue =1524
| last1 =Stephenson
| first1 =F. R.
| last2 =Morrison
| first2 =L. V.
| last3 =Whitrow
| first3 =G. J.
| date =1984
| bibcode =1984RSPTA.313...47S
|pages = 47–70
| doi =10.1098/rsta.1984.0082
|s2cid = 120566848
}} +2.40 ms/century divided by 36525 days.</ref> and in gradual increase of the radius of Moon's orbit, at about 3.82&nbsp;centimeters per year.<ref>{{cite journal
|url=http://physics.ucsd.edu/~tmurphy/apollo/doc/Dickey.pdf |archive-url=https://ghostarchive.org/archive/20221009/http://physics.ucsd.edu/~tmurphy/apollo/doc/Dickey.pdf |archive-date=2022-10-09 |url-status=live
|title=Lunar Laser Ranging: A Continuing Legacy of the Apollo Program
|journal = Science
|volume = 265
|issue=5171
|pages=482–90, see 486
|author =Dickey, J. O.
|date=1994
|display-authors=etal
|doi=10.1126/science.265.5171.482
|pmid=17781305|bibcode = 1994Sci...265..482D |s2cid=10157934
}}</ref>

[[File:PrecessionOfATop.svg|thumb|The [[torque]] caused by the two opposing forces '''F'''<sub>g</sub> and −'''F'''<sub>g</sub> causes a change in the angular momentum '''L''' in the direction of that torque (since torque is the time derivative of angular momentum). This causes the [[Spinning top|top]] to [[precess]].]]

The conservation of angular momentum explains the angular acceleration of an [[Ice skating|ice skater]] as they bring their arms and legs close to the vertical axis of rotation. By bringing part of the mass of their body closer to the axis, they decrease their body's moment of inertia. Because angular momentum is the product of [[moment of inertia]] and [[angular velocity]], if the angular momentum remains constant (is conserved), then the angular velocity (rotational speed) of the skater must increase.

The same phenomenon results in extremely fast spin of compact stars (like [[white dwarf]]s, [[neutron star]]s and [[black hole]]s) when they are formed out of much larger and slower rotating stars.

Conservation is not always a full explanation for the dynamics of a system but is a key constraint. For example, a [[spinning top]] is subject to gravitational torque making it lean over and change the angular momentum about the [[nutation]] axis, but neglecting friction at the point of spinning contact, it has a conserved angular momentum about its spinning axis, and another about its [[precession]] axis. Also, in any [[planetary system]], the planets, star(s), comets, and asteroids can all move in numerous complicated ways, but only so that the angular momentum of the system is conserved.

[[Noether's theorem]] states that every [[conservation law]] is associated with a [[symmetry]] (invariant) of the underlying physics. The symmetry associated with conservation of angular momentum is [[rotational invariance]]. The fact that the physics of a system is unchanged if it is rotated by any angle about an axis implies that angular momentum is conserved.<ref>{{cite book| title=The classical theory of fields|series=Course of Theoretical Physics|first1= L. D. |last1=Landau| first2= E. M. |last2=Lifshitz|publisher=Oxford, Butterworth–Heinemann|year= 1995| isbn =978-0-7506-2768-9}}</ref>

=== Relation to Newton's second law of motion ===

While angular momentum total conservation can be understood separately from [[Newton's laws of motion]] as stemming from [[Noether's theorem]] in systems symmetric under rotations, it can also be understood simply as an efficient method of calculation of results that can also be otherwise arrived at directly from Newton's second law, together with laws governing the forces of nature (such as Newton's third law, [[Maxwell's equations]] and [[Lorentz force]]). Indeed, given initial conditions of position and velocity for every point, and the forces at such a condition, one may use Newton's second law to calculate the second derivative of position, and solving for this gives full information on the development of the physical system with time.<ref>Tenenbaum, M., & Pollard, H. (1985). Ordinary differential equations en elementary textbook for students of mathematics. Engineering and the Sciences.</ref> Note, however, that this is no longer true in [[quantum mechanics]], due to the existence of [[Spin (physics)|particle spin]], which is an angular momentum that cannot be described by the cumulative effect of point-like motions in space.

