Question

5-4. Electrons in the 2-mile linear accelerator of SLAC (Stanford Linear Accelerator Center) reach a final velocity of about 0.999 999 9997c. (a) What is the length of the accelerator to such an electron? (b) Therefore how much time would it take to Problems 61 reach the end of the accelerator to such an electron, assuming the velocity stays constant? (c) How long would it take the electron to make the trip as seen by Stanford?

Answer 1

Given the velocity of electron is

Given the velocity of electron is nu = 0.9999999997 c The gamma factor is gamma = 1/squareroot 1 - nu^2/c^2 = 40824.8 The electron in the moving frame will see a contraction in length. A)Length of the accelerator to the electron is L = L_0/gamma = 2 miles The answer in meters is L = 4.899 times 10^-5 times 1609.34 meters = 0.0788 m b) Time taken to reach the end of the accelerator as seen by the electron is t = L/nu = 0.0788 m/0.9999999997 times 3 times 10^8 = 2.628 times 10^-8 s c) Time taken as seen by Stanford is t_0 = gamma t = 40824.8 times 2.628 times 10^-8 = 1.073 times 10^-5 s

The gamma factor is

$\gamma =\frac{1}{\sqrt{1-v^2/c^2}}=40824.8$

The electron in the moving frame will see a contraction in length.

a) Length of the accelerator to the electron is

$L=\frac{L_0}{\gamma}=\frac{2\,miles}{40824.8}=4.899\times 10^{-5}\,miles$

The answer in meters is

$L=4.899\times 10^{-5}\times 1609.34 \, \,meters=0.0788\,\,m$

b) Time taken to reach the end of the accelerator as seen by the electron is

$t=L/v=\frac{0.0788\,\,m}{0.9999999997\times 3\times 10^8}=2.628\times 10^{-8}\,s$

c) Time taken as seen by Stanford is

$t_o=\gamma t=40824.8\times 2.628\times 10^{-8}=1.073\times 10^{-5}\,s$