Angular momentum

 ( Mechanical Quantities ) 

Angular momentum (sometimes called moment of momentum or rotational momentum) is the rotational analog of Momentum. It is an important physical quantity because it is a Conservation law – the total angular momentum of a closed system remains constant. Angular momentum has both a direction and a magnitude, and both are conserved. Bicycles and motorcycles, , Rifling, and owe their useful properties to conservation of angular momentum. Conservation of angular momentum is also why form spirals and have high rotational rates. In general, conservation limits the possible motion of a system, but it does not uniquely determine it.

The three-dimensional angular momentum for a point particle is classically represented as a pseudovector , the cross product of the particle's position vector (relative to some origin) and its momentum vector; the latter is in Newtonian mechanics. Unlike linear momentum, angular momentum depends on where this origin is chosen, since the particle's position is measured from it.

Angular momentum is an extensive quantity; that is, the total angular momentum of any composite system is the sum of the angular momenta of its constituent parts. For a continuous rigid body or a fluid, the total angular momentum is the volume integral of angular momentum density (angular momentum per unit volume in the limit as volume shrinks to zero) over the entire body.

Similar to conservation of linear momentum, where it is conserved if there is no external force, angular momentum is conserved if there is no external torque. Torque can be defined as the rate of change of angular momentum, analogous to force. The net external torque on any system is always equal to the total torque on the system; the sum of all internal torques of any system is always 0 (this is the rotational analogue of Newton's third law of motion). Therefore, for a closed system (where there is no net external torque), the total torque on the system must be 0, which means that the total angular momentum of the system is constant.

The change in angular momentum for a particular interaction is called angular impulse, sometimes twirl.

The angular momentum $L$ of a uniform rigid sphere rotating around its axis, instead, is given by $$L\; =\; \backslash frac\{4\}\{5\}\; \backslash pi\; M\; f\; r^2$$ where $M$ is the sphere's mass, $f$ is the frequency of rotation and $r$ is the sphere's radius.

Thus, for example, the orbital angular momentum of the Earth with respect to the Sun is about 2.66 × 10⁴⁰ kg⋅m²⋅s⁻¹, while its rotational angular momentum is about 7.05 × 10³³ kg⋅m²⋅s⁻¹.

In the case of a uniform rigid sphere rotating around its axis, if, instead of its mass, its density is known, the angular momentum $L$ is given by $$L\; =\; \backslash frac\{16\}\{15\}\; \backslash pi^2\; \backslash rho\; f\; r^5$$ where $\backslash rho$ is the sphere's density, $f$ is the frequency of rotation and $r$ is the sphere's radius.

In the simplest case of a spinning disk, the angular momentum $L$ is given by $$L\; =\; \backslash pi\; M\; f\; r^2$$ where $M$ is the disk's mass, $f$ is the frequency of rotation and $r$ is the disk's radius.

If instead the disk rotates about its diameter (e.g. coin toss), its angular momentum $L$ is given by $$L\; =\; \backslash frac\{1\}\{2\}\; \backslash pi\; M\; f\; r^2$$

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, should be referred to as the angular momentum relative to that center.

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

the product of the radius of rotation and the linear momentum of the particle $p\; =\; mv$, where $v=\; r\backslash omega$ is the 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: $$L\; =\; rmv\_\backslash perp,$$ where $v\_\backslash perp\; =\; v\backslash sin(\backslash theta)$ is the perpendicular component of the motion. Expanding, $L\; =\; rmv\backslash sin(\backslash theta),$ rearranging, $L\; =\; r\backslash sin(\backslash theta)mv,$ and reducing, angular momentum can also be expressed, $$L\; =\; r\_\backslash perp\; mv,$$ where $r\_\backslash perp\; =\; r\backslash sin(\backslash theta)$ is the length of the moment arm, a line dropped perpendicularly from the origin onto the path of the particle. It is this definition, , to which the term moment of momentum refers.

And the potential energy is $$U\; =\; 0.$$

Then the Lagrangian is $$\backslash mathcal\{L\}\backslash left(\backslash phi,\; \backslash dot\{\backslash phi\}\backslash right)\; =\; T\; -\; U\; =\; \backslash tfrac\{1\}\{2\}mr^2\; \backslash dot\{\backslash phi\}^2.$$

The generalized momentum "canonically conjugate to" the coordinate $\backslash phi$ is defined by $$p\_\backslash phi\; =\; \backslash frac\{\backslash partial\; \backslash mathcal\{L\}\}\{\backslash partial\; \backslash dot\{\backslash phi\}\}\; =\; mr^2\; \backslash dot\{\backslash phi\}\; =\; I\backslash omega\; =\; L.$$

• $I\; =\; r^2m$ is the moment of inertia for a Point particle,
• $\backslash boldsymbol\{\backslash omega\}=\backslash frac\{\backslash mathbf\{r\}\backslash times\backslash mathbf\{v\}\}\{r^2\}$ is the orbital angular velocity of the particle about the origin,
• $\backslash mathbf\{r\}$ is the position vector of the particle relative to the origin, and $r=\backslash left\backslash vert\backslash mathbf\{r\}\backslash right\backslash vert$,
• $\backslash mathbf\{v\}$ is the linear velocity of the particle relative to the origin, and
• $m$ is the mass of the particle.

This can be expanded, reduced, and by the rules of vector calculus, rearranged: $$\backslash begin\{align\}$$

 \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},
     

By defining a unit vector $\backslash mathbf\{\backslash hat\{u\}\}$ perpendicular to the plane of angular displacement, a scalar angular speed $\backslash omega$ results, where $$\backslash omega\backslash mathbf\{\backslash hat\{u\}\}\; =\; \backslash boldsymbol\{\backslash omega\},$$ and $$\backslash omega\; =\; \backslash frac\{v\_\backslash perp\}\{r\},$$ where $v\_\backslash perp$ is the perpendicular component of the motion, as above.

The two-dimensional scalar equations of the previous section can thus be given direction: $$\backslash begin\{align\}$$

 \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}},
     

In the spherical coordinate system the angular momentum vector expresses as

$\backslash mathbf\{L\}\; =$

 m \mathbf{r} \times \mathbf{v} = m r^2 \left(\dot\theta\,\hat{\boldsymbol\varphi} - \dot\varphi \sin\theta\,\mathbf{\hat{\boldsymbol\theta}}\right).
     

