Conservation of angular momentum

General considerationsedit

A rotational analog of 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." 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).

Seen another way, a rotational analogue of Newton's first law of motion might be written, "A rigid body continues in a state of uniform rotation unless acted by an external influence." Thus with no external influence to act upon it, the original angular momentum of the system remains constant.

The conservation of angular momentum is used in analyzing 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 radius vector, and none is perpendicular to the radius. Mathematically, torque ${\displaystyle \mathit{\tau <!--\; \tau \; -->}=\mathbf{r}\times <!--\; \times \; -->\mathbf{F}=\mathbf{0},}$ because in this case ${\displaystyle \mathbf{r}}$ and ${\displaystyle \mathbf{F}}$ are parallel vectors. Therefore, the angular momentum of the body about the center is constant. This is the case with gravitational attraction in the orbits of planets and satellites, 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 Earth–Moon system results in the transfer of angular momentum from Earth to Moon, due to 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, and in gradual increase of the radius of Moon's orbit, at about 3.82 centimeters per year.

The conservation of angular momentum explains the angular acceleration of an ice skater as she brings her arms and legs close to the vertical axis of rotation. By bringing part of the mass of her body closer to the axis, she decreases her 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 dwarfs, neutron stars and black holes) when they are formed out of much larger and slower rotating stars. Decrease in the size of an object n times results in increase of its angular velocity by the factor of n2.

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.

Relation to Newton's second law of motionedit

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. Note, however, that this is no longer true in quantum mechanics, due to the existence of particle spin, which is 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 skater is pulling in her/his 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:

${\displaystyle 0=dL=d(I\cdot <!--\; \cdot \; -->\mathrm{\omega <!--\; \omega \; -->})=dI\cdot <!--\; \cdot \; -->\mathrm{\omega <!--\; \omega \; -->}+I\cdot <!--\; \cdot \; -->d\mathrm{\omega <!--\; \omega \; -->}}$

Using this, we see that the change requires an energy of:

${\displaystyle dE=d\left(\frac{1}{2}I\cdot <!--\; \cdot \; -->{\mathrm{\omega <!--\; \omega \; -->}}^2\right)=\frac{1}{2}dI\cdot <!--\; \cdot \; -->{\mathrm{\omega <!--\; \omega \; -->}}^2+I\cdot <!--\; \cdot \; -->\mathrm{\omega <!--\; \omega \; -->}\cdot <!--\; \cdot \; -->d\mathrm{\omega <!--\; \omega \; -->}=-<!--\; -\; -->\frac{1}{2}dI\cdot <!--\; \cdot \; -->{\mathrm{\omega <!--\; \omega \; -->}}^2}$

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:

${\displaystyle -<!--\; -\; -->r\cdot <!--\; \cdot \; -->{\mathrm{\omega <!--\; \omega \; -->}}^2}$

Let us observe a point of mass m, whose position vector relative to the center of motion is parallel 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:

${\displaystyle -<!--\; -\; -->m\cdot <!--\; \cdot \; -->z\cdot <!--\; \cdot \; -->{\mathrm{\omega <!--\; \omega \; -->}}^2}$

Thus the work required for moving this point to a distance dz farther from the center of motion is:

${\displaystyle dW=-<!--\; -\; -->m\cdot <!--\; \cdot \; -->z\cdot <!--\; \cdot \; -->{\mathrm{\omega <!--\; \omega \; -->}}^2\cdot <!--\; \cdot \; -->dz=-<!--\; -\; -->m\cdot <!--\; \cdot \; -->{\mathrm{\omega <!--\; \omega \; -->}}^2\cdot <!--\; \cdot \; -->d\left(\frac{1}{2}{z}^2\right)}$

For a non-pointlike body one must integrate over this, with m replaced by the mass density per unit z. This gives:

${\displaystyle dW=-<!--\; -\; -->\frac{1}{2}dI\cdot <!--\; \cdot \; -->{\mathrm{\omega <!--\; \omega \; -->}}^2}$

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 her/his 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.

In Lagrangian formalismedit

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, ${\displaystyle L_z}$, the angular momentum around the z axis, is:

${\displaystyle L_z=\frac{\mathrm{\partial <!--\; \partial \; -->}\mathcal{L}}{\mathrm{\partial <!--\; \partial \; -->}{\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathrm{\theta <!--\; \theta \; -->}}}_z}}$

where ${\displaystyle \mathcal{L}}$ is the Lagrangian and ${\displaystyle {\mathrm{\theta <!--\; \theta \; -->}}_z}$ is the angle around the z axis.

Note that ${\displaystyle {\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathrm{\theta <!--\; \theta \; -->}}}_z}$, the time derivative of the angle, is the angular velocity ${\displaystyle {\mathrm{\omega <!--\; \omega \; -->}}_z}$. 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:

${\displaystyle \underset{i}{\sum <!--\; \sum \; -->}\frac{1}{2}{m}_i{v_T}_i^2=\underset{i}{\sum <!--\; \sum \; -->}\frac{1}{2}{m}_i(x_i^2+y_i^2){{{\mathrm{\omega <!--\; \omega \; -->}}_z}_i}^2}$

where the subscript i stands for the i-th body, and m, v_T and ω_z 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:

${\displaystyle \frac{1}{2}\int <!--\; \int \; -->\mathrm{\rho <!--\; \rho \; -->}(x,y,z)(x_i^2+y_i^2){{{\mathrm{\omega <!--\; \omega \; -->}}_z}_i}^2=\frac{1}{2}{I_z}_i{{{\mathrm{\omega <!--\; \omega \; -->}}_z}_i}^2}$

where I_z is the moment of inertia around the z-axis.

