1. Given a Routhian $R(r,\dot{r},\phi,p_{\phi})$, how do you derive Lagrange's equation for $r$? Do you just solve the following for $r$? $$\frac{d}{dt}\frac{∂R}{∂\dot{r}}-\frac{∂R}{∂r}=0.$$

  2. And as a related question, what is the motivation for using a Routhian?


2 Answers 2


No. The coordinate $r$ stil follows the Euler-Lagrange equation, but $\phi$ and $p_\phi$ follow the Hamilton equations. But these are trivial, which is the whole point of the Routhian. The motivation is that the Routhian isn't really $R(r, \dot r, \phi, p_\phi)$ but just $R(r, \dot r)$ with a constant parameter $p_\phi$. $\phi$ isn't a coordinate because by definition, it was a cyclic coordinate in the Lagrangian and so it doesn't appear in the Routhian either. The momentum $p_\phi$, meanwhile, is conserved so it's really just some constant. With both of these conditions, we can simply take them both out, call $p_\phi$ a constant, and we wind up with a problem with 1 fewer effective dimensions. The same would be true using the complete Hamiltonian, but sometimes Lagrangians are easier to work with for the "hard" part of the problem (the non-cyclic coordinates), and the Routhian lets you stay "Lagrangian" for the hard part.

Here is a concrete example. Take the Lagrangian describing a harmonic potential in polar coordinates:

$$ \mathcal L = \frac{1}{2} m \left( \dot r^2 + r^2 \dot \phi^2 \right) - \frac{1}{2} k r^2 $$

Since $\phi$ does not appear in the Lagrangian, it is a cyclic coordinate. From the Euler-Lagrange equation

$$ \frac{d}{dt} \frac{\partial \mathcal L}{\partial \dot \phi} = \frac{\partial \mathcal L}{\partial \phi} = 0 $$

it follows that

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

is a conserved quantity. It would be nice to take $\dot\phi$ out of the set of coordinates and just replace it with the constant $p_\phi$. Then we'd essentially have a Lagrangian whose only coordinates are $r$ and $\dot r$, just with a constant parameter $p_\phi$. The way to do this correctly is to make the Routhian

$$ \mathcal R = \mathcal L - p_\phi \dot \phi = \frac{1}{2} m \dot r^2 - \frac{1}{2} k r^2- \frac{p_\phi^2}{2 mr^2} $$

(Aside: Try solving $\dot \phi$ in terms of $p_\phi$ and substitute into the Lagrangian. You'll get something very different that will lead to very wrong results. If you want to remove the cyclic coordinate, you must do a Legendre transform w.r.t. that coordinate.).

Since the Legendre transform has been done w.r.t. $\theta$ and $\phi$, these coordinates follow Hamilton's equations (with an appropriate change of sign). But these is trivial:

$$ \dot \phi = - \frac{\partial \mathcal R}{\partial p_\phi} = \frac{p_\phi}{mr^2}, \quad \dot p_\phi = \frac{\partial \mathcal R}{\partial \phi} = 0 $$

What we've achieved here is essentially separation of variables. $\phi$ has its own equation of motion, which involves $r$, but the equation of motion of $r$ is independent of $\phi$. We can solve the equation of motion of $r$, which is an effectively one-dimensional problem, and then go back to find out how $\phi$ evolves.

The effective one-dimensional problem has an effective potential

$$ V(r) = \frac{1}{2} k r^2 + \frac{p_\phi^2}{2mr^2} $$

This extra term, which blows up at small $r$, is called the centrifugal barrier, and accounts for the fact that angular momentum conservation makes it harder / impossible for the particle in question to reach the origin. The coordinates $r$ and $\dot r$ didn't have a Legendre transform applied, so they still follow the Euler-Lagrange equation

$$ \frac{d}{dt} \frac{\partial \mathcal R}{\partial \dot r} = \frac{\partial \mathcal R}{\partial r} \Rightarrow m \ddot r = -kr + \frac{p_\phi^2}{3mr^3} $$

A follow up question might be "Yes, its convenient to not stay with a full Lagrangian, but why not just go full Hamiltonian?" The answer here probably depends on the context of the specific problem. With the Routhian, we have a 2nd order differential equation to solve, and then an integral to find the e.o.m. of $\phi$. With a full Hamiltonian, we'd have a system of two 1st order differential equations, followed by the same integral. One might be easier to solve, or better computationally, etc.

  1. Setting. Imagine that the configuration space consists of, say, both small & capital generalized positions $q^j$ and $Q^J$, with corresponding velocities $v^j$ and $V^J$, and momenta $p_j$ and $P_J$, respectively.

  2. Routhian. The Routhian $$\begin{align}R(q,Q,v,P,t) ~=~&V^JP_J-L(q,Q,v,V,t)\cr ~=~& H(q,Q,p,P,t)-v^jp_j \end{align} \tag{R}$$ is a hybrid between [and a partial velocity-momentum Legendre transformation away from] the Lagrangian $$L(q,Q,v,V,t)\tag{L}$$ and the Hamiltonian $$H(q,Q,p,P,t),\tag{H}$$ such that the small velocity variables $v^j$ and the capital momentum variables $P_J$ are kept.

  3. Action principle. The Routhian equations are the Euler-Lagrange (EL) equations for the Routhian action $$S_R[q,Q,v,P]~=~\int \! dt~L_R(q,Q,\dot{q},\dot{Q},P,t), \tag{SR}$$ with Routhian Lagrangian $$ L_R (q,Q,\dot{q},\dot{Q},P,t) ~:=~\dot{Q}^JP_J-R(q,Q,\dot{q},P,t) ,\tag{LR}$$ which leads to Lagrange equations for the small variables and Hamilton's equations for the capital variables.

  4. The motivation is to harvest the usual benefits from both the Lagrangian and Hamiltonian sides. Examples:

    • If the capital position variables $Q^J$ are cyclic variables [which means that $L$, $R$ and $H$ do not depend on $Q^J$], then the capital momentum variables $P_J$ are constants of motion. We can therefore demote the dynamical variables $(Q^J,P_K)$ to external parameters of the model. An action principle for the remaining dynamical variables (i.e. the small variables) is then given by the time integral of (minus) the Routhian $$-\int \! dt~R(q,\dot{q},t). $$ For a simple application, see e.g. this Phys.SE post.

    • If the Lagrangian $L$ depends affinely on $v^j$ and non-affinely on $V^J$, the Faddeev-Jackiw method to first-order formulations yields the Routhian action $S_R$.

  • $\begingroup$ Notes for later: In case of cyclic variables $Q_J$, we might want to impose BC $P_f=P_i={\rm const}$, and therefore change the Lagrangian $L\to L-\frac{d(P_J Q^J)}{dt}$, and action $S\to S-P_J (Q_f^J-Q_i^J)$. Similarly for the Hamiltonian Lagrangian $L_H\to L_H-\frac{d(P_J Q^J)}{dt}=-\dot{P}_J Q^J-H$, and Hamiltonian action $S_H\to S_H-P_J (Q_f^J-Q_i^J)=\int (p_j\mathrm{d}q^j-Q^J\mathrm{d}P_J-H\mathrm{d}t)$. $\endgroup$
    – Qmechanic
    Aug 30, 2021 at 6:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.