Imagine an object with internal particulate velocities $v_{i, internal}$ relative to the center of mass frame for each mass $m_i$ in two situations: one in the center of mass frame, and one where the center of mass is moving with some linear velocity $v_{cm}$.

Naturally, in the center of mass frame, the total kinetic energy of the system of the object is equal to the internal kinetic energy of the object: i.e. $\frac{1}{2} \sum m_i \cdot v_{i, internal}^2$.

Now, in the frame of a moving center of mass, tantamount to boosting each particle's velocity to $v_{i, internal} + v_{cm}$, the total kinetic energy would thus be $\frac{1}{2} \sum m_i \cdot (v_{i, internal} + v_{cm})^2 = \frac{1}{2} \sum m_i \cdot [v_{i, internal}^2 + 2 \cdot v_{i, internal} \cdot v_{cm} + v_{cm}^2]$, which, if separated into the additive terms $K_{cm}$ and $K_{int}$ such that $K_{cm} = \frac{1}{2} Mv_{cm}^2$, implies $K_{int} = \frac{1}{2} \sum m_i \cdot [v_{i, internal}^2 + 2 \cdot v_{i, internal} \cdot v_{cm}]$, as increases with $v_{cm}$.

Is this an accurate description, or does some symmetry I'm not seeing cancel the pesky middle term when summed over all particles (e.g. as it's cancelled for rotational kinetic energy for rotations about the center of mass in the center of mass frame by that rotational symmetry of the object)?


1 Answer 1


I have just realized that, by virtue of it being the center of mass frame (i.e. $p_{cm} = 0$), the sum of mass weighted internal velocities of the particles is ergo, by definition, zero ($\sum m_i \cdot v_{i,internal} = \sum p_i = 0$), so the $2 \cdot v_{cm} \cdot \sum m_i \cdot v_{i, internal}]$ term necessarily equals zero. It indeed cancels when summing over all particles when defined from the center of mass as such, hence leaving $K_{int}$ to be equal to $K_{int} = \frac{1}{2} \sum m_i \cdot v_{i, internal}^2$. Therefore, internal kinetic energy, naturally, does not increase with the velocity of the center of mass.


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