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The Lagrangian of the two-body problem in terms of polar coordinate $r$ and $\phi$ is $$\mathcal{L}=\frac{1}{2} \mu (\dot{r}^2+r^2\dot{\phi}^2)-U(r)$$ And the Lagrange equation corresponding to $\phi$ is a statement of conservation of angular momentum $$\frac{\partial \mathcal{L}}{\partial \dot{\phi}}=\mu r^2 \dot{\phi}=const=l$$ If I use this equation to eliminate the $\dot{\phi}$ in the first equation, I'll have $$\mathcal{L}=\frac{1}{2} \mu \dot{r}^2+\frac{l^2}{2\mu r^2}-U(r)$$ We can define $U_{eff}(r)=U(r)-\frac{l^2}{2\mu r^2}$ so that $\mathcal{L}=\frac{1}{2} \mu \dot{r}^2-U_{eff}(r)$

However, in Taylor's book Classical Mechanics, he first looks at the Lagrange equation corresponding to $r$ (or radial equation) $$\mu r \dot{\phi^2}-\frac{dU}{dr}=\mu \ddot{r}$$ Then define centrifugal force as the first term in LHS $$F_{cf}=\mu r \dot{\phi}^2= \frac{l^2}{\mu r^3}=-\frac{d}{dr}(\frac{l^2}{2 \mu r^2})= -\frac{d U_{cf}}{dr}$$ Return to the radial equation, we have $$\mu \ddot{r}=-\frac{d}{dr}[U(r)+ \frac{l^2}{2 \mu r^2}]=-\frac{d}{dr} U_{eff}(r)$$ It seems that $U_{eff}(r)=U(r)+\frac{l^2}{2 \mu r^2}$ in Taylor's book, instead of $U_{eff}(r)=U(r)-\frac{l^2}{2\mu r^2}$ as we got previously.

I'm quite confused that why I get a minus sign in the definition of $U_{eff}(r)$ if I look directly at the Lagrangian, using the statement of conservation of angular momentum. I wonder I have this minus sign because I change the form of Lagrangian too early but I can't figure out why I can't do it.


marked as duplicate by Qmechanic Jul 31 '17 at 8:49

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