Question

A relativistic train of proper length 200 m approaches a tunnel of the same proper length, at a relative speed of 0.900c. A paint bomb in the engine room is set to explode (and cover everyone with blue paint) when the front of the train passes the far end of the tunnel (event FF). However, when the rear car passes the near end of the tunnel (event RN), a device in that car is set to send a signal to the engine room to deactivate the bomb. Train view: (a) What is the tunnel length? (b) Which event occurs first, FF or RN? (c) What is the time between those events? (d) Does the paint bomb explode? Tunnel view: (e) What is the train length? (f) Which event occurs first? (g) What is the time between those events? (h) Does the paint bomb explode? If your answers to (d) and (h) differ, you need to explain the paradox, because either the engine room is covered with blue paint or not; you cannot have it both ways. If your answers are the same, you need to explain why?

Short answer

(a) The tunnel length is $87.2\text{m}$.

(b) The FF event occurs first.

(c) The time between events is $4.18\times {10}^{-7}\text{s}$.

(d) The paint bomb will explode.

(e) The train length is $87.2\text{m}$.

(f) The RN event occurs first in tunnel view.

(g) The time between events is $4.18\times {10}^{-7}\text{s}$.

(h) The paint bomb will not explode.

Step-by-step solution

Identification of given data

The relative speed of relativistic train is $v_A=0.900c$

The proper length of relativistic train is $L=200\text{m}$

The length of tunnel relative to train is found by using the formula for length contraction. The time for far end and rear end is found by considering the difference in proper length and relative length of tunnel.

Determination of tunnel length (a)

The tunnel length is given as:

$L_t=L\sqrt{1-{\left(\frac{v}{c}\right)}^2}$

Here, $c$ is the speed of light and its value is $3\times {10}^8\text{m}/\text{s}$.

Substitute all the values in the above equation.

$$\begin{gathered} L_t=\left(200\text{m}\right)\sqrt{1-{\left(\frac{0.900c}{c}\right)}^2} \\ =87.2\text{m} \end{gathered}$$

Therefore, the tunnel length is $87.2\text{m}$.

Determination of occurrence of event (b)

The tunnel length is shorter than the proper length of relativistic train so the front of train crosses far end of tunnel before rear end crosses the rear end of tunnel.

Therefore, the FF event occurs first.

Determination of time between events (c)

The time between events is given as:

$t_e=\frac{L-L_t}{v}$

Substitute all the values in the above equation.

$$\begin{gathered} t_e=\frac{200\text{m}-87.2\text{m}}{\left(0.900c\right)\left(\frac{3\times {10}^8\text{m}/\text{s}}{c}\right)} \\ =4.18\times {10}^{-7}\text{s} \end{gathered}$$

Therefore, the time between events is $4.18\times {10}^{-7}\text{s}$.

Identification for explosion of bomb (d)

In the tunnel view FF event occurs first before RN so the paint bomb will explode because paint bomb is set to explode as it passes far end of tunnel.

Therefore, the paint bomb will explode.

Determination of train length (e)

The train length is given as:

$L_T=L\sqrt{1-{\left(\frac{v}{c}\right)}^2}$

Substitute all the values in the above equation.

$$\begin{gathered} L_T=\left(200\text{m}\right)\sqrt{1-{\left(\frac{0.900c}{c}\right)}^2} \\ =87.2\text{m} \end{gathered}$$

Therefore, the train length is $87.2m$.

Determination of occurrence of event in tunnel view (f)

The train length is shorter than the proper length of relativistic train so the rear end of train crosses rear end of tunnel before front end crosses the front end of tunnel.

Therefore, the RN event occurs first in tunnel view.

Determination of time between events (g)

The time between events is given as:

$T=\frac{L-L_T}{v}$

Substitute all the values in the above equation.

role="math" localid="1663151209123" $$\begin{gathered} T=\frac{200\text{m}-87.2\text{m}}{\left(0.900c\right)\left(\frac{3\times {10}^8\text{m}/\text{s}}{c}\right)} \\ =4.18\times {10}^{-7}\text{s} \end{gathered}$$

Therefore, the time between events is $4.18\times {10}^{-7}\text{s}$.

Determination of time between events (h)

The time between events in tunnel view is given as:

$t=\frac{L}{v}$

Substitute all the values in the above equation.

$$\begin{gathered} t=\frac{200\text{m}}{\left(0.900c\right)\left(\frac{3\times {10}^8\text{m}/\text{s}}{c}\right)} \\ =7.41\times {10}^{-7}\text{s} \end{gathered}$$

The time for occurrence of events in tunnel view is more than the time for events in train view so paint bomb will not explode because time for crossing front end of tunnel is more and rear end crosses rear end before front end.