Lorentz Contraction and Energy

[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.

Homework Statement

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?

Homework Equations

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

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

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?

 

[Orodruin]

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.

 

[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.