Proof of König's theorem

Problem

Prove König’s theorem for the kinetic energy of a two-particle system: the total kinetic energy is the sum of the kinetic energy of the centre of mass and the kinetic energy of the two particles relative to the centre of mass.

Solution

Setup. We write the velocity of each particle as $\vec{v}_i = \vec{v}_\text{CM} + \vec{v}_i'$, with $\vec{v}_i'$ the velocity relative to the CM ($i = 1, 2$). The total kinetic energy is $E_\text{cin,tot} = \sum_{i=1}^{2} \tfrac{1}{2}m_i\,|\vec{v}_\text{CM} + \vec{v}_i'|^2$

Expanding the square. Using $|\vec{v}_\text{CM} + \vec{v}_i'|^2 = v_\text{CM}^2 + 2\,\vec{v}_\text{CM}\cdot\vec{v}_i' + v_i'^2$: $E_\text{cin,tot} = \tfrac{1}{2}M_\text{tot}v_\text{CM}^2 + \vec{v}_\text{CM}\cdot\!\sum_i m_i\vec{v}_i' + \sum_i \tfrac{1}{2}m_i v_i'^2$

The mixed term vanishes. By definition of the CM, the momentum in the CM frame is zero: $\sum_i m_i\vec{v}_i' = m_1\vec{v}_1' + m_2\vec{v}_2' = \vec{0}$ so the middle term disappears. Only the two sought terms remain: $\ev{E_\text{cin,tot} = \tfrac{1}{2}M_\text{tot}v_\text{CM}^2 + \sum_i \tfrac{1}{2}m_i v_i'^2 = E_\text{cin,CM} + E_\text{cin,int}}$

Links

Topics: Centre of mass Concepts: Kinetic energy · Internal energy Skills: Symbolic setup