In general, conservation laws do not hold whenever the center of mass of the system is moving?

Question

I am currently studying Classical Mechanics, fifth edition, by Kibble and Berkshire. Problem 3 of chapter 1 is as follows:

Consider a system of three particles, each of mass $m$, whose motion is described by (1.9). If particles 2 and 3, even though not rigidly bound together, are regarded as forming a composite body of mass $2m$ located at the mid-point $\mathbf{r} = \dfrac{1}{2} (\mathbf{r}_2 + \mathbf{r}_3)$, find the equations describing the motion of the two-body system comprising particle 1 and the composite body (2+3). What is the force on the composite body due to particle 1? Show that the equations agree with (1.7). When the masses are unequal, what is the correct definition of the position of the composite (2+3) that will make (1.7) still hold?

I was unsure about this part:

When the masses are unequal, what is the correct definition of the position of the composite (2+3) that will make (1.7) still hold?

The answer is said to be

$$\mathbf{r} = \dfrac{m_2 \mathbf{r}_2 + m_3 \mathbf{r}_3}{m_2 + m_3}.$$

(1.7) is as follows:

$$m_1 \mathbf{a}_1 = -m_2 \mathbf{a}_2$$

To try and understand how this could be done, I recently asked this question. Thanks to user Ja72's comments, I was able to do further research and learned that this is actually the center of mass:

https://en.wikipedia.org/wiki/Two-body_problem#Center_of_mass_motion_(1st_one-body_problem)

https://en.wikipedia.org/wiki/Center_of_mass#Barycentric_coordinates

http://hyperphysics.phy-astr.gsu.edu/hbase/cm.html

The representations shown in these links reminded me of the law of conservation of momentum:

$$m_1\mathbf{v}_1 + m_2 \mathbf{v}_2 = m_1 \mathbf{v}_1^\prime + m_2 \mathbf{v}_2^\prime.$$

I then wondered: Does the law of conservation of momentum also hold for position and acceleration? Because, if it does, then it seems to me that we can represent the problem as follows:

$$m_1 \mathbf{r}_2 + m_2 \mathbf{r}_3 = (m_1 + m_2) \mathbf{r}^\prime \\ \Rightarrow \mathbf{r}^\prime = \dfrac{m_1 \mathbf{r}_2 + m_2 \mathbf{r}_3}{m_1 + m_2},$$

where $(m_1 + m_2)$ is the composite mass. This seems to be of the form that we are looking for. Furthermore, unlike user Ja72's answer here, it uses basic equations of classical mechanics that are discussed in chapter 1, which means that it is more likely to be the solution method that the authors were intending the reader to use to solve this problem.

I had the question of whether the law of conservation of momentum also holds for position and acceleration answered here by the user Dale. They said that conservation of position is not valid, and that, in general, conservation laws do not hold whenever the center of mass of the system is moving. So how do I reconcile this with the fact that using the conservation laws in terms of position seems to get us the correct solution of $\mathbf{r} = \dfrac{m_2 \mathbf{r}_2 + m_3 \mathbf{r}_3}{m_2 + m_3}$? Is it because the center of mass is actually not moving in this case? Or is it just a coincidence? I would greatly appreciate it if people would please take the time to clarify this.

Answer 1

Let us start by defining the center of mass of a 2-particles system as

$$\textbf{r}={m_1\textbf{r}_1+m_2\textbf{r}_2 \over m_1+m_2}$$

now, if there is no external force on the system (two isolated particles) then the only forces acting on the two masses are whatever force is acting between $m_1$ and $m_2$, which we call $\textbf{F}_{12}$ (acting on $m_1$ and due to $m_2$) and $\textbf{F}_{21}$ (acting on $m_2$ and due to $m_1$).

