Spherical pendulum

Spherical pendulum: angles and velocities

Consider the spherical pendulum, a mass m (known as a "pendulum bob") attached to a rigid rod of length l of negligible mass, subject to a local gravitational field g. The system rotates with angular velocity dφ/dt which is not constant. The angle between the rod and vertical is θ and is not constant.

The Lagrangian is^{[nb 3]}

${\displaystyle L(\theta ,{\dot {\theta }},{\dot {\phi }})={\frac {m\ell ^{2}}{2}}({\dot {\theta }}^{2}+\sin ^{2}\theta {\dot {\phi }}^{2})+mg\ell \cos \theta \,,}$

and φ is the cyclic coordinate for the system with constant momentum

${\displaystyle p_{\phi }={\frac {\partial L}{\partial {\dot {\phi }}}}=m\ell ^{2}\sin ^{2}\theta {\dot {\phi }}\,.}$

which again is physically the angular momentum of the system about the vertical. The angle θ and angular velocity dφ/dt vary with time, but the angular momentum is constant. The Routhian is

${\displaystyle {\begin{aligned}R(\theta ,{\dot {\theta }})&=p_{\phi }{\dot {\phi }}-L\\&=p_{\phi }{\dot {\phi }}-{\frac {m\ell ^{2}}{2}}{\dot {\theta }}^{2}-{\frac {p_{\phi }{\dot {\phi }}}{2}}-mg\ell \cos \theta \\&={\frac {p_{\phi }{\dot {\phi }}}{2}}-{\frac {m\ell ^{2}}{2}}{\dot {\theta }}^{2}-mg\ell \cos \theta \\&={\frac {p_{\phi }^{2}}{2m\ell ^{2}\sin ^{2}\theta }}-{\frac {m\ell ^{2}}{2}}{\dot {\theta }}^{2}-mg\ell \cos \theta \end{aligned}}}$

The θ equation is found from the Lagrangian equations

${\displaystyle {\frac {d}{dt}}{\frac {\partial R}{\partial {\dot {\theta }}}}={\frac {\partial R}{\partial \theta }}\quad \Rightarrow \quad -m\ell ^{2}{\ddot {\theta }}=-{\frac {p_{\phi }^{2}\cos \theta }{m\ell ^{2}\sin ^{3}\theta }}+mg\ell \sin \theta \,,}$

or simplifying by introducing the constants

${\displaystyle a={\frac {p_{\phi }^{2}}{m^{2}\ell ^{4}}}\,,\quad b={\frac {g}{\ell }}\,,}$

gives

${\displaystyle {\ddot {\theta }}=a{\frac {\cos \theta }{\sin ^{3}\theta }}-b\sin \theta \,.}$

This equation resembles the simple nonlinear pendulum equation, because it can swing through the vertical axis, with an additional term to account for the rotation about the vertical axis (the constant a is related to the angular momentum p_φ).

Applying the Lagrangian approach there are two nonlinear coupled equations to solve.

The θ equation is

${\displaystyle {\frac {d}{dt}}{\frac {\partial L}{\partial {\dot {\theta }}}}={\frac {\partial L}{\partial \theta }}\quad \Rightarrow \quad m\ell ^{2}{\ddot {\theta }}=m\ell ^{2}\sin \theta \cos \theta {\dot {\phi }}^{2}-mg\ell \sin \theta \,,}$

and the φ equation is

${\displaystyle {\frac {d}{dt}}{\frac {\partial L}{\partial {\dot {\phi }}}}={\frac {\partial L}{\partial \phi }}\quad \Rightarrow \quad 2\sin \theta \cos \theta {\dot {\theta }}{\dot {\phi }}+\sin ^{2}\theta {\ddot {\phi }}=0\,.}$

Heavy symmetrical top

Heavy symmetric top in terms of the Euler angles

The heavy symmetrical top of mass M has Lagrangian^[7][8]

