Main article: Lagrangian mechanics
Routinely, in order to write down the kinetic T = 1 2 m v 2 {\displaystyle T={\tfrac {1}{2}}mv^{2}} and potential V {\displaystyle V} parts of the Lagrangian L = T − V {\displaystyle L=T-V} in arbitrary generalized coordinates the position of the mass is expressed along Cartesian axes. Here, following the conventions shown in the diagram,
Next, time derivatives of these coordinates are taken, to obtain velocities along the axes
Thus,
and
The Lagrangian, with constant parts removed, is1
The Euler–Lagrange equation involving the polar angle θ {\displaystyle \theta }
gives
When ϕ ˙ = 0 {\displaystyle {\dot {\phi }}=0} the equation reduces to the differential equation for the motion of a simple gravity pendulum.
Similarly, the Euler–Lagrange equation involving the azimuth ϕ {\displaystyle \phi } ,
The last equation shows that angular momentum around the vertical axis, | L z | = l sin θ × m l sin θ ϕ ˙ {\displaystyle |\mathbf {L} _{z}|=l\sin \theta \times ml\sin \theta \,{\dot {\phi }}} is conserved. The factor m l 2 sin 2 θ {\displaystyle ml^{2}\sin ^{2}\theta } will play a role in the Hamiltonian formulation below.
The second order differential equation determining the evolution of ϕ {\displaystyle \phi } is thus
The azimuth ϕ {\displaystyle \phi } , being absent from the Lagrangian, is a cyclic coordinate, which implies that its conjugate momentum is a constant of motion.
The conical pendulum refers to the special solutions where θ ˙ = 0 {\displaystyle {\dot {\theta }}=0} and ϕ ˙ {\displaystyle {\dot {\phi }}} is a constant not depending on time.
Main article: Hamiltonian mechanics
The Hamiltonian is
where conjugate momenta are
In terms of coordinates and momenta it reads
H = [ 1 2 m l 2 θ ˙ 2 + 1 2 m l 2 sin 2 θ ϕ ˙ 2 ] ⏟ T + [ − m g l cos θ ] ⏟ V = P θ 2 2 m l 2 + P ϕ 2 2 m l 2 sin 2 θ − m g l cos θ {\displaystyle H=\underbrace {\left[{\frac {1}{2}}ml^{2}{\dot {\theta }}^{2}+{\frac {1}{2}}ml^{2}\sin ^{2}\theta {\dot {\phi }}^{2}\right]} _{T}+\underbrace {{\bigg [}-mgl\cos \theta {\bigg ]}} _{V}={P_{\theta }^{2} \over 2ml^{2}}+{P_{\phi }^{2} \over 2ml^{2}\sin ^{2}\theta }-mgl\cos \theta }
Hamilton's equations will give time evolution of coordinates and momenta in four first-order differential equations
Momentum P ϕ {\displaystyle P_{\phi }} is a constant of motion. That is a consequence of the rotational symmetry of the system around the vertical axis.[dubious – discuss]
Trajectory of the mass on the sphere can be obtained from the expression for the total energy
by noting that the horizontal component of angular momentum L z = m l 2 sin 2 θ ϕ ˙ {\displaystyle L_{z}=ml^{2}\sin ^{2}\!\theta \,{\dot {\phi }}} is a constant of motion, independent of time.2 This is true because neither gravity nor the reaction from the sphere act in directions that would affect this component of angular momentum.
Hence
which leads to an elliptic integral of the first kind3 for θ {\displaystyle \theta }
and an elliptic integral of the third kind for ϕ {\displaystyle \phi }
The angle θ {\displaystyle \theta } lies between two circles of latitude,4 where
Landau, Lev Davidovich; Evgenii Mikhailovich Lifshitz (1976). Course of Theoretical Physics: Volume 1 Mechanics. Butterworth-Heinenann. pp. 33–34. ISBN 0750628960. 0750628960 ↩