# PlanetPhysics/Angular Momentum of a System of Particles

The angular momentum $\mathbf {p} =m\left[\mathbf {r} {\dot {\mathbf {r} }}\right]$ of a particle is equal to $2m$ times the `areal' velocity of the radius vector. The vector sum $\sum m_{k}\left[\mathbf {r} _{k}{\dot {\mathbf {r} }}_{k}\right]$, called the total angular momentum of a system of particles , is one of the most important quantities in mechanics. We shall now investigate the properties of this quantity. For this purpose we multiply the equations of motion for a particle vectorially by $\mathbf {r} _{k}$ and sum over all the particles. The result is
   $\sum _{k}m_{k}\left[\mathbf {r} _{k}{\frac {d^{2}\mathbf {r} _{k}}{dt^{2}}}\right]=\sum _{k}\left[\mathbf {r} _{k}\mathbf {F} _{k}\right]+\sum _{k}\sum _{j}\left[\mathbf {r} _{k}\mathbf {F} _{jk}\right]$
The left member represents the time derivative of the quantity $\sum m_{k}\left[\mathbf {r} _{k}{\dot {\mathbf {r} }}_{k}\right]$, that is, of the total angular momentum. Further, the vector product of the radius vector to the point of application of a force by the force vector is called the moment of the force $\mathbf {F} _{k}$. We denote it by $\mathbf {M} _{k}$. The magnitude of $\mathbf {M} _{k}$ corresponds to the product of force by lever arm, in relation to turning about $O$. The total moment of the external forces is given by $\sum \left[\mathbf {r} _{k}\mathbf {F} _{k}\right]$. The second term in the right member of (5), which represents the resultant of the moments of the internal forces, vanishes if the internal forces between two particles have the direction of the line joining the particles, i.e. if the forces are central. Thus, since $\mathbf {F} _{jk}=-\mathbf {F} _{kj}$, we have for any pair of particles
   $\left[\mathbf {r} _{k}\mathbf {F} _{jk}\right]+\left[\mathbf {r} _{j}\mathbf {F} _{kj}\right]=\left[\left(\mathbf {r} _{k}-\mathbf {r} _{j}\right)\mathbf {F} _{jk}\right]$
But the vector product on the right vanishes, , since we are assuming that $\mathbf {F} _{jk}$ is in the direction of $\mathbf {r} _{k}-\mathbf {r} _{j}$. There remains, therefore,
   ${\frac {d}{dt}}\sum _{k}\left[\mathbf {r} _{k}{\dot {\mathbf {r} }}_{k}\right]={\frac {d\mathbf {P} }{dt}}=\sum _{k}\left[\mathbf {r} _{k}{\dot {\mathbf {F} }}_{k}\right]=\mathbf {M}$
For a \htmladdnormallink{system {http://planetphysics.us/encyclopedia/SimilarityAndAnalogousSystemsDynamicAdjointnessAndTopologicalEquivalence.html} of particles in which the forces between any two particles are in the direction of the line joining these particles, the rate of change of the total angular momentum is equal to the sum of the moments of the applied forces}.
The limitation made above is actually of little importance. From considerations of symmetry, it is difficult to imagine a force acting between two points which does not coincide in direction with the line joining them, for there is no other pre-eminent direction. If the Biot-Savart law seems an exception, it must be remembered that this law deals with the force between a magnetpole and an elementary segment (i.e. not a point or particle) of an Electrical Conductor.
In the particular case in which there are no external forces acting, or if the total moment of the forces vanishes, then, according to equation (7), the total angular momentum of the system remains constant. In this form the theorem explains a large variety of phenomena of everyday life, e.g. the method by which a child sets a swing in motion, the ability of a falling cat to right itself before landing, the familiar turntable experiments, etc. This law finds one of its most beautiful applications in explaining the Einstein-de Haas Effect in Magnetism.
In general, the value of the total angular momentum depends upon the choice of the reference point $O$. If, however, the center of gravity of the system is at rest, this quantity becomes independent of the choice of $O$. If we denote the radius vector to a new center $O^{\prime }$ by $r_{0}$ and a radius vector emanating from $O^{\prime }$ by $\mathbf {r} _{k}^{\prime }$, then
$\mathbf {r_{k}} =\mathbf {r_{0}} +\mathbf {r} _{k}^{\prime }$
The angular momentum, referred to $O$, is
$\mathbf {P} =\sum _{k}m_{k}\left[\mathbf {r_{k}} {\dot {\mathbf {r_{k}} }}\right]$
and referred to $O^{\prime }$
$\mathbf {P^{\prime }} =\sum _{k}m_{k}\left[\mathbf {r} _{k}^{\prime }{\dot {\mathbf {r} _{k}^{\prime }}}\right]$
If we put $\mathbf {r_{0}} +\mathbf {r} _{k}^{\prime }$ for $\mathbf {r} _{k}$ in $P$, we have, on account of ${\dot {\mathbf {r_{0}} }}=0$,
$\mathbf {P} =\sum _{k}m_{k}\left[\left(\mathbf {r_{0}} +\mathbf {r} _{k}^{\prime }\right){\dot {\mathbf {r} _{k}^{\prime }}}\right]=\left[\mathbf {r} _{0}\sum _{k}m_{k}{\dot {\mathbf {r} _{k}^{\prime }}}\right]+\mathbf {P} ^{\prime }$
But the first term vanishes if
$\sum m_{k}{\frac {d\mathbf {r} _{k}^{\prime }}{dt}}=0$
i.e. if the center of gravity is at rest.