# Euler-Lagrange Equations for Two Body Problem

• _Kenny_
In summary, the conversation discusses the use of Lagrangian Mechanics and the two-body problem for a radial force. The Lagrangian of the system is given, and the individual is trying to find the Euler-Lagrange Equations for r_1 and r_2. They are confused about taking the derivative of the potential portion with respect to either r_1 or r_2. It is suggested to change coordinates to center of mass coordinates to simplify the problem. However, the individual wants to know what the Euler-Lagrange Equations would be if they remained with non-reduced coordinates. A discussion ensues about using Cartesian coordinates and the correct way to compute the derivative of the potential.
_Kenny_

## Homework Statement

I'm trying to do a little review of Lagrangian Mechanics through studying the two-body problem for a radial force. I have the Lagrangian of the system $$L=\frac{1}{2}m_1\dot{\vec{r_1}}^{2}+\frac{1}{2}m_2\dot{\vec{r_2}}^{2}-V(|{\vec{r_1}-\vec{r_2}}|)$$. Now I'm trying to find the Euler-Lagrange Equations for $r_1$ and $r_2$ but I'm confused about taking the derivative of the potential portion with respect to either $r_1$ or $r_2$. Please call me stupid and then tell me why I'm being stupid here.

## Homework Equations

$$L=\frac{1}{2}m_1\dot{\vec{r_1}}^{2}+\frac{1}{2}m_2\dot{\vec{r_2}}^{2}-V(|{\vec{r_1}-\vec{r_2}}|)$$

$$\frac{dL}{dq}=\frac{d}{dt}\frac{dL}{d\dot{q}}$$

[/B]

## The Attempt at a Solution

$$\frac{\partial L}{\partial r_1}=-\frac{\partial V(|{\vec{r_1}-\vec{r_2}}|)}{\partial r_1}=...?$$[/B]

You're a stupid because you're doing it in terms of $\vec r_1$ and $\vec r_2$. Change coordinates to $\vec R=\frac{m_1\vec r_1+m_2\vec r_2}{m_1+m_2}$ and $\vec r=\vec r_1-\vec r_2$(centre of mass coordinates) and you'll see that there is only one variable here!

In central force problems, it is useful to instead use the coordinates:
\begin{align} \vec{r} &= \vec{r}_2-\vec{r}_1\\ \vec{R} &= \frac{m_1\vec{r}_1 + m_2\vec{r}_2}{M} \end{align}
The reason is that with these coordinates the Lagrangian reduces to something that looks like a single body with a "reduced mass" $\mu$:
$$L=\frac{1}{2}\underbrace{\frac{m_1m_2}{m_1+m_2}}_{\mu}\dot{\vec{r}}^2 + \underbrace{\frac{1}{2}M\dot{\vec{R}}^2}_{0} - U(\vec{r}) \ ,$$

I'm not switching to the easier coordinates just yet, although I do know they simplify things greatly. What would the Euler-Lagrange Equations be if I remained with the non-reduced coordinates for right now?

_Kenny_ said:
I'm not switching to the easier coordinates just yet, although I do know they simplify things greatly. What would the Euler-Lagrange Equations be if I remained with the non-reduced coordinates for right now?
In contrast to what you may think, its not only a bit harder, its much harder. Because $|\vec r_1-\vec r_2|$ depends on all six components of the position vectors and so you should consider all the components in both the kinetic and potential terms. Just expand the position vectors w.r.t. their Cartesian components. So you'll have 6 EL equations.

_Kenny_ said:
$$\frac{\partial L}{\partial r_1}=-\frac{\partial V(|{\vec{r_1}-\vec{r_2}}|)}{\partial r_1}=...?$$

This is not entirely correct. ##\vec r_1## was a the position of a body. What is ##r_1##? Usually, that denotes the magnitude of the corresponding vector. You could use this as one of the generalized coordinates, but then you would need two other for the body, and what would those be?

I interpret your messages as if you really want to use Cartesian coordinates, in which case it is not ##r_1## that you should be using. It is ##r_{11}, \ r_{12}, \ r_{13} ##. Then you need to compute, for example, $$\partial V(|\vec r_1 - \vec r_2|) \over \partial r_{11} .$$ The question is how?

Observe that ##V(s)## is a scalar function of one scalar argument, so it has some derivative ##V'(s)##. Then you could use the chain rule and obtain $${\partial V(|\vec r_1 - \vec r_2|) \over \partial r_{11} } = V'(|\vec r_1 - \vec r_2|) {\partial |\vec r_1 - \vec r_2| \over \partial r_{11} }.$$

Can you continue from here?

## 1. What is the Euler-Lagrange equation for the two body problem?

The Euler-Lagrange equation for the two body problem is a mathematical equation that describes the motion of two interacting bodies under the influence of Newtonian gravity. It is derived from the principle of least action and is used to find the equations of motion for the two bodies.

## 2. How is the Euler-Lagrange equation derived for the two body problem?

The Euler-Lagrange equation for the two body problem is derived by setting up the action integral, which is the difference between the kinetic and potential energies of the system, and then applying the principle of least action. This results in a set of differential equations known as the Euler-Lagrange equations.

## 3. What are the assumptions made in the Euler-Lagrange equation for the two body problem?

The Euler-Lagrange equation for the two body problem assumes that the two bodies are point masses, that the gravitational force between them is the only force acting on them, and that the gravitational potential is a function of the distance between the two bodies.

## 4. How are the two body problem and the Euler-Lagrange equation related to each other?

The two body problem is a physical problem that involves determining the motion of two interacting bodies. The Euler-Lagrange equation is a mathematical tool used to solve this problem by finding the equations of motion for the two bodies. The two are related in that the Euler-Lagrange equation is derived specifically for the two body problem.

## 5. What are the applications of the Euler-Lagrange equation for the two body problem?

The Euler-Lagrange equation for the two body problem has many applications in physics and engineering. It is commonly used in celestial mechanics to study the motion of celestial bodies such as planets and satellites. It is also used in the design and control of spacecraft and other aerospace systems.

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