# Homework Help: Lorentz Contraction and Energy

1. Oct 19, 2014

### fdsa1234

I already know the solution to this problem, but I'm not sure exactly why it works out the way it does, so I'm looking for an explanation.

1. The problem statement, all variables and given/known data
A particle accelerator accelerates electrons at 40 GeV in a pipe 2 miles (3218.69 metres) long, but only a few cm wide. How long is the accelerator in the rest frame of an electron with the given energy?

2. Relevant equations
$L' = L*\sqrt{(1 - V^2)}$

$E = \frac{m}{\sqrt{(1 - V^2)}}$

3. The attempt at a solution
L' is the Lorentz contracted length of the accelerator in the electron's rest frame; using the two equations with 0.51 MeV as the mass of the electron, I get $\sqrt{(1 - V^2)} = \frac{m}{E} = \frac{(0.51 MeV)}{40 GeV} = 1.2 * 10^-5$

Then, $L' = 1.2 * 10^-5 * 3218.69 = 4 cm$

This is the correct solution. My question is, why are the $\sqrt{(1 - V^2)}$ terms not $\sqrt{(1 - \frac{V^2}{c^2})}$? I thought that the Lorentz contraction equation is $L' = L*\gamma$, where $\gamma$ is $\sqrt{(1 - \frac{V^2}{c^2})}$. What's the explanation?

2. Oct 19, 2014

### Orodruin

Staff Emeritus
It is very common in relativity to use units where c = 1. This explains why it does not appear in your expressions for length contraction as well as in your expression for the total energy.

3. Oct 19, 2014

### fdsa1234

Ah, so $E = \frac{mc^2}{\sqrt{1 - v^2/c^2}}$ simply becomes $E = \frac{m}{\sqrt{1 - V^2}}$. Now I see it. Thank you.