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{{short description|Physical quantity}}

factoids
IÏ‰ {{=}} r Ã— p}}}}{{Classical mechanics|cTopic=Fundamental concepts}}In physics, angular momentum (rarely, moment of momentum or rotational momentum) is the rotational equivalent of linear momentum. It is an important quantity in physics because it is a conserved quantityâ€”the total angular momentum of a closed system remains constant.In three dimensions, the angular momentum for a point particle is a pseudovector r Ã— p, the cross product of the particle's position vector r (relative to some origin) and its momentum vector; the latter is p = mv in Newtonian mechanics. This definition can be applied to each point in continua like solids or fluids, or physical fields. Unlike momentum, angular momentum does depend on where the origin is chosen, since the particle's position is measured from it.Just like for angular velocity, there are two special types of angular momentum: the spin angular momentum and the orbital angular momentum. The spin angular momentum of an object is defined as the angular momentum about its centre of mass coordinate. The orbital angular momentum of an object about a chosen origin is defined as the angular momentum of the centre of mass about the origin. The total angular momentum of an object is the sum of the spin and orbital angular momenta. The orbital angular momentum vector of a particle is always parallel and directly proportional to the orbital angular velocity vector Ï‰ of the particle, where the constant of proportionality depends on both the mass of the particle and its distance from origin. However, the spin angular momentum of the object is proportional but not always parallel to the spin angular velocity Î©, making the constant of proportionality a second-rank tensor rather than a scalar.Angular momentum is additive; the total angular momentum of any composite system is the (pseudo) vector sum of the angular momenta of its constituent parts. For a continuous rigid body, the total angular momentum is the volume integral of angular momentum density (i.e. angular momentum per unit volume in the limit as volume shrinks to zero) over the entire body.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; in other words, the sum of all internal torques of any system is always 0 (this is the rotational analogue of Newton's Third Law). 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 conservation of angular momentum helps explain many observed phenomena, for example the increase in rotational speed of a spinning figure skater as the skater's arms are contracted, the high rotational rates of neutron stars, the Coriolis effect, and the precession of gyroscopes. In general, conservation does limit the possible motion of a system, but does not uniquely determine what the exact motion is.In quantum mechanics, angular momentum (like other quantities) is expressed as an operator, and its one-dimensional projections have quantized eigenvalues. 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, it turns out that the notion of a quantum particle literally "spinning" about an axis does not exist. Nevertheless, elementary particles still possess a spin angular momentum, but this angular momentum does not correspond to spinning motion in the ordinary sense.BOOK, Understanding the Properties of Matter, 2nd, illustrated, revised, Michael, de Podesta, CRC Press, 2002, 978-0-415-25788-6, 29,weblink Extract of page 29

## In classical mechanics

### Definition

#### Orbital angular momentum in two dimensions

File:Ang mom 2d.png|right|thumb|Velocity of the particle m with respect to the origin O can be resolved into components parallel to (vâˆ¥) and perpendicular to (vâŠ¥) the radius vector r. The angular momentum of m is proportional to the perpendicular componentperpendicular componentAngular momentum is a vector quantity (more precisely, a pseudovector) that represents the product of a body's rotational inertia and rotational velocity (in radians/sec) about a particular axis. However, if the particle's trajectory lies in a single plane, it is sufficient to discard the vector nature of angular momentum, and treat it as a scalar (more precisely, a pseudoscalar).BOOK, Wilson, E. B., Linear Momentum, Kinetic Energy and Angular Momentum, The American Mathematical Monthly, XXII, Ginn and Co., Boston, in cooperation with University of Chicago, et al., 1915page= 190, Google books, Angular momentum can be considered a rotational analog of linear momentum. Thus, where linear momentum p is proportional to mass m and linear speed {{nowrap|v,}}
p = mv,
angular momentum L is proportional to moment of inertia I and angular speed omega measured in radians per second.BOOK, Worthington, Arthur M., Dynamics of Rotation, Longmans, Green and Co., London, 1906page= 21, Google books,
L = Iomega.
Unlike mass, which depends only on amount of matter, moment of inertia is also dependent on the position of the axis of rotation and the shape of the matter. Unlike linear speed, which does not depend upon the choice of origin, angular velocity is always measured with respect to a fixed origin. Therefore, strictly speaking, L should be referred to as the angular momentum relative to that center.BOOK, Taylor, John R., Classical Mechanics, University Science Books, Mill Valley, CA, 2005, 978-1-891389-22-1, 90, Because I = r^2m for a single particle and omega = frac{v}{r} for circular motion, angular momentum can be expanded, L = r^2 m cdot frac{v}{r}, and reduced to,
L = rmv,
the product of the radius of rotation r and the linear momentum of the particle p = mv, where v in this case is the equivalent linear (tangential) speed at the radius (= romega).This simple analysis can also apply to non-circular motion if only the component of the motion which is perpendicular to the radius vector is considered. In that case,
L = rmv_perp,
where v_perp = vsin(theta) is the perpendicular component of the motion. Expanding, L = rmvsin(theta), rearranging, L = rsin(theta)mv, and reducing, angular momentum can also be expressed,
L = r_perp mv,
where r_perp = rsin(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, {{math|(length of moment arm)Ã—(linear momentum)}} to which the term moment of momentum refers.BOOK, Dadourian, H. M., Analytical Mechanics for Students of Physics and Engineering, D. Van Nostrand Company, New York, 1913page= 266, Google books,

