# Definition of effective potential energy arising from Lagrangian

In Lagrangian and Hamiltonian mechanics, it's common to define part of the kinetic energy as the "effective potential energy", but I am unclear on which expression we define this from. If we look at the Lagrangian and identify the part of the kinetic energy dependent on the generalized coordinate and not the velocity as part of the effective potential, we get a different sign for one of the terms in $V_{eff}$ than if we do the same thing when looking at the Hamiltonian.

From my experience, in scleronomic systems where the Hamiltonian is the total energy and constant, it is correct to identify the effective potential from the Hamiltonian. When energy is not conserved, we choose to identify the effective potential from the Lagrangian.

They are always different in sign. From this, I conclude that in the first case, it is only correct to take it out of the Hamiltonian and its wrong to do it the other way, and vice versa in the second case I described.

My question is, what is the difference that causes all of this? How is this choice the result of the need for the negative spatial derivative of the effective potential to equal the effective force felt by the mass?

• Commented Feb 23, 2017 at 23:26

To my knowledge, you define effective potentials when you have a conserved quantity in your system so that the motion takes place in less dimensions than that are available in Lagrangian.

From this point of view, what you call as the effective potential is the potential seen in the reduced dimensions. That is why the correct way to identify the effective potential is to examine the equation of motion, not the Lagrangian which lives in a higher dimension.

The simplest example may be a harmonic oscillator in two dimension: $$\mathcal{L}=\frac{1}{2}\left(\dot{r}^2+r^2\dot{\theta}^2\right)-\frac{1}{2}kr^2$$ where $k$ is the spring constant devided by mass. Clearly the Lagrangian lives in two dimensions, however, we have a conserved quantity, that is $$\frac{dl}{dt}=0\quad,\quad l=r^2\dot{\theta}$$ Hence the motion takes place in one dimension with the following EOM: $$\ddot{r}=-kr+\frac{l^2}{r^3}$$ But this EOM could have been obtained from one dimensional Lagrangian $$\mathcal{L}=\frac{1}{2}\dot{r}^2-\left(\frac{1}{2}kr^2+\frac{l^2}{2r^2}\right)$$ hence the effective potential is $$V_\text{eff}=\frac{1}{2}kr^2+\frac{l^2}{2r^2}$$

The reason you get a wrong minus sign in the effective potential if you insert $\dot{\theta}=lr^{-2}$ into the original Lagrangian is simple: This condition is valid in EOM only, which is merely a solution to $\delta\mathcal{L}=0$. You cannot expect $\mathcal{L}$, living in 2 dimensions, to be unmodified when you apply a $1d$ constraint to it.

Long story short: Whether it is Lagrangian formalism or any other one, what you call as the effective potential is the potential which is related to the motion in reduced dimensions. Well at least, these are the only cases that I am aware of.

• Thanks very much for your reply.I understand what you said about the effective potential having to satisfy the EOM rather than be simply pulled out of a 2D lagrangian, I am however unclear on when I can sub in θ˙=lr^-2 into my Lagrangian as I have seen questions in which this is done (where the time derivative of a holonomic constraint is subbed in). If I understand correctly you're saying to always identify v(eff) from the energy expression as in the case of a harmonic oscillator in which energy is conserved. Commented Feb 24, 2017 at 9:14
• However I have seen questions in which the energy is not conserved and the constraint is rheonomic, and because of these differences they chose to identify V(eff) out of the Lagrangian. under what circumstances can I assume that taking it out of the Lagrangian would satisfy the EOM? Commented Feb 24, 2017 at 9:14
• Firstly, I am not saying to always identify V(eff) from Hamiltonian: All I am saying that V(eff) should be identified in the relevant Lagrangian. You can identify V(eff) from Lagrangian as well, but that Lagrangian should be in the same dimensions with the motion. In above example, I used Lagrangian to identify V(eff), but 1d Lagrangian not 2d. In above example, we did not have a constraint: $l=r^2\dot{\theta}$ is not a constraint, but an EOM. You can impose constraints in Lagrangian, but not EOM's. I hope this clarifies :) Commented Feb 24, 2017 at 23:23

To complement Soner's answer, I want to clarify why we cannot naively simply plug in $$\dot{\theta}=l/mr^2$$ in the Lagrangian. This is discussed in a different context in exercise 1.5 of Henneaux, Marc, and Claudio Teitelboim. Quantization of Gauge Systems. Princeton University Press, 1992.

First of all, let us recall that if we want to obtain the equations of motion for an orbit starting at $$(r_i,\theta_i)$$ and finishing at $$(r_f,\theta_f)$$, we want to minimize the action $$S(r,\theta)=\int_{t_i}^{t_f}\text{d}t\left(\frac{1}{2}m\dot{r}^2+\frac{1}{2}mr^2\dot{\theta}^2-U(r)\right)$$ on the space $$\mathcal{E}$$ of trajectories $$r=r(t)$$ and $$\theta=\theta(t)$$ satisfying $$r(t_i)=r_i$$, $$\theta(t_i)=\theta_i$$, $$r(t_f)=r_f$$, and $$\theta(t_f)=\theta_f$$. What one can show, say from Noether's theorem, is that any minima of this action satisfies that $$l=mr^2\dot{\theta}$$ is a constant. This would imply that at the minima of the action, $$\dot{\theta}=l/mr^2$$ for some constant $$l$$. By integrating both sides, we see that this constant, by virtue of the boundary conditions of the space we are working with, is, in fact, a functional of the trajectory $$r=r(t)$$ $$l(r)=\frac{m\Delta\theta}{\int_{t_0}^{t_f}\text{d}t\,\frac{1}{r^2}},\quad\Delta\theta=\theta_f-\theta_i.$$ You see, the difficulty here is one of language. While $$l$$ is indeed a constant over time, it is not a constant on the space of trajectories $$\mathcal{E}$$.