Many problems in physics involve matter in motion about some certain point in space, be it in actual rotation about it, or simply moving past it, where it is desired to know what effect the moving matter has on the point—can it exert energy upon it or perform work about it? Energy, the ability to do work, can be stored in matter by setting it in motion—a combination of its inertia and its displacement. Inertia is measured by its mass, and displacement by its velocity. Their product, is the matter's momentum. Referring this momentum to a central point introduces a complication: the momentum is not applied to the point directly. For instance, a particle of matter at the outer edge of a wheel is, in effect, at the end of a lever of the same length as the wheel's radius, its momentum turning the lever about the center point. This imaginary lever is known as the moment arm. It has the effect of multiplying the momentum's effort in proportion to its length, an effect known as a moment. Hence, the particle's momentum referred to a particular point, is the angular momentum, sometimes called, as here, the moment of momentum of the particle versus that particular center point. The equation $L\; =\; rmv$ combines a moment (a mass $m$ turning moment arm $r$) with a linear (straight-line equivalent) speed $v$. Linear speed referred to the central point is simply the product of the distance $r$ and the angular speed $\backslash omega$ versus the point: $v=r\backslash omega,$ another moment. Hence, angular momentum contains a double moment: $L\; =\; rmr\; \backslash omega.$ Simplifying slightly, $L\; =\; r^2\; m\backslash omega,$ the quantity $r^2m$ is the particle's moment of inertia, sometimes called the second moment of mass. It is a measure of rotational inertia.

The above analogy of the translational momentum and rotational momentum can be expressed in vector form:

• $\backslash mathbf\; p\; =\; m\backslash mathbf\; v$ for linear motion
• $\backslash mathbf\; L\; =\; I\backslash boldsymbol\backslash omega$ for rotation

The direction of momentum is related to the direction of the velocity for linear movement. The direction of angular momentum is related to the angular velocity of the rotation.

Because moment of inertia is a crucial part of the spin angular momentum, the latter necessarily includes all of the complications of the former, which is calculated by multiplying elementary bits of the mass by the squares of their distances from the center of rotation.

For a rigid body, for instance a wheel or an asteroid, the orientation of rotation is simply the position of the rotation axis versus the matter of the body. It may or may not pass through the center of mass, or it may lie completely outside of the body. For the same body, angular momentum may take a different value for every possible axis about which rotation may take place. It reaches a minimum when the axis passes through the center of mass.

For a collection of objects revolving about a center, for instance all of the bodies of the Solar System, the orientations may be somewhat organized, as is the Solar System, with most of the bodies' axes lying close to the system's axis. Their orientations may also be completely random.

In brief, the more mass and the farther it is from the center of rotation (the longer the moment arm), the greater the moment of inertia, and therefore the greater the angular momentum for a given angular velocity. In many cases the moment of inertia, and hence the angular momentum, can be simplified by,

Similarly, for a Point particle $m$ the moment of inertia is defined as, $$I=r^2m$$where $r$ is the radius of the point mass from the center of rotation, and for any collection of particles $m\_i$ as the sum, $$\backslash sum\_i\; I\_i\; =\; \backslash sum\_i\; r\_i^2m\_i\; .$$

Angular momentum's dependence on position and shape is reflected in its units versus linear momentum: kg⋅m²/s or N⋅m⋅s for angular momentum versus kg⋅m/s or N⋅s for linear momentum. When calculating angular momentum as the product of the moment of inertia times the angular velocity, the angular velocity must be expressed in radians per second, where the radian assumes the dimensionless value of unity. (When performing dimensional analysis, it may be productive to use which treats radians as a base unit, but this is not done in the International system of units). The units if angular momentum can be interpreted as torque⋅time. An object with angular momentum of can be reduced to zero angular velocity by an angular impulse of .

The plane perpendicular to the axis of angular momentum and passing through the center of mass

Because the moment of inertia is $mr^2$, it follows that $\backslash frac\{dI\}\{dt\}\; =\; 2mr\backslash frac\{dr\}\{dt\}\; =\; 2rp\_$, and $\backslash frac\{d\backslash mathbf\{L\}\}\{dt\}\; =\; I\backslash frac\{d\backslash boldsymbol\{\backslash omega\}\}\{dt\}\; +\; 2rp\_\backslash boldsymbol\{\backslash omega\},$ which, reduces to $$\backslash boldsymbol\{\backslash tau\}\; =\; I\backslash boldsymbol\{\backslash alpha\}\; +\; 2rp\_\backslash boldsymbol\{\backslash omega\}.$$ This is the rotational analog of Newton's second law. Note that the torque is not necessarily proportional or parallel to the angular acceleration (as one might expect). The reason for this is that the moment of inertia of a particle can change with time, something that cannot occur for ordinary mass.

+2.40 ms/century divided by 36525 days. and in gradual increase of the radius of Moon's orbit, at about 3.82 centimeters per year.
     

As an example, consider decreasing of the moment of inertia, e.g. when a figure skating 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: $$0\; =\; dL\; =\; d\; (I\backslash cdot\; \backslash omega)\; =\; dI\; \backslash cdot\; \backslash omega\; +\; I\; \backslash cdot\; d\backslash omega$$

Using this, we see that the change requires an energy of: $$dE\; =\; d\; \backslash left(\backslash tfrac\{1\}\{2\}\; I\backslash cdot\; \backslash omega^2\backslash right)\; =\; \backslash tfrac\{1\}\{2\}\; dI\; \backslash cdot\; \backslash omega^2\; +\; I\; \backslash cdot\; \backslash omega\; \backslash cdot\; d\backslash omega\; =\; -\backslash tfrac\{1\}\{2\}\; dI\; \backslash cdot\; \backslash omega^2$$ 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: $$-r\backslash cdot\; \backslash omega^2$$

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: $$-m\backslash cdot\; z\backslash cdot\; \backslash omega^2$$

Thus the work required for moving this point to a distance dz farther from the center of motion is: $$dW\; =\; -m\backslash cdot\; z\backslash cdot\; \backslash omega^2\backslash cdot\; dz\; =\; -m\backslash cdot\; \backslash omega^2\backslash cdot\; d\backslash left(\backslash tfrac\{1\}\{2\}\; z^2\backslash right)$$

For a non-pointlike body one must integrate over this, with m replaced by the mass density per unit z. This gives: $$dW\; =\; -\; \backslash tfrac\{1\}\{2\}dI\; \backslash cdot\; \backslash omega^2$$ 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 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.

