Energy of relativistic particle in LHC

[Saitama]

Homework Statement

(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.

Homework Equations

The Attempt at a Solution

The particle revolves in a circular path, hence
[tex]\frac{mv^2}{R}=qvB[/tex]
[tex]mv=eRB[/tex]
(e is the charge of particle)
Since p=mv (momentum)
[tex]p=eRB[/tex]
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!

 

[mfb]

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 :(.

 

[Saitama]

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 :(.

Thanks a lot mfb for the help!

 

[rude man]

1. I don't understand how the mass of proton is not necessary for the problem. Is my expression for p correct? p=eRB

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

 

[Nickie]

Your expression for momentum is correct. The mass of the proton is not necessary because the energy of a relativistic particle is given by the equation E=√(p^2c^2+m^2c^4), where p is the momentum, c is the speed of light, and m is the mass. In this problem, we are only concerned with the momentum of the particle, so the mass is not needed. Additionally, the energy of a particle in a circular path is given by E=γmc^2, where γ is the Lorentz factor. Since the particle is moving at a high speed in the LHC, we can assume that γ is close to 1, and therefore the mass term can be neglected. Thus, the mass of the proton is not necessary for calculating the energy of the relativistic particle in the LHC.