#### Scalarâ€”angular momentum from Lagrangian mechanics

Another approach is to define angular momentum as the conjugate momentum (also called canonical momentum) of the angular coordinate phi expressed in the Lagrangian of the mechanical system. Consider a mechanical system with a mass m constrained to move in a circle of radius a in the absence of any external force field. The kinetic energy of the system is
T = frac{1}{2}ma^2 omega^2 = frac{1}{2}ma^2 dot{phi}^2.
And the potential energy is
U = 0.
Then the Lagrangian is
mathcal{L}left(phi, dot{phi}right) = T - U = frac{1}{2}ma^2 dot{phi}^2.
The generalized momentum "canonically conjugate to" the coordinate phi is defined by
p_phi = frac{partial mathcal{L}}{partial dot{phi}} = ma^2 dot{phi} = Iomega = L.

#### Orbital angular momentum in three dimensions

File:Torque animation.gif|frame|right|Relationship between force (F), torque (Ï„), momentum (p), and angular momentum (L) vectors in a rotating system. r is the position vector.]]To completely define orbital angular momentum in three dimensions, it is required to know the rate at which the position vector sweeps out angle, the direction perpendicular to the instantaneous plane of angular displacement, and the mass involved, as well as how this mass is distributed in spaceBOOK, Watson, W., General Physics, Longmans, Green and Co., New York, 1912page= 33, Google books, . By retaining this vector nature of angular momentum, the general nature of the equations is also retained, and can describe any sort of three-dimensional motion about the center of rotation â€“ circular, linear, or otherwise. In vector notation, the orbital angular momentum of a point particle in motion about the origin can be expressed as:
mathbf{L} = Iboldsymbol{omega},
where
I = r^2m is the moment of inertia for a point mass, boldsymbol{omega}=frac{mathbf{r}timesmathbf{v}}{r^2} is the orbital angular velocity in radians/sec (units 1/sec) of the particle about the origin, mathbf{r} is the position vector of the particle relative to the origin, r=leftvertmathbf{r}rightvert, 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 algebra, rearranged:
begin{align}
mathbf{L} &= left(r^2mright)left(frac{mathbf{r}timesmathbf{v}}{r^2}right)
&= mleft(mathbf{r}timesmathbf{v}right)
&= mathbf{r}times mmathbf{v}
&= mathbf{r}timesmathbf{p},
end{align}which is the cross product of the position vector mathbf{r} and the linear momentum mathbf{p} = mmathbf{v} of the particle. By the definition of the cross product, the mathbf{L} vector is perpendicular to both mathbf{r} and mathbf{p}. It is directed perpendicular to the plane of angular displacement, as indicated by the right-hand rule â€“ so that the angular velocity is seen as counter-clockwise from the head of the vector. Conversely, the mathbf{L} vector defines the plane in which mathbf{r} and mathbf{p} lie.By defining a unit vector mathbf{hat{u}} perpendicular to the plane of angular displacement, a scalar angular speed omega results, where
omegamathbf{hat{u}} = boldsymbol{omega}, and omega = frac{v_perp}{r}, where v_perp is the perpendicular component of the motion, as above.
The two-dimensional scalar equations of the previous section can thus be given direction:
begin{align}
mathbf{L} &= Iboldsymbol{omega}
&= Iomegamathbf{hat{u}}
&= left(r^2mright)omegamathbf{hat{u}}
&= rmv_perp mathbf{hat{u}}
&= r_perp mvmathbf{hat{u}},
end{align}and mathbf{L} = rmvmathbf{hat{u}} for circular motion, where all of the motion is perpendicular to the radius r.

