If you have a conserved quantity, why can't you use it to eliminate a variable in the Lagrangian?

Question

Suppose, for example, we take a particle in polar coordinates $(r, \theta)$ with a central force, so $U = U(r).$ The Lagrangian is $$\mathcal{L} = \dfrac12 m (\dot{r}^2 + (r\dot{\theta})^2) - U(r).$$

The corresponding equation of motion for $r$ is $$m\ddot{r} = m\dot{\theta}^2r - \dfrac{\mathrm{d}U}{\mathrm{d}r}.$$

Using angular momentum conservation, $L = mr^2 \dot{\theta}$ is constant, so we can use it to eliminate $\dot{\theta}$ and obtain $$m\ddot{r} = \dfrac{L^2}{mr^3} - \dfrac{\mathrm{d}U}{\mathrm{d}r}.$$

However, suppose we return to the original statement of the Lagrangian and eliminate $\dot{\theta}$ there. Then we'd have $$\mathcal{L} = \dfrac12 m\left(\dot{r}^2 + \dfrac{L^2}{m^2r^2}\right) - \dfrac{\mathrm{d}U}{\mathrm{d}r}.$$

Applying the Euler-Lagrange equation $\dfrac{\mathrm{d}}{\mathrm{d}t}\dfrac{\partial \mathcal{L}}{\partial \dot{r}} = \dfrac{\partial \mathcal{L}}{\partial r}$ to the above, $$m\ddot{r} = -\dfrac{L^2}{mr^3}-\dfrac{\mathrm{d}U}{\mathrm{d}r}.$$

This is wrong; we can't eliminate $\dot{\theta}$ in the Lagrangian this way before applying the Euler-Lagrange equations, but why not?

Answer 1

You can eliminate coordinates. You are just not allowed to eliminate velocities. If you want to eliminate a velocity, you have to go over to the Routhian. Landau Lifshitz has a discussion in paragraph 41.

The idea is that you want to replace the velocity $\dot \theta$ by the conserved quantity which is the associated generalized momentum $L$. In order to do that you have to perform a (partial) Legendre transform (essentially going over to the Hamiltonian with respect to $\dot \theta$). So for fixed anuglar momentum $L$, we have the Routhian (note that LL use a different sign convention) $$\mathcal{R}= \mathcal{L}- \dot\theta L =\frac{m}{2} \dot r^2 -\frac{L^2}{2m r^2} - U(r) =\frac{m}{2} \dot r^2 - U_\text{eff}(r) \,.$$ So the correct Lagrangian is given by $\mathcal{R}$ the effect of the angular coordinate is the centrifugal potential $L^2/2mr^2$.