The length of a spaceship is measured to be exactly half its rest length. (a) To three significant figures, what is the speed parameter $\mathit{\beta }$of the spaceship relative to the observer’s frame? (b) By what factor do the spaceship’s clocks run slow relative to clocks in the observer’s frame?

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

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

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

Solve the above equation further,

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

Thus, the speed parameter is $\beta =0.8660$.

(b)

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

Here, we calculated that $\beta =0.8660$. So, the factor can be calculated as follows:

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

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