91Ó°ÊÓ

The length of an object in terms of the speed parameter is given by $L=L_0\sqrt{1-{\mathrm{β}}^2}$ . Here, $\mathrm{β}$is the speed parameter, $L_0$ is the rest length.

Given that the length of the spaceship is half of its rest length. So, we have $L=\frac{L_0}{2}$.

Substitute the known values in the above formula and solve for $\mathrm{β}$as follows:

$$\begin{gathered} L=L_0\sqrt{1-{\mathrm{β}}^2} \\ \frac{L_0}{2}=L_0\sqrt{1-{\mathrm{β}}^2} \end{gathered}$$

$$\begin{gathered} \frac{1}{2}=\sqrt{1-{\mathrm{β}}^2} \\ \frac{1}{4}=1-{\mathrm{β}}^2 \end{gathered}$$

Solve the above equation further,

$$\begin{gathered} {\mathrm{β}}^2=1-\frac{1}{4} \\ {\mathrm{β}}^2=\frac{3}{4} \\ \mathrm{β}=0.8660 \end{gathered}$$

Thus, the speed parameter is $\mathrm{β}=0.8660$.

(b)

The factor by which the spaceship’s clocks run slow relative to the clocks in the observer’s frame is equal to $\mathrm{γ}=\frac{1}{\sqrt{1-{\mathrm{β}}^2}}$.

Here, we calculated that $\mathrm{β}=0.8660$. So, the factor can be calculated as follows:

$$\begin{gathered} \mathrm{γ}=\frac{1}{\sqrt{1-{\mathrm{β}}^2}} \\ \mathrm{γ}=\frac{1}{\sqrt{1-{\left(0.8660\right)}^2}} \\ \mathrm{γ}=\frac{1}{0.5} \\ \mathrm{γ}=2.00 \end{gathered}$$

Thus, by a factor of$2.00$ , the spaceship’s clocks run slow relative to clocks in the observer’s frame.