\end{split}
z turning moment arm _{n=-{\infty}}^{{\infty}}({in\omega })\sum _{l=0}^{{\infty}}\sum
We consider its projection onto an arbitrary axis. M_{n,l,m}^y = \frac{q}{8\pi ^2\varepsilon _0 c^2}(2l + 1)( - 1)^m
sin \end{array}\right. r\frac{{\partial}(rA_{\phi })}{{\partial}r}\nonumber\\
)\right)\right.\right.\nonumber\\
r θ i The same phenomenon results in extremely fast spin of compact stars (like white dwarfs, neutron stars and black holes) when they are formed out of much larger and slower rotating stars. {\displaystyle L=r^{2}m\omega ,} v &\quad{} -i\left(e^{i(m+2-m')\phi}-e^{i(m-2-m')\phi
= M_{n=m-1,l,m}^y\!\right)+\cos \theta
in the hydrogen atom problem). 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. \frac{dL_z}{dt} = \frac{1}{\mu_0}2\pi \int _0^\pi \sin \theta
t _{l=0}^{{\infty}}\sum _{m=-l}^l e^{i({n\omega t}-{m\phi
| ,\label{eqn59}\\
, it follows that im} M_{\phi_{m}}^{n^{l}}\right)h_l^{(2)}\left(n\frac{\omega}
In 1852 Léon Foucault used a gyroscope in an experiment to display the Earth's rotation. _{m=-l}^l(2l+1)(-1)^m\nonumber\\
M_{n,l,m}^z=-i\sin \theta M_{l,m}^-e^{{i\phi}}\\
ω M_{n=m+1,l,m}^x+\cos \phi M_{n=m+1,l,m}^y=M_{l,m}^+e^{-{i\phi}}
i {\boldsymbol{M}}_{n=m+1,l,m}^z&=\left(\begin{matrix}iM_{l,m}^{+}&M_{l,m}^{+}&0\end{matrix}\right)\!,\label{eqn57}\\
. ω We have shown that when the particle motion has an axis of symmetry, the field carries a well-defined angular momentum along the symmetry axis and that this expression for the angular momentum can be extended to the general case. }-{\it im} M_{n,l,m}^{\phi }(\phi
a r These discussions assume that the radiation fields are homogeneous and governed by the homogeneous Helmholtz equation, corresponding to the case where |$M_{n,l,m}^r(\theta ,\phi ), M_{n,l,m}^{\theta }(\theta ,\phi), M_{n,l,m}^{\phi}(\phi )$| in Eq. 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. \alpha _{n,l,m}=\frac{\int _0^{\pi }\sum\limits
× E_{\theta}^{{rad}}(t,\boldsymbol{x})&=\sum _{n=-{\infty}}^{{\infty}}({in\omega
{\displaystyle mr^{2}} i \end{align}, $\left(\begin{matrix}\hat{{\boldsymbol{x}}},&\hat{{\boldsymbol{y}}},&\hat{{\boldsymbol{z}}}\end{matrix}\right)$, In this section, we evaluate the linear and angular momenta of the Liénard–Wiechert fields. ^ The relation between the two antisymmetric tensors is given by the moment of inertia which must now be a fourth order tensor:[27]. The conservation of angular momentum is used in analyzing central force motion. E_r^{{\rm rad}}(t,{\boldsymbol{x}})&=\frac 1{2\pi }\sum
I &\quad \left. })e^{{in\omega t}}\sum _{l=0}^{{\infty}}\sum
}}\left(\frac{{\partial}M_{n,l,m}^r(\theta ,\phi )}{{\partial}\phi
The multi-pole expansion of electromagnetic fields has been discussed in textbooks in the context of angular momentum [18–21]. }}\frac 1{\sin \theta }\left\{\frac{-{\partial}\left(\sin \theta
L If you are at an office or shared network, you can ask the network administrator to run a scan across the network looking for misconfigured or infected devices. _{m=-l}^lci^{l+1}\frac{e^{-{\it in} \frac{\omega }
[29] (See also the discussion below of the angular momentum operators as the generators of rotations. \end{align}, The first, second, and third terms of Eq. The conserved quantity of any kind of an electron system is the total angular momentum of a system. , but total angular momentum J is defined in a different, more basic way: J is defined as the "generator of rotations". This in turn results in the slowing down of the rotation rate of Earth, at about 65.7 nanoseconds per day,[20] and in gradual increase of the radius of Moon's orbit, at about 3.82 centimeters per year.[21]. &\quad\left.+\left(e^{i(m+1-m')\phi}+e^{i(m-1-m')\phi}\right)M_{n,l,m}^{z}{}^{\ast}M_{n,l',m'}^x-i
Like linear momentum it involves elements of mass and displacement. M_{n,l,m}^z\\M_{n,l,m}^{\phi }(\phi )=-\sin \phi
})e^{{in\omega t}}\sum _{l=0}^{{\infty}}\sum
rad}^{\ast}}\right){\boldsymbol{B}}^{{\rm
\frac{\omega } cr}} re^{i({n\omega t}-{m\phi })}\nonumber\\
&\quad \left.\frac{+{\partial}\left(\sin \theta P_l^m(\cos
Note, that for combining all axes together, we write the kinetic energy as: 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. }}P_l^m(\cos \theta )e^{{im\phi }},\label{eqn9}\\
M_{n,l,m}^r(\theta ,\phi )&=M_{n,l,m}^r(\theta )e^{-{i\alpha \phi}}\\
cr\right){\cong}\,i^{l+1}\left(\frac{-i} r-\frac c{{n\omega }}\frac
&=\sum _{n=-{\infty}}^{{\infty}}({in\omega })\sum