# Conservation of Energy of the Center of Mass

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## Summary:

Why is the energy of the center of mass of a system of particles conserved?
In classical mechanics, the energy of a system of particles (say with 2 particles) in an external field is given by
$$E=\frac{1}{2}m_1|\vec{v}_1|^2+\frac{1}{2}m_2|\vec{v}_2|^2+V(\vec{r}_1)+V(\vec{r}_2)+V'(|\vec{r}_2-\vec{r}_1|)$$
Where V is the potential energy of the external field, and V' is the energy of interaction between the two particles. It is well known that ##\frac{dE}{dt}=0##, or that energy is conserved.

However, the energy of the center of mass is surprisingly also conserved
$$E'=\frac{1}{2}(m_1+m_2)(\frac{m_1\vec{v}_1+m_2\vec{v}_2}{m_1+m_2})^2+V(\frac{m_1\vec{r}_1+m_2\vec{r}_2}{m_1+m_2})$$
$$\frac{dE'}{dt}=0$$

This makes sense as macroscopically, everything is just made out of tiny particles and if this identity doesn't hold there would be no concept of a macroscopic object being treated as a particle.

However, is there any proof that the energy of the center of mass is conserved, based directly on the fact that the energy of the system is conserved? What are the conditions? For example, if the interactions between the particles in the system could not be described by a potential ##V'(|\vec{r}_2-\vec{r}_1|)##, such as in the case of magnetic forces, would this still hold? Must ##V## be linear in the coordinates?

In addition, does this hold for other quantities? For example, if every particle ##i## had a quantity ##\Omega_i(\vec{r}_i,\vec{p}_i)## associated with it such that ##\frac{d}{dt}(\sum_i \Omega_i(\vec{r}_i,\vec{p}_i))=0##, would ##\frac{d}{dt}\Omega(\vec{r}_{cm},\vec{p}_{cm})=0##?

Thanks!