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In classical mechanics,Routh's procedure orRouthian mechanics is a hybrid formulation ofLagrangian mechanics andHamiltonian mechanics developed byEdward John Routh. Correspondingly, theRouthian is thefunction which replaces both theLagrangian andHamiltonian functions. Although Routhian mechanics is equivalent to Lagrangian mechanics and Hamiltonian mechanics, and introduces no new physics, it offers an alternative way to solve mechanical problems.

The Routhian, like the Hamiltonian, can be obtained from aLegendre transform of the Lagrangian, and has a similar mathematical form to the Hamiltonian, but is not exactly the same. The difference between the Lagrangian, Hamiltonian, and Routhian functions are their variables. For a given set ofgeneralized coordinates representing thedegrees of freedom in the system, the Lagrangian is a function of the coordinates and velocities, while the Hamiltonian is a function of the coordinates and momenta.

The Routhian differs from these functions in that some coordinates are chosen to have corresponding generalized velocities, the rest to have corresponding generalized momenta. This choice is arbitrary, and can be done to simplify the problem. It also has the consequence that theRouthian equations are exactly the Hamiltonian equations for some coordinates and corresponding momenta, and the Lagrangian equations for the rest of the coordinates and their velocities. In each case the Lagrangian and Hamiltonian functions are replaced by a single function, the Routhian. The full set thus has the advantages of both sets of equations, with the convenience of splitting one set of coordinates to the Hamilton equations, and the rest to the Lagrangian equations.

In the case of Lagrangian mechanics, thegeneralized coordinatesq₁,q₂, ... and the corresponding velocitiesdq₁/dt,dq₂/dt, ..., and possibly time^{[nb 1]}t, enter the Lagrangian,

${\displaystyle L(q_1,q_2,\dots ,{\dot{q}}_1,{\dot{q}}_2,\dots ,t),{\dot{q}}_i=\frac{d{q}_i}{dt},}$

where the overdots denotetime derivatives.

In Hamiltonian mechanics, the generalized coordinatesq₁,q₂, ... and the corresponding generalized momentap₁,p₂, ..., and possibly time, enter the Hamiltonian,

${\displaystyle H(q_1,q_2,\dots ,p_1,p_2,\dots ,t)=\sum _i{\dot{q}}_i{p}_i-L(q_1,q_2,\dots ,{\dot{q}}_1(p_1),{\dot{q}}_2(p_2),\dots ,t),p_i=\frac{\mathrm{\partial }L}{\mathrm{\partial }{\dot{q}}_i},}$

where the second equation is the definition of the generalized momentump_i corresponding to the coordinateq_i (partial derivatives are denoted using∂). The velocitiesdq_i/dt are expressed as functions of their corresponding momenta by inverting their defining relation. In this context,p_i is said to be the momentum "canonically conjugate" toq_i.

The Routhian is intermediate betweenL andH; some coordinatesq₁,q₂, ...,q_n are chosen to have corresponding generalized momentap₁,p₂, ...,p_n, the rest of the coordinatesζ₁,ζ₂, ...,ζ_s to have generalized velocitiesdζ₁/dt,dζ₂/dt, ...,dζ_s/dt, and time may appear explicitly;^[1][2]

where again the generalized velocitydq_i/dt is to be expressed as a function of generalized momentump_i via its defining relation. The choice of whichn coordinates are to have corresponding momenta, out of then +s coordinates, is arbitrary.

The above is used byLandau and Lifshitz, andGoldstein. Some authors may define the Routhian to be the negative of the above definition.^[3]

Given the length of the general definition, a more compact notation is to use boldface fortuples (or vectors) of the variables, thusq = (q₁,q₂, ...,q_n),ζ = (ζ₁,ζ₂, ...,ζ_s),p = (p₁,p₂, ...,p_n), anddζ/dt = (dζ₁/dt,dζ₂/dt, ...,dζ_s/dt), so that

