Question

Sophie Zabar, clairvoyante, cried out in pain at precisely the instant her twin brother, 500km away, hit his thumb with a hammer. A skeptical scientist observed both events (brother’s accident, Sophie’s cry) from an airplane traveling at$\frac{\mathbf{12}}{\mathbf{13}}\mathit{c}$ to the right (Fig. 12.19). Which event occurred first, according to the scientist? How much earlier was it, in seconds?

Short answer

Sophie’s cry occurred $4\times {10}^{-3}$s earlier with respect to scientist.

Step-by-step solution

Expression for the time dilation:

Write the expression for the time dilation.

$\mathbf{Y}\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{1}\mathbf{-}{\displaystyle \frac{{\mathbf{v}}^{\mathbf{2}}}{{\mathbf{c}}^2}}}}$ …… (1)

Here, v is the velocity of an airplane, and c is the speed of light.

Determine the required time to occur an event:

Consider brother’s accident be at the origin time, i.e., zero in both the frames. With respect to frame S, Sophie’s co-ordinates will be,

$x=500\mathrm{km}\mathrm{and}t=0.$

With respect to the scientist frame of reference, Sophie’s coordinates using Lorentz transformation equation will be,

$\overline{\mathrm{t}}=\gamma \left(\begin{array}{c}t+\frac{vx}{c^2}\end{array}\right)$

As$t=0$

Hence, the equation becomes,

$\overline{\mathrm{t}}=-\frac{\gamma vx}{c^2}$ …… (2)

Substitute the value of $v=\frac{12}{13}c$in equation (1).

$$\begin{gathered} \mathrm{\gamma }=\frac{1}{\sqrt{1-{\displaystyle \frac{{\left({\displaystyle \frac{12}{13}}\mathrm{c}\right)}^2}{{\mathrm{c}}^2}}}} \\ \mathrm{\gamma }=\frac{1}{\sqrt{{\displaystyle \frac{169-144}{169}}}} \\ \mathrm{\gamma }=\frac{1}{\sqrt{{\displaystyle \frac{25}{169}}}} \\ \mathrm{\gamma }=\frac{13}{5} \end{gathered}$$

Substitute the value of y, v and x in equation (2).

$$\begin{gathered} \overline{\mathrm{t}}=-\frac{\left(\begin{array}{c}{\displaystyle \frac{13}{5}}\end{array}\right)\left(\begin{array}{c}{\displaystyle \frac{12}{13}}c\end{array}\right)\left(500\mathrm{km}\times {\displaystyle \frac{{10}^3\mathrm{m}}{1\mathrm{km}}}\right)}{c^2} \\ \overline{t}=-\frac{12\times {10}^5\mathrm{m}}{\left(\begin{array}{c}3\times {10}^8\mathrm{m}/\mathrm{s}\end{array}\right)} \\ \overline{t}=-4\times {10}^{-3}s \end{gathered}$$

Therefore, Sophie’s cry occurred $4\times {10}^{-3}$s earlier with respect to the scientist.