# Calculate relativistic boost to COM frame from two arbitary velocities?

Looking in Goldstein's book, there doesn't seem to be a standard formula to calculate the COM frame velocity for two particles, from their relativistic velocities in the lab frame, although it is done for the case where one particle is initially at rest. I find this a glaring omission and would like to know if there is a general formula for two relativistic particles moving along the x axis of the lab frame.

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The center of mass 4-momentum is the sum of the 4-momenta of the particles (no vector symbol or index, but the v's are four-component vectors) using the masses as the weights:

$$P_\mathrm{CM} = m_1 v_1 + m_2 v_2$$

The length of this is the mass of the combined system, (mostly minus metric)

$$M^2 = |P|^2 = m_1^2 + m_2^2 + 2m_1 m_2 v_1 \cdot v_2$$

The four-velocity of the center of mass is then

$$v_\mathrm{CM} = {m_1v_1 + m_2 v_2 \over M}$$

and the three velocity is given by the ratio of the space-components of the four vector to the time component:

$$v^0_\mathrm{CM} = {m_1\gamma_1 + m_2 \gamma_2 \over M}$$

So that the center of mass velocity is:

$$\vec{v}_\mathrm{CM} = {m_1\gamma_1 \vec{v}_1 + m_2\gamma_2 \vec{v}_2 \over m_1\gamma_1 + m_2\gamma_2}$$

or the weighted average of the velocities using the relativistic mass (the energy). This formula usually appears with energy letters replacing mass letters:

$$\vec{v}_\mathrm{CM} = { E_1 \vec{v}_1 + E_2 \vec{v}_2 \over E_1 + E_2}$$

Where m_1 and m_2 are the masses, $v_1$ and $v_2$ are the 4-velocities, $E_1$ and $E_2$ are the energies, $\gamma_1 = {1\over \sqrt{1-|\vec v_1|^2}}$ and similarly for $\gamma_2$.

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cool... how would you find the frame where the sum of the velocities of the particles are zero? –  Larry Harson May 28 '12 at 20:34
@LarryHarson: Pretend their masses are equal. –  Ron Maimon May 28 '12 at 21:38
Your derivation is true for N particles and I was thinking of the three-velocities rather than the four-velocities of N particles. For two it's easy since their speeds have to be the same and their four-velocities cancelling implies the same with their three-velocites. –  Larry Harson May 28 '12 at 22:33
This easy isn't it? You just calculate the total momentum ($m_1v_1 + m_2v_2$). The velocity of the COM frame is then simply this momentum divided by the total mass i.e.
$$v_{com} = \frac{m_1v_1 + m_2v_2}{m_1 + m_2}$$