[NOTES/CM-02008] Eliminating Cyclic Coordnates

The Routhian

The Routhian is a kind  of partial Legendre transform of Lagrangian. In the previous section a Legendre transform was used to pass Lagrangian to a Hamiltonian to description. In this process all the generalized velocities were eliminated in favour of canonical momenta. It possible to do this only for some of the generalized velocities. Let the set of generalized coordinates be denoted as \(q, \xi\) and Lagrangian be a function of  \(q,\xi, \dot{q}, \dot{\xi}\). Define the canonical momenta as usual by
\begin{equation}\label{EQ01} p_k  = \frac{\partial L}{\dot\partial {q}_k} \end{equation} 
We introduce Routhian by doing a Legendre transform  from \(\dot q_k\)  to \(p_k\).The Routhian is defined by \begin{equation}\label{EQ02} R(q, p, \xi, \dot {\xi} ) = \sum_k p_k \dot {q}_k - L \end{equation} In an alternate form of dynamics, the canonical momenta take over  the role played by velocities and Hamiltonian becomes central quantity  which governs the dynamics.The EOM can be written in an alternate form called the Hamiltonian EOM. In the Hamiltonian dynamics the velocities are eliminated in favour of canonical momenta. Thus the canonical momenta, in this example, coincide with components of momentum $m\dot{\vec r}$ and Hamiltonian is equal to the energy.

However, it must be remarked that {\it the canonical momenta are not always equal to `ordinary' momenta and Hamiltonian need not be a sum of} $K.E+P.E$. This is the case when the system is described, for example, by a velocity dependent generalized potential. Motion of  a charged particle in external magnetic field constitutes an example of this type where the canonical momentum is not equal to ordinary momentum.  This is the case when the system is described, for example, by a velocity dependent generalized potential. Motion of  a charged particle in external magnetic field constitutes an example of this type where the canonical momentum is not equal to ordinary momentum.

Using Routhian to Eliminate Cyclic Degrees of Freedom
Let us note that a cyclic coordinate is absent and that the canonically conjugate momentum is a   constant, so it should be possible to eliminate these degrees of freedom completely. That this is indeed possible can be seen by doing a 'partial Legendre transform' of the Lagrangian we arrive at a description in terms of canonical momentum which is a constant of motion. The process described above leads to using  Routhian having just the remaining generalised coordinates and  velocities and constants.