Routhian mechanics

Edward John Routh, 1831–1907

In classical mechanics, Routh's procedure or Routhian mechanics is a hybrid formulation of Lagrangian mechanics and Hamiltonian mechanics developed by Edward John Routh. Correspondingly, the Routhian is the function which replaces both the Lagrangian and Hamiltonian 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.

Definitions

The Routhian, like the Hamiltonian, can be obtained from a Legendre 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 of generalized coordinates representing the degrees 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 the Routhian 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, the generalized coordinates Template:Math, ... and the corresponding velocities Template:Math, and possibly time Template:Math, enter the Lagrangian,

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

where the overdots denote time derivatives.

In Hamiltonian mechanics, the generalized coordinates Template:Math and the corresponding generalized momenta Template:Math and possibly time, enter the Hamiltonian,

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

where the second equation is the definition of the generalized momentum Template:Math corresponding to the coordinate Template:Math (partial derivatives are denoted using Template:Math). The velocities Template:Math are expressed as functions of their corresponding momenta by inverting their defining relation. In this context, Template:Math is said to be the momentum "canonically conjugate" to Template:Math.

The Routhian is intermediate between Template:Math and Template:Math; some coordinates Template:Math are chosen to have corresponding generalized momenta Template:Math, the rest of the coordinates Template:Math to have generalized velocities Template:Math, and time may appear explicitly;

Routhian (Template:Math degrees of freedom)

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

where again the generalized velocity Template:Math is to be expressed as a function of generalized momentum Template:Math via its defining relation. The choice of which Template:Math coordinates are to have corresponding momenta, out of the Template:Math coordinates, is arbitrary.

The above is used by Landau and Lifshitz, and Goldstein. Some authors may define the Routhian to be the negative of the above definition.

Given the length of the general definition, a more compact notation is to use boldface for tuples (or vectors) of the variables, thus Template:Math, Template:Math, Template:Math, and Template:Math, so that

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

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

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

Equations of motion

For reference, the Euler-Lagrange equations for Template:Math degrees of freedom are a set of Template:Math coupled second order ordinary differential equations in the coordinates

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

where Template:Math, and the Hamiltonian equations for Template:Math degrees of freedom are a set of Template:Math coupled first order ordinary differential equations in the coordinates and momenta

${\dot{q}}_i=\frac{\partial H}{\partial p_i},{\dot{p}}_i=-\frac{\partial H}{\partial q_i}.$

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

Two degrees of freedom

Consider the case of a system with two degrees of freedom, Template:Math and Template:Math, with generalized velocities Template:Math and Template:Math, and the Lagrangian is time-dependent. (The generalization to any number of degrees of freedom follows exactly the same procedure as with two). The Lagrangian of the system will have the form

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

The differential of Template:Math is

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

Now change variables, from the set (Template:Math, Template:Math, Template:Math, Template:Math) to (Template:Math, Template:Math, Template:Math, Template:Math), simply switching the velocity Template:Math to the momentum Template:Math. This change of variables in the differentials is the Legendre transformation. The differential of the new function to replace Template:Math will be a sum of differentials in Template:Math, Template:Math, Template:Math, Template:Math, and Template:Math. Using the definition of generalized momentum and Lagrange's equation for the coordinate Template:Math:

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

we have

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

and to replace Template:Math by Template:Math, recall the product rule for differentials, and substitute

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

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

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

Introducing the Routhian

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

where again the velocity Template:Math is a function of the momentum Template:Math, we have

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

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

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

Comparing the coefficients of the differentials Template:Math, Template:Math, Template:Math, Template:Math, and Template:Math, the results are Hamilton's equations for the coordinate Template:Math,

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

and Lagrange's equation for the coordinate Template:Math

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

which follow from

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

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 of Template:Math and Template:Math are negatives

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

Any number of degrees of freedom

For Template:Math coordinates as defined above, with Routhian

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

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 of Template:Math with respect to the coordinates Template:Math and Template:Math, momenta Template:Math, and velocities Template:Math, where Template:Math, and Template:Math. The derivatives are

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

$\frac{\partial R}{\partial p_i}={\dot{q}}_i$

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

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

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

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 of Template:Math) gives the Lagrangian equations. The fifth is just the same relation between time partial derivatives as before. To summarize

Routhian equations of motion (n + s degrees of freedom)

${\dot{q}}_i=\frac{\partial R}{\partial p_i},{\dot{p}}_i=-\frac{\partial R}{\partial q_i},$

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

The total number of equations is Template:Math, there are Template:Math Hamiltonian equations plus Template:Math Lagrange equations.