Let us know take the second derivative of ${\textbf{r}}$ $$\ddot{\textbf{r}} = {m_1\ddot{\textbf{r}}_1+m_2\ddot{\textbf{r}}_2 \over m_1+m_2}$$

and using $\textbf{F}=m\textbf{a}$ for each particle

$$\ddot{\textbf{r}} = {m_1\ddot{\textbf{r}}_1+m_2\ddot{\textbf{r}}_2 \over m_1+m_2} = {\textbf{F}_{12}+{\textbf{F}}_{21} \over m_1+m_2} $$

Now, for Newton's third law we must have $\textbf{F}_{12}=-\textbf{F}_{21}$ (action/reaction pair) so that

$$\ddot{\textbf{r}}=0$$

meaning that the center of mass of the two particles $\textbf{r}$ does not accelerate i.e. it goes at constant speed (or is not moving at all). This is analogous to say that, in the absence of external forces, the center of mass moves at constant speed i.e. it conserves momentum as we can multiply $\textbf{r}$ with $(m_1+m_2)$ and then derivate it two times again to get

$$(m_1+m_2)\ddot{\textbf{r}} = {d\over dt}\left((m_1+m_2)\dot{\textbf{r}}\right)={d\over dt}\left( (m_1+m_2)\textbf{v} \right)={d\over dt} \textbf{p}=0$$

where we used that $\ddot{\textbf{r}}=0$ and $\textbf{p}=(m_1+m_2)\textbf{v}$ to get that the total momentum $\textbf{p}$ is conserved.

To sum up so far: if we define a position $$\textbf{r}={m_1\textbf{r}_1+m_2\textbf{r}_2 \over m_1+m_2}$$ we get that without external forces we can think of $\textbf{r}$ as the coordinate of a particle with mass $m_1+m_2$ such that this mass follows Newton's law i.e. momentum conservation. This can be proven to be true also in the presence of external forces under given conditions (e.g., central or constant forces).

In which case $$(m_1+m_2)\ddot{\textbf{r}}=\textbf{F}_{external}$$

Your equation $$(m_1+m_2)\textbf{r}=m_1\textbf{r}_1+m_2\textbf{r}_2$$ simply is the definition of the center of mass rewritten. It does not mean that there is "conservation of position" in general, it just means that, under certain conditions (no external forces or central/constant forces) then the system can be rewritten in a simpler way. It does not mean that $\textbf{r}$ is conserved: actually, $\textbf{r}$ can change if there are external forces or if the system was moving in some direction at the beginning (then it will keep moving with constant speed even without external forces). But it is true that (except some weird cases) that equality will always "work" in the sense that by that definition you can "summarize" the system in a one-particle system.

What is instead always conserved (in the absence of external forces) is the total momentum $\textbf{p}$: that will be constant!

In other words, $$\textbf{p}=(m_1+m_2)\dot{\textbf{r}}\ne 0$$ i.e., $\textbf{r}$ is not conserved (its derivative is not zero, it is $\textbf{p}$).

But rather, the derivative of $\textbf{p}$ is more interesting:

$$\dot{\textbf{p}}=(m_1+m_2)\ddot{\textbf{r}}= \textbf{F}_{ext}$$ where $\textbf{F}_{ext}$ is the total external force!

If $\textbf{F}_{ext}=0$ (no external force) and $\textbf{v}_0=0$ (no initial velocity) than yes: the center of mass does not move (it is "conserved", but that is because $\textbf{p}=0$).

If $\textbf{F}_{ext}=0$ (no external force) and $\textbf{0}=\textbf{v}_0$ (some initial velocity) the the center of mass will move at constant speed $\textbf{v}_0$ (because $\textbf{p}\ne 0$ but it instead there is some "initial momentum". So now $\textbf{r}$ is not conserved, but will still be given by the weighted sum of the two positions!).

If $\textbf{F}_{ext}\ne 0$ (external forces but constant/central/etc) then the center of mass will accelerate as if it were a single particle with total mass $m_1+m_2$.

Re-summing up, $$(m_1+m_2)\textbf{r}=m_1\textbf{r}_1+m_2\textbf{r}_2$$

is not a conservation law: it is a definition! And it is a definition that allows to simplify calculations because all internal forces disappear and one is only left with external ones. Plus, $\textbf{r}$ can be used as a "single" coordinate to describe the system. So you can always write

$$(m_1+m_2)\textbf{r}=m_1\textbf{r}_1+m_2\textbf{r}_2$$

but its behavior will depend on external forces and initial conditions.

Answer 2

Conservation of position is not invalid, it's meaningless. To get the velocity and acceleration of the center of mass, take the first and second derivative of the formula for the position. In the absence of external forces, the momentum is conserved in any inertial frame of reference. (The velocity of the center of mass is different but constant in each.) In a collision, energy is conserved only if the collision is fully elastic.