Solving two body central force motion using Lagrangian

[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?

 

[kuruman]

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 $$\mathbf{ R}_{cm}=\frac{m_1\mathbf{ r}_1+m_2\mathbf{ r}_2}{m_1+m_2}.$$

 

[deuteron]

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

 

[kuruman]

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.

 

[deuteron]

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 $\dot {\vec r}_{cm}=\text{const.}$

 

[kuruman]

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 $\mathbf{r}_i=x_i~\mathbf{\hat x}+y_i~\mathbf{\hat y}+z_i~\mathbf{\hat z}.$

 

[deuteron]

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]

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"?

 

[kuruman]

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 $\vec r_r.$

 

[kuruman]

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.

 

[deuteron]

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:

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

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

 

[kuruman]

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. $$\frac{d}{dt}\frac{\partial \vec {r}_{cm}}{\partial{\vec {\dot r}_{cm}}}-\frac{\partial \mathcal L}{\partial \vec r_{cm}}=0\rightarrow
\begin{cases}
\frac{d}{dt}\frac{\partial \mathcal{L}}{\partial{{\dot x}_{cm}}}-\frac{\partial \mathcal L}{\partial x_{cm}} =0\\
\frac{d}{dt}\frac{\partial \mathcal{L}}{\partial{{\dot y}_{cm}}}-\frac{\partial \mathcal L}{\partial y_{cm}} =0 \\
\frac{d}{dt}\frac{\partial \mathcal{L}}{\partial{{\dot z}_{cm}}}-\frac{\partial \mathcal L}{\partial z_{cm}} =0
\end{cases}$$