Center-of-momentum frame

In physics, a center-of-momentum frame (zero-momentum frame, or COM frame) of a system is any inertial frame in which the center of mass of the system is at rest (has zero velocity and hence momentum). Note that the center of momentum of a system is not a location, but rather defines a particular inertial frame (a velocity and a direction). Thus "center of momentum" already means "center-of-momentum frame" and is a short form of this phrase.

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

In the centre of momentum frame, the total linear momentum of the system is zero - by definition. Also, the total energy of the system is the minimal energy as seen from all possible 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:

$m = E/c^2. \,\!$

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

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

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:

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

2-body problem

An example of the usage of this frame is given below – in a two-body collision, not neccersarily 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 respectivley. 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 quantites):

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

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):

$\begin{align} \frac{{\rm d}\bold{R}}{{\rm d}t} & = \frac{{\rm d}}{{\rm d}t}\left(\frac{m_1\bold{r}_1+m_2\bold{r}_2}{m_1+m_2} \right) \\ & = \frac{m_1\bold{u}_1 + m_2\bold{u}_2 }{m_1+m_2} \\ & = \bold{V} \\ \end{align} \,\!$

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

$m_1\bold{u}_1^\prime + m_2\bold{u}_2^\prime = \boldsymbol{0}$

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

$\bold{V} = \frac{\bold{p}_1 + \bold{p}_2}{m_1+m_2} = \frac{m_1\bold{u}_1 + m_2\bold{u}_2}{m_1+m_2}\,\!$

and in the COM frame, where we assert definitivley that the total momenta of the particles, p₁' and p₂', vanishes:

$\bold{p}_1^\prime + \bold{p}_2^\prime = m_1\bold{u}_1^\prime + m_2\bold{u}_2^\prime = \boldsymbol{0}$

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 the velocity of the COM frame can be removed from the calculation using that frame, so the momenta of the particles in the COM frame can expressed in terms of the quantities in the lab frame (i.e. the given initial values):

$\begin{align} \bold{p}_1^\prime & = m_1\bold{u}_1^\prime \\ & = m_1 \left( \bold{u}_1 - \bold{V} \right) = \frac{m_1m_2}{m_1+m_2} \left( \bold{u}_1 - \bold{u}_2 \right) \\ & = -m\bold{u}_2^\prime = -\bold{p}_2^\prime \\ \end{align} \,\!$

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

$\Delta\bold{u} = \bold{u}_1 - \bold{u}_2$

and the 2-body reduced mass is;

$\mu = \frac{m_1m_2}{m_1+m_2} \,\!$

so the momenta of the particles compactly reduce to

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

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 calulation 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:

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

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

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

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

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

This equation does not imply that

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

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:

$\begin{align} \bold{P} & = \bold{p}_1 + \bold{p}_2 \\ & = (m_1 + m_1)\bold{V} \\ & = M\bold{V} \end{align}\,\!$

Simalar analysis to the above obtains;

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

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

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

See also

• Laboratory frame of reference