In the case where we have cyclic coordinates, it is possible to use a method developed by Routh to eliminate them from the Lagrangian completely. Supposed that the Lagrangian has \(n\) degrees of freedom and the first \(k\) coordinates are cyclic. Thus:
\begin{align}
\die{L}{\dot{q}_i}=\beta_i \;\;\;\text{(i=1…k)}
\tag{4.40}\end{align}
where the \(\beta_i\) are really just the generalizes momenta, but we use the \(\beta_i\) to distinguish the constant momenta. We first rewrite the Langrangian, \(L\), by using the \(\beta_i\) to replace the \(k\) velocities corresponding to cylclic coordinates, \(\dot{q}_i\). We then define the “Routhian” as:
\begin{align}
R&\equiv L-\sum_{i=1}^k\beta_i\dot{q}_i\\
&=R(q_i,\dot{q}_i,\beta_i, t)
\tag{4.41}\end{align}
The Routhian thus depends on \(n-k\) of the \(q_i\) (and their velocities) in \(L\), as well as \(k\) quantities \(\beta_i\) (in the sum), and time. Now consider the variation of the Routhian:
\begin{align}
\delta R=\sum_{i=k+1}^{n}\die{R}{q_i}\delta q_i+\sum_{i=k+1}^{n}\die{R}{\dot{q}_i}\delta \dot{q}_i+\sum_{i=1}^k\die{R}{\beta_i}\delta \beta_i+\die{R}{t}
\tag{4.42}\end{align}
Using the definition of the Routhian, this must also equal:
\begin{align}
\delta R=&\delta\left(L-\sum_{i=1}^k\beta_i\dot{q}_i \right)\\
=&\sum_{i=k+1}^{n}\die{L}{q_i}\delta q_i+\sum_{i=k+1}^{n}\die{L}{\dot{q}_i}\delta \dot{q}_i+\die{L}{t}-\delta\left(\sum_{i=1}^k\beta_i\dot{q}_i\right)\\
=&\sum_{i=k+1}^{n}\die{L}{q_i}\delta q_i+\sum_{i=k+1}^{n}\die{L}{\dot{q}_i}\delta \dot{q}_i+\die{L}{t}-\sum_{i=1}^k\left(\beta_i\delta\dot{q}_i+\dot{q}_i\delta\beta_i\right)\\
=&\sum_{i=k+1}^{n}\die{L}{q_i}\delta q_i+\sum_{i=k+1}^{n}\die{L}{\dot{q}_i}\delta \dot{q}_i+\die{L}{t}-\sum_{i=1}^k\die{L}{\dot{q}_i}\delta\dot{q}_i-\sum_{i=1}^k\dot{q}_i\delta\beta_i
\tag{4.43}\end{align}
However, the second last term is zero, because we have explicitly rewritten the Lagrangian to remove the first \(k\) of the \(\dot{q}_i\). We can thus compare the two versions of the equations for \(\delta R\) and match each term:
\begin{align}
\die{L}{q_i}&=\die{R}{q_i}\\
\die{L}{\dot{q}_i}&=\die{R}{\dot{q}_i}\\
\dot{q}_i&=-\die{R}{\beta_i}\;\;\;\text{(i=1…k)}\\
\die{L}{t}&=\die{R}{t}
\tag{4.44}\end{align}
Given these equivalences, it is clear that the Routhian will also follow the Lagrange equations:
\begin{align}
\frac{d}{dt}\die{R}{\dot{q}_i}-\die{R}{q_i}=0\;\;\;(i=k+1\dots n)
\tag{4.45}\end{align}
where we have now effectively reduced the number of degrees of freedom in the problem, since the Routhian only has \(n-k\) coordinates. The equations of motion are thus given by:
\begin{align}
\dot{q}_i&=-\die{R}{\beta_i}\;\;\;\text{(i=1…k)}\\
\frac{d}{dt}\die{R}{\dot{q}_i}-\die{R}{q_i}&=0\;\;\;(i=k+1\dots n)
\tag{4.46}\end{align}
The procedure for solving a problem with the Routhian is:
Write the Lagrange and identify the cyclic coordinates
Calculate the \(\beta_i\) and use them to eliminate the corresponding \(\dot{q}_i\) in the Lagrangian
Write the Routhian
The equations of motion for the \(k\) velocities from the cyclic variables are easily written in terms of the \(\beta_i\)
The equations of motion for the remaining \(n-k\) coordinates are obtained by applying the usual Lagrange equations to \(R\).
Example 4-10
Use the Routhian to describe the motion of a particle in a gravitational field
The Lagrangian is given by:
\begin{align*}
L=\frac{1}{2}m(\dot{x}^2+\dot{y}^2+\dot{z}^2)-mgz
\end{align*}
where the coordinates \(x\) and \(y\) are cyclic. We have 3 degrees of freedom and 2 cyclic coordinates. We thus have:
\begin{align*}
\die{L}{\dot{x}}&=m\dot{x}=\beta_x\\
\die{L}{\dot{y}}&=m\dot{y}=\beta_y
\end{align*}
We can thus eliminate \(\dot{x}\) and \(\dot{y}\) from the Lagrangian, in favour of the \(\beta_i\):
\begin{align*}
L=\frac{1}{2}m\left(\frac{\beta_x^2}{m^2}+\frac{\beta_y^2}{m^2}+\dot{z}^2\right)-mgz
\end{align*}
The Routhian is thus:
\begin{align*}
R&=\frac{1}{2}m\left(\frac{\beta_x^2}{m^2}+\frac{\beta_y^2}{m^2}+\dot{z}^2\right)-mgz-\beta_x\dot{x}-\beta_y\dot{y}\nonumber\\
&=\frac{1}{2}m\left(\frac{\beta_x^2}{m^2}+\frac{\beta_y^2}{m^2}+\dot{z}^2\right)-mgz-\frac{\beta_x^2}{m}-\frac{\beta_y^2}{m}\nonumber\\
&=\frac{1}{2}m\dot{z}^2-mgz-\frac{\beta_x^2}{2m}-\frac{\beta_y^2}{2m}
\end{align*}
and it does not depend on the cyclic coordinates or their velocities.
We can then write the equations of motion:
\begin{align*}
\dot{x}&=-\die{R}{\beta_x}=\frac{\beta_x}{m}\to x(t)=x_0+\frac{\beta_x}{m}t\\
\dot{y}&=-\die{R}{\beta_y}=\frac{\beta_y}{m}\to y(t)=y_0+\frac{\beta_y}{m}t\\
\ddot{z}&=-g
\end{align*}