Under the transformation, $x\_i\; \backslash rightarrow\; x\_i\; +\; \backslash delta\; x\_i$, the action becomes: $$S\backslash left(x\_\{i\}+\backslash delta;t\_\{1\},t\_\{2\}\backslash right)=\backslash !\backslash int\_\{t\_\{1\}\}^\{t\_\{2\}\}d\; t\backslash left(\backslash frac\{1\}\{2\}m\backslash frac\{d(x\_\{i\}+\backslash delta\; x\_\{i\})\}\{d\; t\}\backslash frac\{d(x\_\{i\}+\backslash delta\; x\_\{i\})\}\{d\; t\}-V(x\_\{i\}+\backslash delta\; x\_\{i\})\backslash right)$$ where we can employ the expansion of the terms up-to first order in $\backslash delta\; x\_i$: giving the following change in action: $$Sx\_\{i\}+\backslash delta\backslash simeq\; Sx\_\{i\}+\backslash int\_\{t\_\{1\}\}^\{t\_\{2\}\}d\; t\backslash ,\backslash delta\; x\_\{i\}\backslash left(-\; \backslash frac\{\backslash partial\; V\}\{\backslash partial\; x\_i\}-m\{\backslash frac\{d^\{2\}x\_\{i\}\}\{d\; t^\{2\}\}\}\backslash right)+m\backslash int\_\{t\_\{1\}\}^\{t\_\{2\}\}d\; t\{\backslash frac\{d\}\{d\; t\}\}\backslash left(\backslash delta\; x\_\{i\}\{\backslash frac\{d\; x\_\{i\}\}\{d\; t\}\}\backslash right).$$

Since all rotations can be expressed as matrix exponential of skew-symmetric matrices, i.e. as $R(\backslash hat\; n,\backslash theta)\; =\; e^\{M\; \backslash theta\}$ where $M$ is a skew-symmetric matrix and $\backslash theta$ is angle of rotation, we can express the change of coordinates due to the rotation $R(\backslash hat\; n,\backslash delta\; \backslash theta\; )$, up-to first order of infinitesimal angle of rotation, $\backslash delta\; \backslash theta$ as: $$\backslash delta\; x\_i\; =\; M\_\{ij\}\; x\_j\; \backslash delta\; \backslash theta\; .$$

Combining the equation of motion and rotational invariance of action, we get from the above equations that:$$0=\backslash delta\; S=\backslash int\_\{t\_\{1\}\}^\{t\_\{2\}\}d\; t\backslash frac\{d\}\{d\; t\}\backslash left(m\backslash frac\{d\; x\_\{i\}\}\{d\; t\}\backslash delta\; x\_\{i\}\backslash right)=\; M\_\{i\; j\}\backslash ,\backslash delta\; \backslash theta\; \backslash ,\; m\; \backslash ,x\_\{j\}\backslash frac\{d\; x\_\{i\}\}\{d\; t\}\backslash Bigg\backslash vert\_\{t\_\{1\}\}^\{t\_\{2\}\}$$Since this is true for any matrix $M\_\{ij\}$ that satisfies $M\_\{ij\}\; =\; -\; M\_\{ji\}\; ,$ it results in the conservation of the following quantity: $$\backslash ell\_\{ij\}(t)\; :=\; m\backslash left(x\_i\; \backslash frac\{dx\_j\}\{dt\}-x\_j\; \backslash frac\{dx\_i\}\{dt\}\backslash right),$$ as $\backslash ell\_\{ij\}(t\_1)=\backslash ell\_\{ij\}(t\_2)$. This corresponds to the conservation of angular momentum throughout the motion.

Note that $\backslash dot\backslash theta\_z$, the time derivative of the angle, is the angular velocity $\backslash omega\_z$. 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: $$\backslash sum\_i\; \backslash tfrac\{1\}\{2\}m\_i\; \{v\_T\}\_i^2\; =\; \backslash sum\_i\; \backslash tfrac\{1\}\{2\}\; m\_i\; \backslash left(x\_i^2\; +\; y\_i^2\backslash right)\; \{\; \{\backslash omega\_z\}\_i\}^2$$ 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: $$\backslash frac\{1\}\{2\}\backslash int\; \backslash rho(x,y,z)\; \backslash left(x\_i^2\; +\; y\_i^2\backslash right)\; \{\; \{\backslash omega\_z\}\_i\}^2\backslash ,dx\backslash ,dy\; =\; \backslash frac\{1\}\{2\}\; \{I\_z\}\_i\; \{\; \{\backslash omega\_z\}\_i\}^2$$ where integration runs over the area of the body,

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 ith object: $$\backslash begin\{align\}$$

 {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
     

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: $$L\_z\; =\; \backslash sum\_i\; \{I\_z\}\_i\; \backslash cdot\; \{\backslash omega\_z\}\_i$$

From Euler–Lagrange equations it then follows that: $$0\; =\; \backslash frac\{\backslash partial\; \backslash cal\{L\}\; \}\{\backslash partial\; \{\; \{\backslash theta\_z\}\_i\}\; \}\; -\; \backslash frac\{d\}\{dt\}\backslash left(\backslash frac\{\backslash partial\; \backslash cal\{L\}\; \}\{\backslash partial\; \{\; \{\backslash dot\backslash theta\_z\}\_i\}\}\backslash right)\; =\; \backslash frac\{\backslash partial\; \backslash cal\{L\}\; \}\{\backslash partial\; \{\; \{\backslash theta\_z\}\_i\}\; \}\; -\; \backslash frac\{d\{L\_z\}\_i\}\{dt\}$$

Since the lagrangian is dependent upon the angles of the object only through the potential, we have: $$\backslash frac\{d\{L\_z\}\_i\}\{dt\}\; =\; \backslash frac\{\backslash partial\; \backslash cal\{L\}\}\{\backslash partial\; \{\; \{\backslash theta\_z\}\_i\}\; \}\; =\; -\backslash frac\{\backslash partial\; V\}\{\backslash partial\; \{\; \{\backslash theta\_z\}\_i\}\; \}$$ which is the torque on the ith 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 $V(\{\backslash theta\_z\}\_i,\; \{\backslash theta\_z\}\_j)\; =\; V(\{\backslash theta\_z\}\_i\; -\; \{\backslash theta\_z\}\_j)$). We therefore get for the total angular momentum: $$\backslash frac\{d\; L\_z\}\{dt\}\; =\; -\backslash frac\{\backslash partial\; V\}\{\backslash partial\; \{\backslash theta\_z\}\; \}\; =\; 0$$ 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.

Hamilton's equations relate the angle around the z-axis to its conjugate momentum, the angular momentum around the same axis: $$\backslash begin\{align\}$$

 \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}
     

The first equation gives $$\{L\_z\}\_i\; =\; \{I\_z\}\_i\; \backslash cdot\; \{\; \{\backslash dot\{\backslash theta\}\_z\}\_i\}\; =\; \{I\_z\}\_i\; \backslash cdot\; \{\backslash omega\_z\}\_i$$

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: $$$$

 E_k = \frac{1}{2}\sum_i \frac
= \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)
     
     
where  pr 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:
     $$E\_k\; =\; \backslash sum\_i\; \backslash left(\backslash frac\{\; \{p\_r\}\_i^2\}\{2m\_i\}\; +\; \backslash frac\{|\{\backslash mathbf\{L\}\_i\}^2\}\{2m\_i\; \{r\_i\}^2\}\backslash right)$$
     

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).
     