We can now proceed keeping this in mind. On the minima, we know that $$\int_{t_0}^{t_f}\text{d}t\,\frac{1}{2}mr^2\dot{\theta}^2=\int_{t_0}^{t_f}\text{d}t\,\frac{l(r)^2}{2mr^2}=\frac{l(r)^2}{2m}\int_{t_0}^{t_f}\text{d}t\,\frac{1}{r^2}=\frac{\Delta\theta}{2}l(r).$$ Therefore, the minima is also obtained by minimizing on $$X$$ the function $$\tilde{S}(r,\theta)=\int_{t_i}^{t_f}\text{d}t\left(\frac{1}{2}m\dot{r}^2+\frac{l(r)^2}{2mr^2}-U(r)\right)=\int_{t_i}^{t_f}\text{d}t\left(\frac{1}{2}m\dot{r}^2-U(r)\right)+\frac{\Delta\theta}{2}l(r)$$ This is a well-posed problem, although it is not one that can be solved using Euler-Lagrange equations. The problem is that this action is not local, since the functional $$l(r)$$ is not the integral of some density. One can however proceed noting that $$\delta l=\frac{2m\Delta\theta}{\left(\int_{t_i}^{t_f}\text{d}t\,\frac{1}{r^2}\right)^2}\int_{t_i}^{t_f}\text{d}t\,\frac{1}{r^3}\delta r=\frac{2l(r)^2}{m\Delta\theta}\int_{t_i}^{t_f}\text{d}t\,\frac{1}{r^3}\delta r.$$ We conclude $$\delta \tilde{S}=\int_{t_i}^{t_f}\text{d}t\,\left(-m\ddot{r}-U'(r)+\frac{l(r)^2}{mr^3}\right)\delta r,$$ which gives the correct equations of motion $$m\ddot{r}=-U'(r)+\frac{l^2}{mr^3}.$$

Now, while this clarifies the subtleties of replacing $$\dot{\theta}=l/mr^2$$ in the action, it doesn't bring any new light on how one can obtain from the action $$\tilde{S}$$ the equivalent action $$S(r)=\int_{t_i}^{t_f}\text{d}t\,\left(\frac{1}{2}m\dot{r}^2-\frac{l^2}{2mr^2}-U(r)\right),$$ for a truly constant $$l$$, of course, other than noting that they both give the same minima. There is another procedure, shedding light into the role of boundary conditions, which does give the correct action.

Since adding boundary terms preserves the equations of motion, we can instead consider the action $$\bar{S}(r,\theta)=\int_{t_i}^{t_f}\text{d}t\left(\frac{1}{2}m\dot{r}^2+\frac{1}{2}mr^2\dot{\theta}^2-U(r)\right)-[mr^2\theta\dot{\theta}]_{t_i}^{t_f}\,=\int_{t_i}^{t_f}\text{d}t\left(\frac{1}{2}m\dot{r}^2+\frac{1}{2}mr^2\dot{\theta}^2-U(r)-\frac{\text{d}}{\text{d}t}(mr^2\theta\dot{\theta})\right).$$ Indeed, the variation of such an action is of the form $$\delta\bar{S}=\int_{t_0}^{t_f}\text{d}t\,\left((\dots)\delta r+(\dots)\delta\theta\right)+\left[(\dots)\delta r+(\dots)\delta{\dot{\theta}}\right]_{t_0}^{t_f}.$$ Thus, the variational problem for $$\bar{S}$$ is only well-posed if we consider spaces of trajectories where we fix boundary conditions for $$\dot{\theta}$$ instead of $$\theta$$. In such spaces, $$l$$ is truly a constant, e.g. $$l=mr_i^2\dot{\theta}_i,$$ and we can consider the equivalent action in which $$\dot{\theta}=l/mr^2$$ $$\bar{S}(r,\theta)=\int_{t_i}^{t_f}\text{d}t\left(\frac{1}{2}m\dot{r}^2+\frac{l^2}{2mr^2}-U(r)\right)-l[\theta]_{t_i}^{t_f}\,=\int_{t_i}^{t_f}\text{d}t\left(\frac{1}{2}m\dot{r}^2+\frac{l^2}{2mr^2}-U(r)-l\dot{\theta}\right)=\int_{t_i}^{t_f}\text{d}t\left(\frac{1}{2}m\dot{r}^2-\frac{l^2}{2mr^2}-U(r)\right)=S(r).$$ In particular, we now obtain the correct effective potential term.