2016-06-12 12:14:47 8 Comments

If there is a system, described by an Lagrangian $\mathcal{L}$ of the form

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

where $T$ is the kinetic energy and $V$ the potential energy, it is also possible to define the total energy $E$ of the system

$$ E = T + V =\frac{m}{2}\left(\dot{r}^2+r^2\dot{\phi}^2\right) - \frac{k}{r}.\tag{2}$$

If the angular momentum $M$ is defined by $$ M = m r^{2} \dot{\phi},\tag{3}$$ then $E$ can be written as

$$ E = \frac{m}{2}\dot{r}^2 + \underbrace{\frac{M^2}{2mr^2}-\frac{k}{r}}_{V_\textrm{eff}\left(r\right)},\tag{4}$$

where the last two terms are written as new, "effective" potential $V_\textrm{eff}\left(r\right)$.

In addition, using the definition of $M$, the Lagrangian $\mathcal{L}$ can be written as

$$ \mathcal{L} = \frac{m}{2}\dot{r}^2 + \underbrace{\frac{M^2}{2mr^2}+\frac{k}{r}}_{-V_\textrm{eff}\left(r\right)},\tag{5}$$

where the sign of $V_\textrm{eff}\left(r\right)$ has been changed, since $\mathcal{L} = T-V$. But from this argument, it appears that there are two different possible ways to construct the same effective potential. This seems to me contradictory. Where is my mistake?

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## 2 comments

## @Qmechanic 2016-06-12 13:27:41

The underlying reason for OP's flawed argument is, that a premature use of EOMs in the stationary action principle $$ S~=~\int\!dt ~L(r,\dot{r};\theta,\dot{\theta}), \qquad L(r,\dot{r};\theta,\dot{\theta})~=~\frac{1}{2}m(\dot{r}^2 +r^2\dot{\theta}^2) -V(r),\tag{A}$$ invalidates the variational principle. Concretely OP is achieving the incorrect Lagrangian (5) by eliminating the angular coordinate from the original Lagrangian (1) via the fact that the angular momentum (3) is a constant of motion (which is the EOM for the angular coordinate).

Now for OP's example, it turns out that a correct reduction can be perform via the Hamiltonian formulation $$ H(r,p_r;\theta,p_{\theta}) ~=~\frac{p_{r}^2}{2m}+ \frac{p_{\theta}^2}{2mr^2} + V(r). \tag{B}$$ Note that the canonical momentum $p_{\theta}$ (which is the angular momentum) is a constant of motion because $\theta$ is a cyclic variable.

We next re-interpret the system (B) in a rotating frame following the particle with fictitious forces and only 1D radial kinematics. The Hamiltonian (B) becomes $$ H(r,p_r)~=~\frac{p_r^2}{2m}+ V_{\rm cf}(r)+ V(r), \tag{C}$$ where $$ V_{\rm cf}(r)~:=~\frac{p_{\theta}^2}{2mr^2} \tag{D}$$ is a centrifugal potential in a 1D radial world, cf. my Phys.SE answer here. Hence the Hamiltonian system (C) has effectively only one radial degree of freedom.

Finally, perform a Legendre transformation on the Hamiltonian (C) to obtain the corresponding 1D Lagrangian system: $$ L(r,\dot{r})~=~\frac{m}{2 }\dot{r}^2\color{Red}{-}V_{\rm cf}(r)-V(r).\tag{E} $$ Note the crucial minus sign marked in red. One may check that the corresponding EL eq. for (E) is the correct EOM.

The Lagrangian (E) is minus Routhian, cf. this Phys.SE post.

## @Blazej 2016-06-12 12:34:46

Effective potential is defined by the formula $E=T_{radial}+V_{eff}(r)$. Your calculation shows that once you make this identification it is not true that $\mathcal L = T_{radial}-V_{eff}$, but that is fine. This happens because centrifugal term (i.e. the one with angular momentum) is really kinetic term and not a true potential. Hence it must enter the Lagrangian with plus sign, unlike the "real" potentials.