Question

Spatial separation between two events. For the passing reference frames of Fig. 37-25, events A and B occur with the following spacetime coordinates: according to the unprimed frame,$\mathbf{(}{\mathbf{x}}_{\mathbf{A}}\mathbf{,}{\mathbf{t}}_{\mathbf{A}}\mathbf{)}$and role="math" localid="1663045013644" $\mathbf{(}{\mathbf{x}}_{\mathbf{B}}\mathbf{,}{\mathbf{t}}_{\mathbf{B}}\mathbf{)}$according to the primed frame,$\mathbf{(}{{\mathbf{x}}^{\mathbf{\text{'}}}}_{\mathbf{A}}\mathbf{,}{{\mathbf{t}}^{\mathbf{\text{'}}}}_{\mathbf{A}}\mathbf{)}$ androle="math" localid="1663045027721" $\mathbf{(}{{\mathbf{x}}^{\mathbf{\text{'}}}}_{\mathbf{B}}\mathbf{,}{{\mathbf{t}}^{\mathbf{\text{'}}}}_{\mathbf{B}}\mathbf{)}$. In the umprimed frame$\mathbf{\Delta t}\mathbf{=}{\mathbf{t}}_{\mathbf{B}}\mathbf{-}{\mathbf{t}}_{\mathbf{A}}\mathbf{=}\mathbf{1}\mathbf{.00}\mathbf{\text{\hspace{0.17em}μs}}$androle="math" localid="1663045143133" $\mathbf{\Delta x}\mathbf{=}{\mathbf{x}}_{\mathbf{B}}\mathbf{-}{\mathbf{x}}_{\mathbf{A}}\mathbf{=}\mathbf{240}\mathbf{\text{\hspace{0.17em}m}}$.(a) Find an expression forin ${\mathbf{\Delta x}}^{\mathbf{\text{'}}}$terms of the speed parameter$\mathbf{\beta }$and the given data. Graph${\mathbf{\Delta x}}^{\mathbf{\text{'}}}$versus$\mathbf{\beta }$for two ranges of$\mathbf{\beta }$: (b)$\mathbf{0}$ to$\mathbf{0.01}$and (c)$\mathbf{0.1}$to $\mathbf{1}$. (d) At what value of$\mathbf{\beta }$is${\mathbf{\Delta x}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{0}$?

Short answer

a) The expression for the${\mathrm{\Delta x}}^{\text{'}}$ is$\frac{\left(240\text{\hspace{0.17em}m}\right)-\left(300\text{\hspace{0.17em}m}\right)\mathrm{\beta }}{\sqrt{1-{\mathrm{\beta }}^2}}$ .

(b) The graph between${\mathrm{\Delta x}}^{\text{'}}$ and$\mathrm{\beta }$ for the value$0$ and $0.01$.

(c)The graph between ${\mathrm{\Delta x}}^{\text{'}}$and$\mathrm{\beta }$ for the value$0.01$ to $1$.

(d) The value of $\mathrm{\beta }$is $0.8$.

Step-by-step solution

Write the given data from the question.

The time difference between the two events,

$\begin{array}{l}\mathrm{\Delta t}={\mathrm{t}}_{\mathrm{B}}-{\mathrm{t}}_{\mathrm{A}}\\ \mathrm{\Delta t}=1\text{\hspace{0.17em}μs}\end{array}$

The distance,

$\begin{array}{l}\mathrm{\Delta x}={\mathrm{x}}_{\mathrm{B}}-{\mathrm{x}}_{\mathrm{A}}\\ \mathrm{\Delta x}=240\text{\hspace{0.17em}m}\end{array}$

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

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

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

The equation to calculate the expression for the${\mathbf{\Delta x}}^{\mathbf{\text{'}}}$ is given as follows.

${\mathbf{\Delta x}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{\gamma }\mathbf{(}\mathbf{\Delta x}\mathbf{-}\mathbf{\beta c\Delta t}\mathbf{)}$ …(ii)

Here, $\mathbf{\gamma }$is the Lorentz factor and$\mathbf{c}$ is the speed of light.

Calculate the expression for Δx'.

(a)

Calculate the expression for${\mathrm{\Delta x}}^{\text{'}}$.

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

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

$\begin{array}{l}{\mathrm{\Delta x}}^{\text{'}}=\frac{1}{\sqrt{1-{\mathrm{\beta }}^2}}(\mathrm{\Delta x}-\mathrm{\beta c\Delta t})\\ {\mathrm{\Delta x}}^{\text{'}}=\frac{\mathrm{\Delta x}-\mathrm{\beta c\Delta t}}{\sqrt{1-{\mathrm{\beta }}^2}}\end{array}$

Substitute$240\text{\hspace{0.17em}m}$ for$\mathrm{\Delta x}$ ,$3\times {10}^8\text{\hspace{0.17em}m}/\text{s}$ for$\mathrm{c}$ and$1\text{\hspace{0.17em}μs}$ for$\mathrm{\Delta t}$ into equation (i).

$\begin{array}{l}{\mathrm{\Delta x}}^{\text{'}}=\frac{240\text{\hspace{0.17em}m}-\mathrm{\beta }\times 3\times {10}^8\text{\hspace{0.17em}m}/\text{s}\times 1\times {10}^{-6}\text{\hspace{0.17em}s}}{\sqrt{1-{\mathrm{\beta }}^2}}\\ {\mathrm{\Delta x}}^{\text{'}}=\frac{\left(240\text{\hspace{0.17em}m}\right)-\left(300\text{\hspace{0.17em}m}\right)\mathrm{\beta }}{\sqrt{1-{\mathrm{\beta }}^2}}\end{array}$ …(iii)

Hence the expression for the${\mathrm{\Delta x}}^{\text{'}}$ is $\frac{\left(240\text{\hspace{0.17em}m}\right)-\left(300\text{\hspace{0.17em}m}\right)\mathrm{\beta }}{\sqrt{1-{\mathrm{\beta }}^2}}$.

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

(b)

The graph between${\mathrm{\Delta x}}^{\text{'}}$ and$\mathrm{\beta }$ is shown below.

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

(c)

The graph between${\mathrm{\Delta x}}^{\text{'}}$ and $\mathrm{\beta }$is shown below.

Calculate the value of the β at which Δx'=0 .

Calculate the value of $\mathrm{\beta }$.

Substitute$0$ for${\mathrm{\Delta x}}^{\text{'}}$ into equation (iii).

$\begin{array}{c}\frac{240-300\mathrm{\beta }}{\sqrt{1-{\mathrm{\beta }}^2}}=0\\ 240-300\mathrm{\beta }=0\\ 300\mathrm{\beta }=240\\ \mathrm{\beta }=0.8\end{array}$

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