Center-of-momentum frame

Two-body problem

An example of the usage of this frame is given below – in a two-body collision, not necessarily elastic (where kinetic energy is conserved). The COM frame can be used to find the momentum of the particles much easier than in a lab frame: the frame where the measurement or calculation is done. The situation is analyzed using Galilean transformations and conservation of momentum (for generality, rather than kinetic energies alone), for two particles of mass m₁ and m₂, moving at initial velocities (before collision) u₁ and u₂ respectively. The transformations are applied to take the velocity of the frame from the velocity of each particle from the lab frame (unprimed quantities) to the COM frame (primed quantities):

${\displaystyle {\mathbf{u}}_1^{\mathrm{\prime }}={\mathbf{u}}_1-\mathbf{V},{\mathbf{u}}_2^{\mathrm{\prime }}={\mathbf{u}}_2-\mathbf{V},}$

where V is the velocity of the COM frame. Since V is the velocity of the COM, i.e. the time derivative of the COM location R (position of the center of mass of the system):

${\displaystyle \begin{array}{cc}\frac{\mathrm{d}\mathbf{R}}{\mathrm{d}t}& =\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{m_1{\mathbf{r}}_1+m_2{\mathbf{r}}_2}{m_1+m_2}\right)\\ & =\frac{m_1{\mathbf{u}}_1+m_2{\mathbf{u}}_2}{m_1+m_2}\\ & =\mathbf{V}\end{array}}$

so at the origin of the COM frame, R' = 0, this implies

${\displaystyle m_1{\mathbf{u}}_1^{\mathrm{\prime }}+m_2{\mathbf{u}}_2^{\mathrm{\prime }}=\mathbf{0}}$

The same results can be obtained by applying momentum conservation in the lab frame, where the momenta are p₁ and p₂:

${\displaystyle \mathbf{V}=\frac{{\mathbf{p}}_1+{\mathbf{p}}_2}{m_1+m_2}=\frac{m_1{\mathbf{u}}_1+m_2{\mathbf{u}}_2}{m_1+m_2}}$

and in the COM frame, where it is asserted definitively that the total momenta of the particles, p₁′ and p₂′, vanishes:

${\displaystyle {\mathbf{p}}_1^{\mathrm{\prime }}+{\mathbf{p}}_2^{\mathrm{\prime }}=m_1{\mathbf{u}}_1^{\mathrm{\prime }}+m_2{\mathbf{u}}_2^{\mathrm{\prime }}=\mathbf{0}.}$

Using the COM frame equation to solve for V returns the lab frame equation above, demonstrating any frame (including the COM frame) may be used to calculate the momenta of the particles. It has been established that the velocity of the COM frame can be removed from the calculation using the above frame, so the momenta of the particles in the COM frame can be expressed in terms of the quantities in the lab frame (i.e. the given initial values):

${\displaystyle \begin{array}{cc}{\mathbf{p}}_1^{\mathrm{\prime }}& =m_1{\mathbf{u}}_1^{\mathrm{\prime }}\\ & =m_1\left({\mathbf{u}}_1-\mathbf{V}\right)=\frac{m_1{m}_2}{m_1+m_2}\left({\mathbf{u}}_1-{\mathbf{u}}_2\right)\\ & =-m_2{\mathbf{u}}_2^{\mathrm{\prime }}=-{\mathbf{p}}_2^{\mathrm{\prime }}\end{array}.}$

Notice that the relative velocity in the lab frame of particle 1 to 2 is

${\displaystyle \mathrm{\Delta }\mathbf{u}={\mathbf{u}}_1-{\mathbf{u}}_2}$

and the 2-body reduced mass is

${\displaystyle \mu =\frac{m_1{m}_2}{m_1+m_2}}$

so the momenta of the particles compactly reduce to

${\displaystyle {\mathbf{p}}_1^{\mathrm{\prime }}=-{\mathbf{p}}_2^{\mathrm{\prime }}=\mu \mathrm{\Delta }\mathbf{u}.}$

This is a substantially simpler calculation of the momenta of both particles; the reduced mass and relative velocity can be calculated from the initial velocities in the lab frame and the masses, and the momentum of one particle is simply the negative of the other. The calculation can be repeated for final velocities v₁ and v₂ in place of the initial velocities u₁ and u₂, since after the collision the velocities still satisfy the above equations:

${\displaystyle \begin{array}{cc}\frac{\mathrm{d}\mathbf{R}}{\mathrm{d}t}& =\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{m_1{\mathbf{r}}_1+m_2{\mathbf{r}}_2}{m_1+m_2}\right)\\ & =\frac{m_1{\mathbf{v}}_1+m_2{\mathbf{v}}_2}{m_1+m_2}\\ & =\mathbf{V}\end{array}}$

so at the origin of the COM frame, R = 0, this implies after the collision

${\displaystyle m_1{\mathbf{v}}_1^{\mathrm{\prime }}+m_2{\mathbf{v}}_2^{\mathrm{\prime }}=\mathbf{0}}$

In the lab frame, the conservation of momentum fully reads:

${\displaystyle m_1{\mathbf{u}}_1+m_2{\mathbf{u}}_2=m_1{\mathbf{v}}_1+m_2{\mathbf{v}}_2=(m_1+m_2)\mathbf{V}}$

This equation does not imply that

${\displaystyle m_1{\mathbf{u}}_1=m_1{\mathbf{v}}_1=m_1\mathbf{V},m_2{\mathbf{u}}_2=m_2{\mathbf{v}}_2=m_2\mathbf{V}}$

instead, it simply indicates the total mass M multiplied by the velocity of the centre of mass V is the total momentum P of the system:

${\displaystyle \begin{array}{cc}\mathbf{P}& ={\mathbf{p}}_1+{\mathbf{p}}_2\\ & =(m_1+m_2)\mathbf{V}\\ & =M\mathbf{V}\end{array}}$

Similar analysis to the above obtains

${\displaystyle {\mathbf{p}}_1^{\mathrm{\prime }}=-{\mathbf{p}}_2^{\mathrm{\prime }}=\mu \mathrm{\Delta }\mathbf{v}=\mu \mathrm{\Delta }\mathbf{u},}$

where the final relative velocity in the lab frame of particle 1 to 2 is

${\displaystyle \mathrm{\Delta }\mathbf{v}={\mathbf{v}}_1-{\mathbf{v}}_2=\mathrm{\Delta }\mathbf{u}.}$