${\displaystyle R(\mathbf{q},\mathit{\zeta },\mathbf{p},\dot{\mathit{\zeta }},t)=\mathbf{p}\cdot \dot{\mathbf{q}}-L(\mathbf{q},\mathit{\zeta },\dot{\mathbf{q}},\dot{\mathit{\zeta }},t),}$

where · is thedot product defined on the tuples, for the specific example appearing here:

${\displaystyle \mathbf{p}\cdot \dot{\mathbf{q}}=\sum _{i=1}^n{p}_i{\dot{q}}_i.}$

For reference, theEuler-Lagrange equations fors degrees of freedom are a set ofs coupled second orderordinary differential equations in the coordinates

${\displaystyle \frac{d}{dt}\frac{\mathrm{\partial }L}{\mathrm{\partial }{\dot{q}}_j}=\frac{\mathrm{\partial }L}{\mathrm{\partial }{q}_j},}$

wherej = 1, 2, ...,s, and theHamiltonian equations forn degrees of freedom are a set of2n coupled first order ordinary differential equations in the coordinates and momenta

${\displaystyle {\dot{q}}_i=\frac{\mathrm{\partial }H}{\mathrm{\partial }{p}_i},{\dot{p}}_i=-\frac{\mathrm{\partial }H}{\mathrm{\partial }{q}_i}.}$

Below, the Routhian equations of motion are obtained in two ways, in the process other useful derivatives are found that can be used elsewhere.

Consider the case of a system with twodegrees of freedom,q andζ, with generalized velocitiesdq/dt anddζ/dt, and the Lagrangian is time-dependent. (The generalization to any number of degrees of freedom follows exactly the same procedure as with two).^[4] The Lagrangian of the system will have the form

${\displaystyle L(q,\zeta ,\dot{q},\dot{\zeta },t)}$

Thedifferential ofL is

${\displaystyle dL=\frac{\mathrm{\partial }L}{\mathrm{\partial }q}dq+\frac{\mathrm{\partial }L}{\mathrm{\partial }\zeta }d\zeta +\frac{\mathrm{\partial }L}{\mathrm{\partial }\dot{q}}d\dot{q}+\frac{\mathrm{\partial }L}{\mathrm{\partial }\dot{\zeta }}d\dot{\zeta }+\frac{\mathrm{\partial }L}{\mathrm{\partial }t}dt.}$

Now change variables, from the set (q,ζ,dq/dt,dζ/dt) to (q,ζ,p,dζ/dt), simply switching the velocitydq/dt to the momentump. This change of variables in the differentials is theLegendre transformation. The differential of the new function to replaceL will be a sum of differentials indq,dζ,dp,d(dζ/dt), anddt. Using the definition of generalized momentum and Lagrange's equation for the coordinateq:

${\displaystyle p=\frac{\mathrm{\partial }L}{\mathrm{\partial }\dot{q}},\dot{p}=\frac{d}{dt}\frac{\mathrm{\partial }L}{\mathrm{\partial }\dot{q}}=\frac{\mathrm{\partial }L}{\mathrm{\partial }q}}$

we have

${\displaystyle dL=\dot{p}dq+\frac{\mathrm{\partial }L}{\mathrm{\partial }\zeta }d\zeta +pd\dot{q}+\frac{\mathrm{\partial }L}{\mathrm{\partial }\dot{\zeta }}d\dot{\zeta }+\frac{\mathrm{\partial }L}{\mathrm{\partial }t}dt}$

and to replacepd(dq/dt) by(dq/dt)dp, recall theproduct rule for differentials,^{[nb 2]} and substitute

${\displaystyle pd\dot{q}=d(\dot{q}p)-\dot{q}dp}$

to obtain the differential of a new function in terms of the new set of variables:

${\displaystyle d(L-p\dot{q})=\dot{p}dq+\frac{\mathrm{\partial }L}{\mathrm{\partial }\zeta }d\zeta -\dot{q}dp+\frac{\mathrm{\partial }L}{\mathrm{\partial }\dot{\zeta }}d\dot{\zeta }+\frac{\mathrm{\partial }L}{\mathrm{\partial }t}dt.}$

