Question

In Fig. 37-9, observer S detects two flashes of light. A big flash occurs at ${\mathbf{x}}_{\mathbf{1}}\mathbf{=}\mathbf{1200}\mathbf{}\mathbf{m}$and, $\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\mu s}$later, a small flash occurs at ${\mathbf{x}}_{\mathbf{2}}\mathbf{=}\mathbf{480}\mathbf{}\mathbf{m}$. As detected by observer S^', the two flashes occur at a single coordinate x^'. (a) What is the speed parameter of S^', and (b) is S^' moving in the positive or negative direction of the x axis? To S^', (c) which flash occurs first and (d) what is the time interval between the flashes?

Short answer

(a) The speed parameter is 0.48.

(b) The S^' frame is moving in the negative x-direction.

(c) Thebigger flash occurs first.

(d) The time interval is $4.39\mathrm{\mu s}$.

Step-by-step solution

Describe the expression for length of contraction

The expression of time dilation is given by,

$\mathbf{∆}{\mathbf{x}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{\gamma }\left(∆x-v∆t\right)$ ……. (1)

Find the speed parameter of

(a)

In the frame S, the bigger flash occurs at x₁ position and the smaller flash occurs at x₂ position .

$∆\mathrm{x}={\mathrm{x}}_1-{\mathrm{x}}_2$

In the frame S^' both the flashes occur at the same time.

$∆{\mathrm{x}}^{\text{'}}=0$

Substitute$∆{\mathrm{x}}^{\text{'}}=0$ and$∆\mathrm{x}={\mathrm{x}}_1-{\mathrm{x}}_2$ in equation (1).

$$\begin{gathered} 0=\mathrm{\gamma }\left({\mathrm{x}}_1-{\mathrm{x}}_2-\mathrm{v}∆\mathrm{t}\right) \\ \mathrm{v}=\frac{{\mathrm{x}}_1-{\mathrm{x}}_2}{∆\mathrm{t}} \end{gathered}$$ …… (2)

Substitute 1200 m for x₁, 480 m for x₂, and$5.00\mathrm{\mu s}$ for$∆\mathrm{t}$ in equation (2).

$\begin{array}{rcl}\mathrm{v}& =& \frac{1200\mathrm{m}-480\mathrm{m}}{\left(500\mathrm{\mu s}\right)\left({\displaystyle \frac{{10}^{-6}\mathrm{s}}{1\mathrm{\mu s}}}\right)}\\ & =& 1.44\times {10}^8\mathrm{m}/\mathrm{s}\end{array}$

The expression to calculate speed parameter is given by,

$\mathrm{\beta }=\frac{\mathrm{v}}{\mathrm{c}}$ ……. (3)

Substitute $1.44\times {10}^8\mathrm{m}/\mathrm{s}$for v, and $3\times {10}^8\mathrm{m}/\mathrm{s}$for c in equation (3).

$\begin{array}{rcl}\mathrm{\beta }& =& \frac{1.44\times {10}^8\mathrm{m}/\mathrm{s}}{3\times {10}^8\mathrm{m}/\mathrm{s}}\\ & =& 0.48\end{array}$

Therefore, the speed parameter is 0.48.

Determine whether S' moving in the positive or negative direction of the x axis

(b)

The speed is positive, so, the direction of S^' frame must be along the negative x-axis. Therefore, the S^' frame is moving in the negative x-direction.

Find the flash that occur first

(c)

The expression to calculate the time dilation is given by,

$\begin{array}{rcl}∆{\mathrm{t}}^{\text{'}}& =& \frac{\left(∆\mathrm{t}-{\displaystyle \frac{\mathrm{v}∆\mathrm{x}}{{\mathrm{c}}^2}}\right)}{\sqrt{1-{\mathrm{\beta }}^2}}\\ & =& \frac{\left(∆\mathrm{t}-{\displaystyle \frac{\mathrm{v}\left({\mathrm{x}}_1-{\mathrm{x}}_2\right)}{{\mathrm{c}}^2}}\right)}{\sqrt{1-{\mathrm{\beta }}^2}}\end{array}$ ……. (4)

Substitute 1200 m for x₁, 480 m for x₂,$5.00\mathrm{\mu s}$for $∆\mathrm{t}$,$1.44\times {10}^8\mathrm{m}/\mathrm{s}$for v,$3\times {10}^8\mathrm{m}/\mathrm{s}$for c, and 0.48 for in equation (4).

$\begin{array}{rcl}∆{\mathrm{t}}^{\text{'}}& =& \frac{\left(5.00\mathrm{\mu s}\left({\displaystyle \frac{{10}^{-6}\mathrm{s}}{1\mathrm{\mu s}}}\right)-{\displaystyle \frac{\left(1.44\times {10}^8\mathrm{m}/\mathrm{s}\right)\left(1200\mathrm{m}-480\mathrm{m}\right)}{{\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)}^2}}\right)}{\sqrt{1-{\left(0.48\right)}^2}}\\ & =& 4.386\times {10}^{-6}\mathrm{s}\left(\frac{1\mathrm{\mu s}}{{10}^{-6}\mathrm{s}}\right)\\ & =& 4.39\mathrm{\mu s}\end{array}$

The time is positive and hence, the bigger flash occurs first.

Therefore, the bigger flash occurs first.

Find the time interval between the flashes

(d)

Substitute 1200 m for x₁, 480 m for x₂,$5.00\mathrm{\mu s}$ for $∆\mathrm{t}$,$1.44\times {10}^8\mathrm{m}/\mathrm{s}$ for v,$3\times {10}^8\mathrm{m}/\mathrm{s}$ for c, and 0.48 for$\mathrm{\beta }$ in equation (4).

$\begin{array}{rcl}∆{\mathrm{t}}^{\text{'}}& =& \frac{\left(5.00\mathrm{\mu s}\left({\displaystyle \frac{{10}^{-6}\mathrm{s}}{1\mathrm{\mu s}}}\right)-{\displaystyle \frac{\left(1.44\times {10}^8\mathrm{m}/\mathrm{s}\right)\left(1200\mathrm{m}-480\mathrm{m}\right)}{{\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)}^2}}\right)}{\sqrt{1-{\left(0.48\right)}^2}}\\ & =& 4.386\times {10}^{-6}\mathrm{s}\left(\frac{1\mathrm{\mu s}}{{10}^{-6}\mathrm{s}}\right)\\ & =& 4.39\mathrm{\mu s}\end{array}$

Therefore, the time interval is $4.39\mathrm{\mu s}$.