Chapter 9, Problem 17P

(a)

To determine

The relative speed between $\text{S}$ and ${\text{S}}^{\prime }$.

Answer to Problem 17P

The relative speed between $\text{S}$ and ${\text{S}}^{\prime }$ is $\underset{\_}{2.50\times {10}^8\text{m}/\text{s}}$.

Explanation of Solution

Write the Lorenz transformation equation for position.

  $\Delta x^{\prime }=\gamma \left(\Delta x-\upsilon \Delta t\right)$        (I)

Here, $\Delta x^{\prime }$ is the distance of separation of two events in frame ${\text{S}}^{\prime }$, $\Delta x$ is the distance of separation in frame $\text{S}$, $\upsilon$ is the speed with which ${\text{S}}^{\prime }$ moving, $\Delta t$ is the change in time, and $\gamma$ is the Lorentz factor.

Conclusion:

Substitute, $0$ for $\Delta x^{\prime }$, $\left(5.00-3.00\right)\text{m}$ for $\Delta x$, $\left(9.00-1.00\right)\times {10}^{-9}\text{s}$ for $\Delta t$ in equation (I) to obtain the value of speed.

  $0=\gamma \left[\left(5.00-3.00\right)\text{m}-\upsilon \left(\left(9.00-1.00\right)\times {10}^{-9}\text{s}\right)\right]$        (II)

$\gamma$ is a constant and not equal to zero. Equate the term in bracket to zero to find the value of $\upsilon$.

  $\begin{array}{c}\left[\left(5.00-3.00\right)\text{m}-\upsilon \left(8.00\times {10}^{-9}\text{s}\right)\right]=0\\ \upsilon =\frac{\left(5.00-3.00\right)\text{m}}{8.00\times {10}^{-9}\text{s}}\\ =2.50\times {10}^8\text{m}/\text{s}\end{array}$

Therefore, the relative speed between $\text{S}$ and ${\text{S}}^{\prime }$ is $\underset{\_}{2.50\times {10}^8\text{m}/\text{s}}$.

(b)

To determine

The location of two flashes with respect to frame ${\text{S}}^{\prime }$.

Answer to Problem 17P

The location of two flashes with respect to frame ${\text{S}}^{\prime }$ is $\underset{\_}{4.98\text{m}}$.

Explanation of Solution

Write the Lorenz transformation equation for position.

  $x^{\prime }=\gamma \left(x-\upsilon t\right)$        (III)

Here, $x^{\prime }$ is the position of flash in ${\text{S}}^{\prime }$, $x$ is the position of flash frame $\text{S}$, $\upsilon$ is the speed with which ${\text{S}}^{\prime }$ moving, $t$ is the time, and $\gamma$ is the Lorentz factor.

Substitute, $\frac{1}{\sqrt{1-\frac{{\upsilon }^2}{c^2}}}$ for $\gamma$ in equation (I).

  $x^{\prime }=\frac{\left(x-\upsilon t\right)}{\sqrt{1-\frac{{\upsilon }^2}{c^2}}}$        (IV)

Conclusion:

Substitute, $2.50\times {10}^8\text{m}/\text{s}$ for $\upsilon$, $3.00\times {10}^8\text{m}/\text{s}$ for $c$, $3.00\text{m}$ for $x$, and $1.00\times {10}^{-9}\text{s}$ for $t$ in equation (I) to obtain the position of red flash in frame ${\text{S}}^{\prime }$.

  $\begin{array}{c}{x^{\prime }}_{\text{R}}=\frac{\left(3.00\text{m}-\left(2.50\times {10}^8\text{m}/\text{s}\right)\left(1.00\times {10}^{-9}\text{s}\right)\right)}{\sqrt{1-\frac{{\left(2.50\times {10}^8\text{m}/\text{s}\right)}^2}{{\left(3.00\times {10}^8\text{m}/\text{s}\right)}^2}}}\\ =4.98\text{m}\end{array}$

Substitute, $2.50\times {10}^8\text{m}/\text{s}$ for $\upsilon$, $3.00\times {10}^8\text{m}/\text{s}$ for $c$, $5.00\text{m}$ for $x$, and $9.00\times {10}^{-9}\text{s}$ for $t$ in equation (I) to obtain the position of blue flash in frame ${\text{S}}^{\prime }$.

  $\begin{array}{c}{x^{\prime }}_{\text{B}}=\frac{\left(5.00\text{m}-\left(2.50\times {10}^8\text{m}/\text{s}\right)\left(9.00\times {10}^{-9}\text{s}\right)\right)}{\sqrt{1-\frac{{\left(2.50\times {10}^8\text{m}/\text{s}\right)}^2}{{\left(3.00\times {10}^8\text{m}/\text{s}\right)}^2}}}\\ =4.977\text{m}\\ \approx 4.98\text{m}\end{array}$

Therefore, the. Location of two flashes with respect to frame ${\text{S}}^{\prime }$ is $\underset{\_}{4.98\text{m}}$.

(c)

To determine

The time at which red flash occur in ${\text{S}}^{\prime }$ frame.

Answer to Problem 17P

The time at which red flash occur in ${\text{S}}^{\prime }$ frame is $\underset{\_}{-1.33\times {10}^8\text{s}}$.

Explanation of Solution

Write the Lorenz transformation equation for time.

  $t=\frac{1}{\sqrt{1-\frac{{\upsilon }^2}{c^2}}}\left(t-\frac{\upsilon }{c^2}x\right)$        (V)

Conclusion:

Substitute, $2.50\times {10}^8\text{m}/\text{s}$ for $\upsilon$, $3.00\times {10}^8\text{m}/\text{s}$ for $c$, $3.00\text{m}$ for $x$, and $1.00\times {10}^{-9}\text{s}$ for $t$ in the equation (V).

  $\begin{array}{c}{t}^{\prime }=\frac{1}{\sqrt{1-\frac{{\left(2.50\times {10}^8\text{m}/\text{s}\right)}^2}{{\left(3.00\times {10}^8\text{m}/\text{s}\right)}^2}}}\left(1.00\times {10}^{-9}\text{s}-\frac{2.50\times {10}^8\text{m}/\text{s}}{{\left(3.00\times {10}^8\text{m}/\text{s}\right)}^2}\left(3.00\text{m}\right)\right)\\ =-1.33\times {10}^{-8}\text{s}\end{array}$

Therefore, the time at which red flash occur in ${\text{S}}^{\prime }$ frame is $\underset{\_}{-1.33\times {10}^8\text{s}}$.