### Center-of-mass frame

In physics, the center-of-momentum frame (zero-momentum frame, or COM frame) of a system is the unique inertial frame in which the center mass of the system is at rest. The total momentum of the system vanishes in this reference frame. The center of momentum of a system is not a location, but usually refers to the coordinate reference frame in which the momenta of a system's components add to zero. Thus "center of momentum" already means "center-of-momentum frame" and is a short form of this phrase.[1]

A special case of the center-of-momentum frame is the center-of-mass frame: an inertial frame in which the center of mass (which is a physical point) is at the origin at all times. In all COM frames, the center of mass is at rest, but it may not necessarily be at rest at the origin of the coordinate system.

## Properties

### General

The center of momentum frame is defined as the inertial frame in which the sum over the linear momentum of each particle vanishes. Let $S$ denote the lab reference system and $S\text{'}$ denote the center of momentum reference frame. Quantities in $S\text{'}$ are denoted by a prime. Using a galilean transformation, the particle velocity in $S\text{'}$ is

$v\text{'} = v - V_c,$

where $V_c = \frac\left\{\sum_i m_i v_i\right\}\left\{\sum_i m_i\right\}$

is the velocity of the mass center. The total momentum in the center-of-mass system then vanishes:

$\sum_\left\{i\right\} p\text{'}_i = \sum_\left\{i\right\} m_i v\text{'}_i = \sum_\left\{i\right\} m_i \left(v_i - V_c\right) = \sum_\left\{i\right\} m_i v_i - \sum_i m_i \frac\left\{\sum_j m_j v_i\right\}\left\{\sum_j m_j\right\} = \sum_i m_i v_i - \sum_j m_j v_j = 0$

Also, the total energy of the system is the minimal energy as seen from all inertial reference frames.

### Special relativity

In relativity, COM frame exists for a massive system. In the COM frame the total energy of the system is the "rest energy", and this quantity (when divided by the factor c2, where c is the speed of light) therefore gives the rest mass (positive invariant mass) of the system:

$m = \frac\left\{E\right\}\left\{c^2\right\}.$

The invariant mass of the system is actually given by the relativistic invariant relation:

$m^2 =\left\left(\frac\left\{E\right\}\left\{c^2\right\}\right\right)^2-\left\left(\frac\left\{p\right\}\left\{c\right\}\right\right)^2 \,\!$

but for zero momentum the momentum term (p/c)2 vanishes, hence the total energy coincides with the rest energy.

Systems which have energy but zero invariant mass (such as photons moving in a single direction, or equivalently, plane electromagnetic waves) do not have COM frames, because there is no frame which they have zero net momentum. Due to the invariance of the speed of light, such massless systems must travel at the speed of light in any frame, and therefore always possess a net momentum-magnitude which is equal to their energy divided by the speed of light:

$p = E/c. \,\!$

## 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):[2]

$\bold\left\{u\right\}_1^\prime = \bold\left\{u\right\}_1 - \bold\left\{V\right\} , \quad \bold\left\{u\right\}_2^\prime = \bold\left\{u\right\}_2 - \bold\left\{V\right\}$

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):[3]

\begin\left\{align\right\}

\frac{{\rm d}t} & = \fracTemplate:\rm d{{\rm d}t}\left(\frac{m_1\bold{r}_1+m_2\bold{r}_2}{m_1+m_2} \right) \\ & = \frac{m_1\bold{v}_1 + m_2\bold{v}_2 }{m_1+m_2} \\ & = \bold{V} \\ \end{align} \,\!

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

$m_1\bold\left\{v\right\}_1^\prime + m_2\bold\left\{v\right\}_2^\prime = \boldsymbol\left\{0\right\}$

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

$m_1\bold\left\{u\right\}_1 + m_2\bold\left\{u\right\}_2 = m_1\bold\left\{v\right\}_1 + m_2\bold\left\{v\right\}_2 = \left(m_1+m_2\right)\bold\left\{V\right\}$

This equation does not imply that

$m_1\bold\left\{u\right\}_1 = m_1\bold\left\{v\right\}_1 = m_1\bold\left\{V\right\}, \quad m_2\bold\left\{u\right\}_2 = m_2\bold\left\{v\right\}_2 = m_2\bold\left\{V\right\}$

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:

\begin\left\{align\right\} \bold\left\{P\right\} & = \bold\left\{p\right\}_1 + \bold\left\{p\right\}_2 \\

& = (m_1 + m_2)\bold{V} \\ & = M\bold{V} \end{align}\,\!

Similar analysis to the above obtains;

$\bold\left\{p\right\}_1^\prime = -\bold\left\{p\right\}_2^\prime = \mu \Delta\bold\left\{v\right\} = \mu \Delta\bold\left\{u\right\} \,\!$

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

$\Delta\bold\left\{v\right\} = \bold\left\{v\right\}_1 - \bold\left\{v\right\}_2 = \Delta\bold\left\{u\right\}.$