91影视

Consider the speed of particle is$蠀=0.94c$.

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

Write the formula of distance travelled by the particle.

$\mathit{v}\mathbf{=}\frac{\mathbf{d}}{\mathbf{t}}$ 鈥︹�� (1)

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

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

$\mathbf{鈭�}{\mathit{t}}_{\mathbf{0}}\mathbf{=}\mathbf{鈭�}\mathit{t}\sqrt{\mathbf{1}\mathbf{-}\frac{{\mathbf{v}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}}$ 鈥︹�� (2)

Here, $\mathbf{鈭�}\mathit{t}$ 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.

$\mathit{t}\mathbf{=}{\mathit{t}}_{\mathbf{0}}\sqrt{\mathbf{1}\mathbf{-}\frac{{\mathbf{v}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}}$ 鈥︹�� (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:

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

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 $鈭�t_0$by employing Eq (2):

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

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

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

Determine theparticle鈥檚 time frame of reference, it survived.

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

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

Frome the particle鈥檚 own time frame of reference. it survived for $0.011渭s$.

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