# Energy of relativistic particle in LHC

Tags:
1. Mar 5, 2013

### Saitama

1. The problem statement, all variables and given/known data
(see attachment)
The proton charge is $1.6 \times 10^{-19} C$ and the speed of light is $3 \times 10^8 m/s$. The proton's mass is not necessary in this problem.

2. Relevant equations

3. The attempt at a solution
The particle revolves in a circular path, hence
$$\frac{mv^2}{R}=qvB$$
$$mv=eRB$$
(e is the charge of particle)
Since p=mv (momentum)
$$p=eRB$$
I don't understand how the mass of proton is not necessary for the problem. Is my expression for p correct?

Any help is appreciated. Thanks!

File size:
54.4 KB
Views:
96
2. Mar 5, 2013

### Staff: Mentor

Your equations do not use the proton mass, how could it be relevant? You can use the approximation $E=pc$ to get the energy, as the mass of the proton is negligible relative to its momentum.

Your formulas are good for nonrelativistic particles only, you cannot use this approximation for protons in the LHC.
The result is the same with relativistic formulas, but that is just a coincidence here.

The sketch has wrong positions for LHCb and ALICE :(.

3. Mar 5, 2013

### Saitama

Thanks a lot mfb for the help!

4. Mar 5, 2013

### rude man

Yes it is, even relativistically, if you remember p is the relativistic momentum = mv = γm0v.