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I have this problem:

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 $\theta$.

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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I think you've been (understandably) misled by the wording of the problem.

...as viewed from the particle B's rest frame...

That's badly written because there are two different reference frames that could reasonably be called "particle B's rest frame": the one in which particle B is at rest before the collision, and the one in which it's at rest after the collision. Assuming there is a nontrivial collision, these are not the same reference frame.

You've expressed $p_B^i$ in the former frame and $p_B^f$ in the latter frame, but what you should do is express them both in the same frame, since in order to apply conservation of momentum, you need every momentum to be expressed in the same reference frame. (Either one should work.) For example, in the frame in which B is at rest before the collision, it won't be at rest after the collision, and so $p_B^f$ in that frame is going to be something other than $(m_B, 0, 0, 0)$.

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    $\begingroup$ Oh, so what they mean is the Lab frame basically? (if I am to choose the B-frame before collision) $\endgroup$ – Alex Marshall Sep 23 '18 at 2:47
  • $\begingroup$ I can't be sure which frame they mean for you to use, but I suspect it is that one. $\endgroup$ – David Z Sep 23 '18 at 3:51
  • $\begingroup$ The question is poorly worded. When I first read it, I interpreted it as the before-collision-particle-B rest frame, but I can't be sure that that is what was intended. $\endgroup$ – Trevor Kafka Sep 23 '18 at 4:12

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