 

 
 
  

  Angular momentum in orbital mechanics
 
 

 
 
 
While in classical mechanics the language of angular momentum can be replaced by Newton's laws of motion, it is particularly useful for motion in central potential such as planetary motion in the solar system. Thus, the orbit of a planet in the solar system is defined by its energy, angular momentum and angles of the orbit major axis relative to a coordinate frame.
     

In astrodynamics and celestial mechanics, a quantity closely related to angular momentum is defined as
Richard H. Battin (1999). 9781563473425, American Institute of Aeronautics and Astronautics, Inc.. ISBN 9781563473425
     $$\backslash mathbf\{h\}\; =\; \backslash mathbf\{r\}\; \backslash times\; \backslash mathbf\{v\},$$
called  specific angular momentum. Note that $\backslash mathbf\{L\}\; =\; m\backslash mathbf\{h\}.$ Mass is often unimportant in orbital mechanics calculations, because motion of a body is determined by gravity. The primary body of the system is often so much larger than any bodies in motion about it that the gravitational effect of the smaller bodies on it can be neglected; it maintains, in effect, constant velocity. The motion of all bodies is affected by its gravity in the same way, regardless of mass, and therefore all move approximately the same way under the same conditions.
     

 

 
 
  

  Solid bodies
 
 

 
Angular momentum is also an extremely useful concept for describing rotating rigid bodies such as a gyroscope or a rocky planet.
For a continuous mass distribution with density function  ρ( r), a differential volume element  dV with position vector  r within the mass has a mass element  dm =  ρ( r) dV. Therefore, the infinitesimal angular momentum of this element is:
     $$d\backslash mathbf\{L\}\; =\; \backslash mathbf\{r\}\backslash times\; dm\; \backslash mathbf\{v\}\; =\; \backslash mathbf\{r\}\backslash times\; \backslash rho(\backslash mathbf\{r\})\; dV\; \backslash mathbf\{v\}\; =\; dV\; \backslash mathbf\{r\}\backslash times\; \backslash rho(\backslash mathbf\{r\})\; \backslash mathbf\{v\}$$
and volume integral this differential over the volume of the entire mass gives its total angular momentum:
     $$\backslash mathbf\{L\}=\backslash int\_V\; dV\; \backslash mathbf\{r\}\backslash times\; \backslash rho(\backslash mathbf\{r\})\; \backslash mathbf\{v\}$$
     

In the derivation which follows, integrals similar to this can replace the sums for the case of continuous mass.
     

 

 
 
  

  Collection of particles
 
 

 
 
 
For a collection of particles in motion about an arbitrary origin, it is informative to develop the equation of angular momentum by resolving their motion into components about their own center of mass and about the origin. Given,
 

  

   $m\_i$ is the mass of particle $i$,
  
  

   $\backslash mathbf\{R\}\_i$ is the position vector of particle $i$ w.r.t. the origin,
  
  

   $\backslash mathbf\{V\}\_i$ is the velocity of particle $i$ w.r.t. the origin,
  
  

   $\backslash mathbf\{R\}$ is the position vector of the center of mass w.r.t. the origin,
  
  

   $\backslash mathbf\{V\}$ is the velocity of the center of mass w.r.t. the origin,
  
  

   $\backslash mathbf\{r\}\_i$ is the position vector of particle $i$ w.r.t. the center of mass,
  
  

   $\backslash mathbf\{v\}\_i$ is the velocity of particle $i$ w.r.t. the center of mass,
  
 

The total mass of the particles is simply their sum,
     $$M=\backslash sum\_i\; m\_i.$$
     

The position vector of the center of mass is defined by,
     
     $$M\backslash mathbf\{R\}=\backslash sum\_i\; m\_i\; \backslash mathbf\{R\}\_i.$$
     

By inspection,
     

     
$\backslash mathbf\{R\}\_i\; =\; \backslash mathbf\{R\}\; +\; \backslash mathbf\{r\}\_i$ and $\backslash mathbf\{V\}\_i\; =\; \backslash mathbf\{V\}\; +\; \backslash mathbf\{v\}\_i.$
     
     

The total angular momentum of the collection of particles is the sum of the angular momentum of each particle,
     

 
Expanding $\backslash mathbf\{R\}\_i$,
     

     
$\backslash begin\{align\}$
     
     
 \mathbf{L} &= \sum_i \left[\left(\mathbf{R} + \mathbf{r}_i\right) \times m_i\mathbf{V}_i \right] \\
            &= \sum_i \left[ \mathbf{R} \times m_i\mathbf{V}_i + \mathbf{r}_i \times m_i\mathbf{V}_i \right]
     
\end{align}
     

Expanding $\backslash mathbf\{V\}\_i$,
     

     
$\backslash begin\{align\}$
     
     
 \mathbf{L} &= \sum_i \left[ \mathbf{R} \times m_i\left(\mathbf{V} + \mathbf{v}_i\right) + \mathbf{r}_i \times m_i(\mathbf{V} + \mathbf{v}_i) \right ]\\
            &= \sum_i \left[ \mathbf{R} \times m_i\mathbf{V} + \mathbf{R} \times m_i\mathbf{v}_i + \mathbf{r}_i \times m_i\mathbf{V} + \mathbf{r}_i \times m_i\mathbf{v}_i \right]\\
            &= \sum_i \mathbf{R} \times m_i\mathbf{V} + \sum_i \mathbf{R} \times m_i\mathbf{v}_i + \sum_i \mathbf{r}_i \times m_i\mathbf{V} + \sum_i \mathbf{r}_i \times m_i\mathbf{v}_i
     
\end{align}
     

It can be shown that (see sidebar),
{
class="toccolours" style="float:right; margin-left:0.5em; margin-right:0.5em; font-size:84%; background:white; color:black; width:30em; max-width:30%;" cellspacing="5"
Prove that $\backslash sum\_i\; m\_i\backslash mathbf\{r\}\_i\; =\; \backslash mathbf\{0\}$
     $$\backslash begin\{align\}$$
           \mathbf{r}_i &= \mathbf{R}_i - \mathbf{R} \\
        m_i\mathbf{r}_i &= m_i\left(\mathbf{R}_i - \mathbf{R}\right) \\
 \sum_i m_i\mathbf{r}_i &= \sum_i m_i\left(\mathbf{R}_i - \mathbf{R}\right)\\
                        &= \sum_i (m_i\mathbf{R}_i - m_i\mathbf{R})\\
                        &= \sum_i m_i\mathbf{R}_i - \sum_i m_i\mathbf{R}\\
                        &= \sum_i m_i\mathbf{R}_i - \left( \sum_i m_i \right) \mathbf{R}\\
                        &= \sum_i m_i\mathbf{R}_i - M\mathbf{R}
     
\end{align}
which, by the definition of the center of mass, is $\backslash mathbf\{0\},$ and similarly for $\backslash sum\_i\; m\_i\; \backslash mathbf\{v\}\_i.$
     