### Discussion

Angular momentum can be described as the rotational analog of linear momentum. Like linear momentum it involves elements of mass and displacement. Unlike linear momentum it also involves elements of position and shape.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 pointâ€”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 motionâ€”a combination of its inertia and its displacement. Inertia is measured by its mass, and displacement by its velocity. Their product,
begin{align}
(text{amount of inertia}) times (text{amount of displacement})&=text{amount of (inertiaâ‹…displacement)}text{mass} times text{velocity} &= text{momentum}m times v &= pend{align}is the matter's momentum.BOOK,weblink Physics: Advanced Course, Barker, George F., 66, Henry Holt and Company, New York, 4thvia= Google Books, 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,
begin{align}
(text{moment arm}) times (text{amount of inertia}) times (text{amount of displacement})&=text{moment of (inertiaâ‹…displacement)}text{length} times text{mass} times text{velocity} &= text{moment of momentum}r times m times v &= Lend{align}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 omega versus the point: v=romega, another moment. Hence, angular momentum contains a double moment: L=rmromega. Simplifying slightly, L=r^2momega, 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.BOOK,weblink Physics: Advanced Course, Barker, George F., 67â€“68, Henry Holt and Company, New York, 4thvia= Google Books, File:Moment of inertia examples.gif|thumb|right|Moment of inertia (shown here), and therefore angular momentum, is different for every possible configuration of mass and axis of 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.BOOK, Oberg, Erik, Machinery's Handbook, 26th, Industrial Press, Inc., New York, 978-0-8311-2625-4display-authors=etal, 143, Therefore, the total moment of inertia, and the angular momentum, is a complex function of the configuration of the matter about the center of rotation and the orientation of the rotation for the various bits.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.BOOK, Watson, W., General Physics, Longmans, Green and Co., New York, 1912page= 34, Google books, It reaches a minimum when the axis passes through the center of mass.BOOK, Kent, William, The Mechanical Engineers' Pocket Book, 9th, John Wiley and Sons, Inc., New York, 1916page= 517, Google books, 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,BOOK, Oberg, Erik, Machinery's Handbook, 26th, Industrial Press, Inc., New York, 978-0-8311-2625-4display-authors=etal, 146,
I=k^2m, where k is the radius of gyration, the distance from the axis at which the entire mass m may be considered as concentrated.
Similarly, for a point mass 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,
sum_i I_i = sum_i r_i^2m_i
Angular momentum's dependence on position and shape is reflected in its units versus linear momentum: kgâ‹…m2/s, Nâ‹…mâ‹…s, or Jâ‹…s for angular momentum versus kgâ‹…m/s or Nâ‹…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 (Dimensional_analysis#Siano's_extension:_orientational_analysis|orientational analysis) which treats radians as a base unit, but this is outside the scope of the International system of units). Angular momentum's units can be interpreted as torqueâ‹…time or as energyâ‹…time per angle. An object with angular momentum of {{nowrap|L Nâ‹…mâ‹…s}} can be reduced to zero rotation (all of the rotational energy can be transferred out of it) by an angular impulse of {{nowrap|L Nâ‹…mâ‹…s}}BOOK, Oberg, Erik, Machinery's Handbook, 26th, Industrial Press, Inc., New York, 978-0-8311-2625-4display-authors=etalL Nâ‹…m}} for one second, or energy of {{nowrap|L J}} for one second.BOOK, Kent, William, The Mechanical Engineers' Pocket Book, 9th, John Wiley and Sons, Inc., New York, 1916page= 527, Google books, The plane perpendicular to the axis of angular momentum and passing through the center of massBOOK, Battin, Richard H., An Introduction to the Mathematics and Methods of Astrodynamics, Revised Edition, American Institute of Aeronautics and Astronautics, Inc., 978-1-56347-342-5, 1999, , p. 97 is sometimes called the invariable plane, because the direction of the axis remains fixed if only the interactions of the bodies within the system, free from outside influences, are considered.BOOK, Rankine, W. J. M., A Manual of Applied Mechanics, 6th, Charles Griffin and Company, London,weblinkpage= 507Invariable plane>invariable plane of the Solar System.

