0
$\begingroup$

Studying particle physics I had a doubt that I can't seem to find a solution:

Is possibile in a particle decay to produce some particles with no momentum?

As an example: take the following decay

$$\Upsilon(4S) \rightarrow B+\overline{B} $$

where the particle $\Upsilon (4S)$ is produced by the collision of an electron an a positron beam with different energies (such as in the BaBar experiment), so that $\Upsilon(4S)$ is produced with some momentum.

Or as another example the three body decay

$$K^+ \rightarrow \pi^0 + e^+ + \nu_e $$

where $K^+$ has some momentum. Would it be possible in the laboratory frame to produce the $\pi^0$, for example, with no momentum?

Improve this question
$\endgroup$
4
  • 3
    $\begingroup$ In some frame, a product with momentum will have none: its rest frame. $\endgroup$
    – Cosmas Zachos
    May 1, 2019 at 17:51
  • $\begingroup$ @CosmasZachos Yes, that's clear. But maybe I didn't make it clear: I want to know if in the laboratory frame could be possible! $\endgroup$
    – Davide Morgante
    May 1, 2019 at 18:09
  • 1
    $\begingroup$ What is so special about the lab frame? go to the rest frame of the pion, Lorentz transform the decaying Kaon, and you have a magic lab frame such that.... $\endgroup$
    – Cosmas Zachos
    May 1, 2019 at 18:22
  • 2
    $\begingroup$ "What is the minimum lab frame momentum of the kaon such that the pion cannot be produced at rest in the lab?" would be a suitable homework question for a upper-division course in particle physics (for graduate students it would be a warm-up question.), and it thinking about it in those terms might give you a hint. $\endgroup$
    – dmckee --- ex-moderator kitten
    May 1, 2019 at 19:29

1 Answer 1

Reset to default
1
$\begingroup$

Suppose you have a proton of energy E hitting another proton. The result is 3 protons and an anti-proton. What is the minimum energy for this to happen?

The answer is 7 times the rest energy of a single proton. But why?

Also consider why pair production doesn't happen in a vacuum even though its always energetically permitted.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thanks, I'll think about that! $\endgroup$
    – Davide Morgante
    May 2, 2019 at 9:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.