91Ó°ÊÓ

The time difference between the two events,

$$\begin{gathered} \mathrm{Δ}t=t_B-t_A \\ \mathrm{Δ}t=1\mathrm{μ}s \end{gathered}$$

The distance,

$$\begin{gathered} \mathrm{Δ}x=x_B-x_A \\ \mathrm{Δ}x=240\text{m} \end{gathered}$$

The expression to calculate the Lorentz factor is given as follows.

$\mathit{γ}\mathbf{=}\sqrt{\frac{\mathbf{1}}{\mathbf{1}\mathbf{-}{\mathbf{β}}^{\mathbf{2}}}}$ …(¾±)

The equation to calculate the expression for theis given as follows.

$\mathit{Δ}\mathit{t}\mathbf{\text{'}}\mathbf{=}\mathit{γ}\left(\mathrm{Δ}t-\frac{\mathrm{β}\mathrm{Δ}x}{c}\right)$ …(¾±¾±)

Here,$\mathrm{β}$is the speed parameter, $\mathrm{γ}$is the Lorentz factor and c is the speed of light.

(a)

Calculate the expression for $\mathrm{Δ}t\text{'}$.

Substitute$\sqrt{\frac{1}{1-{\mathrm{β}}^2}}$for$\mathrm{γ}$into equation (ii).

$\mathrm{Δ}t\text{'}=\sqrt{\frac{1}{1-{\mathrm{β}}^2}}\left(\mathrm{Δ}t-\frac{\mathrm{β}\mathrm{Δ}x}{c}\right)$

Substitute $240mfor∆x,1\mathrm{μ}sfor\mathrm{Δ}tand3×{10}^{8}m/sforc$r into above equation.

$$\begin{gathered} \mathrm{Δ}t\text{'}=\sqrt{\frac{1}{1-{\mathrm{β}}^2}}\left(1×{10}^{-6}\text{s}-\frac{\mathrm{β}×240\text{m}}{3×{10}^8}\right) \\ \mathrm{Δ}t\text{'}=\frac{1}{\sqrt{1-{\mathrm{β}}^2}}\left(1×{10}^{-6}\text{s}-80×{10}^{-8}\text{s}\mathrm{β}\right) \\ \mathrm{Δ}t\text{'}=\frac{1}{\sqrt{1-{\mathrm{β}}^2}}\left(1×{10}^{-6}\text{s}-0.80×{10}^{-6}\text{s}\mathrm{β}\right) \\ \mathrm{Δ}t\text{'}=\frac{1-0.80\mathrm{β}}{\sqrt{1-{\mathrm{β}}^2}} \end{gathered}$$ …(¾±¾±¾±)

Hence the expression for$\mathrm{Δ}t\text{'}$ is,$\frac{1-0.80\mathrm{β}}{\sqrt{1-{\mathrm{β}}^2}}\text{s}$ .

(b)

The graph between $\mathrm{Δ}t\text{'}and\mathrm{β}$is shown below.

(c)

The graph between$\mathrm{Δ}t\text{'}$ and $\mathrm{β}$is shown below

(d)

To calculate the value of the take the derivative of the equation (iii),

$$\begin{gathered} \frac{d\left(\mathrm{Δ}t\text{'}\right)}{d\mathrm{β}}=\frac{-0.8\sqrt{1-{\mathrm{β}}^2}-\left(1-0.8\mathrm{β}\right)\left(-2\mathrm{β}\right)\left(\frac{1}{2\sqrt{1-{\mathrm{β}}^2}}\right)}{\left(1-{\mathrm{β}}^2\right)} \\ \frac{d\left(\mathrm{Δ}t\text{'}\right)}{d\mathrm{β}}=\frac{\mathrm{β}\left(1-0.8\mathrm{β}\right)-0.8\left(1-{\mathrm{β}}^2\right)}{{\left(1-{\mathrm{β}}^2\right)}^{3/2}} \\ \frac{d\left(\mathrm{Δ}t\text{'}\right)}{d\mathrm{β}}=\frac{\mathrm{β}-0.8}{{\left(1-{\mathrm{β}}^2\right)}^{3/2}} \end{gathered}$$

Substitute 0 for $\frac{d\left(\mathrm{Δ}t\text{'}\right)}{d\mathrm{β}}$into above equation.

$$\begin{gathered} 0=\frac{\mathrm{β}-0.8}{{\left(1-{\mathrm{β}}^2\right)}^{3/2}} \\ \mathrm{β}-0.8=0 \\ \mathrm{β}=0.8 \end{gathered}$$

Hence the value of the$\mathrm{β}$ is 0.8 .

(e)

Calculate the minimum value of $\mathrm{Δ}t\text{'}$.

Substitute 0.8 for $\mathrm{β}$into equation (iii).

$$\begin{gathered} \mathrm{Δ}t\text{'}=\frac{1-0.80×0.8}{\sqrt{1-{0.8}^2}} \\ \mathrm{Δ}t\text{'}=\frac{1-0.64}{0.6} \\ \mathrm{Δ}t\text{'}=\frac{0.36}{0.6} \\ \mathrm{Δ}t\text{'}=0.6\mathrm{μ}s \end{gathered}$$

Hence the minimum value of$\mathrm{Δ}t\text{'}$ isrole="math" localid="1663062673499" $0.6\mathrm{μ}\text{s}$ .

Calculate the velocity,

$v=\frac{\mathrm{Δ}x}{\mathrm{Δ}t}$

Substitute $240mfor\mathrm{Δ}xand1\mathrm{μ}sfor\mathrm{Δ}t$into above equation.

$$\begin{gathered} v=\frac{240\text{m}}{1×{10}^{-6}\text{s}} \\ v=240×{10}^{6}m/s \\ v=2.4×{10}^{8}m/s \end{gathered}$$

Since the velocity $2.4×{10}^{8}m/s$is less than the speed of the light therefore event A can cause the event B.