# Solving two body central force motion using Lagrangian

• deuteron
deuteron
Homework Statement
Find the equation of motion
Relevant Equations
##\mathcal L= T-U##
For the central force ##F=-\nabla U(r_r)## where ##\vec r_r=\vec r_1-\vec r_2##, and ##\vec r_1## and ##\vec r_2## denote the positions of the masses, we get the following kinetic energy using the definition of center of mass ##\vec r_{cm}= \frac{m_1\vec r_1+m_2\vec r_2}{m_1+m_2}##:

$$T= \frac 1 2 (m_1+m_2) \dot{\vec r}_{cm}^2 + \frac 12 \frac{m_1m_2}{m_1+m_2} \dot {\vec r}_r^2$$

We can write the ##\dot {\vec r}## terms in polar coordinates as:

$$\dot{\vec r} = \dot r^2 \vec e_r+ r^2\dot\theta^2\vec e_\theta$$

However, then we get the following equation of mass for the center of mass:

$$\frac{\partial\mathcal L}{\partial r_{cm}}=(m_1+m_2)r_{cm}\dot\theta_{cm}= \frac d {dt}\frac {\partial\mathcal L}{\partial\dot r_{cm}}= M\ddot r_{cm}$$
$$\frac {\partial\mathcal L}{\partial\theta_{cm}}=0=\frac d {dt} \frac {\partial\mathcal L}{\partial \dot \theta_{cm}}= (m_1+m_2) \ddot\theta_{cm}$$

from which we get:

$$\dot\theta_{cm}=\text{const.}$$
$$M\ddot r _{cm} = M r_{cm}\dot\theta_{cm}$$

and if we don't decide ##\vec r_{cm}## to be the origin, which I don't think we *have* to do, then ##\ddot r_{cm}## has a value, which I am not really sure is true. What am I doing wrong above?

Last edited:
PhDeezNutz
What generalized coordinates are you using? Remember that you need 3 coordinates per mass which means 6 for this case. This ##r_{cm}= \frac{m_1r_1+m-2r_2}{m_1+m_2}## is not the definition of the center of mass. It is the vector equation $$\mathbf{ R}_{cm}=\frac{m_1\mathbf{ r}_1+m_2\mathbf{ r}_2}{m_1+m_2}.$$

kuruman said:
What generalized coordinates are you using? Remember that you need 3 coordinates per mass which means 6 for this case. This ##r_{cm}= \frac{m_1r_1+m-2r_2}{m_1+m_2}## is not the definition of the center of mass. It is the vector equation $$\mathbf{ R}_{cm}=\frac{m_1\mathbf{ r}_1+m_2\mathbf{ r}_2}{m_1+m_2}.$$
Since in central force problems angular momentum is conserved and therefore the masses move in the plane perpendicular to the angular momentum we only need 2 coordinates per mass, in this case total 4 coordinates, which I started with Cartesian and then changed to polar coordinates at the line "We can write the ##\dot{\vec r}## in˙ terms in polar coordinates as:"

Thanks for the warning tho, I fixed the notation with the vector symbols

The statement of the problem is just "Find the equation of motion." Of what? I don't understand ##\theta_{cm}##. If you have two interacting masses, the CM coordinates separate out and you are left with a Lagrangian in relative coordinates. If you have two interacting masses in a central potential then ##\theta_{cm}## makes sense, but you have to choose your origin at the force center.

Please write down a precise statement of the problem that you are considering. Also, it would help if you provided a diagram defining your coordinates.

kuruman said:
The statement of the problem is just "Find the equation of motion." Of what? I don't understand ##\theta_{cm}##. If you have two interacting masses, the CM coordinates separate out and you are left with a Lagrangian in relative coordinates. If you have two interacting masses in a central potential then ##\theta_{cm}## makes sense, but you have to choose your origin at the force center.

