Decomposing Lagrangian into CM and relative parts with presence of uniform gravitational field

Question

Most problems concerning two-body motion (using Lagrangian methods) often only consider the motion of two particles subject to no external forces. However, the Lagrangian should be decomposable into the center of mass (CM) and relative parts even if our closed system (momentum of the CM is constant) of two interacting masses is producing an attractive gravitational field.

The Lagrangian that describes the two-body system with no attractive gravitational field is

$${\cal L}=\frac{1}{2}M\dot{\vec{R}}^2 +\left(\frac{1}{2}\mu\dot{\vec{r}}^2-U(r)\right)={\cal L}_{\rm CM}+{\cal L}_{\rm rel}$$

If we then considered an attractive gravitational field, how would this change the above Lagrangian?

My thought process is that the CM term will not change, since the gravitational field is attractive w.r.t the two masses, thereby implying that only the terms containing r will be affected by the field.

The gravitational field between masses 1 and 2 is simply described as conservative and central: $\vec{g}$.

How would I rewrite the Lagrangian with this g-field now in play?

Answer 1

It's not clear to me from your question if you are interested in the gravitational force that two particles exert on each other, or on the gravitational force the two particles experienced if they move in an external field. The way you handle these two cases is different, so below I outline how you can handle both.

If you are interested in the gravitational force that the two particles exert on each other, then you simply add a gravitational term in the potential \begin{equation} U(r) = U_{\rm non-gravity}(r) + U_{\rm gravity}(r) \end{equation} where \begin{equation} U_{\rm gravity}(r) = -\frac{G m_1 m_2}{r} \end{equation} The potential only appears in the relative term of the Lagrangian.

If you are interested in the particles moving in an external field, then...

• ...if the external field is a constant in magnitude and direction, one strategy is to take advantage of the equivalence principle to cancel this force. Move into a freely falling frame that is accelerating with the same magnitude and opposite direction as the gravitational acceleration, and the gravitational force will be canceled by the fictitious force from being in a non-inertial reference frame.
• ...if the external field is not constant (or if you just don't like using the equivalence principle), then you can write the Lagrangian in terms of the coordinates of the two particles $x_1$ and $x_2$ (without going to the center of mass and relative coordinates) as \begin{equation} \mathcal{L} = \frac{1}{2} m_1 \dot{x}_1^2 + \frac{1}{2} m_2 \dot{x}_2^2 - U(|\vec{x}_1 - \vec{x}_2|) - U_{\rm g, ext}(\vec{x}_1, \vec{x}_2) \end{equation} where $U_{\rm g, ext}(\vec{x}_1, \vec{x}_2)$ describes the potential energy due to the location of the point masses in the external gravitational field. For example, if you had a mass $M$ placed at a position $\vec{R}$, then the external potential energy would be \begin{equation} U_{\rm g, ext}(\vec{x}_1, \vec{x}_2) = - \frac{G M m_1}{|\vec{R}-\vec{r}_1|} - \frac{G M m_2}{|\vec{R}-\vec{r}_2|} \end{equation} As another example, you could consider the two particles in a uniform gravitational field where the magnitude of the gravitational acceleration is $g$, in which case \begin{equation} U_{\rm g, ext}(\vec{x}_1, \vec{x}_2) = m_1 g h_1 + m_2 g h_2 \end{equation} where $h_1$ is the height of particle 1 (aka the displacement from particle 1 to wherever your surface of zero potential is), and $h_2$ is the height of particle 2. Regardless of the specific form of $U_{\rm g, ext}$, since the external gravitational potential breaks translation invariance, momentum is not conserved, so the split into center of mass and relative coordinates may not be a useful thing to do in this case.

Answer 2

two-body motion

first you transfer the absolute coordinates $~r_1~,r_2~$ to the relative coordinate $~r~$ and the center of mass coordinate $~R_c~$

I use this two equations

$$R_c=\frac{m_1\,r_1+m_2\,r_2}{m_1+m_2}\\ r=r_1-r_2$$

the solution for $~r_1~,r_2~$ is: $$r_1=\frac{R_c+r\,m_2}{m_1+m_2}\\ r_2=\frac{R_c-r\,m_1}{m_1+m_2}$$

from here you obtain the kinetic energy and the potential energy

$$T=\frac{m_1}{2}\dot r_1^2+\frac{m_2}{2}\dot r_2^2= \frac 12 M\,\dot R_c^2+\frac 12 \frac{m_1\,m_2}{m_1+m_2}\dot r^2$$

$$U(r)=-\frac{m_1\,m_2\,G}{r}$$

the potential energy in case of the gravitation vector $~\vec g~$ is:

$$U=-m_1\,g\,r_1-m_2\,g\,r_2=-M\,g\,R_c$$

EOM's

$$\ddot R_c=-g\\ \ddot r =0\quad \Rightarrow\\ \ddot r_1=-g\\ \ddot r_2=-g$$