# Homework Help: Question in special relativity

1. Jun 27, 2009

### m.medhat

Hello,
I have a two question in relativity from the book”university physics” and the answers of these two questions was found in the end of this book , but the answer of the book do not correspond my answer .
The questions are :-
Kaon production . in high-energy physics , new particles can be created by collisions of fast-moving projectile particles with stationary particles . some of the kinetic energy of the incident particle is used to create the mass of the new particle . A proton – proton collision can result in the creation of a negative kaon (K-) and a positive kaon (K+) :-
Calculate the minimum kinetic energy of the incident proton that will allow this reaction to occur if the second (target) proton is initially at rest . the rest energy of each kaon is 493.7 MeV . the rest energy of each proton is 938.3 MeV . (Hint :- it is useful here to work in the frame in which the total momentum is zero . but note that here the lorentz transformation must be used to relate the velocities in the laboratory frame to those in the zero-total-momentum frame .
The answer of the book = 2494 MeV
While my solution was :-
The kinetic energy for proton equal to the rest energy for new two particles .
The kinetic energy for proton = 493.7+493.7 = 987.4 MeV
Which solution is correct ?why?
The next question is :-
Lorentz transformation for acceleration . find the lorentz transformation equation for velocity . let frame S’ have a constant x-component of velocity (u) relative to frame S . an object moves relative to frame S along the x-axis with instantaneous velocity (vx) and instantaneous acceleration (ax) . show that its instantaneous acceleration in frame S’ is

a’x = ax [(1-v2/c2)^3/2] [(1-uvx/c2)^-3]

(hint :- express the acceleration in S’ as a’x = dv’x / dt’ )
and show that the acceleration in frame S can be expressed as

ax = a’x [(1-v2/c2)^3/2] [(1+uv’x/c2)^-3]

please I want the solution be with steps and fine details .

2. Jun 27, 2009

### Staff: Mentor

Realize that there are two protons and two kaons. Once you find the kinetic energy of the protons in the zero-momentum frame, you need to transform back into the lab frame. Hint: Find the speed of the protons in the zero-momentum frame, then find their speed relative to each other.

3. Jun 29, 2009

### m.medhat

very thanks for reply but i still do not understand , please i want a complete solution for the two questions

4. Jun 29, 2009

### dx

Giving comlpete solutions is against PF rules.

5. Jul 1, 2009

### m.medhat

i am sorry i don't know that , please i want any help tolve these two questions

6. Jul 1, 2009

### queenofbabes

Do you know how to set up the zero momemtum frame properly? You don't seem to have done so, why not try again and let us see how you've done it?

7. Jul 2, 2009

### m.medhat

i do not know how to set up the zero momentum frame properly (i want a help)

8. Jul 2, 2009

### Staff: Mentor

The zero momentum frame is the frame in which the protons approach each other with the same speed. After the collision, the protons are at rest. Hint: What happened to the proton's kinetic energy?

9. Jul 4, 2009

### m.medhat

i thought that protons kinetic energy will be converted to rest energy for two kaons . but i still don't know how to solve the question (please i want help)

10. Jul 4, 2009

### queenofbabes

You think correctly, but since you still seem clueless regarding the zero momentum frame(ZMF) , here are some steps:

1) The zero momentum frame moves at a certain constant velocity with respect to the lab frame. Maybe call it v'. Remember within the ZMF both protons are moving towards each other at the same speed (and hence no net momentum). **So what is the relation between v' and v, the velocity of the original proton in the lab frame?**

2) You correctly deduced KE is converted to rest mass energy. Note that this refers to KE *in the ZMF!!* Make sure you understand why! After the collision, everything is at rest.

3) Work backwards to deduce v', and hence KE of original proton.