Introducing the Routhian

${\displaystyle R(q,\zeta ,p,\dot{\zeta },t)=p\dot{q}(p)-L}$

where again the velocitydq/dt is a function of the momentump, we have

${\displaystyle dR=-\dot{p}dq-\frac{\mathrm{\partial }L}{\mathrm{\partial }\zeta }d\zeta +\dot{q}dp-\frac{\mathrm{\partial }L}{\mathrm{\partial }\dot{\zeta }}d\dot{\zeta }-\frac{\mathrm{\partial }L}{\mathrm{\partial }t}dt,}$

but from the above definition, the differential of the Routhian is

${\displaystyle dR=\frac{\mathrm{\partial }R}{\mathrm{\partial }q}dq+\frac{\mathrm{\partial }R}{\mathrm{\partial }\zeta }d\zeta +\frac{\mathrm{\partial }R}{\mathrm{\partial }p}dp+\frac{\mathrm{\partial }R}{\mathrm{\partial }\dot{\zeta }}d\dot{\zeta }+\frac{\mathrm{\partial }R}{\mathrm{\partial }t}dt.}$

Comparing the coefficients of the differentialsdq,dζ,dp,d(dζ/dt), anddt, the results areHamilton's equations for the coordinateq,

${\displaystyle \dot{q}=\frac{\mathrm{\partial }R}{\mathrm{\partial }p},\dot{p}=-\frac{\mathrm{\partial }R}{\mathrm{\partial }q},}$

andLagrange's equation for the coordinateζ

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

which follow from

${\displaystyle \frac{\mathrm{\partial }L}{\mathrm{\partial }\zeta }=-\frac{\mathrm{\partial }R}{\mathrm{\partial }\zeta },\frac{\mathrm{\partial }L}{\mathrm{\partial }\dot{\zeta }}=-\frac{\mathrm{\partial }R}{\mathrm{\partial }\dot{\zeta }},}$

and taking the total time derivative of the second equation and equating to the first. Notice the Routhian replaces the Hamiltonian and Lagrangian functions in all the equations of motion.

The remaining equation states the partial time derivatives ofL andR are negatives

${\displaystyle \frac{\mathrm{\partial }L}{\mathrm{\partial }t}=-\frac{\mathrm{\partial }R}{\mathrm{\partial }t}.}$

Forn +s coordinates as defined above, with Routhian

${\displaystyle R(q_1,\dots ,q_n,{\zeta }_1,\dots ,{\zeta }_s,p_1,\dots ,p_n,{\dot{\zeta }}_1,\dots ,{\dot{\zeta }}_s,t)=\sum _{i=1}^n{p}_i{\dot{q}}_i(p_i)-L}$

the equations of motion can be derived by a Legendre transformation of this Routhian as in the previous section, but another way is to simply take the partial derivatives ofR with respect to the coordinatesq_i andζ_j, momentap_i, and velocitiesdζ_j/dt, wherei = 1, 2, ...,n, andj = 1, 2, ...,s. The derivatives are

${\displaystyle \frac{\mathrm{\partial }R}{\mathrm{\partial }{q}_i}=-\frac{\mathrm{\partial }L}{\mathrm{\partial }{q}_i}=-\frac{d}{dt}\frac{\mathrm{\partial }L}{\mathrm{\partial }{\dot{q}}_i}=-{\dot{p}}_i}$

${\displaystyle \frac{\mathrm{\partial }R}{\mathrm{\partial }{p}_i}={\dot{q}}_i}$

${\displaystyle \frac{\mathrm{\partial }R}{\mathrm{\partial }{\zeta }_j}=-\frac{\mathrm{\partial }L}{\mathrm{\partial }{\zeta }_j},}$

${\displaystyle \frac{\mathrm{\partial }R}{\mathrm{\partial }{\dot{\zeta }}_j}=-\frac{\mathrm{\partial }L}{\mathrm{\partial }{\dot{\zeta }}_j},}$

${\displaystyle \frac{\mathrm{\partial }R}{\mathrm{\partial }t}=-\frac{\mathrm{\partial }L}{\mathrm{\partial }t}.}$

The first two are identically the Hamiltonian equations. Equating the total time derivative of the fourth set of equations with the third (for each value ofj) gives the Lagrangian equations. The fifth is just the same relation between time partial derivatives as before. To summarize^[5]

The total number of equations is2n +s, there are2n Hamiltonian equations pluss Lagrange equations.