     
$\backslash sum\_i\; m\_i\backslash mathbf\{r\}\_i\; =\; \backslash mathbf\{0\}$ and $\backslash sum\_i\; m\_i\backslash mathbf\{v\}\_i\; =\; \backslash mathbf\{0\},$
     
therefore the second and third terms vanish,
     

     
$\backslash mathbf\{L\}\; =\; \backslash sum\_i\; \backslash mathbf\{R\}\; \backslash times\; m\_i\backslash mathbf\{V\}\; +\; \backslash sum\_i\; \backslash mathbf\{r\}\_i\; \backslash times\; m\_i\backslash mathbf\{v\}\_i\; .$
     
     

The first term can be rearranged,
     

     
$\backslash sum\_i\; \backslash mathbf\{R\}\; \backslash times\; m\_i\backslash mathbf\{V\}\; =\; \backslash mathbf\{R\}\; \backslash times\; \backslash sum\_i\; m\_i\backslash mathbf\{V\}\; =\; \backslash mathbf\{R\}\; \backslash times\; M\backslash mathbf\{V\},$
     
and total angular momentum for the collection of particles is finally,
     
     

 
The first term is the angular momentum of the center of mass relative to the origin. Similar to  , below, it is the angular momentum of one particle of mass  M at the center of mass moving with velocity  V. The second term is the angular momentum of the particles moving relative to the center of mass, similar to , below. The result is general—the motion of the particles is not restricted to rotation or revolution about the origin or center of mass. The particles need not be individual masses, but can be elements of a continuous distribution, such as a solid body.
     

Rearranging equation () by vector identities, multiplying both terms by "one", and grouping appropriately,
     $$\backslash begin\{align\}$$
 \mathbf{L} &= M(\mathbf{R} \times \mathbf{V}) + \sum_i \left[m_i\left(\mathbf{r}_i \times \mathbf{v}_i\right)\right], \\
            &= \frac{R^2}{R^2}M\left(\mathbf{R} \times \mathbf{V}\right) + \sum_i \left[ \frac{r_i^2}{r_i^2}m_i\left(\mathbf{r}_i \times \mathbf{v}_i\right)\right] , \\
            &= R^2M \left( \frac{\mathbf{R} \times \mathbf{V} }{R^2} \right) + \sum_i \left[ r_i^2 m_i \left( \frac{\mathbf{r}_i \times \mathbf{v}_i}{r_i^2} \right) \right] , \\
     
\end{align}
gives the total angular momentum of the system of particles in terms of moment of inertia $I$ and angular velocity $\backslash boldsymbol\{\backslash omega\}$,
     

 
 
 

 
 
  

  Single particle case
 
 

 
In the case of a single particle moving about the arbitrary origin,
     $$\backslash begin\{align\}$$
 \mathbf{r}_i &= \mathbf{v}_i = \mathbf{0}, \\
   \mathbf{r} &= \mathbf{R}, \\
   \mathbf{v} &= \mathbf{V}, \\
            m &= M,
     
\end{align}
     $$\backslash sum\_i\; \backslash mathbf\{r\}\_i\; \backslash times\; m\_i\backslash mathbf\{v\}\_i\; =\; \backslash mathbf\{0\},$$
     $$\backslash sum\_i\; I\_i\backslash boldsymbol\{\backslash omega\}\_i\; =\; \backslash mathbf\{0\},$$ and equations () and () for total angular momentum reduce to,
     $$\backslash mathbf\{L\}\; =\; \backslash mathbf\{R\}\; \backslash times\; m\backslash mathbf\{V\}\; =\; I\_R\backslash boldsymbol\{\backslash omega\}\_R.$$
     

 

 
 
  

  Case of a fixed center of mass
 
 

 
For the case of the center of mass fixed in space with respect to the origin,
     $$\backslash mathbf\{V\}\; =\; \backslash mathbf\{0\},$$
     $$\backslash mathbf\{R\}\; \backslash times\; M\backslash mathbf\{V\}\; =\; \backslash mathbf\{0\},$$
     $$I\_R\backslash boldsymbol\{\backslash omega\}\_R\; =\; \backslash mathbf\{0\},$$ and equations () and () for total angular momentum reduce to,
     $$\backslash mathbf\{L\}\; =\; \backslash sum\_i\; \backslash mathbf\{r\}\_i\; \backslash times\; m\_i\backslash mathbf\{v\}\_i\; =\; \backslash sum\_i\; I\_i\backslash boldsymbol\{\backslash omega\}\_i.$$
     

 
 

 
 
  

  Angular momentum in general relativity
 
 

 
In modern (20th century) theoretical physics, angular momentum (not including any intrinsic angular momentum – see below) is described using a different formalism, instead of a classical pseudovector. In this formalism, angular momentum is the 2-form Noether charge associated with rotational invariance. As a result, angular momentum is generally not conserved locally for general , unless they have rotational symmetry;
     
 (1973). 9780521099066, Cambridge University Press. ISBN 9780521099066 whereas globally the notion of angular momentum itself only makes sense if the spacetime is asymptotically flat.
      If the spacetime is only axially symmetric like for the Kerr metric, the total angular momentum is not conserved but $p\_\{\backslash phi\}$ is conserved which is related to the invariance of rotating around the symmetry-axis, where note that $p\_\{\backslash phi\}=g\_\{\backslash phi\; \backslash mu\}p^\{\backslash mu\}=mg\_\{\backslash mu\; \backslash phi\}\; dX^\{\backslash mu\}/d\backslash tau$ where $g\_\{\backslash mu\backslash nu\}$ is the metric, $m=\backslash sqrt$ is the rest mass, $dX^\{\backslash mu\}/d\backslash tau$ is the four-velocity, and $X^\{\backslash mu\}=(t,r,\backslash theta,\backslash phi)$ is the four-position in spherical coordinates.
     

