Elastic collision in one of the particles' frame

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SUMMARY

The discussion centers on calculating the change in particle A's 4-momentum in the rest frame of particle B during an elastic collision. The initial 4-momentum of particle A is given as (E, p, 0, 0), while particle B is at rest with 4-momentum (m_B, 0, 0, 0). The final 4-momenta are expressed as (E_f, p_f cos(θ), p_f sin(θ), 0) for particle A and (m_B', 0, 0, 0) for particle B. The key takeaway is that total momentum is conserved across the system, not the individual momenta of the particles.

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  • Understanding of 4-momentum in special relativity
  • Knowledge of conservation laws in physics
  • Familiarity with elastic collisions
  • Basic grasp of scattering angles and their implications
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  • Study the principles of 4-momentum conservation in elastic collisions
  • Learn how to calculate scattering angles in particle physics
  • Explore the implications of relativistic effects on momentum
  • Investigate the role of energy conservation in particle interactions
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Homework Statement


A particle B is standing still while another one, A, is moving towards it with initial 4-momentum ##(E,p,0,0)##. Calculate the change in particle A's 4-momentum as viewed from the particle B's rest frame, in terms of the initial energy E and the scattering angle ##\theta##.

Homework Equations

The Attempt at a Solution


I am a bit confused about the 4-momentum conservation. Initially we have ##p^i_A=(E,p,0,0)## and ##p^i_B=(m_B,0,0,0)## finally we should have ##p^f_A=(E_f, p_f cos(\theta),p_f sin(\theta),0)## and ##p^f_B=(m_B,0,0,0)##. To get the change in momentum I would do ##p^f_A-p^i_A##. But the total momentum should be conserved in any frame, but I am not sure how does that work here. In order to conserve it, we would need ##E=E_f## and ##\theta=0## but then the problem would be trivial and also physically you can obviously have angles other than 0. What am I doing wrong?
 
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Particle B will move after the collision.
 
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Likes   Reactions: maughanster
mfb gave a good answer. After the collision, particle B is not stationary. Maybe this already clear, but the TOTAL momentum is conserved, not necessarily the momentum of each individual particle.
 

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