${\displaystyle L(\theta ,{\dot {\theta }},{\dot {\psi }},{\dot {\phi }})={\frac {I_{1}}{2}}({\dot {\theta }}^{2}+{\dot {\phi }}^{2}\sin ^{2}\theta )+{\frac {I_{3}}{2}}({\dot {\psi }}^{2}+{\dot {\phi }}^{2}\cos ^{2}\theta )+I_{3}{\dot {\psi }}{\dot {\phi }}\cos \theta -Mg\ell \cos \theta }$

where ψ, φ, θ are the Euler angles, θ is the angle between the vertical z-axis and the top's z′-axis, ψ is the rotation of the top about its own z′-axis, and φ the azimuthal of the top's z′-axis around the vertical z-axis. The principal moments of inertia are I₁ about the top's own x′ axis, I₂ about the top's own y′ axes, and I₃ about the top's own z′-axis. Since the top is symmetric about its z′-axis, I₁ = I₂. Here the simple relation for local gravitational potential energy V = Mglcosθ is used where g is the acceleration due to gravity, and the centre of mass of the top is a distance l from its tip along its z′-axis.

The angles ψ, φ are cyclic. The constant momenta are the angular momenta of the top about its axis and its precession about the vertical, respectively:

${\displaystyle p_{\psi }={\frac {\partial L}{\partial {\dot {\psi }}}}=I_{3}{\dot {\psi }}+I_{3}{\dot {\phi }}\cos \theta }$

${\displaystyle p_{\phi }={\frac {\partial L}{\partial {\dot {\phi }}}}={\dot {\phi }}(I_{1}\sin ^{2}\theta +I_{3}\cos ^{2}\theta )+I_{3}{\dot {\psi }}\cos \theta }$

From these, eliminating dψ/dt:

${\displaystyle p_{\phi }-p_{\psi }\cos \theta =I_{1}{\dot {\phi }}\sin ^{2}\theta }$

we have

${\displaystyle {\dot {\phi }}={\frac {p_{\phi }-p_{\psi }\cos \theta }{I_{1}\sin ^{2}\theta }}\,,}$

and to eliminate dφ/dt, substitute this result into p_ψ and solve for dψ/dt to find

${\displaystyle {\dot {\psi }}={\frac {p_{\psi }}{I_{3}}}-\cos \theta \left({\frac {p_{\phi }-p_{\psi }\cos \theta }{I_{1}\sin ^{2}\theta }}\right)\,.}$

The Routhian can be taken to be

${\displaystyle R(\theta ,{\dot {\theta }})=p_{\psi }{\dot {\psi }}+p_{\phi }{\dot {\phi }}-L={\frac {1}{2}}(p_{\psi }{\dot {\psi }}+p_{\phi }{\dot {\phi }})-{\frac {I_{1}{\dot {\theta }}^{2}}{2}}+Mg\ell \cos \theta }$

and since

${\displaystyle {\frac {p_{\phi }{\dot {\phi }}}{2}}={\frac {p_{\phi }^{2}}{2I_{1}\sin ^{2}\theta }}-{\frac {p_{\psi }p_{\phi }\cos \theta }{2I_{1}\sin ^{2}\theta }}\,,}$

${\displaystyle {\frac {p_{\psi }{\dot {\psi }}}{2}}={\frac {p_{\psi }^{2}}{2I_{3}}}-{\frac {p_{\psi }p_{\phi }\cos \theta }{2I_{1}\sin ^{2}\theta }}+{\frac {p_{\psi }^{2}\cos ^{2}\theta }{2I_{1}\sin ^{2}\theta }}}$

we have

${\displaystyle R={\frac {p_{\psi }^{2}}{2I_{3}}}+{\frac {p_{\psi }^{2}\cos ^{2}\theta }{2I_{1}\sin ^{2}\theta }}+{\frac {p_{\phi }^{2}}{2I_{1}\sin ^{2}\theta }}-{\frac {p_{\psi }p_{\phi }\cos \theta }{I_{1}\sin ^{2}\theta }}-{\frac {I_{1}{\dot {\theta }}^{2}}{2}}+Mg\ell \cos \theta \,.}$