Since the Lagrangian has the same units asenergy, the units of the Routhian are also energy. InSI units this is theJoule.

Taking the total time derivative of the Lagrangian leads to the general result

${\displaystyle \frac{\mathrm{\partial }L}{\mathrm{\partial }t}=\frac{d}{dt}\left(\sum _{i=1}^n{\dot{q}}_i\frac{\mathrm{\partial }L}{\mathrm{\partial }{\dot{q}}_i}+\sum _{j=1}^s{\dot{\zeta }}_j\frac{\mathrm{\partial }L}{\mathrm{\partial }{\dot{\zeta }}_j}-L\right).}$

If the Lagrangian is independent of time, the partial time derivative of the Lagrangian is zero,∂L/∂t = 0, so the quantity under the total time derivative in brackets must be a constant, it is the total energy of the system^[6]

${\displaystyle E=\sum _{i=1}^n{\dot{q}}_i\frac{\mathrm{\partial }L}{\mathrm{\partial }{\dot{q}}_i}+\sum _{j=1}^s{\dot{\zeta }}_j\frac{\mathrm{\partial }L}{\mathrm{\partial }{\dot{\zeta }}_j}-L.}$

(If there are external fields interacting with the constituents of the system, they can vary throughout space but not time). This expression requires the partial derivatives ofL with respect toall the velocitiesdq_i/dt anddζ_j/dt. Under the same condition ofR being time independent, the energy in terms of the Routhian is a little simpler, substituting the definition ofR and the partial derivatives ofR with respect to the velocitiesdζ_j/dt,

${\displaystyle E=R-\sum _{j=1}^s{\dot{\zeta }}_j\frac{\mathrm{\partial }R}{\mathrm{\partial }{\dot{\zeta }}_j}.}$

Notice only the partial derivatives ofR with respect to the velocitiesdζ_j/dt are needed. In the case thats = 0 and the Routhian is explicitly time-independent, thenE =R, that is, the Routhian equals the energy of the system. The same expression forR in whens = 0 is also the Hamiltonian, so in allE =R =H.

If the Routhian has explicit time dependence, the total energy of the system is not constant. The general result is

${\displaystyle \frac{\mathrm{\partial }R}{\mathrm{\partial }t}={\displaystyle \frac{d}{dt}}\left(R-\sum _{j=1}^s{\dot{\zeta }}_j\frac{\mathrm{\partial }R}{\mathrm{\partial }{\dot{\zeta }}_j}\right),}$

which can be derived from the total time derivative ofR in the same way as forL.

Often the Routhian approach may offer no advantage, but one notable case where this is useful is when a system hascyclic coordinates (also called "ignorable coordinates"), by definition those coordinates which do not appear in the original Lagrangian. The Lagrangian equations are powerful results, used frequently in theory and practice, since the equations of motion in the coordinates are easy to set up. However, if cyclic coordinates occur there will still be equations to solve for all the coordinates, including the cyclic coordinates despite their absence in the Lagrangian. The Hamiltonian equations are useful theoretical results, but less useful in practice because coordinates and momenta are related together in the solutions - after solving the equations the coordinates and momenta must be eliminated from each other. Nevertheless, the Hamiltonian equations are perfectly suited to cyclic coordinates because the equations in the cyclic coordinates trivially vanish, leaving only the equations in the non cyclic coordinates.

The Routhian approach has the best of both approaches, because cyclic coordinates can be split off to the Hamiltonian equations and eliminated, leaving behind the non cyclic coordinates to be solved from the Lagrangian equations. Overall fewer equations need to be solved compared to the Lagrangian approach.

