# What does this Euler-Lagrange equation represent?

I have the following Lagrangian:

$$L=\frac{\mu}{2}\left(\dot{r}^2+r^2\dot\phi^2\right)-U(r),$$

The Euler-Lagrange equations are thus:

$$\frac{d}{dt}\left(\mu r^2\dot\phi\right)=0$$ $$\frac{d}{dt}(\mu \dot r)=\mu r\dot{\phi}^2-\frac{\partial}{\partial r}U(r).$$

I am trouble understanding what each Euler-Lagrange equation represent:

For example the first one: $$\frac{d}{dt}\left(\mu r^2\dot\phi\right)=0$$

What does $$\mu r^2\dot\phi\$$ means?

Would it be the angular momentum?

• All symbols are undefined. What does $\mu$ mean, and r, $\phi$, U(r)? And what again is the definition of angular momentum? Mar 5, 2021 at 11:49
• Mar 5, 2021 at 15:51

## 2 Answers

Yes, the term $$\mu r^2\dot{\phi}$$ represents the angular momentum, and the first Euler equation telling you that the angular momentum is the constant of motion for the given problem. The second Euler equation is just the radial force equation. If you can try to write newton's equation of motion like $$F_r=m(\ddot{r}-r\dot{\theta}^2)\ \ \text{and}\ \ \ F_\theta=m(r\ddot{\theta}+2\dot{r}\dot{\theta})$$ Then you can compare it to the Euler equation to get the meaning of the terms.

• You are making assumptions. Reasonable assumptions , but still. We don't know that $\phi$ is even an angle. All symbols in the question are undefined. Mar 5, 2021 at 11:50
• Yes! I can understand. OP didn't specify the problem, It has just given Langrangian and you can not interpret anything from that. Except for what @AFG has done. Mar 5, 2021 at 12:08

It is the generalized momentum conjugate to $$\phi$$, defined as

$$p_\phi=\frac{\partial L}{\partial \dot{\phi}}.$$

With your choice of generalized coordinates, $$r$$ and $$\phi$$, it is the angular momentum.

As $$\phi$$ itself does not appear in the Lagrangian, it is said to be an $$\textit{ignorable}$$ coordinate, so its conjugate momentum is a conserved quantity. This comes from Euler-Lagrange equations, $$\frac{d}{dt}\Big(\frac{\partial L}{\partial \dot{q}_i}\Big)=\frac{\partial L}{\partial q_i},$$ which can be rewritten as $$\frac{dp_i}{dt}=\frac{\partial L}{\partial q_i}$$ with the definition above. You can see that if the Lagrangian does not explicitey depend on a certain generalized coordinate $$q_i$$, its conjugate momentum $$p_i$$ is conserved.