Please please help me out. This is a question about time dilation. Thanks!?!

Question

Provide the time dilation equation found in Section 15.4 of the text. Explain each step of the derivation.

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Answer 1 · 2017-11-16T00:43:45

Let #t_"moving" = # the time that the observer in the moving vehicle measures.

Let #t_"stationary" = # the time that the stationary observer measures

Let #c =# the speed of light

Let #v =# the speed of the vehicle

As the beam is travels toward and strikes the first mirror, the moving observer sees the light-pulse travel a distance

#a = ct_"moving"#

The stationary observer sees the vehicle move to the right a distance

#b = vt_"stationary"#

The stationary observer sees the light-pulse travel diagonally a distance

#d = ct_"stationary"#.

We know that the distances, a, b, and d form a right triangle with d being the hypotenuse:

#d^2 = a^2 + b^2#

Substitute for a, b, and d:

#(ct_"stationary")^2 = (ct_"moving")^2 + (vt_"stationary")^2#

Subtract #(vt_"stationary")^2# from both sides:

#(ct_"stationary")^2-(vt_"stationary")^2 = (ct_"moving")^2#

Factor out #t_"stationary"^2# and distribute the squares:

#(c^2-v^2)t_"stationary"^2 = c^2t_"moving"^2#

Divide both sides by #(c^2-v^2)#:

#t_"stationary"^2 = (c^2t_"moving"^2)/(c^2-v^2)#

Multiply the right side by 1 in the form of #(1/c^2)/(1/c^2)#:

#t_"stationary"^2 = t_"moving"^2/(1-v^2/c^2)#

Use the square root on both sides:

#t_"stationary" = t_"moving"/sqrt(1-v^2/c^2)#

The above is one form; the other form is:

#t_"moving" = t_"stationary"sqrt(1-v^2/c^2)#