Please write down a precise statement of the problem that you are considering. Also, it would help if you provided a diagram defining your coordinates.
The equation of motion of the masses that are interacting under the central potential. ##\theta_{cm}## refers to the angle that the position vector ##r_{cm}## makes with the ##x-##axis of the chosen coordinate frame (polar coordinates). There is not one force center since both the bodies exert central force on each other. This is a basic Kepler two body problem, I just mathematically can't get to the statement that ##\dot {\vec r}_{cm}=\text{const.}##

Then why do you even bother with ##\theta_{cm}##? Define $$\mathbf{ R}=\frac{m_1\mathbf{ r}_1+m_2\mathbf{ r}_2}{m_1+m_2}~;~~~\mathbf{r}=\mathbf{r}_2-\mathbf{r}_1$$solve for ##\mathbf{r}_2## and ##\mathbf{r}_1## in terms of ##\mathbf{R}## and ##\mathbf{r}## and substitute in $$T=\frac{1}{2}m_1 \mathbf{\dot{r}_1}^2+\frac{1}{2}m_2 \mathbf{\dot{r}_2}^2.$$ The motion of the CM should separate out and result in ##\mathbf{\dot R}=\rm{const.}## Use the Cartesian representation ##\mathbf{r}_i=x_i~\mathbf{\hat x}+y_i~\mathbf{\hat y}+z_i~\mathbf{\hat z}.##

kuruman said:
Then why do you even bother with ##\theta_{cm}##? Define $$\mathbf{ R}=\frac{m_1\mathbf{ r}_1+m_2\mathbf{ r}_2}{m_1+m_2}~;~~~\mathbf{r}=\mathbf{r}_2-\mathbf{r}_1$$solve for ##\mathbf{r}_2## and ##\mathbf{r}_1## in terms of ##\mathbf{R}## and ##\mathbf{r}## and substitute in $$T=\frac{1}{2}m_1 \mathbf{\dot{r}_1}^2+\frac{1}{2}m_2 \mathbf{\dot{r}_2}^2.$$ The motion of the CM should separate out and result in ##\mathbf{\dot R}=\rm{const.}## Use the Cartesian representation ##\mathbf{r}_i=x_i~\mathbf{\hat x}+y_i~\mathbf{\hat y}+z_i~\mathbf{\hat z}.##
yes i already did that and started the question after arriving at ##T= \frac 12 (m_1+m_2) \dot{\vec r}_{cm}^2 + \frac 12 \mu \dot{\vec r}_r^2##

After that, I wrote ##\vec r_{cm}## and ##\vec r_r## in terms of polar coordinates, and then applied Lagrangian equations of motion as you'll see in my question.

After applying Lagrange, I got a non-zero expression for the angular and radial parts of the center of mass, which should not have happened, that's why I think I did something wrong, but I can't figure out which step is the wrong one.

deuteron said:
yes i already did that and started the question after arriving at ##T= \frac 12 (m_1+m_2) \dot{\vec r}_{cm}^2 + \frac 12 \mu \dot{\vec r}_r^2##

After that, I wrote ##\vec r_{cm}## and ##\vec r_r## in terms of polar coordinates, and then applied Lagrangian equations of motion as you'll see in my question.

After applying Lagrange, I got a non-zero expression for the angular and radial parts of the center of mass, which should not have happened, that's why I think I did something wrong, but I can't figure out which step is the wrong one.
Can you explain how you get ##\dot R=\text{const.}## and what you mean with "the motion of the CM should separate out"?

deuteron said:
yes i already did that and started the question after arriving at ##T= \frac 12 (m_1+m_2) \dot{\vec r}_{cm}^2 + \frac 12 \mu \dot{\vec r}_r^2##
OK, then your Lagrangian is $$\mathcal{L}=\frac 12 (m_1+m_2) \dot{\vec r}_{cm}^2 +\frac 12 \mu \dot{\vec r}_r^2-V(r_r)$$ For generalized coordinate ##\vec {r}_{cm}## three equations of motion are obtained from $$\frac{d}{dt}\frac{\partial \mathcal{L}}{\partial{\vec {\dot r}_{cm}}}-\frac{\partial \mathcal L}{\partial \vec r_{cm}}=0.$$What do you get? Repeat for generalized coordinate ##\vec r_r.##

Last edited:
deuteron said:
Can you explain how you get ##\dot R=\text{const.}## and what you mean with "the motion of the CM should separate out"?
If you obtain the equation of motion for ##\mathbf{R}=\vec r_{cm}## as I outlined in post #9, you will get ##(m_1+m_2)\ddot{\mathbf R}=0## which implies ##\dot{\mathbf R}=\text{const.}## You can actually do that in your head by just looking at the Lagrangian in post #9.