The Routhian formulation is useful for systems withcyclic coordinates, because by definition those coordinates do not enterL, and henceR. The corresponding partial derivatives ofL andR with respect to those coordinates are zero, which equates to the corresponding generalized momenta reducing to constants. To make this concrete, if theq_i are all cyclic coordinates, and theζ_j are all non cyclic, then

${\displaystyle \frac{\mathrm{\partial }L}{\mathrm{\partial }{q}_i}={\dot{p}}_i=-\frac{\mathrm{\partial }R}{\mathrm{\partial }{q}_i}=0\Rightarrow p_i={\alpha }_i,}$

where theα_i are constants. With these constants substituted into the Routhian,R is a function of only the non cyclic coordinates and velocities (and in general time also)

${\displaystyle R({\zeta }_1,\dots ,{\zeta }_s,{\alpha }_1,\dots ,{\alpha }_n,{\dot{\zeta }}_1,\dots ,{\dot{\zeta }}_s,t)=\sum _{i=1}^n{\alpha }_i{\dot{q}}_i({\alpha }_i)-L({\zeta }_1,\dots ,{\zeta }_s,{\dot{q}}_1({\alpha }_1),\dots ,{\dot{q}}_n({\alpha }_n),{\dot{\zeta }}_1,\dots ,{\dot{\zeta }}_s,t),}$

The2n Hamiltonian equation in the cyclic coordinates automatically vanishes,

${\displaystyle {\dot{q}}_i=\frac{\mathrm{\partial }R}{\mathrm{\partial }{\alpha }_i}=f_i({\zeta }_1(t),\dots ,{\zeta }_s(t),{\dot{\zeta }}_1(t),\dots ,{\dot{\zeta }}_s(t),{\alpha }_1,\dots ,{\alpha }_n,t),{\dot{p}}_i=-\frac{\mathrm{\partial }R}{\mathrm{\partial }{q}_i}=0,}$

and thes Lagrangian equations are in the non cyclic coordinates

${\displaystyle \frac{d}{dt}\frac{\mathrm{\partial }R}{\mathrm{\partial }{\dot{\zeta }}_j}=\frac{\mathrm{\partial }R}{\mathrm{\partial }{\zeta }_j}.}$

Thus the problem has been reduced to solving the Lagrangian equations in the non cyclic coordinates, with the advantage of the Hamiltonian equations cleanly removing the cyclic coordinates. Using those solutions, the equations for${\displaystyle {\dot{q}}_i}$can be integrated to compute${\displaystyle q_i(t)}$.

If we are interested in how the cyclic coordinates change with time, the equations for the generalized velocities corresponding to the cyclic coordinates can be integrated.

Routh's procedure does not guarantee the equations of motion will be simple, however it will lead to fewer equations.

One general class of mechanical systems with cyclic coordinates are those withcentral potentials, because potentials of this form only have dependence on radial separations and no dependence on angles.

Consider a particle of massm under the influence of a central potentialV(r) inspherical polar coordinates(r,θ,φ)

${\displaystyle L(r,\dot{r},\theta ,\dot{\theta },\dot{\varphi })=\frac{m}{2}({\dot{r}}^2+r^2{\dot{\theta }}^2+r^2{\sin }^2\theta {\dot{\phi }}^2)-V(r).}$

Noticeφ is cyclic, because it does not appear in the Lagrangian. The momentum conjugate toφ is the constant

${\displaystyle p_{\varphi }=\frac{\mathrm{\partial }L}{\mathrm{\partial }\dot{\varphi }}=m{r}^2{\sin }^2\theta \dot{\varphi },}$

in whichr anddφ/dt can vary with time, but the angular momentump_φ is constant. The Routhian can be taken to be

We can solve forr andθ using Lagrange's equations, and do not need to solve forφ since it is eliminated by Hamiltonian's equations. Ther equation is

