Question

Question: Temporal separation between two events. Events and occur with the following spacetime coordinates in the reference frames of Fig. 37-25: according to the unprimed frame,andaccording to the primed frame, $\left(x_A,t_A\right)\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\left(x_B,t_B\right)$. In the unprimed frame $\mathbf{∆}\mathit{t}\mathbf{=}{\mathit{t}}_{\mathbf{B}}\mathbf{-}{\mathit{t}}_{\mathbf{A}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathit{\mu }\mathit{s}$and $\mathit{\Delta }\mathit{x}\mathbf{=}{\mathit{x}}_{\mathbf{B}}\mathbf{-}{\mathit{x}}_{\mathbf{A}}\mathbf{=}\mathbf{240}\mathbf{}\mathbf{\text{m}}$. (a) Find an expression for $\mathit{\Delta }\mathit{t}\mathbf{\text{'}}$in terms of the speed parameter$\mathit{\beta }$and the given data. Graph $\mathit{\Delta }\mathit{t}\mathbf{\text{'}}$ versus $\mathit{\beta }$for the following two ranges of $\mathit{\beta }$: (b) 0 to 0.01and 0.1 (c) 0.1 to 1. (d) At what value of $\mathit{\beta }$is $\mathit{\Delta }\mathit{t}\mathbf{\text{'}}$minimum and (e) what is that minimum? (f) Can one of these events cause the other? Explain.

Short answer

Answer

(a) The expression for$\Delta t\text{'}$is$\frac{1-0.80\beta }{\sqrt{1-{\beta }^2}}\mu \text{s}$.

(b) The graph between$\Delta t\text{'}$and$\beta$for the value 0 and 0.01

.

(c)The graph between$\Delta t\text{'}$ and$\beta$ for the value 0.01 to 1.

(d) The value of the $\beta$is 0.8.

(e) The minimum value of$\Delta t\text{'}$ is$0.6\mu s$ .

(f) The event Acan cause event B .

Step-by-step solution

Write the given data from the question.

The time difference between the two events,

$$\begin{gathered} \Delta t=t_B-t_A \\ \Delta t=1\mu s \end{gathered}$$

The distance,

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

The formula to calculate the expression for and graphs between and speed parameter.

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

$\mathit{\gamma }\mathbf{=}\sqrt{\frac{\mathbf{1}}{\mathbf{1}\mathbf{-}{\mathbf{\beta }}^{\mathbf{2}}}}$ …(i)

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

$\mathit{\Delta }\mathit{t}\mathbf{\text{'}}\mathbf{=}\mathit{\gamma }\left(\Delta t-\frac{\beta \Delta x}{c}\right)$ …(ii)

Here,$\beta$is the speed parameter, $\gamma$is the Lorentz factor and c is the speed of light.

Calculate the expression forΔt'  .

(a)

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

Substitute$\sqrt{\frac{1}{1-{\beta }^2}}$for$\gamma$into equation (ii).

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

Substitute $240mfor∆x,1\mu sfor\Delta tand3\times {10}^{8}m/sforc$r into above equation.

$$\begin{gathered} \Delta t\text{'}=\sqrt{\frac{1}{1-{\beta }^2}}\left(1\times {10}^{-6}\text{s}-\frac{\beta \times 240\text{m}}{3\times {10}^8}\right) \\ \Delta t\text{'}=\frac{1}{\sqrt{1-{\beta }^2}}\left(1\times {10}^{-6}\text{s}-80\times {10}^{-8}\text{s}\beta \right) \\ \Delta t\text{'}=\frac{1}{\sqrt{1-{\beta }^2}}\left(1\times {10}^{-6}\text{s}-0.80\times {10}^{-6}\text{s}\beta \right) \\ \Delta t\text{'}=\frac{1-0.80\beta }{\sqrt{1-{\beta }^2}} \end{gathered}$$ …(iii)

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

Draw the graph between  Δt' and β  for the value 0 to 0.01.

(b)

The graph between $\Delta t\text{'}and\beta$is shown below.

Draw the graph between  Δt' and β  for the value 0.1  to 1.

(c)

The graph between$\Delta t\text{'}$ and $\beta$is shown below

Calculate the value of β at which Δt' is minimum.

(d)

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

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

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

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

Hence the value of the$\beta$ is 0.8 .

Calculate the minimum value of Δt' at β=0.8

(e)

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

Substitute 0.8 for $\beta$into equation (iii).

$$\begin{gathered} \Delta t\text{'}=\frac{1-0.80\times 0.8}{\sqrt{1-{0.8}^2}} \\ \Delta t\text{'}=\frac{1-0.64}{0.6} \\ \Delta t\text{'}=\frac{0.36}{0.6} \\ \Delta t\text{'}=0.6\mu s \end{gathered}$$

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

8: Determine that can one of these events cause another event

Calculate the velocity,

$v=\frac{\Delta x}{\Delta t}$

Substitute $240mfor\Delta xand1\mu sfor\Delta t$into above equation.

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

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