Question

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?

Short answer

1. The value of $\beta$ is $0.8660$.
2. By a factor of $2.00$, the spaceship’s clocks run slow relative to clocks in the observer’s frame.

Step-by-step solution

The speed parameter (a) 

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 $\beta =0.8660$.

Solution of part (b)

(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.