${\displaystyle \frac{d}{dt}\frac{\mathrm{\partial }R}{\mathrm{\partial }\dot{r}}=\frac{\mathrm{\partial }R}{\mathrm{\partial }r}\Rightarrow -m\ddot{r}=-\frac{p_{\varphi }^2}{m{r}^3{\sin }^2\theta }-mr{\dot{\theta }}^2+\frac{\mathrm{\partial }V}{\mathrm{\partial }r},}$

and theθ equation is

${\displaystyle \frac{d}{dt}\frac{\mathrm{\partial }R}{\mathrm{\partial }\dot{\theta }}=\frac{\mathrm{\partial }R}{\mathrm{\partial }\theta }\Rightarrow -m(2r\dot{r}\dot{\theta }+r^2\ddot{\theta })=-\frac{p_{\varphi }^2\cos \theta }{m{r}^2{\sin }^3\theta }.}$

The Routhian approach has obtained two coupled nonlinear equations. By contrast the Lagrangian approach leads tothree nonlinear coupled equations, mixing in the first and second time derivatives ofφ in all of them, despite its absence from the Lagrangian.

Ther equation is

${\displaystyle \frac{d}{dt}\frac{\mathrm{\partial }L}{\mathrm{\partial }\dot{r}}=\frac{\mathrm{\partial }L}{\mathrm{\partial }r}\Rightarrow m\ddot{r}=mr{\dot{\theta }}^2+mr{\sin }^2\theta {\dot{\varphi }}^2-\frac{\mathrm{\partial }V}{\mathrm{\partial }r},}$

theθ equation is

${\displaystyle \frac{d}{dt}\frac{\mathrm{\partial }L}{\mathrm{\partial }\dot{\theta }}=\frac{\mathrm{\partial }L}{\mathrm{\partial }\theta }\Rightarrow 2r\dot{r}\dot{\theta }+r^2\ddot{\theta }=r^2\sin \theta \cos \theta {\dot{\varphi }}^2,}$

theφ equation is

${\displaystyle \frac{d}{dt}\frac{\mathrm{\partial }L}{\mathrm{\partial }\dot{\varphi }}=\frac{\mathrm{\partial }L}{\mathrm{\partial }\varphi }\Rightarrow 2r\dot{r}{\sin }^2\theta \dot{\varphi }+2r^2\sin \theta \cos \theta \dot{\theta }\dot{\varphi }+r^2{\sin }^2\theta \ddot{\varphi }=0.}$

Consider thespherical pendulum, a massm (known as a "pendulum bob") attached to a rigid rod of lengthl of negligible mass, subject to a local gravitational fieldg. The system rotates with angular velocitydφ/dt which isnot constant. The angle between the rod and vertical isθ and isnot constant.

The Lagrangian is^{[nb 3]}

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

andφ is the cyclic coordinate for the system with constant momentum

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

which again is physically the angular momentum of the system about the vertical. The angleθ and angular velocitydφ/dt vary with time, but the angular momentum is constant. The Routhian is

Theθ equation is found from the Lagrangian equations

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

or simplifying by introducing the constants

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

gives

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

This equation resembles the simple nonlinearpendulum equation, because it can swing through the vertical axis, with an additional term to account for the rotation about the vertical axis (the constanta is related to the angular momentump_φ).

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

Theθ equation is

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

and theφ equation is

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

The heavysymmetrical top of massM has Lagrangian^[7][8]

${\displaystyle L(\theta ,\dot{\theta },\dot{\psi },\dot{\varphi })=\frac{I_1}{2}({\dot{\theta }}^2+{\dot{\varphi }}^2{\sin }^2\theta )+\frac{I_3}{2}({\dot{\psi }}^2+{\dot{\varphi }}^2{\cos }^2\theta )+I_3\dot{\psi }\dot{\varphi }\cos \theta -Mg\ell \cos \theta }$

whereψ,φ,θ are theEuler angles,θ is the angle between the verticalz-axis and the top'sz′-axis,ψ is the rotation of the top about its ownz′-axis, andφ the azimuthal of the top'sz′-axis around the verticalz-axis. The principalmoments of inertia areI₁ about the top's ownx′ axis,I₂ about the top's owny′ axes, andI₃ about the top's ownz′-axis. Since the top is symmetric about itsz′-axis,I₁ =I₂. Here the simple relation for localgravitational potential energyV =Mglcosθ is used whereg is the acceleration due to gravity, and the centre of mass of the top is a distancel from its tip along itsz′-axis.

