**1. The problem statement, all variables and given/known data**

Two particles with masses ##m_1## and ##m_2## scatter elastically. Show that in the centre of momentum frame ##\vec{p_1}## and ##\vec{p_2}##, the 3-momenta before collision and ##\vec{q_1}## and ##\vec{q_2}##, the 3-momenta after the collision, all lie on a circle in 3-momenta space.

**2. Relevant equations**

4-momenta of particles, energy conservation, equation of a circle.

**3. The attempt at a solution**

In the C.O.M frame of the system, the 4 momenta look like $$\left(\frac{E_1}{c}, \vec{p_1}\right)\,\,\text{for particle 1}\,\,\,\,\text{and}\,\,\,\,\left(\frac{E_2}{c}, \vec{p_2}\right)\,\,\,\,\text{for particle 2}.$$ In this frame both particles have the same speed, but moving in the opposite direction. So, I think we can write ##\vec{p_1} = \gamma(v_1)m_1 \vec{v_1}## and ##\vec{p_2} = \gamma(-v_1)m_2 \vec{-v_1}##, where ##\gamma(-v_1) = \gamma(v_1)##.

The 4 momenta after the collision (in this frame) is similarly of the form $$\left(\frac{E_1′}{c}, \vec{q_1}\right)\,\,\,\text{for particle 1}\,\,\,\text{and}\,\,\,\left(\frac{E_2′}{c}, \vec{q_2}\right)\,\,\,\text{for particle 2},$$ where ##E_i^2 = |\vec{p_i}|^2c^2 + m_i^2 c^4## and ##E_i’^2 = |\vec{q_i}|^2c^2 + m_i^2c^4##. For an elastic scattering, ##E_1 + E_2 = E_1′ + E_2’##.

I am wondering if I had to introduce the explicit form for the 3 momenta. I could write an explicit form for the 3 momenta of the particles after the collision and use my energy equation to obtain the velocities after the collision in terms of ##v_1## (Both particles would possess same speed after collision since we are always in centre of mass frame, I think). But that seems like a messy approach. Is there a better way?

Many thanks.

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