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When studying central forces it is possible to propose the Lagrangian:

$$ L = T-U=\frac{1}{2}m \dot{r}^2+\frac{1}{2}mr^2 \dot{\theta}^2 - U(r)$$

Then we can solve the equation of motions for $\theta$: $$ \frac{d}{dt}\frac{\partial}{\partial \dot{\theta}}L-\frac{\partial}{\partial \theta}L=0=\frac{d}{dt}\left( mr^2\dot{\theta}\right) \rightarrow mr^2\dot{\theta} = constant = l$$

Then $\dot{\theta} = l/mr^2$

Why can't I take this equality and put it into the Lagrangian $\mathcal{L}$ before solving the equations of motion for $r$? So:

$$ L = \frac{1}{2}m \dot{r}^2+\frac{l^2}{2 mr^2} - U(r) $$

$$ \frac{d}{dt}\frac{\partial}{\partial \dot{r}}L-\frac{\partial}{\partial r}L=0\rightarrow\frac{d}{dt}(mr)+\frac{l^2}{mr^3}+\frac{\partial}{\partial r}U(r)=0$$

The result would be obviously different if I solve the equations for $r$ and then substitute $\dot{\theta} = l/mr^2$:

$$L = \frac{1}{2}m \dot{r}^2+\frac{1}{2}mr^2 \dot{\theta}^2 - U(r)$$

$$\frac{d}{dt}\frac{\partial}{\partial \dot{r}}L-\frac{\partial}{\partial r}L=0\rightarrow\frac{d}{dt}(mr)-\frac{l^2}{mr^3}+\frac{\partial}{\partial r}U(r)=0$$

In this case, the difference is the sign on the angular momentum term. But in general, it is not the same at all. Why is solving the Euler-Lagrange Equation first the right approach?


marked as duplicate by Qmechanic classical-mechanics Jul 10 at 8:52

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I think that your mistake is treating the $\theta$ as generalized coordinate, which is not any more. You can get the equation of motion either with method I or II.

the generalized coordinate is in both method $r$



equation of motion


with: $\dot{\theta}=\frac{l}{m\,r^2}$

$$m\,\ddot{r}=m\,r\,\left( \frac{l}{m\,r^2}\right)^2-\frac{d}{dr}U(r)$$

$$\ddot{r}=-{\frac {{l}^{2}}{{m}^{2}{r}^{3}}}-{\frac {{\frac {d}{dr}}U \left( r \right) }{m}} $$


with: $\dot{\theta}=\frac{l}{m\,r^2}$



equation of motion

$$\ddot{r}=-{\frac {{l}^{2}}{{m}^{2}{r}^{3}}}-{\frac {{\frac {d}{dr}}U \left( r \right) }{m}} $$

  • $\begingroup$ I think in methd II you are missing a (-) sign in the equation of motion. In the first term of the right side. $\endgroup$ – IvanMartinez Jul 10 at 14:36
  • $\begingroup$ But the equation are the same as in method I? $\endgroup$ – Eli Jul 11 at 7:15

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