#### Angular momentum and torque

{{See also|Torque#Definition and relation to angular momentum}}Newton's second law of motion can be expressed mathematically,
mathbf{F} = mmathbf{a},
or force = mass Ã— acceleration. The rotational equivalent for point particles may be derived as follows:
mathbf{L} = Iboldsymbol{omega}
which means that the torque (i.e. the time derivative of the angular momentum) is
boldsymbol{tau} = frac{dI}{dt}boldsymbol{omega} + Ifrac{dboldsymbol{omega}}{dt}.
Because the moment of inertia is mr^2, it follows that frac{dI}{dt} = 2mrfrac{dr}{dt} = 2rp_{||}, and frac{dmathbf{L}}{dt} = Ifrac{dboldsymbol{omega}}{dt} + 2rp_{||}boldsymbol{omega}, which, reduces to
boldsymbol{tau} = Iboldsymbol{alpha} + 2rp_{||}boldsymbol{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.

### Angular momentum in orbital mechanics

In astrodynamics and celestial mechanics, a massless (or per unit mass) angular momentum is definedBOOK, Battin, Richard H., An Introduction to the Mathematics and Methods of Astrodynamics, Revised Edition, American Institute of Aeronautics and Astronautics, Inc., 978-1-56347-342-5, 1999, 115,
mathbf{h} = mathbf{r} times mathbf{v},
called specific angular momentum. Note that mathbf{L} = mmathbf{h}. Mass is often unimportant in orbital mechanics calculations, because motion is defined by gravity. The primary body of the system is often so much larger than any bodies in motion about it that the smaller bodies have a negligible gravitational effect on it; it is, in effect, stationary. All bodies are apparently attracted by its gravity in the same way, regardless of mass, and therefore all move approximately the same way under the same conditions.

### Solid bodies

For a continuous mass distribution with density function Ï(r), a differential volume element dV with position vector r within the mass has a mass element dm = Ï(r)dV. Therefore, the infinitesimal angular momentum of this element is:
dmathbf{L} = mathbf{r}times dm mathbf{v} = mathbf{r}times rho(mathbf{r}) dV mathbf{v} = dV mathbf{r}times rho(mathbf{r}) mathbf{v}
and integrating this differential over the volume of the entire mass gives its total angular momentum:
mathbf{L}=int_V dV mathbf{r}times rho(mathbf{r}) mathbf{v}
In the derivation which follows, integrals similar to this can replace the sums for the case of continuous mass.