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

${\displaystyle p_{\psi }=\frac{\mathrm{\partial }L}{\mathrm{\partial }\dot{\psi }}=I_3\dot{\psi }+I_3\dot{\varphi }\cos \theta }$

${\displaystyle p_{\varphi }=\frac{\mathrm{\partial }L}{\mathrm{\partial }\dot{\varphi }}=\dot{\varphi }(I_1{\sin }^2\theta +I_3{\cos }^2\theta )+I_3\dot{\psi }\cos \theta }$

From these, eliminatingdψ/dt:

${\displaystyle p_{\varphi }-p_{\psi }\cos \theta =I_1\dot{\varphi }{\sin }^2\theta }$

we have

${\displaystyle \dot{\varphi }=\frac{p_{\varphi }-p_{\psi }\cos \theta }{I_1{\sin }^2\theta },}$

and to eliminatedφ/dt, substitute this result intop_ψ and solve fordψ/dt to find

${\displaystyle \dot{\psi }=\frac{p_{\psi }}{I_3}-\cos \theta \left(\frac{p_{\varphi }-p_{\psi }\cos \theta }{I_1{\sin }^2\theta }\right).}$

The Routhian can be taken to be

${\displaystyle R(\theta ,\dot{\theta })=p_{\psi }\dot{\psi }+p_{\varphi }\dot{\varphi }-L=\frac{1}{2}(p_{\psi }\dot{\psi }+p_{\varphi }\dot{\varphi })-\frac{I_1{\dot{\theta }}^2}{2}+Mg\ell \cos \theta }$

and since

${\displaystyle \frac{p_{\varphi }\dot{\varphi }}{2}=\frac{p_{\varphi }^2}{2I_1{\sin }^2\theta }-\frac{p_{\psi }{p}_{\varphi }\cos \theta }{2I_1{\sin }^2\theta },}$

${\displaystyle \frac{p_{\psi }\dot{\psi }}{2}=\frac{p_{\psi }^2}{2I_3}-\frac{p_{\psi }{p}_{\varphi }\cos \theta }{2I_1{\sin }^2\theta }+\frac{p_{\psi }^2{\cos }^2\theta }{2I_1{\sin }^2\theta }}$

we have

${\displaystyle R=\frac{p_{\psi }^2}{2I_3}+\frac{p_{\psi }^2{\cos }^2\theta }{2I_1{\sin }^2\theta }+\frac{p_{\varphi }^2}{2I_1{\sin }^2\theta }-\frac{p_{\psi }{p}_{\varphi }\cos \theta }{I_1{\sin }^2\theta }-\frac{I_1{\dot{\theta }}^2}{2}+Mg\ell \cos \theta .}$

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

${\displaystyle R=\frac{1}{2I_1{\sin }^2\theta }\left[p_{\psi }^2{\cos }^2\theta +p_{\varphi }^2-2p_{\psi }{p}_{\varphi }\cos \theta \right]-\frac{I_1{\dot{\theta }}^2}{2}+Mg\ell \cos \theta }$

The equation of motion forθ is, by direct calculation,

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

${\displaystyle -I_1\ddot{\theta }=-\frac{\cos \theta }{I_1{\sin }^3\theta }\left[p_{\psi }^2{\cos }^2\theta +p_{\varphi }^2-\frac{p_{\psi }{p}_{\varphi }}{2}\cos \theta \right]+\frac{1}{2I_1{\sin }^2\theta }\left[-2p_{\psi }^2\cos \theta \sin \theta +\frac{p_{\psi }{p}_{\varphi }}{2}\sin \theta \right]-Mg\ell \sin \theta ,}$

or by introducing the constants

${\displaystyle a=\frac{p_{\psi }^2}{I_1^2},b=\frac{p_{\varphi }^2}{I_1^2},c=\frac{p_{\psi }{p}_{\varphi }}{2I_1^2},k=\frac{Mg\ell }{I_1},}$

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 tothree nonlinear coupled equations to solve, despite the absence of the coordinatesψ andφ in the Lagrangian.

