Fewmet
- 406
- 38
Homework Statement
The tunnel of the Large Hadron Collider is 27.00 km long. The LHC accelerates protons to a speed of 0.999997828 times the speed of light. How long would it take the proton to pass through the tunnel according to an observer at rest with respect to the building? To an observer moving with the proton?
∆x=27.00 km = 9.00 x 10-5light seconds
v=0.999997848c
Homework Equations
Δx′=γ(Δx−vΔt)
Δt′=γ(Δt−vΔx/c2)
L=L0/γ
ΔT=γΔT0
The Attempt at a Solution
For an observer at rest with respect to the building
vav=Δx/Δt
∆t=9.00 x 10-5 seconds
I think that part is fine. I tried two different approaches to finding the time in the reference frame of the proton, and got two different answers.
First approach
Treat this as two events:
Event 1) at t=0 and x=0 the proton is at one end of the collider. Then
t' = 0 s and
x' = 0 light seconds
event 2) at t= 9.00 x 10-5 seconds and x= 9.00 x 10-5 light seconds the proton is at the other end of the collider. Using the Lorentz transformations with v= 0.999997824 c, I get
x'= 9.39 x 10-8 light seconds
t'=9.39 x 10-8 seconds
So ∆t = 9.30 x 10-8 seconds - 0 seconds = 9.30 x 10-8 seconds.
Second Approach
From the perspective of the proton, the collider moves towards it at 0.999997824 c. The collider has a rest length of 9.00 x 10-5 light seconds, so an observer moving with the proton sees L, where
L=L0/γ = 9.00 x 10-5 light seconds * 0.00208614
=1.88 x 10-7 light seconds.
A body of that length moves past the observer in time t:
t =L/v= 1.88 x 10-7 light seconds/0.999997824 c =1.88 x 10-7 seconds.
The second approach is getting me twice the time of the first approach. I also see the same discrepancy in the relativistic lengths. I know I am overlooking something fundamental, but cannot see what.