### Collection of particles

#### Center of mass

(File:Ang mom vector diagram.png|thumb|The angular momentum of the particles i is the sum of the cross products R Ã— MV + Î£ri Ã— mivi.)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, mathbf{R}_i is the position vector of particle i vs the origin, mathbf{V}_i is the velocity of particle i vs the origin, mathbf{R} is the position vector of the center of mass vs the origin, mathbf{V} is the velocity of the center of mass vs the origin, mathbf{r}_i is the position vector of particle i vs the center of mass, mathbf{v}_i is the velocity of particle i vs the center of mass,
The total mass of the particles is simply their sum,
M=sum_i m_i.
The position vector of the center of mass is defined by,BOOK
, Wilson
, E. B.
, Linear Momentum, Kinetic Energy and Angular Momentum
, The American Mathematical Monthly
, XXII
, Ginn and Co., Boston, in cooperation with University of Chicago, et al.
, 1915
,
Mmathbf{R}=sum_i m_i mathbf{R}_i.
By inspection,
mathbf{R}_i = mathbf{R} + mathbf{r}_i and mathbf{V}_i = mathbf{V} + mathbf{v}_i.
The total angular momentum of the collection of particles is the sum of the angular momentum of each particle,{{Equation box 1 |indent=: |equation =mathbf{L} = sum_i left( mathbf{R}_i times m_i mathbf{V}_i right)     ({{EquationRef|1}}) }}Expanding mathbf{R}_i,
begin{align}
mathbf{L} &= sum_i left[left(mathbf{R} + mathbf{r}_iright) times m_imathbf{V}_i right]
&= sum_i left[ mathbf{R} times m_imathbf{V}_i + mathbf{r}_i times m_imathbf{V}_i right]
end{align}Expanding mathbf{V}_i,
begin{align}
mathbf{L} &= sum_i left[ mathbf{R} times m_ileft(mathbf{V} + mathbf{v}_iright) + mathbf{r}_i times m_i(mathbf{V} + mathbf{v}_i) right ]
&= sum_i left[ mathbf{R} times m_imathbf{V} + mathbf{R} times m_imathbf{v}_i + mathbf{r}_i times m_imathbf{V} + mathbf{r}_i times m_imathbf{v}_i right]
&= sum_i mathbf{R} times m_imathbf{V} + sum_i mathbf{R} times m_imathbf{v}_i + sum_i mathbf{r}_i times m_imathbf{V} + sum_i mathbf{r}_i times m_imathbf{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 sum_i m_imathbf{r}_i = mathbf{0}begin{align}
mathbf{r}_i &= mathbf{R}_i - mathbf{R}
m_imathbf{r}_i &= m_ileft(mathbf{R}_i - mathbf{R}right)
sum_i m_imathbf{r}_i &= sum_i m_ileft(mathbf{R}_i - mathbf{R}right)
&= sum_i (m_imathbf{R}_i - m_imathbf{R})
&= sum_i m_imathbf{R}_i - sum_i m_imathbf{R}
&= sum_i m_imathbf{R}_i - left( sum_i m_i right) mathbf{R}
&= sum_i m_imathbf{R}_i - Mmathbf{R}
end{align}which, by the definition of the center of mass, is mathbf{0}, and similarly for sum_i m_imathbf{v}_i.
sum_i m_imathbf{r}_i = mathbf{0} and sum_i m_imathbf{v}_i = mathbf{0},
therefore the second and third terms vanish,
mathbf{L} = sum_i mathbf{R} times m_imathbf{V} + sum_i mathbf{r}_i times m_imathbf{v}_i .
The first term can be rearranged,
sum_i mathbf{R} times m_imathbf{V} = mathbf{R} times sum_i m_imathbf{V} = mathbf{R} times Mmathbf{V},
and total angular momentum for the collection of particles is finally,BOOK
, Wilson
, E. B.
, Linear Momentum, Kinetic Energy and Angular Momentum
, The American Mathematical Monthly
, XXII
, Ginn and Co., Boston, in cooperation with University of Chicago, et al.
, 1915
, {{Equation box 1 |indent=: |equation =mathbf{L} = mathbf{R} times Mmathbf{V} + sum_i mathbf{r}_i times m_imathbf{v}_i     ({{EquationRef|2}}) }}The first term is the angular momentum of the center of mass relative to the origin. Similar to Single particle, 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 Fixed center of mass, below. The result is generalâ€”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 ({{EquationNote|2}}) by vector identities, multiplying both terms by "one", and grouping appropriately,
begin{align}
mathbf{L} &= M(mathbf{R} times mathbf{V}) + sum_i left[m_ileft(mathbf{r}_i times mathbf{v}_iright)right],
&= frac{R^2}{R^2}Mleft(mathbf{R} times mathbf{V}right) + sum_i left[ frac{r_i^2}{r_i^2}m_ileft(mathbf{r}_i times mathbf{v}_iright)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 boldsymbol{omega},{{Equation box 1 |indent=: |equation =mathbf{L} = I_Rboldsymbol{omega}_R + sum_i I_iboldsymbol{omega}_i.     ({{EquationRef|3}}) }}

#### Single particle

In the case of a single particle moving about the arbitrary origin,
mathbf{r}_i = mathbf{v}_i = mathbf{0}, mathbf{r} = mathbf{R}, mathbf{v} = mathbf{V}, m = M, sum_i mathbf{r}_i times m_imathbf{v}_i = mathbf{0}, sum_i I_iboldsymbol{omega}_i = mathbf{0}, and equations ({{EquationNote|2}}) and ({{EquationNote|3}}) for total angular momentum reduce to, mathbf{L} = mathbf{R} times mmathbf{V} = I_Rboldsymbol{omega}_R.