The first term is constant, and can be ignored since only the derivatives of R will enter the equations of motion. The simplified Routhian, without loss of information, is thus

${\displaystyle R={\frac {1}{2I_{1}\sin ^{2}\theta }}\left[p_{\psi }^{2}\cos ^{2}\theta +p_{\phi }^{2}-2p_{\psi }p_{\phi }\cos \theta \right]-{\frac {I_{1}{\dot {\theta }}^{2}}{2}}+Mg\ell \cos \theta }$

The equation of motion for θ is, by direct calculation,

${\displaystyle {\frac {d}{dt}}{\frac {\partial R}{\partial {\dot {\theta }}}}={\frac {\partial R}{\partial \theta }}\quad \Rightarrow \quad }$

${\displaystyle -I_{1}{\ddot {\theta }}=-{\frac {\cos \theta }{I_{1}\sin ^{3}\theta }}\left[p_{\psi }^{2}\cos ^{2}\theta +p_{\phi }^{2}-{\frac {p_{\psi }p_{\phi }}{2}}\cos \theta \right]+{\frac {1}{2I_{1}\sin ^{2}\theta }}\left[-2p_{\psi }^{2}\cos \theta \sin \theta +{\frac {p_{\psi }p_{\phi }}{2}}\sin \theta \right]-Mg\ell \sin \theta \,,}$

or by introducing the constants

${\displaystyle a={\frac {p_{\psi }^{2}}{I_{1}^{2}}}\,,\quad b={\frac {p_{\phi }^{2}}{I_{1}^{2}}}\,,\quad c={\frac {p_{\psi }p_{\phi }}{2I_{1}^{2}}}\,,\quad k={\frac {Mg\ell }{I_{1}}}\,,}$

a simpler form of the equation is obtained

${\displaystyle {\ddot {\theta }}={\frac {\cos \theta }{\sin ^{3}\theta }}(a\cos ^{2}\theta +b-c\cos \theta )+{\frac {1}{2\sin \theta }}(2a\cos \theta -c)+k\sin \theta \,.}$

Although the equation is highly nonlinear, there is only one equation to solve for, it was obtained directly, and the cyclic coordinates are not involved.

By contrast, the Lagrangian approach leads to three nonlinear coupled equations to solve, despite the absence of the coordinates ψ and φ in the Lagrangian.

The θ equation is

${\displaystyle {\frac {d}{dt}}{\frac {\partial L}{\partial {\dot {\theta }}}}={\frac {\partial L}{\partial \theta }}\quad \Rightarrow \quad I_{1}{\ddot {\theta }}=(I_{1}-I_{3}){\dot {\phi }}^{2}\sin \theta \cos \theta -I_{3}{\dot {\psi }}{\dot {\phi }}\sin \theta +Mg\ell \sin \theta \,,}$

the ψ equation is

${\displaystyle {\frac {d}{dt}}{\frac {\partial L}{\partial {\dot {\psi }}}}={\frac {\partial L}{\partial \psi }}\quad \Rightarrow \quad {\ddot {\psi }}+{\ddot {\phi }}\cos \theta -{\dot {\phi }}{\dot {\theta }}\sin \theta =0\,,}$

and the φ equation is

${\displaystyle {\frac {d}{dt}}{\frac {\partial L}{\partial {\dot {\phi }}}}={\frac {\partial L}{\partial \phi }}\quad \Rightarrow \quad {\ddot {\phi }}(I_{1}\sin ^{2}\theta +I_{3}\cos ^{2}\theta )+{\dot {\phi }}(I_{1}-I_{3})2\sin \theta \cos \theta {\dot {\theta }}+I_{3}{\ddot {\psi }}\cos \theta -I_{3}{\dot {\psi }}\sin \theta {\dot {\theta }}=0\,,}$

Classical charged particle in a uniform magnetic field

Classical charged particle in uniform B field, using cylindrical coordinates. Top: If the radial coordinate r and angular velocity dθ/dt vary, the trajectory is a helicoid with varying radius but uniform motion in the z direction. Bottom: Constant r and dθ/dt means a helicoid with constant radius.

