(a) What’s the percent error introduced when you use Galileo’s rule, instead of Einstein’s, with${\mathbf{v}}_{\mathbf{AB}}\mathbf{=}\mathbf{5}\mathbf{}\mathbf{mi}\mathbf{/}\mathbf{h}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{v}}_{\mathbf{BC}}\mathbf{=}\mathbf{60}\mathbf{}\mathbf{mi}\mathbf{/}\mathbf{h}$and?

(b) Suppose you could run at half the speed of light down the corridor of a train going three-quarters the speed of light. What would your speed be relative to the ground?

(c) Prove, using Eq. 12.3, that if${\mathbf{v}}_{\mathbf{AB}}\mathbf{<}\mathbf{c}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{v}}_{\mathbf{BC}}\mathbf{}\mathbf{<}\mathbf{c}\mathbf{}\mathbf{then}\mathbf{}{\mathbf{v}}_{\mathbf{AC}}\mathbf{<}\mathbf{c}$Interpret this result.

Write the expression for Einstein’s addition rule.

${\mathit{v}}_{\mathbf{A}\mathbf{C}}\mathbf{=}\frac{{\mathbf{v}}_{\mathbf{A}\mathbf{B}}\mathbf{+}{\mathbf{v}}_{\mathbf{B}\mathbf{C}}}{\mathbf{1}\mathbf{+}\left({\displaystyle \frac{v_{AB}{v}_{BC}}{C^2}}\right)}$ …… (1)

Write the expression for Galileo’s addition rule.

${\mathit{v}}_{\mathbf{A}\mathbf{C}}\mathbf{=}{\mathit{v}}_{\mathbf{A}\mathbf{B}}\mathbf{+}{\mathit{v}}_{\mathbf{B}\mathbf{C}}$ …… (2)

(a)

Write the expression for the velocity of A with respect to C (according to Einstein).

$v_{\mathrm{E}}=\frac{{\mathrm{v}}_{\mathrm{AB}}+{\mathrm{v}}_{\mathrm{BC}}}{1+\left({\displaystyle \frac{{\mathrm{v}}_{\mathrm{AB}}{\mathrm{v}}_{\mathrm{BC}}}{{\mathrm{C}}^2}}\right)}$

Substitute the value of equation (2) in the above expression.

$$\begin{gathered} v_{\mathrm{E}}=\frac{{\mathrm{v}}_{\mathrm{G}}}{1+\left({\displaystyle \frac{{\mathrm{v}}_{\mathrm{AB}}{\mathrm{v}}_{\mathrm{BC}}}{{\mathrm{C}}^2}}\right)} \\ {v}_E=v_G{\left[1+\left(\frac{v_{AB}{v}_{BC}}{C^2}\right)\right]}^{-1} \end{gathered}$$

Use binomial expansion and neglect the higher-order terms.

$v_{\mathrm{E}}=v_{G}1-\left(\frac{{\mathrm{v}}_{\mathrm{AB}}{\mathrm{v}}_{\mathrm{BC}}}{{\mathrm{C}}^2}\right)$

Subtract the above equation from.$v_G$

$$\begin{gathered} v_G-v_E=v_G-v_G\left[1-\left(\frac{v_{AB}{v}_{BC}}{C^2}\right)\right] \\ {v}_G-v_E=v_G\left(\frac{v_{AB}{v}_{BC}}{C^2}\right) \\ \frac{v_G-v_E}{v_G}=\frac{v_{AB}{v}_{BC}}{C^2} \end{gathered}$$

For the percent error, the above equation becomes,

$\frac{v_G-v_E}{v_G}=\frac{v_{AB}{v}_{BC}}{C^2}\times 100$

Substitute all the values of $v_{AB}$and $v_{BC}$in the above expression.

$$\begin{gathered} \frac{v_G-v_E}{v_G}=\frac{\left(5\mathrm{mi}/\mathrm{h}\right)\left(60\mathrm{mi}/\mathrm{h}\right)}{{\left(3\times {10}^8\mathrm{m}/\mathrm{s}\times {\displaystyle \frac{2.237\mathrm{mi}/\mathrm{h}}{1\mathrm{m}/\mathrm{s}}}\right)}^2}\times 100 \\ \frac{v_G-v_E}{v_G}=\frac{300}{{\left(6.7\times {10}^8\mathrm{m}/\mathrm{s}\right)}^2}\times 100 \\ \frac{v_G-v_E}{v_G}=6.7\times {10}^{-16}\times 100 \\ \frac{v_G-v_E}{v_G}=6.7\times {10}^{-14} \end{gathered}$$

Therefore, the percent error is$6.7\times {10}^{-14}$.

(b)

Based on the given problem, the value of$v_{AB}$and will be,

and .

$v_{AB}=\frac{c}{2}\mathrm{and}{\mathrm{v}}_{\mathrm{BC}}=\frac{3\mathrm{c}}{4}.$

Substitute the value of$v_{AB}$and$v_{BC}$ in equation (1) to calculate the velocity of a person relative to the ground.

$$\begin{gathered} v_{AC}\frac{\left({\displaystyle \frac{c}{2}}\right)+\left({\displaystyle \frac{3c}{4}}\right)}{1+{\displaystyle \frac{\left({\displaystyle \frac{c}{2}}\right)\left({\displaystyle \frac{3c}{4}}\right)}{c^2}}} \\ {v}_{AC}=\frac{1.25c}{1+{\displaystyle \frac{3}{8}}} \\ {v}_{AC}=\frac{10c}{11} \end{gathered}$$

Therefore, the speed of a person relative to the ground is.

(c)

Let’s assume,

$$\begin{gathered} \beta =\frac{v_{AC}}{c} \\ {\beta }_1=\frac{v_{AB}}{c} \\ {\beta }_2=\frac{v_{BC}}{c} \end{gathered}$$

Substitute the value of$v_{AB}$and$v_{BC}$in equation (1).

$$\begin{gathered} \beta c=\frac{c{\beta }_1+c{\beta }_2}{1+{\displaystyle \frac{\left(c{\beta }_1\right)\left(c{\beta }_2\right)}{c^2}}} \\ \beta =\frac{{\beta }_1+{\beta }_2}{1+{\beta }_1{\beta }_2} \end{gathered}$$

Squaring both sides,

$$\begin{gathered} {\beta }^2=\frac{{\left({\beta }_1+{\beta }_2\right)}^2}{{\left(1+{\beta }_1{\beta }_2\right)}^2} \\ {\beta }^2=\frac{{\beta }_1^2+{\beta }_2^2+2{\beta }_1{\beta }_2}{1+{\beta }_1^2+{\beta }_2^2+2{\beta }_1{\beta }_2} \\ {\beta }^2=1-\frac{\left(1-{\beta }_1^2\right)\left(1-{\beta }_2^2\right)}{{\left(1+{\beta }_1{\beta }_2\right)}^2} \\ {\beta }^2=1-∆ \end{gathered}$$

From the above obtained value,${\beta }^2$should be less than 1. Hence,

$$\begin{gathered} \left|\beta \right|<1 \\ \left|\frac{v_{AC}}{C}\right|<1 \\ {v}_{AC}<1 \end{gathered}$$

The above result implies that if the person’s relative speed to the train and the relative speed of a train with respect to the ground is less than c, the person’s speed relative to the ground will also be less than c. So, it is impossible to travel with the speed of light or greater than it.

Therefore, the result is obtained as $v_{AC}<C$and interpreted as if the person’s relative speed to the train and the relative speed of a train with respect to the ground is less than c, the person’s speed relative to the ground will also be less than c. So, it is impossible to travel with the speed of light or greater than it.