Theθ equation is

${\displaystyle \frac{d}{dt}\frac{\mathrm{\partial }L}{\mathrm{\partial }\dot{\theta }}=\frac{\mathrm{\partial }L}{\mathrm{\partial }\theta }\Rightarrow I_1\ddot{\theta }=(I_1-I_3){\dot{\varphi }}^2\sin \theta \cos \theta -I_3\dot{\psi }\dot{\varphi }\sin \theta +Mg\ell \sin \theta ,}$

theψ equation is

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

and theφ equation is

${\displaystyle \frac{d}{dt}\frac{\mathrm{\partial }L}{\mathrm{\partial }\dot{\varphi }}=\frac{\mathrm{\partial }L}{\mathrm{\partial }\varphi }\Rightarrow \ddot{\varphi }(I_1{\sin }^2\theta +I_3{\cos }^2\theta )+\dot{\varphi }(I_1-I_3)2\sin \theta \cos \theta \dot{\theta }+I_3\ddot{\psi }\cos \theta -I_3\dot{\psi }\sin \theta \dot{\theta }=0,}$

Consider a classicalcharged particle of massm andelectric chargeq in a static (time-independent) uniform (constant throughout space)magnetic fieldB.^[9] The Lagrangian for a charged particle in a generalelectromagnetic field given by themagnetic potentialA andelectric potential${\displaystyle \varphi }$ is

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

It is convenient to usecylindrical 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,${\displaystyle \varphi =0}$, and we can choose the axial gauge for the magnetic potential

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

and the Lagrangian is

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

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 andz. The canonical momenta conjugate toθ andz are the constants

${\displaystyle p_{\theta }=\frac{\mathrm{\partial }L}{\mathrm{\partial }\dot{\theta }}=m{r}^2\dot{\theta }+\frac{qB{r}^2}{2},p_z=\frac{\mathrm{\partial }L}{\mathrm{\partial }\dot{z}}=m\dot{z},}$

so the velocities are

${\displaystyle \dot{\theta }=\frac{1}{m{r}^2}\left(p_{\theta }-\frac{qB{r}^2}{2}\right),\dot{z}=\frac{p_z}{m}.}$

The angular momentum about thez axis isnotp_θ, but the quantitymr²dθ/dt, which is not conserved due to the contribution from the magnetic field. The canonical momentump_θ is the conserved quantity. It is still the case thatp_z is the linear or translational momentum along thez axis, which is also conserved.

The radial componentr and angular velocitydθ/dt can vary with time, butp_θ is constant, and sincep_z is constant it followsdz/dt is constant. The Routhian can take the form

where in the last line, thep_z²/2m term is a constant and can be ignored without loss of continuity. The Hamiltonian equations forθ andz automatically vanish and do not need to be solved for. The Lagrangian equation inr

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

is by direct calculation

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

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

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

the differential equation is

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

To see howz changes with time, integrate the momenta expression forp_z above

${\displaystyle z=\frac{p_z}{m}t+c_z,}$

wherec_z is an arbitrary constant, the initial value ofz to be specified in theinitial conditions.

The motion of the particle in this system ishelicoidal, 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 onr,dr/dt,θ,dθ/dt, will determine if the trajectory of the particle has a constantr or varyingr. If initiallyr is nonzero butdr/dt = 0, whileθ anddθ/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. Ifr is constant, the angular velocity is also constant according to the conservedp_θ.

With the Lagrangian approach, the equation forr would includedθ/dt which has to be eliminated, and there would be equations forθ andz to solve for.

Ther equation is

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

theθ equation is

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

and thez equation is

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

Thez equation is trivial to integrate, but ther andθ equations are not, in any case the time derivatives are mixed in all the equations and must be eliminated.

• Calculus of variations
• Phase space
• Configuration space
• Many-body problem
• Rigid body mechanics