Zipping round and round

From the classical theory we know that the orbit of a particle circling around in a magnetic field satisfies

$R = \frac{m v}{q B}$

Special theory of relativity gives a simple correction to this result:

$R = \frac{\gamma m v}{q B}$

Where $\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ .

The relativistic energy can be calculated as $E = \gamma m c^2$ . By taking $v \approx c$ we can substitute $\gamma m v = \gamma m c = q B R$ and thus $E = q B R c$ .

p=mV=qBr p=57.6E-16 p2c2=2.9E-12 m2c4=0.00000002E-12 E=sqrt[(2.9-0.00000002)E-12] E=1.7E-6J

The relation between the radius of the LHC and the momentum of the proton is given by ( $\omega$ is the angular velocity and $v$ is the velocity of the proton):

$\vec{F}=\frac{d\vec{p}}{dt} \Rightarrow |e|vB = p\omega \Rightarrow \frac{v}{\omega} = R = \frac{p}{|e|B}$

Since the speed of the particle is very close to the speed of light, from the equation of relativistic momentum, one gets $p >> m_pc$ and:

$E=\sqrt{p^2c^2+m_p^2c^4} \approx pc = |e|BRc=1.73*10^{-6}~ J$ .

We don't really need to know the exact value of proton's mass, since at high speed, the behavior of proton is the same as that of a massless particle!!!