#### Fixed center of mass

For the case of the center of mass fixed in space with respect to the origin,
mathbf{V} = mathbf{0}, mathbf{R} times Mmathbf{V} = mathbf{0}, I_Rboldsymbol{omega}_R = mathbf{0}, and equations ({{EquationNote|2}}) and ({{EquationNote|3}}) for total angular momentum reduce to, mathbf{L} = sum_i mathbf{r}_i times m_imathbf{v}_i = sum_i I_iboldsymbol{omega}_i.
{{clear}}

## Angular momentum (modern definition)

File:Angular momentum bivector and pseudovector.svg|275px|thumb|The 3-angular momentum as a bivector (plane element) and axial vectoraxial vectorIn modern (20th century) theoretical physics, angular momentum (not including any intrinsic angular momentum â€“ 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 not conserved for general curved spacetimes, unless it happens to be asymptotically rotationally invariant.{{citation needed|date=May 2013}}In classical mechanics, the angular momentum of a particle can be reinterpreted as a plane element:
mathbf{L} = mathbf{r} wedge mathbf{p} ,,
in which the exterior product âˆ§ replaces the cross product Ã— (these products have similar characteristics but are nonequivalent). This has the advantage of a clearer geometric interpretation as a plane element, defined from the x and p vectors, and the expression is true in any number of dimensions (two or higher). In Cartesian coordinates:
begin{align}
mathbf{L} &= left(xp_y - yp_xright)mathbf{e}_x wedge mathbf{e}_y + left(yp_z - zp_yright)mathbf{e}_y wedge mathbf{e}_z + left(zp_x - xp_zright)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,.
The angular velocity can also be defined as an antisymmetric second order tensor, with components Ï‰ij. The relation between the two antisymmetric 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. {{ISBN|978-0-486-63612-2}}.
L_{ij} = I_{ijkell} omega_{kell} ,.
Again, this equation in L and Ï‰ as tensors is true in any number of dimensions. This equation also appears in the geometric algebra formalism, in which L and Ï‰ 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 antisymmetric tensor of second order:
M_{alphabeta} = X_alpha P_beta - X_beta P_alpha
in the language of four-vectors, namely the four position X and the four momentum P, and absorbs the above L together with the motion of the centre of mass of the particle.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 centre of mass is for the system.

## In quantum mechanics

Angular momentum in quantum mechanics differs in many profound respects from angular momentum in classical mechanics. In relativistic quantum mechanics, it differs even more, in which the above relativistic definition becomes a tensorial operator.

### Spin, orbital, and total angular momentum

File:Classical angular momentum.svg|350px|thumb|Angular momenta of a classical object.{{ubl| Left: "spin" angular momentum S is really orbital angular momentum of the object at every point.| Right: extrinsic orbital angular momentum L about an axis.Top: the moment of inertia tensor I and angular velocity Ï‰ (L is not always parallel to Ï‰).FEYNMAN'S LECTURES ON PHYSICS (VOLUME 2)AUTHOR2=R.B. LEIGHTON PUBLISHER=ADDISONâ€“WESLEYPAGES=31â€“7, 978-0-201-02117-2, Bottom: momentum p and its radial position r from the axis. The total angular momentum (spin plus orbital) is J. For a quantum particle the interpretations are different; spin (physics)>particle spinspin (physics)>particle spinThe classical definition of angular momentum as mathbf{L} = mathbf{r}timesmathbf{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.{{harvnb|Hall|2013}} 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. Almost all elementary particles have spin. 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, for example electrons have "spin 1/2" (this actually means "spin Ä§/2") while photons have "spin 1" (this actually means "spin Ä§").Finally, there is total angular momentum J, which combines both the spin and orbital angular momentum of all particles and fields. (For one particle, J = L + S.) Conservation of angular momentum applies to J, but not to L or S; for example, the spinâ€“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 fractional values.JOURNAL, 10.1126/sciadv.1501748, 28861467, There are many ways to spin a photon: Half-quantization of a total optical angular momentum, Science Advances, 2, 4, e1501748, 2016, Ballantine, K. E., Donegan, J. F., Eastham, P. R., 2016SciA....2E1748B, 5565928,