Energy

Since the Lagrangian has the same units as energy, the units of the Routhian are also energy. In SI units this is the Joule.

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

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

If the Lagrangian is independent of time, the partial time derivative of the Lagrangian is zero, Template:Math, so the quantity under the total time derivative in brackets must be a constant, it is the total energy of the system

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

(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 of Template:Math with respect to all the velocities Template:Math and Template:Math. Under the same condition of Template:Math being time independent, the energy in terms of the Routhian is a little simpler, substituting the definition of Template:Math and the partial derivatives of Template:Math with respect to the velocities Template:Math,

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

Notice only the partial derivatives of Template:Math with respect to the velocities Template:Math are needed. In the case that Template:Math and the Routhian is explicitly time-independent, then Template:Math, that is, the Routhian equals the energy of the system. The same expression for Template:Math in when Template:Math is also the Hamiltonian, so in all Template:Math.

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

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

which can be derived from the total time derivative of Template:Math in the same way as for Template:Math.

Cyclic coordinates

Often the Routhian approach may offer no advantage, but one notable case where this is useful is when a system has cyclic 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 with cyclic coordinates, because by definition those coordinates do not enter Template:Math, and hence Template:Math. The corresponding partial derivatives of Template:Math and Template:Math with respect to those coordinates are zero, which equates to the corresponding generalized momenta reducing to constants. To make this concrete, if the Template:Math are all cyclic coordinates, and the Template:Math are all non cyclic, then

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

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

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

The Template:Math Hamiltonian equation in the cyclic coordinates automatically vanishes,

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

and the Template:Math Lagrangian equations are in the non cyclic coordinates

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

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 ${\dot{q}}_i$can be integrated to compute $q_i(t)$.

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.

Examples

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

Central potential in spherical coordinates

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

Consider a particle of mass Template:Math under the influence of a central potential Template:Math in spherical polar coordinates Template:Math

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

Notice Template:Math is cyclic, because it does not appear in the Lagrangian. The momentum conjugate to Template:Math is the constant

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

in which Template:Math and Template:Math can vary with time, but the angular momentum Template:Math is constant. The Routhian can be taken to be

$\begin{array}{cc}R(r,\dot{r},\theta ,\dot{\theta })& =p_{\varphi }\dot{\varphi }-L\\ & =p_{\varphi }\dot{\varphi }-\frac{m}{2}{\dot{r}}^2-\frac{m}{2}{r}^2{\dot{\theta }}^2-\frac{p_{\varphi }\dot{\varphi }}{2}+V(r)\\ & =\frac{p_{\varphi }\dot{\varphi }}{2}-\frac{m}{2}{\dot{r}}^2-\frac{m}{2}{r}^2{\dot{\theta }}^2+V(r)\\ & =\frac{p_{\varphi }^2}{2m{r}^2{\sin }^2\theta }-\frac{m}{2}{\dot{r}}^2-\frac{m}{2}{r}^2{\dot{\theta }}^2+V(r).\end{array}$

We can solve for Template:Math and Template:Math using Lagrange's equations, and do not need to solve for Template:Math since it is eliminated by Hamiltonian's equations. The Template:Math equation is

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

and the Template:Math equation is

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

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

The Template:Math equation is

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

the Template:Math equation is

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

the Template:Math equation is

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

Symmetric mechanical systems

Spherical pendulum

Spherical pendulum: angles and velocities

Consider the spherical pendulum, a mass Template:Math (known as a "pendulum bob") attached to a rigid rod of length Template:Math of negligible mass, subject to a local gravitational field Template:Math. The system rotates with angular velocity Template:Math which is not constant. The angle between the rod and vertical is Template:Math and is not constant.

The Lagrangian is

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

and Template:Math is the cyclic coordinate for the system with constant momentum

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

which again is physically the angular momentum of the system about the vertical. The angle Template:Math and angular velocity Template:Math vary with time, but the angular momentum is constant. The Routhian is

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

The Template:Math equation is found from the Lagrangian equations

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

or simplifying by introducing the constants

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

gives

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

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 Template:Math is related to the angular momentum Template:Math).

