Question

asked Feb 23, 2020 123k views

1 Answer

Xeph · Answer 1 · 2020-02-27T15:54:04+0000

a. 37500; 299,792,457.9 m/s

The formula for the length contraction for a particle moving close to the speed of light is:

$L' = (L)/(\gamma)$

where

L' is the length observed by the particle moving

L is the length observed by an observer at rest

In this problem, we have

L = 3 km = 3000 m is the length of the SLAC measured by an observer at rest

L' = 8cm = 0.08 m is the length measured by the electrons moving

Substituting into the formula, we find the gamma factor

$\gamma = (L)/(L')=(3000 m)/(0.08 m)=37,500$

The formula for the gamma factor is

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

where v is the electron's speed and c is the speed of light. Re-arranging the equation, we find v:

$1-(v^2)/(c^2)=(1)/(\gamma^2)\\v=c \sqrt{ 1-(1)/(\gamma^2)}=(299 792 458 m/s)\sqrt{1-(1)/((37500)^2)}=299,792,457.9 m/s$

b.
$1.0\cdot 10^(-5)s$

For an observer at rest in the laboratory, the electron is moving at a speed of

v = 299,792,457.9 m/s

and it covers a total distance of

L = 3000 m

which is the length of the SLAC measured by the observer. Therefore, the time it takes for the electron to travel down the tube is

$t=(L)/(v)=(3000 m)/(299,792,457.9 m/s)=1.0\cdot 10^(-5)s$

c.
$2.67\cdot 10^(-10)s$

From the electron's point of view, the length of the SLAC is actually contracted, so the electron "sees" a total distance to cover of

L' = 0.08 m

And this means that the total time of travel of the electron, in its frame of reference will be shorter; in particular it is given by the formula:

$t' = (t)/(\gamma)$

where

$t=1.0\cdot 10^(-5)s$ is the time measured by the observer at rest

$\gamma=37500$

Substituting,

$t' = (1.0\cdot 10^(-5)s)/(37500)=2.67\cdot 10^(-10)s$

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