### Quantization

In quantum mechanics, angular momentum is quantized â€“ that is, it cannot vary continuously, but only in "quantum leaps" between certain allowed values. For any system, the following restrictions on measurement results apply, where hbar is the reduced Planck constant and hat{n} is any Euclidean vector such as x, y, or z:{| class="wikitable"If you measurement in quantum mechanics>measureâ€¦| The result can be...| L_hat{n}| ldots, -2hbar, -hbar, 0, hbar, 2hbar, ldots| S_hat{n} or J_hat{n}| ldots, -frac{3}{2}hbar, -hbar, -frac{1}{2}hbar, 0, frac{1}{2}hbar, hbar, frac{3}{2}hbar, ldots| begin{align}
&L^2
={} &L_x^2 + L_y^2 + L_z^2
end{align}| left[hbar^2 n(n + 1)right], where n = 0, 1, 2, ldots| S^2 or J^2| left[hbar^2 n(n + 1)right], where n = 0, frac{1}{2}, 1, frac{3}{2}, ldotsFile:Circular Standing Wave.gif|thumb|right|In this standing wave on a circular string, the circle is broken into exactly 8 wavelengthwavelength(There are additional restrictions as well, see angular momentum operator for details.)The reduced Planck constant hbar is tiny by everyday standards, about 10âˆ’34 J s, 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 electron shells and subshells in chemistry is significantly affected by the quantization of angular momentum.Quantization of angular momentum was first postulated by Niels Bohr in his Bohr model of the atom and was later predicted by Erwin SchrÃ¶dinger in his SchrÃ¶dinger equation.

### Uncertainty

In the definition mathbf{L}=mathbf{r}timesmathbf{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 commute, for example L_xL_y 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: mathbf{L}=mathbf{r}timesmathbf{p}, but total angular momentum J is defined in a different, more basic way: J is defined as the "generator of rotations".WEB,weblink Robert Grayson Littlejohn, Lecture notes on rotations in quantum mechanics, Robert, Littlejohn, 13 Jan 2012, Physics 221B Spring 2011, 2011, More specifically, J is defined so that the operator
R(hat{n},phi) equiv expleft(-frac{i}{hbar}phi, mathbf{J}cdot hat{mathbf{n}}right)
is the rotation operator that takes any system and rotates it by angle phi about the axis hat{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 Lie algebras and Lie groups 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.

## In electrodynamics

{{See also|Momentum#Particle in field|l1=Momentum (Particle in field)}}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 Ã— 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)
mathbf{p} = mmathbf{v} = mathbf{P} - e 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
mathbf{K}= mathbf{r} times ( mathbf{P} - emathbf{A} )
The interplay with quantum mechanics is discussed further in the article on canonical commutation relations.

## In optics

In classical Maxwell electrodynamics the Poynting vectoris a linear momentum density of electromagnetic field.JOURNAL, 10.1088/0953-4075/41/10/101001, Angular momentum of photons and phase conjugation, Journal of Physics B: Atomic, Molecular and Optical Physics, 41, 10, 101001, 2008, Okulov, A Yu, 0801.2675, 2008JPhB...41j1001O,
mathbf{S}(mathbf{r}, t) = epsilon_0 c^2 mathbf{E}(mathbf{r}, t) times mathbf{B}(mathbf{r}, t).
The angular momentum density vector mathbf{L}(mathbf{r}, t) is given by a vector productas in classical mechanics:JOURNAL,weblink A.Y., Okulov, Optical and Sound Helical structures in a Mandelstam â€“ Brillouin mirror, JETP Letters, 88, 8, 561â€“566, 2008, Russian, 10.1134/s0021364008200046, 2008JETPL..88..487O,
mathbf{L}(mathbf{r}, t) = mathbf{r} times mathbf{S}(mathbf{r}, t).
The above identities are valid locally, i.e. in each space point mathbf{r} in a given moment t.

## History

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.BOOK, Newton, Isaac, The Mathematical Principles of Natural Philosophy, 322, H. D. Symonds, London, Andrew Motte, translator, 1803chapter=Axioms; or Laws of Motion, Law I, Google books,
He did not further investigate angular momentum directly in the Principia,
From such kind of reflexions also sometimes arise the circular motions of bodies about their own centres. 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.

