# Is the force equation derived from the Euler–Lagrange equivalent to the force equation derived from energy conservation?

I'm looking at the problem of a mass moving under the affect of a central potential $V(r)$. I can get find the force equation in two ways, and it seems to me like I'm getting two different equations... Why is that?

The force equation from energy conservation:

$$\frac{1}{2}m(\dot{r}^2+r^2\dot{\theta}^2)+V(r)=E_0$$ Where the first term is the kinetic energy, the second is the potential energy of the central force, $E_0$ is the conserved energy, and we work in polar coordinates where the force points to the origin. If we take a time derivative and divide the result by $\dot{r}$ we get the force equation:

$$m(\dot{r}\ddot{r}+r\dot{r}\dot{\theta}^2+r^2\dot{\theta}\ddot{\theta})+\frac{\partial V(r)}{\partial r}\dot{r}=0$$ $$m\ddot{r}+mr\dot{\theta}^2+m\frac{r^2}{\dot{r}}\dot{\theta}\ddot{\theta}+\frac{\partial V(r)}{\partial r}=0$$

On the other hand we may write the Lagrangian: $$L=\frac{1}{2}m(\dot{r}^2+r^2\dot{\theta}^2)-V(r)$$ And use the Euler–Lagrange equations to find the force equation: $$Euler–Lagrange:\frac{d}{dt}\frac{\partial L}{\partial \dot{q_i}}-\frac{\partial L}{\partial q_i}=0$$ Where $q_i$ are the coordinates: $r,\theta$. We get: $$m\ddot{r}+2mr\dot{r}\dot{\theta}+mr^2\ddot{\theta}-mr\dot{\theta}^2+\frac{\partial V}{\partial r}=0$$

Which is not the same as the force equation we got from the conservation of energy!

** edit: Notice that I have a mistake in the last equation. Euler–Lagrange is a set of two equations, and I thought it was a summation over i. that is not true, obviously, from unit considerations.... I'm keeping the mistake in the question in case someone else does the same mistake **

• Kyle, I don't understand your question. – Adi Ro May 9 '17 at 11:52
• @Kyle The chain rule. $\partial_tV(r)=\dot{r}V'(r)$. – Bob Knighton May 9 '17 at 11:57
• Ah, nevermind. I was being foolish. – Kyle Kanos May 9 '17 at 11:58

## 1 Answer

This is true, you don't get the same equation. This is because you still haven't derived all of the equations your system satisfies. For instance, let's look at the $\theta$ Euler-Lagrange equation. It reads

$$\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial\dot{\theta}}=0$$

Which implies that the quantity $\ell\equiv\partial L/\partial\dot{\theta}=mr^2\dot{\theta}$ is conserved (and is indeed the angular momentum of the system). Using the equation $\dot{\ell}=0$ as well as $\dot{E}=0$ should give you agreement with the Lagrangian equations of motion. Indeed

$$\dot{\ell}=0\longleftrightarrow mr^2\ddot{\theta}\dot{\theta}=-2mr\dot{r}\dot{\theta}^2$$

Which should fix your discrepancy. Moreover, the $r$ Euler-Lagrange equation should give you

$$m\ddot{r}-mr\dot{\theta}^2+V'(r)=0$$

Which, along with the $\dot{\ell}=0$ constraint gives you the conservation of energy equation, naturallyv (I'm honestly unsure where you got the Euler-Lagrange equations in your answer, since dimensional analysis tells you that some of the terms don't even have the same units).

If you want to deduce the equations of motion of your system using conservation methods, you need to make sure you are using every conserved quantity, or else you are underdetermining the motion of your system. The Lagrangian method magically takes care of this for you, by identifying which quantities are conserved simply based on symmetries via the Noether method.

I hope this helped!