Consider a classical charged particle of mass m and electric charge q in a static (time-independent) uniform (constant throughout space) magnetic field B.^[9] The Lagrangian for a charged particle in a general electromagnetic field given by the magnetic potential A and electric potential ${\displaystyle \phi }$ is

${\displaystyle L={\frac {m}{2}}{\dot {\mathbf {r} }}^{2}-q\phi +q{\dot {\mathbf {r} }}\cdot \mathbf {A} \,,}$

It is convenient to use cylindrical coordinates (r, θ, z), so that

${\displaystyle {\dot {\mathbf {r} }}=\mathbf {v} =(v_{r},v_{\theta },v_{z})=({\dot {r}},r{\dot {\theta }},{\dot {z}})\,,}$

${\displaystyle \mathbf {B} =(B_{r},B_{\theta },B_{z})=(0,0,B)\,.}$

In this case of no electric field, the electric potential is zero, ${\displaystyle \phi =0}$, and we can choose the axial gauge for the magnetic potential

${\displaystyle \mathbf {A} ={\frac {1}{2}}\mathbf {B} \times \mathbf {r} \quad \Rightarrow \quad \mathbf {A} =(A_{r},A_{\theta },A_{z})=(0,Br/2,0)\,,}$

and the Lagrangian is

${\displaystyle L(r,{\dot {r}},{\dot {\theta }},{\dot {z}})={\frac {m}{2}}({\dot {r}}^{2}+r^{2}{\dot {\theta }}^{2}+{\dot {z}}^{2})+{\frac {qBr^{2}{\dot {\theta }}}{2}}\,.}$

Notice this potential has an effectively cylindrical symmetry (although it also has angular velocity dependence), since the only spatial dependence is on the radial length from an imaginary cylinder axis.

There are two cyclic coordinates, θ and z. The canonical momenta conjugate to θ and z are the constants

${\displaystyle p_{\theta }={\frac {\partial L}{\partial {\dot {\theta }}}}=mr^{2}{\dot {\theta }}+{\frac {qBr^{2}}{2}}\,,\quad p_{z}={\frac {\partial L}{\partial {\dot {z}}}}=m{\dot {z}}\,,}$

so the velocities are

${\displaystyle {\dot {\theta }}={\frac {1}{mr^{2}}}\left(p_{\theta }-{\frac {qBr^{2}}{2}}\right)\,,\quad {\dot {z}}={\frac {p_{z}}{m}}\,.}$

The angular momentum about the z axis is not p_θ, but the quantity mr²dθ/dt, which is not conserved due to the contribution from the magnetic field. The canonical momentum p_θ is the conserved quantity. It is still the case that p_z is the linear or translational momentum along the z axis, which is also conserved.

The radial component r and angular velocity dθ/dt can vary with time, but p_θ is constant, and since p_z is constant it follows dz/dt is constant. The Routhian can take the form

