Collisions in the centre of mass frame

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SUMMARY

In the centre of mass frame, the angle of deflection in a collision differs from that in the lab frame due to the relative motion of the frames. The transformation between these frames is achieved using Galilean transformations, where the velocity components of a particle are adjusted based on the relative velocity of the frames. Specifically, if a particle has velocity components (ux, uy, uz) in frame S, then in frame S', moving at velocity v in the +x direction, the components become (ux-v, uy, uz). The angles can be calculated using the tangent function: tanθ = uy/ux for frame S and tanθ' = uy/(ux-v) for frame S'.

PREREQUISITES
  • Understanding of Galilean transformations
  • Familiarity with the concept of centre of mass frame
  • Basic knowledge of trigonometry, specifically tangent functions
  • Concept of velocity components in different reference frames
NEXT STEPS
  • Study Galilean transformations in detail
  • Learn about the centre of mass frame in particle physics
  • Explore the derivation of angles of deflection in various reference frames
  • Investigate the implications of relativistic transformations on collision angles
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Physics students, researchers in particle physics, and anyone interested in understanding collision dynamics in different reference frames.

Lucy Yeats
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I've just found out that in the centre of mass frame, the angle of deflection in a collision is different from in the lab frame.

I vaguely understand why: if the frame you viewed the particles in was also moving but only horizontally it would make their horizontal movement appear to decrease while their vertical movement would stay constant, which would seem to decrease the angle.

I have no idea how you would go about finding angles of deflection in the centre of mass frame. Could someone help me derive/ tell me a formula for doing so?
 
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The angle is a free parameter. The issue is transforming between the frames.
 
How would I go about transforming between frames?

Thanks for helping! :-)
 
Call one set of axes S. Let another set of axes, S', be co-incident with S. Let S' now move steadily in the +x direction, relative to S. Now suppose there's a particle moving with velocity components ux, uy, uz as described on the S axes. On the S' axes the components will be (ux-v), uy, uz. This is a galilean (non-relativistic) transform.

From the components you can find the direction cosines of the velocity vectors in the two frames. If the particle is moving in, say, just the x and y directions then it's even easier: in S, tanθ = uy/ux, whereas in S', tanθ' = uy/(ux-v)
 
Ah, I think I get it now- thanks.
 
Good! Despite my forgetting to say that v was the velocity of the S' frame relative to the S!
 

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