# 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.

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 $\gamma_2$.

• 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.

$$v_{com} = \frac{m_1v_1 + m_2v_2}{m_1 + m_2}$$

• Yes, but I was thinking of a relativistic boost – Physiks lover Apr 26 '12 at 10:41
• Oops, I didn't read the title properly. Can you give me the details of the Goldstein book and a page number and I'll have a look. As far as I'm aware 4-momentum is conserved so the same argument will apply though you'll end up with a more complicated formula. – John Rennie Apr 27 '12 at 14:02
• it's towards the end of the the relativity chapter which talks about the relativistic mechanics of colliding particles. Yes, it goes into the conservation of the total 4-momentum, but says nothing about a general formula for the COM frame – Physiks lover Apr 27 '12 at 16:45
• I don't have the book. Can you give me a link to it on Amazon or whatever so I know which book I'm looking for. – John Rennie Apr 27 '12 at 17:06
• @JohnRennie: The formula you give only works if m is the (ostensibly deprecated) relativistic mass. This is not an invariant of the particle, it changes with boost, and it is a stand-in for the energy, so it is best to express the formula using energies. – Ron Maimon May 27 '12 at 3:48

I was looking for an answer to this with just frame velocities, no particles. Here's what I tried:

Let $v_1$ and $v_2$ be the velocities of two frames moving along the $x$-axis relative to lab frame $F$. Another frame $F'$ is moving similarly with velocity $u$, and it sees the prior velocities as $$v_1' = \frac{v_1-u}{1-v_1u} \quad\text{and}\quad v_2' = \frac{v_2-u}{1-v_2u}.$$ in units where $c=1$. Define the velocity $u$ to be the case where $F'$ sees each frame moving in opposite directions at the same speed: \begin{align} v_1' &= -v_2' \\ \frac{v_1-u}{1-v_1u} &= -\frac{v_2-u}{1-v_2u} \end{align} This leads to the quadratic equation $$u^2(v_1+v_2) - 2u(1+v_1v_2) + (v_1+v_2) = 0$$ with solutions $$u_\pm = \frac{(1+v_1v_2)\pm\sqrt{(1-v_1^2)(1-v_2^2)}}{(v_1+v_2)}$$ Treating the 2nd order terms $v_1^2$ and $v_2^2$ as small quantities , and ignoring the 4th order $v_1^2v_2^2$, the approximation to the negative solution $u_-$ gives the average $$u \approx \frac{v_1+v_2}{2}.$$ You can check that $u_+$ is just the inverse, such that $u_+ u_- = 1$. So $$\boxed{u = \frac{(c^2+v_1v_2) - \sqrt{(c^2-v_1^2)(c^2-v_2^2)}}{(v_1+v_2)}.}$$ in regular units.