In classical mechanics, the angular momentum of a particle can be reinterpreted as a plane element:
     $$\backslash mathbf\{L\}\; =\; \backslash mathbf\{r\}\; \backslash wedge\; \backslash mathbf\{p\}\; \backslash ,,$$
in which the exterior product (∧) replaces the cross product (×) (these products have similar characteristics but are nonequivalent). This has the advantage of a clearer geometric interpretation as a plane element, defined using the vectors  x and  p, and the expression is true in any number of dimensions. In Cartesian coordinates:
     $$\backslash begin\{align\}$$
 \mathbf{L} &= \left(xp_y - yp_x\right)\mathbf{e}_x \wedge \mathbf{e}_y + \left(yp_z - zp_y\right)\mathbf{e}_y \wedge \mathbf{e}_z + \left(zp_x - xp_z\right)\mathbf{e}_z \wedge \mathbf{e}_x\\
            &= L_{xy}\mathbf{e}_x \wedge \mathbf{e}_y + L_{yz}\mathbf{e}_y \wedge \mathbf{e}_z + L_{zx}\mathbf{e}_z \wedge \mathbf{e}_x \,,
     
\end{align}
or more compactly in index notation:
     $$L\_\{ij\}\; =\; x\_i\; p\_j\; -\; x\_j\; p\_i\backslash ,.$$
     

The angular velocity can also be defined as an anti-symmetric second order tensor, with components  ωij. The relation between the two anti-symmetric tensors is given by the moment of inertia which must now be a fourth order tensor:Synge and Schild, Tensor calculus, Dover publications, 1978 edition, p. 161. .
     $$L\_\{ij\}\; =\; I\_\{ijk\backslash ell\}\; \backslash omega\_\{k\backslash ell\}\; \backslash ,.$$
     

Again, this equation in  L and  ω as tensors is true in any number of dimensions. This equation also appears in the geometric algebra formalism, in which  L and  ω are bivectors, and the moment of inertia is a mapping between them.
     

In relativistic mechanics, the relativistic angular momentum of a particle is expressed as an anti-symmetric tensor of second order:
     $$M\_\{\backslash alpha\backslash beta\}\; =\; X\_\backslash alpha\; P\_\backslash beta\; -\; X\_\backslash beta\; P\_\backslash alpha$$
in terms of , namely the four-position  X and the four-momentum  P, and absorbs the above  L together with the moment of mass, i.e., the product of the relativistic mass of the particle and its center of mass, which can be thought of as describing the motion of its center of mass, since mass–energy is conserved.
     

In each of the above cases, for a system of particles the total angular momentum is just the sum of the individual particle angular momenta, and the center of mass is for the system.
     

 

 
 
  

  Angular momentum in quantum mechanics
 
 

 
 
In quantum mechanics, angular momentum (like other quantities) is expressed as an operator, and its one-dimensional projections have point spectrum. Angular momentum is subject to the Heisenberg uncertainty principle, implying that at any time, only one projection (also called "component") can be measured with definite precision; the other two then remain uncertain. Because of this, the axis of rotation of a quantum particle is undefined. Quantum particles  do possess a type of non-orbital angular momentum called "spin", but this angular momentum does not correspond to a spinning motion.
 (2025). 9780415257886, CRC Press. . ISBN 9780415257886 In relativistic quantum mechanics the above relativistic definition becomes a tensorial operator.
     

 

 
 
  

  Spin, orbital, and total angular momentum
 
 

 
 
 
[[File:Classical angular momentum.svg|upright=1.25|thumb|Angular momenta of a  classical object.
]]
     

The classical definition of angular momentum as $\backslash mathbf\{L\}\; =\; \backslash mathbf\{r\}\backslash times\backslash mathbf\{p\}$ can be carried over to quantum mechanics, by reinterpreting  r as the quantum position operator and  p as the quantum momentum operator.  L is then an operator, specifically called the  orbital angular momentum operator. The components of the angular momentum operator satisfy the commutation relations of the Lie algebra SO(3). Indeed, these operators are precisely the infinitesimal action of the rotation group on the quantum Hilbert space. Section 17.3 (See also the discussion below of the angular momentum operators as the generators of rotations.)
     

However, in quantum physics, there is another type of angular momentum, called  spin angular momentum, represented by the spin operator  S. Spin is often depicted as a particle literally spinning around an axis, but this is a misleading and inaccurate picture: spin is an intrinsic property of a particle, unrelated to any sort of motion in space and fundamentally different from orbital angular momentum. All elementary particles have a characteristic spin (possibly zero),
 (2025). 9789813237070, World Scientific. . ISBN 9789813237070
 and almost all elementary particles have nonzero spin.
 (2025). 9780387271279, Springer Science & Business Media. . ISBN 9780387271279 For example  have "spin 1/2" (this actually means "spin ħ/2"),  have "spin 1" (this actually means "spin ħ"), and Pion have spin 0.
 (1998). 9780521565837, Cambridge University Press. . ISBN 9780521565837
     

Finally, there is total angular momentum  J, which combines both the spin and orbital angular momentum of all particles and fields. (For one particle, .) Conservation of angular momentum applies to  J, but not to  L or  S; for example, the spin–orbit interaction allows angular momentum to transfer back and forth between  L and  S, with the total remaining constant. Electrons and photons need not have integer-based values for total angular momentum, but can also have half-integer values.
     

In molecules the total angular momentum  F is the sum of the rovibronic (orbital) angular momentum  N, the electron spin angular momentum  S, and the nuclear spin angular momentum  I. For electronic singlet states the rovibronic angular momentum is denoted  J rather than  N. As explained by Van Vleck,
      the components of the molecular rovibronic angular momentum referred to molecule-fixed axes have different commutation relations from those for the components about space-fixed axes.
     

 

 
 
  

  Quantization
 
 

 
 
In quantum mechanics, angular momentum is quantized – that is, it cannot vary continuously, but only in "Quantum number" between certain allowed values. For any system, the following restrictions on measurement results apply, where $\backslash hbar$ is the reduced Planck constant and $\backslash hat\; n$ is any Euclidean vector such as x, y, or z:
     
If you measure...
The result can be...
$L\_\backslash hat\{n\}$
$\backslash ldots,\; -2\backslash hbar,\; -\backslash hbar,\; 0,\; \backslash hbar,\; 2\backslash hbar,\; \backslash ldots$
$S\_\backslash hat\{n\}$ or $J\_\backslash hat\{n\}$
$\backslash ldots,\; -\backslash frac\{3\}\{2\}\backslash hbar,\; -\backslash hbar,\; -\backslash frac\{1\}\{2\}\backslash hbar,\; 0,\; \backslash frac\{1\}\{2\}\backslash hbar,\; \backslash hbar,\; \backslash frac\{3\}\{2\}\backslash hbar,\; \backslash ldots$
$\backslash begin\{align\}$
     &L^2 \\
 ={} &L_x^2 + L_y^2 + L_z^2
     
\end{align}
$\backslash left\backslash hbar^2$, where $n\; =\; 0,\; 1,\; 2,\; \backslash ldots$
$S^2$ or $J^2$
$\backslash left\backslash hbar^2$, where $n\; =\; 0,\; \backslash tfrac\{1\}\{2\},\; 1,\; \backslash tfrac\{3\}\{2\},\; \backslash ldots$
The reduced Planck constant $\backslash hbar$ is tiny by everyday standards, about 10−34 Joule-second, and therefore this quantization does not noticeably affect the angular momentum of macroscopic objects. However, it is very important in the microscopic world. For example, the structure of  and subshells in chemistry is significantly affected by the quantization of angular momentum.
     

