How much total kinetic energy is needed in this case?

Click For Summary
SUMMARY

The discussion centers on calculating the total kinetic energy required to produce a new particle X through electron-positron collisions in a particle accelerator. The resting mass of particle X is given as 210 MeV/c², while the resting mass of electrons and positrons is 0.511 MeV/c². Participants suggest using Einstein's equation E=mc² and the center of mass (CM) system for calculations, along with Lorentz transformations to transition to the laboratory frame. The conversation emphasizes the importance of understanding relativistic mass and energy conservation in particle physics.

PREREQUISITES
  • Understanding of Einstein's equation E=mc²
  • Familiarity with relativistic mass and energy concepts
  • Knowledge of center of mass (CM) system in particle physics
  • Basic principles of Lorentz transformations
NEXT STEPS
  • Study relativistic energy-momentum relations, specifically E² = (pc)² + (m₀c²)²
  • Learn about Lorentz transformations and their applications in particle physics
  • Explore conservation of momentum in high-energy collisions
  • Investigate the implications of particle mass in different reference frames
USEFUL FOR

Students and educators in physics, particularly those focusing on particle physics and relativistic mechanics, will benefit from this discussion.

eri3an
Messages
4
Reaction score
0
I have a problem with an exercise from school. could anyone help me?

Let say that you are searching for a new particel X by making collison between positrones and elektrones in an particelaccelerator. you guess that the new particles resting mass is 210 MeV/c^2 and you hope to produce it by:

(e+) + (e–) ---> X + photons

a) Let us say that the electron (e-) is in resting state in the laboratory system. How much energy is needed to accerate positrons (e+) to such a velocity for the reaction to take place?

b)On less energy demanding way of doing it is to let e+ och e– collide with each other from opposite directions with the same velocity How much total kinetic energy is needed in this case?

resting mass for an electrone or positron is 0.511 MeV/c^2

please I'm not interested in just getting an answer how do i solve it? I've tried by deriving from the E(a) + E(b) = E(c) + E(d) and tried to go using conservation of momentum but it doesn't seem to work :( Maybe I'm out on bad ice?
 
Last edited:
Physics news on Phys.org
Hi eri3an,
you must remember Einstein's equation : E = mc^2, where m is the mass of the particle when it is moving.
 
eri3an said:
I have a problem with an exercise from school. could anyone help me?

Let say that you are searching for a new particel X by making collison between positrones and elektrones in an particelaccelerator. you guess that the new particles resting mass is 210 MeV/c^2 and you hope to produce it by:

(e+) + (e–) ---> X + photons

a) Let us say that the electron (e-) is in resting state in the laboratory system. How much energy is needed to accerate positrons (e+) to such a velocity for the reaction to take place?

b)On less energy demanding way of doing it is to let e+ och e– collide with each other from opposite directions with the same velocity How much total kinetic energy is needed in this case?

resting mass for an electrone or positron is 0.511 MeV/c^2

please I'm not interested in just getting an answer how do i solve it? I've tried by deriving from the E(a) + E(b) = E(c) + E(d) and tried to go using conservation of momentum but it doesn't seem to work :( Maybe I'm out on bad ice?

I suggest you to study the problem in the CM (centre of mass) system.
Then you can pass to the system of the laboratory by Lorentz transformations.
 
borisleprof said:
Hi eri3an,
you must remember Einstein's equation : E = mc^2, where m is the mass of the particle when it is moving.

i just though this statement could use some elaboration:
E=\gamma m_0c^2=\sqrt{m_0^2c^4+p^2c^2}

i think using the real mass, i mean the thing you divide the force with to get the accelaration is better (and that's what borisleprof did).
but in most textbooks, they talk about the mass you measure at the rest frame...
 
Last edited:

Similar threads

  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 10 ·
Replies
10
Views
6K
  • · Replies 20 ·
Replies
20
Views
9K
  • · Replies 54 ·
2
Replies
54
Views
5K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
29
Views
8K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 21 ·
Replies
21
Views
5K
  • · Replies 2 ·
Replies
2
Views
11K
  • · Replies 0 ·
Replies
0
Views
2K