${\displaystyle {\begin{aligned}R(r,{\dot {r}})&=p_{\theta }{\dot {\theta }}+p_{z}{\dot {z}}-L\\&=p_{\theta }{\dot {\theta }}+p_{z}{\dot {z}}-{\frac {m}{2}}{\dot {r}}^{2}-{\frac {p_{\theta }{\dot {\theta }}}{2}}-{\frac {p_{z}{\dot {z}}}{2}}-{\frac {1}{2}}qBr^{2}{\dot {\theta }}\\[6pt]&=(p_{\theta }-qBr^{2}){\frac {\dot {\theta }}{2}}-{\frac {m}{2}}{\dot {r}}^{2}+{\frac {p_{z}{\dot {z}}}{2}}\\[6pt]&={\frac {1}{2mr^{2}}}\left(p_{\theta }-qBr^{2}\right)\left(p_{\theta }-{\frac {qBr^{2}}{2}}\right)-{\frac {m}{2}}{\dot {r}}^{2}+{\frac {p_{z}^{2}}{2m}}\\[6pt]&={\frac {1}{2mr^{2}}}\left(p_{\theta }^{2}-{\frac {3}{2}}qBr^{2}+{\frac {(qB)^{2}r^{4}}{2}}\right)-{\frac {m}{2}}{\dot {r}}^{2}\end{aligned}}}$

where in the last line, the p_z²/2m term is a constant and can be ignored without loss of continuity. The Hamiltonian equations for θ and z automatically vanish and do not need to be solved for. The Lagrangian equation in r

${\displaystyle {\frac {d}{dt}}{\frac {\partial R}{\partial {\dot {r}}}}={\frac {\partial R}{\partial r}}}$

is by direct calculation

${\displaystyle -m{\ddot {r}}={\frac {1}{2m}}\left[{\frac {-2}{r^{3}}}\left(p_{\theta }^{2}-{\frac {3}{2}}qBr^{2}+{\frac {(qB)^{2}r^{4}}{2}}\right)+{\frac {1}{r^{2}}}(-3qBr+2(qB)^{2}r^{3})\right]\,,}$

which after collecting terms is

${\displaystyle m{\ddot {r}}={\frac {1}{2m}}\left[{\frac {2p_{\theta }^{2}}{r^{3}}}-(qB)^{2}r\right]\,,}$

and simplifying further by introducing the constants

${\displaystyle a={\frac {p_{\theta }^{2}}{m^{2}}}\,,\quad b=-{\frac {(qB)^{2}}{2m^{2}}}\,,}$

the differential equation is

${\displaystyle {\ddot {r}}={\frac {a}{r^{3}}}+br}$

To see how z changes with time, integrate the momenta expression for p_z above

${\displaystyle z={\frac {p_{z}}{m}}t+c_{z}\,,}$

where c_z is an arbitrary constant, the initial value of z to be specified in the initial conditions.

The motion of the particle in this system is helicoidal, with the axial motion uniform (constant) but the radial and angular components varying in a spiral according to the equation of motion derived above. The initial conditions on r, dr/dt, θ, dθ/dt, will determine if the trajectory of the particle has a constant r or varying r. If initially r is nonzero but dr/dt = 0, while θ and dθ/dt are arbitrary, then the initial velocity of the particle has no radial component, r is constant, so the motion will be in a perfect helix. If r is constant, the angular velocity is also constant according to the conserved p_θ.

With the Lagrangian approach, the equation for r would include dθ/dt which has to be eliminated, and there would be equations for θ and z to solve for.

The r equation is

${\displaystyle {\frac {d}{dt}}{\frac {\partial L}{\partial {\dot {r}}}}={\frac {\partial L}{\partial r}}\quad \Rightarrow \quad m{\ddot {r}}=mr{\dot {\theta }}^{2}+qBr{\dot {\theta }}\,,}$

the θ equation is

${\displaystyle {\frac {d}{dt}}{\frac {\partial L}{\partial {\dot {\theta }}}}={\frac {\partial L}{\partial \theta }}\quad \Rightarrow \quad m(2r{\dot {r}}{\dot {\theta }}+r^{2}{\ddot {\theta }})+qBr{\dot {r}}=0\,,}$

and the z equation is

${\displaystyle {\frac {d}{dt}}{\frac {\partial L}{\partial {\dot {z}}}}={\frac {\partial L}{\partial z}}\quad \Rightarrow \quad m{\ddot {z}}=0\,.}$

The z equation is trivial to integrate, but the r and θ equations are not, in any case the time derivatives are mixed in all the equations and must be eliminated.