91Ó°ÊÓ

Space pilot Mavis zips past Stanley at a constant speed relative to him of 0.800c. Mavis and Stanley start timers at zero when the front of Mavis’s ship is directly above Stanley. When Mavis reads 5.00 s on her timer, she turns on a bright light under the front of her spaceship. (a) Use the Lorentz coordinate transformation derived in Example 37.6 to calculate x and t as measured by Stanley for the event of turning on the light. (b) Use the time dilation formula, Eq. (37.6), to calculate the time interval between the two events (the front of the spaceship passing overhead and turning on the light) as measured by Stanley. Compare to the value off you calculated in part (a). (c) Multiply the time interval by Mavis’s speed, both as measured by Stanley, to calculate the distance she has traveled as measured by him when the light turns on. Compare to the value of x you calculated in part (a).

Step by step solution

01

Formulas used to solve the question

Gamma:

$\mathit{γ}\mathbf{=}\sqrt{\mathbf{1}\mathbf{-}\frac{{\mathbf{v}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}}$

Time interval:

$\mathbf{∆}\mathit{t}\mathbf{=}\frac{\mathbf{∆}{\mathbf{t}}_{\mathbf{0}}}{\sqrt{\mathbf{1}\mathbf{-}{\displaystyle \frac{{\mathbf{u}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}}}}$

02

Determine the value of x and t

Stanely in frame S, Mavis in frame $\mathit{S}\mathbf{\text{'}}$.

${\mathit{t}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{5}\mathit{s}\mathbf{,}{\mathit{x}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{0}$(the light is at zero x coordinate in mavis frame)

Now,

$\mathit{x}\mathbf{=}\mathit{γ}\mathbf{(}{\mathit{x}}^{\mathbf{\text{'}}}\mathbf{+}\mathit{u}{\mathit{t}}^{\mathbf{\text{'}}}\mathbf{)}$

Here,

$\mathit{γ}\mathbf{=}\sqrt{\mathbf{1}\mathbf{-}\frac{{\left(0.800c\right)}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{67}$

So,

$\mathbf{x}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{67}\mathbf{(}\mathbf{0}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{8}\mathbf{c}\mathbf{*}\mathbf{5}\mathbf{)}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{003}\mathbf{*}{\mathbf{10}}^{\mathbf{9}}\mathbf{m}$

And

$\mathit{t}\mathbf{=}{\mathit{t}}^{\mathbf{\text{'}}}\mathit{y}\mathbf{=}\mathbf{5}\mathbf{*}\mathbf{1}\mathbf{.}\mathbf{67}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{33}\mathit{s}$

03

Determine the time interval between the two events

Using the time dilation formula, the proper time is$\mathit{Δ}{\mathit{t}}_{\mathbf{0}}\mathbf{=}\mathbf{5}\mathit{s}$ in frame of mavis.

$\mathbf{∆}\mathit{t}\mathbf{=}\frac{\mathbf{5}}{\sqrt{\mathbf{1}\mathbf{-}{\displaystyle \frac{{\left(0.8c\right)}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}}}}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{333}\mathit{s}$(same result as part (a))

04

Determine the distance travelled when the light turns on

$\mathit{v}\mathbf{*}\mathit{t}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{800}\mathit{c}\mathbf{*}\mathbf{8}\mathbf{.}\mathbf{33}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{800}\mathbf{*}\mathbf{3}\mathbf{*}{\mathbf{10}}^{\mathbf{8}}\mathbf{*}\mathbf{8}\mathbf{.}\mathbf{33}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{003}\mathbf{*}{\mathbf{10}}^{\mathbf{9}}\mathit{m}$

(same result as part(a))

Thus, the value of x and t is$\mathbf{2}\mathbf{.}\mathbf{003}\mathbf{*}{\mathbf{10}}^{\mathbf{9}}\mathit{m}$ and8.33 s respectively. The time interval between the two events is 8.33 s. The distance travelled when the light turns on is $\mathbf{2}\mathbf{.}\mathbf{003}\mathbf{*}{\mathbf{10}}^{\mathbf{9}}\mathit{m}$.

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