As an example, consider decreasing of the [[moment of inertia]], e.g. when a [[figure skating|figure skater]] is pulling in their hands, speeding up the circular motion. In terms of angular momentum conservation, we have, for angular momentum ''L'', moment of inertia ''I'' and angular velocity ''ω'':
<math display="block"> 0 = dL = d (I\cdot \omega) = dI \cdot \omega + I \cdot d\omega </math>

Using this, we see that the change requires an energy of:
<math display="block"> dE = d \left(\tfrac{1}{2} I\cdot \omega^2\right) = \tfrac{1}{2} dI \cdot \omega^2 + I \cdot \omega \cdot d\omega = -\tfrac{1}{2} dI \cdot \omega^2</math>
so that a decrease in the moment of inertia requires investing energy.

This can be compared to the work done as calculated using Newton's laws. Each point in the rotating body is accelerating, at each point of time, with radial acceleration of:
<math display="block">-r\cdot \omega^2</math>

Let us observe a point of mass ''m'', whose position vector relative to the center of motion is perpendicular to the z-axis at a given point of time, and is at a distance ''z''. The [[centripetal force]] on this point, keeping the circular motion, is:
<math display="block">-m\cdot z\cdot \omega^2</math>

Thus the work required for moving this point to a distance ''dz'' farther from the center of motion is:
<math display="block">dW = -m\cdot z\cdot \omega^2\cdot dz = -m\cdot \omega^2\cdot d\left(\tfrac{1}{2} z^2\right)</math>

For a non-pointlike body one must integrate over this, with ''m'' replaced by the mass density per unit ''z''. This gives:
<math display="block">dW = - \tfrac{1}{2}dI \cdot \omega^2</math>
which is exactly the energy required for keeping the angular momentum conserved.

Note, that the above calculation can also be performed per mass, using [[kinematics]] only. Thus the phenomena of figure skater accelerating tangential velocity while pulling their hands in, can be understood as follows in layman's language: The skater's palms are not moving in a straight line, so they are constantly accelerating inwards, but do not gain additional speed because the accelerating is always done when their motion inwards is zero. However, this is different when pulling the palms closer to the body: The acceleration due to rotation now increases the speed; but because of the rotation, the increase in speed does not translate to a significant speed inwards, but to an increase of the rotation speed.

=== Stationary-action principle ===
In classical mechanics it can be shown that the rotational invariance of action functionals implies conservation of angular momentum. The action is defined in classical physics as a functional of positions, <math>x_i (t)</math> often represented by the use of square brackets, and the final and initial times. It assumes the following form in cartesian coordinates:<math display="block">S\left([x_{i}];t_{1},t_{2}\right)\equiv\int_{t_{1}}^{t_{2}}d t\left(\frac{1}{2}m\frac{d x_{i}}{d t}\ \frac{d x_{i}}{d t}-V(x_{i})\right)</math>where the repeated indices indicate summation over the index. If the action is invariant of an infinitesimal transformation, it can be mathematically stated as: <math display="inline">\delta S = S\left([x_{i}+\delta x_i];t_{1},t_{2}\right)-S\left([x_{i}];t_{1},t_{2}\right) =0</math>.

Under the transformation, <math>x_i \rightarrow x_i + \delta x_i </math>, the action becomes:
<math display="block">S\left([x_{i}+\delta x_i];t_{1},t_{2}\right)=\!\int_{t_{1}}^{t_{2}}d t\left(\frac{1}{2}m\frac{d(x_{i}+\delta x_{i})}{d t}\frac{d(x_{i}+\delta x_{i})}{d t}-V(x_{i}+\delta x_{i})\right)</math>
where we can employ the expansion of the terms up-to first order in <math display="inline">\delta x_i</math>:
<math display="block">\begin{align}
\frac{d(x_i+\delta x_i)}{d t} \frac{d( x_{i}+\delta x_{i})}{ d t } &\simeq\frac{d x_{i}}{d t} \frac{d x_{i}}{d t}-2\frac{d^{2}x_{i}}{d t^{2}}\delta x_{i}+2\frac{d}{d t}\left(\delta x_{i}\frac{d x_{i}}{d t}\right)\\
V(x_{i}+\delta x_{i}) & \simeq V(x_{i})+\delta x_{i}\frac{\partial V}{\partial x_i}\\
\end{align}</math>giving the following change in action:
<math display="block">S[x_{i}+\delta x_{i}]\simeq S[x_{i}]+\int_{t_{1}}^{t_{2}}d t\,\delta x_{i}\left(- \frac{\partial V}{\partial x_i}-m{\frac{d^{2}x_{i}}{d t^{2}}}\right)+m\int_{t_{1}}^{t_{2}}d t{\frac{d}{d t}}\left(\delta x_{i}{\frac{d x_{i}}{d t}}\right).</math>

