An experimenter determines that a particle created at one end of the laboratory apparatus moved at 0.94c and survived for $\mathbf{0}\mathbf{.}\mathbf{032}\mathbf{}\mathit{\mu }\mathit{s}$decaying just as it reached the other end. (a) According to the experimenter. How far did the particle move? (b) In its own frame of reference, how long did the particle survive? (c) According to the particle, what was the length of the laboratory apparatus?

Consider the speed of particle is$\upsilon =0.94c$.

Consider a time to survive at $t=0.032\mu s$.

Write the formula of distance travelled by the particle.

$\mathit{v}\mathbf{=}\frac{\mathbf{d}}{\mathbf{t}}$ …… (1)

Here, is distance travelled by particle, v is speed of particle and t is time to survive.

Write the formula of particle’s time frame of reference, it survived.

$\mathbf{∆}{\mathit{t}}_{\mathbf{0}}\mathbf{=}\mathbf{∆}\mathit{t}\sqrt{\mathbf{1}\mathbf{-}\frac{{\mathbf{v}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}}$ …… (2)

Here, $\mathbf{∆}\mathit{t}$ is time dilation, v is speed of particle and c is speed of light.

Write the formula of length of the laboratory apparatus according to the particle.

$\mathit{t}\mathbf{=}{\mathit{t}}_{\mathbf{0}}\sqrt{\mathbf{1}\mathbf{-}\frac{{\mathbf{v}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}}$ …… (3)

Here, ${\mathit{t}}_{\mathbf{0}}$ is reference length, v is speed of particle and c is speed of light.

The connection between space and time can be used to characterise velocity:

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

Due to relativistic effects, time relative to an object moves more quickly when it is moving very quickly. The time-dilation equation helps explain this:

$∆t=\frac{∆t_0}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}}$

The following equation may be used to show how a relativistic effect causes an object's length to contract:

$t=t_0\sqrt{1-\frac{v^2}{c^2}}$

Using Eq. (1), we can calculate the particle's distance travelled given the particle's duration and speed:

Substitute 0.94c for v and 0.032 for t into equation (1).

$$\begin{gathered} 0.94\mathrm{c}=\frac{\mathrm{d}}{\left(0.032.{10}^{-6}\right)} \\ \mathrm{d}=\left(0.94\mathrm{c}\right).\left(0.032.{10}^{-6}\right) \\ \mathrm{d}=\left(0.94.3.0.{10}^{-6}\right).\left(0.032.{10}^{-6}\right) \\ \mathrm{d}=9.02\mathrm{m} \end{gathered}$$

Therefore, the distance travelled by the particle is 9.02 m.

Next, we calculate the particle's survival period in relation to its own frame of reference. We ascertain the expression for $∆t_0$by employing Eq (2):

$$\begin{gathered} ∆t=\frac{∆t_0}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}} \\ ∆t_0=∆t\sqrt{1-\frac{v^2}{c^2}} \end{gathered}$$

Set role="math" localid="1660040336736" $∆t=t$, determine the time the particle survives using the derived expression for $∆t_0$:

Let $∆t=t$. The resulting equation for $∆t_0$is used to calculate the duration of the particle's existence:

Determine theparticle’s time frame of reference, it survived.

Substitute $\left(0.032.{10}^{-6}\right)$for $∆t$and $\left(0.94c\right)$ for vinto equation (2).

$$\begin{gathered} ∆t_0=\left(0.032.{10}^{-6}\right)\sqrt{1-\frac{{\left(0.94c\right)}^2}{c^2}} \\ =\left(0.032.{10}^{-6}\right)\sqrt{1-0.{94}^2} \\ =0.011\mu s \end{gathered}$$

Frome the particle’s own time frame of reference. it survived for $0.011\mu s$.

Take the particle's computed distance travelled as the reference length, $t_0$. Using (3), one can determine how far the particle has travelled.

Substitute $\left(9.02\right)$for $t_0$and $\left(0.94c\right)$for v into equation (3).

$$\begin{gathered} t=\left(9.02\right)\sqrt{1-\frac{{\left(0.94c\right)}^2}{c^2}} \\ =3.08m \end{gathered}$$

According to the particle, the laboratory equipment has a length of 3.08.m