Thus, assuming the potential energy does not depend on ω_z (this assumption may fail for electromagnetic systems), we have the angular momentum of the i-th object:

We have thus far rotated each object by a separate angle; we may also define an overall angle θ_z by which we rotate the whole system, thus rotating also each object around the z-axis, and have the overall angular momentum:

${\displaystyle L_z=\underset{i}{\sum <!--\; \sum \; -->}{I_z}_i\cdot <!--\; \cdot \; -->{{\mathrm{\omega <!--\; \omega \; -->}}_z}_i}$

From Euler-Lagrange equations it then follows that:

${\displaystyle 0=\frac{\mathrm{\partial <!--\; \partial \; -->}\mathcal{L}}{\mathrm{\partial <!--\; \partial \; -->}{{\mathrm{\theta <!--\; \theta \; -->}}_z}_i}-<!--\; -\; -->\frac{d}{dt}\left(\frac{\mathrm{\partial <!--\; \partial \; -->}\mathcal{L}}{\mathrm{\partial <!--\; \partial \; -->}{{\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathrm{\theta <!--\; \theta \; -->}}}_z}_i}\right)=\frac{\mathrm{\partial <!--\; \partial \; -->}\mathcal{L}}{\mathrm{\partial <!--\; \partial \; -->}{{\mathrm{\theta <!--\; \theta \; -->}}_z}_i}-<!--\; -\; -->\frac{d{L_z}_i}{dt}}$

Since the lagrangian is dependent upon the angles of the object only through the potential, we have:

${\displaystyle \frac{d{L_z}_i}{dt}=\frac{\mathrm{\partial <!--\; \partial \; -->}\mathcal{L}}{\mathrm{\partial <!--\; \partial \; -->}{{\mathrm{\theta <!--\; \theta \; -->}}_z}_i}=-<!--\; -\; -->\frac{\mathrm{\partial <!--\; \partial \; -->}V}{\mathrm{\partial <!--\; \partial \; -->}{{\mathrm{\theta <!--\; \theta \; -->}}_z}_i}}$

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_z (thus it may depend on the angles of objects only through their differences, in the form ${\displaystyle V({{\mathrm{\theta <!--\; \theta \; -->}}_z}_i,{{\mathrm{\theta <!--\; \theta \; -->}}_z}_j)=V({{\mathrm{\theta <!--\; \theta \; -->}}_z}_i-<!--\; -\; -->{{\mathrm{\theta <!--\; \theta \; -->}}_z}_j)}$). We therefore get for the total angular momentum:

${\displaystyle \frac{d{L}_z}{dt}=-<!--\; -\; -->\frac{\mathrm{\partial <!--\; \partial \; -->}V}{\mathrm{\partial <!--\; \partial \; -->}{\mathrm{\theta <!--\; \theta \; -->}}_z}=0}$

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 degrees of freedom, also specification of three rotational degrees of freedom; however these cannot be defined as rotations around the Cartesian axes (see Euler angles). This caveat is reflected in quantum mechanics in the non-trivial commutation relations of the different components of the angular momentum operator.

In Hamiltonian formalismedit

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:

${\displaystyle \frac{1}{2}{I_z}_i{{{\mathrm{\omega <!--\; \omega \; -->}}_z}_i}^2=\frac{{{L_z}_i}^2}{2{I_z}_i}}$

which is analogous to the energy dependence upon momentum along the z-axis, ${\displaystyle \frac{{{p_z}_i}^2}{{2m}_i}}$.

Hamilton's equations relate the angle around the z-axis to its conjugate momentum, the angular momentum around the same axis:

The first equation gives

${\displaystyle {L_z}_i={I_z}_i\cdot <!--\; \cdot \; -->{{\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathrm{\theta <!--\; \theta \; -->}}}_z}_i={I_z}_i\cdot <!--\; \cdot \; -->{{\mathrm{\omega <!--\; \omega \; -->}}_z}_i}$

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:

${\displaystyle E_k=\frac{1}{2}\underset{i}{\sum <!--\; \sum \; -->}\frac{|{\mathbf{p}}_{\mathbf{i}}{\mathbf{|}}^{\mathbf{2}}}{2m_i}=\underset{i}{\sum <!--\; \sum \; -->}\left(\frac{{{p_r}_i}^2}{2m_i}+\frac{1}{2}{{\mathbf{L}}_{\mathbf{i}}}^{\mathsf{\text{T}}}{I_i}^{-<!--\; -\; -->1}{\mathbf{L}}_{\mathbf{i}}\right)}$

where p_r is the momentum in the radial direction, and the moment of inertia is a 3-dimensional matrix; bold letters stand for 3-dimensional vectors.

For point-like bodies we have:

${\displaystyle E_k=\underset{i}{\sum <!--\; \sum \; -->}\left(\frac{{{p_r}_i}^2}{2m_i}+\frac{|{\mathbf{L}}_{\mathbf{i}}{|}^2}{2m_i{r_i}^2}\right)}$

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).