### The Law of Areas

#### Newton's derivation

(File:Newton area law derivation.gif|frame|right|Newton's derivation of the area law using geometric means.)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 {{math|r}}, and that the heights of the triangles are proportional to the perpendicular component of velocity {{math|vâŠ¥}}. Hence, if the area swept per unit time is constant, then by the triangular area formula {{math|{{sfrac|1|2}}(base)(height)}}, the product {{math|(base)(height)}} and therefore the product {{math|rvâŠ¥}} are constant: if {{math|r}} and the base length are decreased, {{math|vâŠ¥}} and height must increase proportionally. Mass is constant, therefore angular momentum {{math|rmvâŠ¥}} is conserved by this exchange of distance and velocity.In the case of triangle SBC, area is equal to {{sfrac|1|2}}(SB)(VC). Wherever C is eventually located due to the impulse applied at B, the product (SB)(VC), and therefore {{math|rmvâŠ¥}} remain constant. Similarly so for each of the triangles.

### 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 WEB,weblink Borrelli, Arianna, 2011, Angular momentum between physics and mathematics, 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.WEB,weblink Euler : Mechanica Vol. 1, Bruce, Ian, 2008, Bernoulli wrote in a 1744 letter of a "moment of rotational motion", possibly the first conception of angular momentum as we now understand it.WEB,weblinkwebsite=The Euler Archive, In 1799, Pierre-Simon Laplace first realized that a fixed plane was associated with rotation â€” 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 LÃ©on 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,"BOOK, Rankine, W. J. M., A Manual of Applied Mechanics, 6th, Charles Griffin and Company, London,weblinkpage= 506, Google books, 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,JOURNAL, Hayward, Robert B., On a Direct Method of estimating Velocities, Accelerations, and all similar Quantities with respect to Axes moveable in any manner in Space with Applications, Transactions of the Cambridge Philosophical Society, 1864, 10, 1bibcode=1864TCaPS..10....1H, 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, JOURNAL, Gompertz, Benjamin, On Pendulums vibrating between Cheeks, The Journal of Science and the Arts, 1818, III, V, 17via=Google books, ; BOOK, Herapath, John, Mathematical Physics, Whittaker and Co., London, 1847, 56via=Google books, 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, JOURNAL, Landen, John, Of the Rotatory Motion of a Body of any Form whatever, Philosophical Transactions, 1785, LXXV, I, 311â€“332url=http://rstl.royalsocietypublishing.org/content/75/311.full.pdf+html,

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## References

• BOOK, Cohen-Tannoudji, Claude, Diu, Bernard, LaloÃ«, Franck, 2006, Quantum Mechanics, John Wiley & Sons, 978-0-471-56952-7, 2 volume set,
• BOOK, E. U., Condon, G. H., Shortley, 1935, The Theory of Atomic Spectra,weblink Cambridge University Press, 978-0-521-09209-8, Especially Chapter 3,
• BOOK, Edmonds, A. R., 1957, Angular Momentum in Quantum Mechanics, Princeton University Press, 978-0-691-07912-7,
• {{citation|first=Brian C.|last=Hall|title=Quantum Theory for Mathematicians|series=Graduate Texts in Mathematics|volume=267 |publisher=Springer|year=2013|isbn=978-0-387-40122-5}}.
• BOOK, Jackson, John David, 1998, Classical Electrodynamics, 3rd, John Wiley & Sons, 978-0-471-30932-1,
• BOOK, Serway, Raymond A., Jewett, John W., Physics for Scientists and Engineers, 6th, Brooks/Cole, 2004, 978-0-534-40842-8,weblink
• BOOK, Thompson, William J., 1994, Angular Momentum: An Illustrated Guide to Rotational Symmetries for Physical Systems, Wiley, 978-0-471-55264-2,
• BOOK, Tipler, Paul, Physics for Scientists and Engineers: Mechanics, Oscillations and Waves, Thermodynamics, 5th, W. H. Freeman, 2004, 978-0-7167-0809-4,
• BOOK, Richard Feynman, Feynman R, Robert B. Leighton, Leighton R, and Matthew Sands, Sands M.,weblink 19â€“4 Rotational kinetic energy, The Feynman Lectures on Physics, online, The Feynman Lectures Website, September 2013,

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