Question

Anna and Bob have identical spaceship 60 m long. The diagram shows Bob’s observations of Anna’s ship, which passes at a speed of $\mathit{c}\mathbf{/}\sqrt{\mathbf{2}}$. Clocks at the back of both ships read 0 just as they pass. Bob is at the center of his ship and at t = 0 on his wrist watch peers at a second clock on Anna’s ship.

(a) What does this clock read?

(b) Later, the back of Anna’s ship passes Bob. At what time does this occur according to Bob?

(c) What will observers in Bob’s frame see on Anna’s two clocks at this time?

(d) Identify two events that show time dilation and two that show length contraction according to Anna.

Short answer

(a) The value of Anna’s clock read at time t, = -100ns.

(b) The value of Bob time occurs later, the back of Anna’s ship passes is t= 141 ns.

(c) The value of position of two clocks according to Bob’s to get the readings on Anna’s frame is $t{\text{'}}_1\approx 100\mathrm{ns}\mathrm{and}\mathrm{t}{\text{'}}_2=0$.

(d) It confirms that the events have been time dilated and length contracted from Anna's perspective by using the correct time and length, together with the length contraction and time dilation formulae.

Step-by-step solution

Write the given data from the question.

Consider the Anna and Bob have identical spaceship 60 m long.

Consider a speed of ships passes at $c/\sqrt{2}$.

Consider a time of ship reads at at t = 0 .

Determine the formula of Anna’s clock read at time, Bob time occurs later, the back of Anna’s ship passes and position of two clocks according to Bob’s to get the readings on Anna’s frame.

Write the formula of Anna’s clock read at time.

$\mathit{t}\mathbf{\text{'}}\mathbf{=}{\mathit{\gamma }}_{\mathbf{v}}\left(t-\frac{vx}{c^2}\right)$ …… (1)

Here, ${\mathit{\gamma }}_{\mathbf{v}}$ is relativistic velocity transformation, t is time taken by ship, v is speed of ship, v is Bob’s position and c speed of light.

Write the formula of Bob time occurs later, the back of Anna’s ship passes.

$\mathit{t}\mathbf{=}\frac{\mathbf{d}}{\mathbf{v}}$ …… (2)

Here, is bob’s position at center andis speed of ship.

Write the formula of position of first clocks according to Bob’s to get the readings on Anna’s frame.

role="math" localid="1660024619938" $\mathit{t}{\mathit{\text{'}}}_{\mathbf{1}}{\mathit{\gamma }}_{\mathbf{v}}\mathbf{(}\mathbf{t}\mathbf{-}\frac{\mathbf{v}{\mathbf{x}}_{\mathbf{1}}}{{\mathbf{c}}^{\mathbf{2}}}\mathbf{)}$ …… (3)

Here, ${\mathit{y}}_{\mathbf{v}}$ is relativistic velocity transformation, t is time taken by ship, v is speed of ship, v is Bob’s position and speed of light.

Write the formula of position of second clocks according to Bob’s to get the readings on Anna’s frame.

$\mathit{t}{\mathit{\text{'}}}_{\mathbf{2}}{\mathit{\gamma }}_{\mathbf{v}}\mathbf{(}\mathbf{t}\mathbf{-}\frac{{\mathbf{vx}}_{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}\mathbf{)}$ …… (4)

Here, ${\mathit{\gamma }}_{\mathbf{v}}$ is relativistic velocity transformation, t is time taken by ship, v is speed of ship, x is Bob’s position and c speed of light.

(a) Determine the value of Anna’s clock read at time.

As shown by the accompanying image, all of Bob's clocks are synchronised at t=0 in his frame${\mathbf{}}^{\mathbf{\left(}\mathbf{S}\mathbf{\right)}}$. Additionally, if we place the origin of the positive$x$at the back of Bob's spaceship and the positive x on the right, Bob is located at x =30 m . With this knowledge in hand, we could now immediately replace in the equation for the Lorentz transformation to obtain the reading of Anna's clock$\left(\mathrm{S}\text{'}\mathrm{frame}\right)$.