Since all rotations can be expressed as [[3D rotation group#Exponential map|matrix exponential]] of skew-symmetric matrices, i.e. as <math>R(\hat n,\theta) = e^{M \theta}</math> where <math>M </math> is a skew-symmetric matrix and <math>\theta </math> is angle of rotation, we can express the change of coordinates due to the rotation <math>R(\hat n,\delta \theta )</math>, up-to first order of infinitesimal angle of rotation, <math>\delta  \theta </math> as:
<math display="block">\delta  x_i  =  M_{ij} x_j \delta \theta . </math>

Combining the equation of motion and '''rotational invariance of action''', we get from the above equations that:<math display="block">0=\delta S=\int_{t_{1}}^{t_{2}}d t\frac{d}{d t}\left(m\frac{d x_{i}}{d t}\delta x_{i}\right)= M_{i j}\,\delta \theta \,  m \,x_{j}\frac{d x_{i}}{d t}\Bigg\vert_{t_{1}}^{t_{2}}</math>Since this is true for any matrix <math>M_{ij} </math> that satisfies <math>M_{ij} = - M_{ji}    , </math> it results in the conservation of the following quantity:
<math display="block">\ell_{ij}(t) := m\left(x_i \frac{dx_j}{dt}-x_j \frac{dx_i}{dt}\right),</math>
as <math>\ell_{ij}(t_1)=\ell_{ij}(t_2)</math>. This corresponds to the conservation of angular momentum throughout the motion.<ref>{{Cite book |last=Ramond |first=Pierre |url=https://books.google.com/books?id=aXr-DwAAQBAJ |title=Field Theory: A Modern Primer |publisher=Routledge |year=2020 |isbn=9780429689017 |edition=2nd}}[https://books.google.com/books?id=aXr-DwAAQBAJ&pg=PA1 Extract of page 1]</ref>

=== 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]].

=== Hamiltonian formalism ===
Equivalently, in [[Hamiltonian mechanics]] the Hamiltonian can be described as a function of the angular momentum. As before, the part of the kinetic energy related to rotation around the z-axis for the ''i''th object is:
<math display="block">\frac{1}{2} {I_z}_i { {\omega_z}_i}^2 = \frac{ { {L_z}_i}^2}{2 {I_z}_i}</math>
which is analogous to the energy dependence upon momentum along the z-axis, <math>\frac{ { {p_z}_i}^2}{ {2m}_i}</math>.

Hamilton's equations relate the angle around the z-axis to its conjugate momentum, the angular momentum around the same axis:
<math display="block">\begin{align}
  \frac{d{\theta_z}_i}{dt} &= \frac{\partial \mathcal{H} }{\partial {L_z}_i} = \frac{ {L_z}_i}{ {I_z}_i} \\
       \frac{d{L_z}_i}{dt} &= -\frac{\partial \mathcal{H} }{\partial {\theta_z}_i} = -\frac{\partial V}{\partial {\theta_z}_i}
\end{align}</math>

The first equation gives
<math display="block">{L_z}_i = {I_z}_i \cdot { {\dot{\theta}_z}_i} = {I_z}_i \cdot {\omega_z}_i</math>

And so we get the same results as in the Lagrangian formalism.

Note, that for combining all axes together, we write the kinetic energy as:
<math display="block">
  E_k = \frac{1}{2}\sum_i \frac{|\mathbf{p}_i|^2}{2m_i}
      = \sum_i \left(\frac{ {p_r}_i^2}{2m_i} + \frac{1}{2} {\mathbf{L}_i}^\textsf{T}{I_i}^{-1} \mathbf{L}_i\right)
</math>
where ''p''<sub>r</sub> is the momentum in the radial direction, and the [[moment of inertia#Inertia matrix in different reference frames|moment of inertia is a 3-dimensional matrix]]; bold letters stand for 3-dimensional vectors.

For point-like bodies we have:
<math display="block">E_k = \sum_i \left(\frac{ {p_r}_i^2}{2m_i} + \frac{|{\mathbf{L}_i}|^2}{2m_i {r_i}^2}\right)</math>

This form of the kinetic energy part of the Hamiltonian is useful in analyzing [[central potential]] problems, and is easily transformed to a [[quantum mechanical]] work frame (e.g. in the [[hydrogen atom]] problem).