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

The Template:Math equation is

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

and the Template:Math equation is

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

Heavy symmetrical top

Heavy symmetric top in terms of the Euler angles

The heavy symmetrical top of mass Template:Math has Lagrangian

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

where Template:Math are the Euler angles, Template:Math is the angle between the vertical Template:Math-axis and the top's Template:Math-axis, Template:Math is the rotation of the top about its own Template:Math-axis, and Template:Math the azimuthal of the top's Template:Math-axis around the vertical Template:Math-axis. The principal moments of inertia are Template:Math about the top's own Template:Math axis, Template:Math about the top's own Template:Math axes, and Template:Math about the top's own Template:Math-axis. Since the top is symmetric about its Template:Math-axis, Template:Math. Here the simple relation for local gravitational potential energy Template:Math is used where Template:Math is the acceleration due to gravity, and the centre of mass of the top is a distance Template:Math from its tip along its Template:Math-axis.

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

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

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

From these, eliminating Template:Math:

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

we have

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

and to eliminate Template:Math, substitute this result into Template:Math and solve for Template:Math to find

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

The Routhian can be taken to be

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

and since

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

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

we have

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

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

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

The equation of motion for Template:Math is, by direct calculation,

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

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

or by introducing the constants

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

a simpler form of the equation is obtained

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

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 Template:Math and Template:Math in the Lagrangian.

The Template:Math equation is

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

the Template:Math equation is

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

and the Template:Math equation is

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

Velocity-dependent potentials

Classical charged particle in a uniform magnetic field

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

Consider a classical charged particle of mass Template:Math and electric charge Template:Math in a static (time-independent) uniform (constant throughout space) magnetic field Template:Math. The Lagrangian for a charged particle in a general electromagnetic field given by the magnetic potential Template:Math and electric potential $\varphi$ is

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

It is convenient to use cylindrical coordinates Template:Math, so that

$\dot{𝐫}=𝐯=(v_r,v_{\theta },v_z)=(\dot{r},r\dot{\theta },\dot{z}),$

$𝐁=(B_r,B_{\theta },B_z)=(0,0,B).$

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

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

and the Lagrangian is

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

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, Template:Math and Template:Math. The canonical momenta conjugate to Template:Math and Template:Math are the constants

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

so the velocities are

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

The angular momentum about the z axis is not Template:Math, but the quantity Template:Math, which is not conserved due to the contribution from the magnetic field. The canonical momentum Template:Math is the conserved quantity. It is still the case that Template:Math is the linear or translational momentum along the z axis, which is also conserved.

The radial component Template:Math and angular velocity Template:Math can vary with time, but Template:Math is constant, and since Template:Math is constant it follows Template:Math is constant. The Routhian can take the form

$\begin{array}{cc}R(r,\dot{r})& =p_{\theta }\dot{\theta }+p_z\dot{z}-L\\ & =p_{\theta }\dot{\theta }+p_z\dot{z}-\frac{m}{2}{\dot{r}}^2-\frac{p_{\theta }\dot{\theta }}{2}-\frac{p_z\dot{z}}{2}-\frac{1}{2}qB{r}^2\dot{\theta }\\ & =(p_{\theta }-qB{r}^2)\frac{\dot{\theta }}{2}-\frac{m}{2}{\dot{r}}^2+\frac{p_z\dot{z}}{2}\\ [6pt]& =\frac{1}{2m{r}^2}(p_{\theta }-qB{r}^2)(p_{\theta }-\frac{qB{r}^2}{2})-\frac{m}{2}{\dot{r}}^2+\frac{p_z^2}{2m}\\ & =\frac{1}{2m{r}^2}(p_{\theta }^2-\frac{3}{2}qB{r}^2+\frac{(qB{)}^2{r}^4}{2})-\frac{m}{2}{\dot{r}}^2\end{array}$

where in the last line, the Template:Math term is a constant and can be ignored without loss of continuity. The Hamiltonian equations for Template:Math and Template:Math automatically vanish and do not need to be solved for. The Lagrangian equation in Template:Math

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

is by direct calculation

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

which after collecting terms is

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

and simplifying further by introducing the constants

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

the differential equation is

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

To see how Template:Math changes with time, integrate the momenta expression for Template:Math above

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

where Template:Math is an arbitrary constant, the initial value of Template:Math 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 Template:Math, Template:Math, Template:Math, Template:Math, will determine if the trajectory of the particle has a constant Template:Math or varying Template:Math. If initially Template:Math is nonzero but Template:Math, while Template:Math and Template:Math are arbitrary, then the initial velocity of the particle has no radial component, Template:Math 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 Template:Math.

With the Lagrangian approach, the equation for Template:Math would include Template:Math which has to be eliminated, and there would be equations for Template:Math and Template:Math to solve for.

The Template:Math equation is

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

the Template:Math equation is

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

and the Template:Math equation is

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

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

See also

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

References

• Template:Cite book
• Template:Cite book
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• Template:Cite book