Quantization of angular momentum was first postulated by Niels Bohr in Bohr model of the atom and was later predicted by Erwin Schrödinger in his Schrödinger equation.
     

 

 
 
  

  Uncertainty
 
 

 
In the definition $\backslash mathbf\{L\}=\backslash mathbf\{r\}\backslash times\backslash mathbf\{p\}$, six operators are involved: The position operators $r\_x$, $r\_y$, $r\_z$, and the momentum operators $p\_x$, $p\_y$, $p\_z$. However, the Heisenberg uncertainty principle tells us that it is not possible for all six of these quantities to be known simultaneously with arbitrary precision. Therefore, there are limits to what can be known or measured about a particle's angular momentum. It turns out that the best that one can do is to simultaneously measure both the angular momentum vector's magnitude and its component along one axis.
     

The uncertainty is closely related to the fact that different components of an angular momentum operator do not commutator, for example $L\_xL\_y\; \backslash neq\; L\_yL\_x$. (For the precise commutation relations, see angular momentum operator.)
     

 

 
 
  

  Total angular momentum as generator of rotations
 
 

 
As mentioned above, orbital angular momentum  L is defined as in classical mechanics: $\backslash mathbf\{L\}=\backslash mathbf\{r\}\backslash times\backslash mathbf\{p\}$, but  total angular momentum  J is defined in a different, more basic way:  J is defined as the "generator of rotations". More specifically,  J is defined so that the operator
     $$R(\backslash hat\{n\},\backslash phi)\; \backslash equiv\; \backslash exp\backslash left(-\backslash frac\{i\}\{\backslash hbar\}\backslash phi\backslash ,\; \backslash mathbf\{J\}\backslash cdot\; \backslash hat\{\backslash mathbf\{n\}\; \}\backslash right)$$
is the rotation operator that takes any system and rotates it by angle $\backslash phi$ about the axis $\backslash hat\{\backslash mathbf\{n\}\}$. (The "exp" in the formula refers to operator exponential.) To put this the other way around, whatever our quantum Hilbert space is, we expect that the rotation group SO(3) will act on it. There is then an associated action of the Lie algebra so(3) of SO(3); the operators describing the action of so(3) on our Hilbert space are the (total) angular momentum operators.
     

The relationship between the angular momentum operator and the rotation operators is the same as the relationship between  and  in mathematics. The close relationship between angular momentum and rotations is reflected in Noether's theorem that proves that angular momentum is conserved whenever the laws of physics are rotationally invariant.
     

 

 
 
  

  Angular momentum in electrodynamics
 
 

 
 
 
When describing the motion of a charged particle in an electromagnetic field, the canonical momentum  P (derived from the Lagrangian for this system) is not gauge invariant. As a consequence, the canonical angular momentum  L =  r ×  P is not gauge invariant either. Instead, the momentum that is physical, the so-called  kinetic momentum (used throughout this article), is (in SI units)
     $$\backslash mathbf\{p\}\; =\; m\backslash mathbf\{v\}\; =\; \backslash mathbf\{P\}\; -\; e\; \backslash mathbf\{A\}$$
where  e is the electric charge of the particle and  A the magnetic vector potential of the electromagnetic field. The gauge-invariant angular momentum, that is  kinetic angular momentum, is given by
     $$\backslash mathbf\{K\}=\; \backslash mathbf\{r\}\; \backslash times\; (\; \backslash mathbf\{P\}\; -\; e\backslash mathbf\{A\}\; )$$
     

The interplay with quantum mechanics is discussed further in the article on canonical commutation relations.
     

 

 
 
  

  Angular momentum in optics
 
 

 
In  classical Maxwell electrodynamics the Poynting vector
is a linear momentum density of electromagnetic field.
     $$\backslash mathbf\{S\}(\backslash mathbf\{r\},\; t)\; =\; \backslash epsilon\_0\; c^2\; \backslash mathbf\{E\}(\backslash mathbf\{r\},\; t)\; \backslash times\; \backslash mathbf\{B\}(\backslash mathbf\{r\},\; t).$$
     

The angular momentum density vector $\backslash mathbf\{L\}(\backslash mathbf\{r\},\; t)$ is given by a vector product
as in classical mechanics:
     $$\backslash mathbf\{L\}(\backslash mathbf\{r\},\; t)\; =\; \backslash epsilon\_0\; \backslash mu\_0\; \backslash mathbf\{r\}\; \backslash times\; \backslash mathbf\{S\}(\backslash mathbf\{r\},\; t).$$
     

The above identities are valid  locally, i.e. in each space point $\backslash mathbf\{r\}$ in a given moment $t$.
     

 

 
 
  

  Angular momentum in nature and the cosmos
 
 

 
 
Tropical cyclones and other related weather phenomena involve conservation of angular momentum in order to explain the dynamics.  Winds revolve slowly around low pressure systems, mainly due to the coriolis effect.  If the low pressure intensifies and the slowly circulating air is drawn toward the center, the molecules must speed up in order to conserve angular momentum.  By the time they reach the center, the speeds become destructive.
     

     Johannes Kepler determined the laws of planetary motion without knowledge of conservation of momentum.  However, not long after his discovery their derivation was determined from conservation of angular momentum.  Planets move more slowly the further they are out in their elliptical orbits, which is explained intuitively by the fact that orbital angular momentum is proportional to the radius of the orbit.  Since the mass does not change and the angular momentum is conserved, the velocity drops.
     

Tidal acceleration is an effect of the tidal forces between an orbiting natural satellite (e.g. the Moon) and the primary planet that it orbits (e.g. Earth).  The gravitational torque between the Moon and the tidal bulge of Earth causes the Moon to be constantly promoted to a slightly higher orbit (~3.8 cm per year) and Earth to be decelerated (by −25.858 ± 0.003″/cy²) in its rotation (the length of the day increases by ~1.7 ms per century, +2.3 ms from tidal effect and −0.6 ms from post-glacial rebound).  The Earth loses angular momentum which is transferred to the Moon such that the overall angular momentum is conserved.
     

 

 
 
  

  Angular momentum in engineering and technology
 
 

 
Examples of using conservation of angular momentum for practical advantage are abundant.  In engines such as steam engines or internal combustion engines, a flywheel is needed to efficiently convert the lateral motion of the pistons to rotational motion.
     

Inertial navigation systems explicitly use the fact that angular momentum is conserved with respect to the inertial frame of space.  Inertial navigation is what enables submarine trips under the polar ice cap, but are also crucial to all forms of modern navigation.
     

     rifle use the stability provided by conservation of angular momentum to be more true in their trajectory.  The invention of rifled firearms and cannons gave their users significant strategic advantage in battle, and thus were a technological turning point in history.
     

