Lagrangian, central forces and conservation of angular momentum [duplicate]

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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 StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jul 10 at 8:52

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 $$r$$

I)

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

equation of motion

$$m\,\ddot{r}=m\,r\,\dot{\theta}^2-\frac{d}{dr}U(r)$$

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

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

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

II)

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

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

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

equation of motion

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

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