The speed of ship will be:

$v=\frac{c}{\sqrt{2}}$

Determine the relativistic velocity transformation.

$$\begin{gathered} {\gamma }_v=\frac{1}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}} \\ =\frac{1}{\sqrt{1-{\displaystyle \frac{1}{2}}}} \\ =\sqrt{2} \end{gathered}$$

Determine the Anna’s clock read at time.

role="math" localid="1660025289346" $$\begin{gathered} \mathrm{t}\text{'}=\sqrt{2}\left(0-\frac{{\displaystyle \frac{\mathrm{c}}{\sqrt{2}}\times 30\mathrm{m}}}{{\mathrm{c}}^2}\right) \\ =-\frac{30\mathrm{m}}{3\times {10}^8\mathrm{m}/\mathrm{s}} \\ =-100\mathrm{ns} \end{gathered}$$

Therefore, the value of Anna’s clock read at time t' = - 100 n .

(c) Determine the value of position of two clocks according to Bob’s to get the readings on Anna’s frame.

Bob's frame of reference will once more be taken into account, but I have worried about the readings in Anna's frame ( S' ) at t = 141 ns. It only needs to determine the two clocks' locations in relation to Bob, and then use our equations for the Lorentz transformation to obtain the readings on Anna's frame.

Determine the position of first clocks according to Bob’s to get the readings on Anna’s frame.

Substitute $\sqrt{2}$for ${\gamma }_v$, 141 nm for t , $\frac{c}{\sqrt{2}}$ for v and 30 m for $x_1$into equation (3)

$$\begin{gathered} t{\text{'}}_1=\sqrt{2}\left(141\mathrm{nm}-\frac{{\displaystyle \frac{c}{\sqrt{2}}}\times 30\mathrm{m}}{c^2}\right) \\ =\left(141\sqrt{2}-100\right)\mathrm{ns} \\ \mathrm{t}\text{'}\approx 100\mathrm{ns} \end{gathered}$$

Now, determine the position of second clocks according to Bob’s to get the readings on Anna’s frame.

Substitute $\sqrt{2}$for , ${\gamma }_v$ for 141 nm, t , $\frac{c}{\sqrt{2}}$for v and 30 m for $x_2$into equation (4).

$$\begin{gathered} t{\text{'}}_2=\sqrt{2}\left(141\mathrm{nm}-\frac{{\displaystyle \frac{c}{\sqrt{2}}\times 30\mathrm{m}}}{c^2}\right) \\ =\sqrt{2}\left(141-141\right)\mathrm{ns} \\ =0 \end{gathered}$$

Therefore, the value of position of two clocks according to Bob’s to get the readings on Anna’s frame is $t{\text{'}}_1\approx 100\mathrm{ns}$and$t{\text{'}}_2=0$ .

(d) Determine the Identify two events that show time dilation and two that show length contraction according to Anna.

We must establish the appropriate time and appropriate length in order to determine the suitable time dilation and appropriate length contraction from Anna's perspective. In order to determine the correct time, we must measure two events that are occurring at the same time. The two events are ${\mathrm{E}}_2$and ${\mathrm{E}}_3$, respectively, in the picture below. Bob's clock, which is located at the same spot, reads 0 and 141 ns , respectively, whereas Anna's front clocks afterwards reads - 100 ns and the back clock reads + 100 ns . The correct time is therefore 141 ns , which is dilated to 200 ns. Let's check this with the use of our time dilation formula.

$\begin{array}{l}{t}_A={\gamma }_v{t}_B\\ =\sqrt{2}\times 141\mathrm{ns}\\ \approx 200\mathrm{ns}\end{array}$

As move on to the second portion of the question, which is once more about the length contraction, we must determine the correct length, or, to put it another way, two events that happened simultaneously from Anna's perspective. You can see from the illustration that the readings on the various clocks in Anna's frame for ${\mathrm{E}}_1$and ${\mathrm{E}}_4$are the same. The distance between Anna's clocks is said to be according to the problem description, but as this is a length contracted observation, the actual length must have been demonstrated.

$$\begin{gathered} L_A={\gamma }_v{L}_B \\ =\sqrt{2}\times 30\mathrm{m} \\ =42.4\mathrm{m} \end{gathered}$$

Additionally, we know that already aware of Bob's spaceship's 60 m length. Anna measures the distance in her frame at the same time and concludes that Bob's spaceship should fit between the two clocks.

$$\begin{gathered} {\mathrm{L}}_{Bob\mathit{-}ship}=\frac{60\mathrm{m}}{\sqrt{2}} \\ =42.4\mathrm{m} \end{gathered}$$

As Anna could see, the length of the spaceship is exactly equal to the separation between the two clocks.

Fig. 1

Therefore, it confirms that the events have been time dilated and length contracted from Anna's perspective by using the correct time and length, together with the length contraction and time dilation formulae.