 

 
 
  

  History
 
 

 
     Isaac Newton, in the  Principia, hinted at angular momentum in his examples of the first law of motion,
A top, whose parts by their cohesion are perpetually drawn aside from rectilinear motions, does not cease its rotation, otherwise than as it is retarded by the air. The greater bodies of the planets and comets, meeting with less resistance in more free spaces, preserve their motions both progressive and circular for a much longer time.
     
He did not further investigate angular momentum directly in the  Principia, saying:
From such kind of reflexions also sometimes arise the circular motions of bodies about their own centers. But these are cases which I do not consider in what follows; and it would be too tedious to demonstrate every particular that relates to this subject.Newton, Axioms; or Laws of Motion, Corollary III
However, his geometric proof of the law of areas is an outstanding example of Newton's genius, and indirectly proves angular momentum conservation in the case of a central force.
     

 

 
 
  

  Law of Areas
 
 

 
 
 
 

 
 
  

  Newton's derivation
 
 

 
 
As a planet orbits the Sun, the line between the Sun and the planet sweeps out equal areas in equal intervals of time. This had been known since Kepler expounded his second law of planetary motion. Newton derived a unique geometric proof, and went on to show that the attractive force of the Sun's gravity was the cause of all of Kepler's laws.
     

During the first interval of time, an object is in motion from point  A to point  B. Undisturbed, it would continue to point  c during the second interval. When the object arrives at  B, it receives an impulse directed toward point  S. The impulse gives it a small added velocity toward  S, such that if this were its only velocity, it would move from  B to  V during the second interval. By the rules of velocity composition, these two velocities add, and point  C is found by construction of parallelogram  BcCV. Thus the object's path is deflected by the impulse so that it arrives at point  C at the end of the second interval. Because the triangles  SBc and  SBC have the same base  SB and the same height  Bc or  VC, they have the same area. By symmetry, triangle  SBc also has the same area as triangle  SAB, therefore the object has swept out equal areas  SAB and  SBC in equal times.
     

At point  C, the object receives another impulse toward  S, again deflecting its path during the third interval from  d to  D. Thus it continues to  E and beyond, the triangles  SAB,  SBc,  SBC,  SCd,  SCD,  SDe,  SDE all having the same area. Allowing the time intervals to become ever smaller, the path  ABCDE approaches indefinitely close to a continuous curve.
     

Note that because this derivation is geometric, and no specific force is applied, it proves a more general law than Kepler's second law of planetary motion. It shows that the Law of Areas applies to any central force, attractive or repulsive, continuous or non-continuous, or zero.
     

 

 
 
  

  Conservation of angular momentum in the law of areas
 
 

 
The proportionality of angular momentum to the area swept out by a moving object can be understood by realizing that the bases of the triangles, that is, the lines from  S to the object, are equivalent to the radius , and that the heights of the triangles are proportional to the perpendicular component of velocity . Hence, if the area swept per unit time is constant, then by the triangular area formula , the product  and therefore the product  are constant: if  and the base length are decreased,  and height must increase proportionally. Mass is constant, therefore angular momentum  is conserved by this exchange of distance and velocity.
     

In the case of triangle  SBC, area is equal to ( SB)( VC). Wherever  C is eventually located due to the impulse applied at  B, the product ( SB)( VC), and therefore  remain constant. Similarly so for each of the triangles.
     

Another areal proof of conservation of angular momentum for any central force uses Mamikon's sweeping tangents theorem.
 (2025). 9781470443351, MAA Press. ISBN 9781470443351
     

 

 
 
  

  After Newton
 
 

 
     Leonhard Euler, Daniel Bernoulli, and Patrick d'Arcy all understood angular momentum in terms of conservation of areal velocity, a result of their analysis of Kepler's second law of planetary motion. It is unlikely that they realized the implications for ordinary rotating matter.see  for an excellent and detailed summary of the concept of angular momentum through history.
     

In 1736 Euler, like Newton, touched on some of the equations of angular momentum in his  Mechanica without further developing them.
     
     

Bernoulli wrote in a 1744 letter of a "moment of rotational motion", possibly the first conception of angular momentum as we now understand it.
     

In 1799, Pierre-Simon Laplace first realized that a fixed plane was associated with rotation—his  invariable plane.
     

     Louis Poinsot in 1803 began representing rotations as a line segment perpendicular to the rotation, and elaborated on the "conservation of moments".
     

In 1852 Léon Foucault used a gyroscope in an experiment to display the Earth's rotation.
     

William J. M. Rankine's 1858  Manual of Applied Mechanics defined angular momentum in the modern sense for the first time:
... a line whose length is proportional to the magnitude of the angular momentum, and whose direction is perpendicular to the plane of motion of the body and of the fixed point, and such, that when the motion of the body is viewed from the extremity of the line, the radius-vector of the body seems to have right-handed rotation.
In an 1872 edition of the same book, Rankine stated that "The term  angular momentum was introduced by Mr. Hayward," probably referring to R.B. Hayward's article  On a Direct Method of estimating Velocities, Accelerations, and all similar Quantities with respect to Axes moveable in any manner in Space with Applications, which was introduced in 1856, and published in 1864. Rankine was mistaken, as numerous publications feature the term starting in the late 18th to early 19th centuries.see, for instance, ;  However, Hayward's article apparently was the first use of the term and the concept seen by much of the English-speaking world. Before this, angular momentum was typically referred to as "momentum of rotation" in English.see, for instance, 
     

 

 
 
  

  See also
 
 

 
 
 
 
 
 

 
 
  

  Further reading
 
 

 
 

  

   
 (2025). 9780471569527, John Wiley & Sons. ISBN 9780471569527
  
  

   
 (2025). 9780521092098, Cambridge University Press. ISBN 9780521092098
  
  

   
A. R. Edmonds (2025). 9780691079127, Princeton University Press. . ISBN 9780691079127
  
  

.
  
  

   
John David Jackson (1998). 9780471309321, John Wiley & Sons. . ISBN 9780471309321
  
  

   
 (2025). 9780534408428, Brooks/Cole. . ISBN 9780534408428
  
  

   
William J. Thompson (1994). 9780471552642, Wiley. ISBN 9780471552642
  
  

   
Paul Tipler (2025). 9780716708094, W. H. Freeman. ISBN 9780716708094
  
 

 

 
 
  

  External links
 
 

 
 
 

  

      "What Do a Submarine, a Rocket and a Football Have in Common? Why the prolate spheroid is the shape for success" (  Scientific American, November 8, 2010)
  
  

      Conservation of Angular Momentum  – a chapter from an online textbook
  
  

      Angular Momentum in a Collision Process – derivation of the three-dimensional case
  
  

     
   
  target="_blank" rel="nofollow"> Angular Momentum and Rolling Motion – more momentum theory
  
 

 
 
 
 
 
Categories: Mechanical Quantities, Rotation, Conservation Laws, Moment, Angular Momentum