The motion of the CM separates out in the sense that the equation of motion for the CM has no ##\mathbf r## in it and the equation of motion for ##\mathbf r## has no ##\mathbf R## in it.

kuruman said:
OK, then your Lagrangian is $$\mathcal{L}=\frac 12 (m_1+m_2) \dot{\vec r}_{cm}^2 +\frac 12 \mu \dot{\vec r}_r^2-V(r_r)$$ For generalized coordinate ##\vec {r}_{cm}## three equations of motion are obtained from $$\frac{d}{dt}\frac{\partial \vec {r}_{cm}}{\partial{\vec {\dot r}_{cm}}}-\frac{\partial \mathcal L}{\partial \vec r_{cm}}=0.$$What do you get? Repeat for generalized coordinate ##\vec r_r.##
But are we allowed to derive ##\mathcal L## with respect to a vector ##\vec r_{cm}##? Instead, if we decompose the vector into its polar coordinates, we get for the Lagrangian

$$\mathcal L= \frac 12 M (\dot r_{cm}^2 + r_{cm}^2 \dot\theta_{cm}^2) + \frac 12\mu (\dot r_r^2 + r_r^2\dot\theta_r^2)-V(r_r)$$

which then gives for the equation of motion:

$$\frac {\partial\mathcal L}{\partial r_{cm}} = M r \dot\theta_{cm}^2$$

$$\frac d {dt}\frac {\partial\mathcal L}{\partial\dot r_{cm}}= M\ddot r_{cm}$$

You missed my point in post#9. I said there are three equations. This means that there are 3 separate equations of motion for the CM in Cartesian coordinates. Forget the polar representation for the CM. It can only buy you grief. $$\frac{d}{dt}\frac{\partial \vec {r}_{cm}}{\partial{\vec {\dot r}_{cm}}}-\frac{\partial \mathcal L}{\partial \vec r_{cm}}=0\rightarrow \begin{cases} \frac{d}{dt}\frac{\partial \mathcal{L}}{\partial{{\dot x}_{cm}}}-\frac{\partial \mathcal L}{\partial x_{cm}} =0\\ \frac{d}{dt}\frac{\partial \mathcal{L}}{\partial{{\dot y}_{cm}}}-\frac{\partial \mathcal L}{\partial y_{cm}} =0 \\ \frac{d}{dt}\frac{\partial \mathcal{L}}{\partial{{\dot z}_{cm}}}-\frac{\partial \mathcal L}{\partial z_{cm}} =0 \end{cases}$$

## What is the two-body central force problem?

The two-body central force problem involves studying the motion of two particles that interact through a force that depends only on the distance between them. This is a fundamental problem in classical mechanics and is essential for understanding systems such as planetary orbits, binary stars, and molecular interactions.

## How do you set up the Lagrangian for the two-body central force problem?

To set up the Lagrangian, you start by expressing the kinetic energy of the system in terms of the relative and center-of-mass coordinates. The potential energy depends only on the distance between the two bodies. The Lagrangian $$L$$ is then given by the difference between the total kinetic energy and the potential energy: $$L = T - V$$, where $$T$$ is the kinetic energy and $$V$$ is the potential energy.

## What is the significance of reducing the two-body problem to a one-body problem?

Reducing the two-body problem to a one-body problem simplifies the analysis significantly. By transforming to the center-of-mass and relative coordinates, the problem can be reduced to studying the motion of a single particle with a reduced mass moving in an effective potential. This reduction makes it easier to solve the equations of motion and understand the dynamics of the system.

## How do you derive the equations of motion from the Lagrangian?

The equations of motion are derived using the Euler-Lagrange equations, which are obtained by taking the partial derivatives of the Lagrangian with respect to the generalized coordinates and their time derivatives. For a coordinate $$q_i$$, the Euler-Lagrange equation is given by $$\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q_i}} \right) - \frac{\partial L}{\partial q_i} = 0$$. Applying this to the Lagrangian of the two-body problem yields the equations of motion for the system.

## What are the conserved quantities in the two-body central force problem?

In the two-body central force problem, the conserved quantities are the total energy, the linear momentum, and the angular momentum. The conservation of these quantities arises from the symmetries of the system: time translation symmetry leads to energy conservation, spatial translation symmetry leads to linear momentum conservation, and rotational symmetry leads to angular momentum conservation.

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