3
$\begingroup$

The problem given is

"A proton of total energy 3GeV makes a head-on collision with a 5GeV electron. Calculate the available energy in the centre-of-mass system to create any new additional particles in the collision"

My attempt to solve this was to calculate $pc$ for the proton using $(E_p^2 - (m_pc^2)^2)^{1/2}$ which evaluated to $2849 MeV$. I then compared invariants in the lab frame $p^\mu = \frac{1}{c}(E_p+E_e,E_e-pc)$ and the centre of mass frame $p'^\mu = \frac{1}{c}(E,0)$, where $E$ is the Energy available for particle creation. This gave $E^2 = (E_p+E_e)^2 - (E_e-pc)^2$, where all of the terms on the right are known. However, the value for $E$ that this expression gave was $7.71 GeV$, this was different from the given value of $6.76 GeV$.

Improve this question
$\endgroup$
3
  • $\begingroup$ What does your quantity $E$ represent? $\endgroup$
    – Brian Moths
    Commented Jan 19, 2016 at 21:14
  • $\begingroup$ Energy available for particle creation. $\endgroup$
    – kylergs
    Commented Jan 19, 2016 at 21:34
  • $\begingroup$ It appears that the problem assumes that both the proton and the electron will appear in the final state. That is $e + p \longrightarrow e' + p' + X$ and you are expected to find the maximum possible mass/energy of $X$. BTW--if you can see why this is "obvious" from the text of your question then you know something about these problems. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Jan 20, 2016 at 4:00

1 Answer 1

Reset to default
1
$\begingroup$

In the expression $E^2 = (E_p+E_e)^2 - (E_e-pc)^2$, $E$ represents total energy in the zero momentum frame. The electron and proton do not annihilate so the energy available for particle creation can be found from the total energy minus the rest energies of the proton and electron. So the energy available for particle creation $E_x = E - m_pc^2 -m_ec^2$.

Improve this answer
$\endgroup$
0

Your Answer

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

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