# Geneva LHC (Homework 5, Physics 226)

4. In the Large Hadron Collider (LHC) outside Geneva, magnetic elds bend the trajectories of protons so that they circulate repeatedly within a 27 km long circular tunnel. Inside the accelerator are two simultaneously counter-circulating beams. At the design energy and luminosity, the protons will be accelerated to 7 TeV and each beam will consist of about 2800 bunches of protons (distributed around the ring), with $10^{11}$ protons/bunch.

(a) What gamma-factor will protons accelerated to the design energy have?

(b) What is the angular velocity !lab of these protons as measured by a stationary clock?

(c) What is the proper-time angular velocity !p of the protons?

(d) What is the magnitude of the 4-acceleration of the protons?

(e) Accelerating any charged particle causes it to emit electromagnetic radiation. The power radiated (i.e., the rate at which energy is lost via electromagnetic radiation) by an accelerating particle with charge q was rst derived by J. Larmor in 1897, and is given by $P_{rad} = (2k_e/3)q^2a^2/c^3$ where $a^2$ is the square of the 4-acceleration. At the design parameters of the LHC, what total power, in watts, will be radiated by the protons circulating around the LHC ring?

a)

Equation (3.3.5) is

$γ=E/(m_0 c^2 )$

For protons $m_0 c^2=938.272 MeV$  so that

$γ=(7*10^{12})/(938.272*10^6 )=7460.5$

b)

$γ^2=1/(1-v^2/c^2)$   so that  $v^2/c^2 =1-1/γ^2$     and $v=c\sqrt{(1-1/γ^2)}$
$ω_{lab}=v/R=c/R \sqrt{(1-1/γ^2)}=(3*10^8)/((27*10^3)/2π)*\sqrt{(1-1/7460.5^2)}=$

$=1.3963*10^5 ( rad/s)$

c)

$ω=2π/T$   so that $T=2π/ω_lab =45.00 μs$

The Lorentz invariant is

$d s^2=-c^2 t^2+d x^2$   in laboratory   and $d s^2=-c^2 τ^2+0$ (in rest system)
So that the proper time (time in rest reference system) is

$c^2 τ^2=c^2 T^2-d x^2=(3*10^8 )^2*(45*10^{-6})^2-(27*10^3)^2=-5.4675*10^8 (s^2)$
$τ^2=-1.8225 s^2$

$ω_p^2=(4π^2)/τ^2 =-21.66 (rad/s)^2$

d)

$p^2=-m^2 c^2$      and $f=p/τ$     so that $f^2=(-m_0^2 c^2)/τ^2$
$a^2=f^2/(m_0^2 )=-c^2/τ^2 =(3*10^8 )^2/1.8225=4.938*10^{16} (m^2/s^4 )$

e)

$P=2/3*k e^2*a^2/c^3 =2/3*9*10^9*(1.6*10^{-19} )^2*(4.938*10^{16})/(3*10^8)^3=$ $=2.81*10^{-37} W/proton$

Total number of protons is

$N=2800*10^{11}$    so that $P_{tot}=7.87*10^{-23} W$