I am stuck on the following question:

A proton of total energy $E$ collides with a proton at rest and creates a pion in addition to the two protons:

$$p+p \rightarrow p+p+ \pi^{0} $$

Following the collision, all three particles move at the same velocity.

Show that $$E = m_p c^2 + \left(2+ \displaystyle \frac{m_{\pi}}{2m_p} \right) m_{\pi}c^2 $$ where $m_p$ and $m_{\pi}$ are the rest masses of the proton and pion respectively.

I have tried conserving energy to get:

$$E + m_p c^2 = 2m_p \gamma (v) c^2 +\gamma (v) m_{\pi} c^2$$

The issue is, what do we do about the $\gamma$ (where $\gamma$ is the Lorentz factor)?

Thanks - all help is appreciated.


Conservation of energy on its own is usually not sufficient for this sort of problems; you also need to work with conservation of momentum, then resort to a number of nasty substitutions and whatnot. Instead, why not go for conservation of 4-momentum?


where $1$ and $2$ label the protons, and $q$ are final state momenta (for simplicity, I've dropped the "$\mu$" indices). Squaring on both sides (or, if using indices, taking the inner products):

$$p_1^2+p_2^2+p_1\cdot p_2=q_1^2+q_2^2+q_\pi^2+2q_1\cdot q_2+2q_1\cdot q_\pi+2q_2\cdot q_\pi.$$

Now remember that $p=(E,\vec{p})$ and so $p^2=E^2-|\vec{p}|^2=m^2$ :

$$2p_1\cdot p_2=2m_p^2+m_\pi^2+4m_pm_\pi.$$

Note that to simplify the RHS I've used the fact that all decay products have the same momentum. To reduce the LHS, I now use the fact that proton $1$ is at rest, i.e. $p_1=(m_p,0)$ : $$2p_1\cdot p_2=2Em_p.$$

Plugging it in the above equation yields the desired result (up to factors of $c=1$ in natural units):


  • $\begingroup$ Thanks for this. I haven't actually learnt about the four-vector. Do you recommend it? Does it make SR problems easier? Any recommended texts on it? $\endgroup$ – PhysicsMathsLove May 13 '17 at 16:23
  • 1
    $\begingroup$ If you're only just starting on SR, then I'd say whatever textbook you're using should really build up to four-vectors: they're the real objects of study, and as you learn about them you should realize how natural it is to work with them. $\endgroup$ – Demosthene May 13 '17 at 16:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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