Question

Bullwinkle in reference frame S^' passes you in reference frame S along the common direction of the x^' and x axes, as in Fig. 37-9. He carries three meter sticks: meter stick 1 is parallel to the x^' axis, meter stick 2 is parallel to the y^' axis, and meter stick 3 is parallel to the z^' axis. On his wristwatch he counts off 15.0 s, which takes 30.0 s according to you. Two events occur during his passage. According to you, event 1 occurs at ${\mathbf{x}}_{\mathbf{1}}\mathbf{=}\mathbf{33}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$and ${\mathbf{t}}_{\mathbf{1}}\mathbf{=}\mathbf{22}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{ns}$, and event 2 occurs at ${\mathbf{x}}_{\mathbf{2}}\mathbf{=}\mathbf{53}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$and ${\mathbf{t}}_{\mathbf{2}}\mathbf{=}\mathbf{62}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{ns}$. According to your measurements, what is the length of (a) meter stick 1, (b) meter stick 2, and (c) meter stick 3? According to Bullwinkle, what are (d) the spatial separation and (e) the temporal separation between events 1 and 2, and (f) which event occurs first?

Short answer

1. The length of meter stick 1 is 0.5 m.
2. The length of meter stick 2 is 1 m.
3. The length of meter stick 3 is 1 m.
4. The spatial separation is 19.216 m.
5. The temporal separation is -35.46 ns.
6. The event 2 is occurred first.

Step-by-step solution

Describe the expression for dilation time

The expression of time dilation is given by,

$\mathbf{∆}\mathbf{t}\mathbf{=}\mathbf{\gamma }\mathbf{∆}{\mathbf{t}}_{\mathbf{o}}$ …… (1)

Find the length of meter stick 1 (a)

Here, the time according to watch is, $∆\mathrm{t}=30\mathrm{s}$, and the time according to Bull winkle is, $∆{\mathrm{t}}_{\mathrm{o}}=15\mathrm{s}$.

From equation (1),

$\begin{array}{rcl}\mathrm{\gamma }& =& \frac{30\mathrm{s}}{15\mathrm{s}}\\ & =& 2\end{array}$

It is known that $\mathrm{\gamma }=\frac{1}{\sqrt{1-{\displaystyle \frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}}}}$then,

$\begin{array}{rcl}\frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}& =& 1-\frac{1}{{\mathrm{\gamma }}^2}\\ \frac{\mathrm{v}}{\mathrm{c}}& =& \sqrt{1-\frac{1}{{\mathrm{\gamma }}^2}}\\ & =& \sqrt{1-\frac{1}{2^2}}\\ \mathrm{v}& =& 0.866\mathrm{c}\end{array}$

The expression to calculate the length is given by,

$\mathrm{L}=\frac{{\mathrm{L}}_{\mathrm{o}}}{\mathrm{\gamma }}$ ……. (2)

Substitute 1 m for L_o, and 2 for $\mathrm{\gamma }$in equation (2).

$\begin{array}{rcl}\mathrm{L}& =& \frac{1\mathrm{m}}{2}\\ & =& 0.5\mathrm{m}\end{array}$

Therefore, the length of meter stick 1 is 0.5 m.

Find the length of meter stick 2 (b)

The meter stick 2 is along y-axis, and the frame S^' is moving along positive x-axis. Therefore, there is no length contraction along y^' direction. So, the length of the meter stick parallel to y^' axis will remain same.

Hence, according to the measurement of length of the meter stick parallel to y^' axis is 1 m.

Find the length of meter stick 3 (c)

There is no length contraction for the meter stick placed parallel to z axis. Thus, according to the measurement of length of the meter stick parallel to z^' axis is 1 m.

Find the spatial separation (d)

The expression to calculate the spatial separation is given by,

$∆{\mathrm{x}}^{\text{'}}=\mathrm{\gamma }(∆\mathrm{x}-\mathrm{v}∆\mathrm{t})$ ……. (3)

Find $∆\mathrm{x}$as follows.

$\begin{array}{rcl}∆\mathrm{x}& =& {\mathrm{x}}_2-{\mathrm{x}}_1\\ & =& 53\mathrm{m}-33\mathrm{m}\\ & =& 20\mathrm{m}\end{array}$

Find $∆\mathrm{t}$as follows.

$\begin{array}{rcl}∆\mathrm{t}& =& {\mathrm{t}}_2-{\mathrm{t}}_1\\ & =& 62\mathrm{ns}-22\mathrm{ns}\\ & =& \left(40\mathrm{ns}\right)\left(\frac{{10}^{-9}\mathrm{s}}{1\mathrm{ns}}\right)\\ & =& 40\times {10}^{-9}\mathrm{s}\end{array}$

Find v as follows.

$\begin{array}{rcl}\mathrm{v}& =& \left(0.866\right)\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)\\ & =& 2.598\times {10}^8\mathrm{m}/\mathrm{s}\end{array}$

Substitute 2 for $\mathrm{\gamma }$, 20 m for $∆\mathrm{x}$, $40\times {10}^{-9}\mathrm{s}$ for $∆\mathrm{t}$, and $2.598\times {10}^8\mathrm{m}/\mathrm{s}$ for v in equation (3).

$\begin{array}{rcl}∆{\mathrm{x}}^{\text{'}}& =& 2\left[20\mathrm{m}-\left(2.598\times {10}^8\mathrm{m}/\mathrm{s}\right)\left(40\times {10}^{-9}\mathrm{s}\right)\right]\\ & =& 2\left(20\mathrm{m}-10.392\mathrm{m}\right)\\ & =& 19.216\mathrm{m}\end{array}$

Therefore, the spatial separation is 19.216 m.

Find the temporal separation between events 1 and 2 (e)

The expression to calculate temporal separation is given by,

$∆{\mathrm{t}}^{\text{'}}=\mathrm{\gamma }\left(∆\mathrm{t}-\mathrm{v}\frac{∆\mathrm{x}}{{\mathrm{c}}^2}\right)$ ……. (4)

Substitute 2 for $\mathrm{\gamma }$, 20 m for $∆\mathrm{x}$, $40\times {10}^{-9}\mathrm{s}$for $∆\mathrm{t}$, $3\times {10}^8\mathrm{m}/\mathrm{s}$for c, and $2.598\times {10}^8\mathrm{m}/\mathrm{s}$for v in equation (4).

$\begin{array}{rcl}∆{\mathrm{t}}^{\text{'}}& =& 2\left[40\times {10}^{-9}\mathrm{s}-\frac{\left(0.866\right)\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)\left(20\mathrm{m}\right)}{{\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)}^2}\right]\\ & =& 2\left[40\times {10}^{-9}\mathrm{s}-5.77\times {10}^{-8}\mathrm{s}\right]\\ & =& -35.46\times {10}^{-9}\mathrm{s}\\ & =& -35.46\mathrm{ns}\end{array}$

Therefore, the temporal separation is -35.46 ns.

Find the event that occur first (f)

The value of temporal separation is negative. Therefore, the event 2 is occurred before event 1 in S^'.