Question: Here we verify the conditions under which in equation (2-33) will be negative. (a) Show that is equivalent to the following:

$\frac{\mathbf{v}}{\mathbf{c}}\mathbf{>}\frac{{\mathbf{u}}_{\mathbf{0}}}{\mathbf{c}}\frac{\mathbf{2}}{\mathbf{1}\mathbf{+}{\mathbf{u}}_{\mathbf{0}}^{\mathbf{2}}\mathbf{/}{\mathbf{c}}^{\mathbf{2}}}$

(b) By construction v, cannot exceed ${\mathit{u}}_{\mathbf{0}}$, for if it did, the information could not catch up with Amy at event 2. Use this to argue that if ${\mathit{u}}_{\mathbf{0}}\mathbf{<}\mathit{c}$, then must be positive for whatever value is allowed to have${\left(x-1\right)}^{\mathbf{2}}\mathbf{⩾}\mathbf{0}$. (c) Using the fact that , show that the right side of the expression in part (a) never exceeds . This confirms that when ${\mathit{u}}_{\mathbf{0}}\mathbf{>}\mathit{c}$v need not exceed to produce a negative$\mathit{t}{\mathbf{\text{'}}}_{\mathbf{3}}$ .

The special relativity mainly portrays the relationship between time and space. Moreover, in this, the physics laws are invariant in the reference frame.

According to the equation (2-33), for getting a condition for $t{\text{'}}_3<0$, the equation of$t{\text{'}}_3$ is expressed as:

$t{\text{'}}_3={\gamma }_v{x}_2\frac{2}{u_0}\left[1-\frac{v}{c}\left(\frac{u_0}{2c}+\frac{c}{2u_0}\right)\right]$

For , $t{\text{'}}_3<0$the above equation will be expressed as:

$$\begin{gathered} 1-\frac{v}{c}\left(\frac{u_0}{2c}+\frac{c}{2u_0}\right)<0 \\ \frac{v}{c}>\frac{1}{\left(\frac{u_0}{2c}+\frac{c}{2u_0}\right)} \end{gathered}$$

Hence, further as:

$\frac{u_0}{2c}+\frac{c}{2u_0}=\frac{u_0^2+c^2}{2c{u}_0}$

The above equation shows that$\frac{v}{c}>\frac{2c{u}_0}{u_0^2+c^2}$ is equal to $\frac{u_0}{c}\frac{2}{1+\frac{u_0^2}{c^2}}$.

Thus, $t{\text{'}}_3<0$is equivalent to the following $\frac{v}{c}>\frac{u_0}{c}\frac{2}{1+u_0^2/c^2}$.

The argument is about the fact that the negative time cannot happen for$u_0<c$ . Here, the information shows that it cannot cross the light’s speed for preserving the events casualty. Hence, for$t{\text{'}}_3<0$ , it is needed to be identified whether it will happen for$u_0<0$ .

$\frac{v}{c}>\frac{u_0}{c}\frac{2}{1+\frac{u_0^2}{c^2}}$

From the above equation, it can be identified that $u_0<c$is mainly equivalent to$\frac{2}{1+\frac{u_0^2}{c^2}}>1$. Hence,$v⩽u_0$and by construction, it can be observed that $\frac{v}{c}$is not greater than$\frac{u_0}{c}$. Moreover, there must be a positive value of v for the values of $t{\text{'}}_3$for which$u_0<c$.

Thus,$u_0<c$if , then$t{\text{'}}_3$should be positive for whatever value is v allowed to have.

For the value $u_0>c$, it is needed to be checked that for a certain velocity$v<c$ , the time taken may be negative which shows that the casualty may not be preserved.

For the value of $u_0>c$, it must be equivalent to ${\left(\frac{u_0}{c}-1\right)}^2$which can be greater than or equal to zero. Hence the equation can be expressed as:

$$\begin{gathered} \frac{u_0^2}{c^2}-2\frac{u_0}{c}+1⩾ \\ 1+\frac{u_0^2}{c^2}⩾2\frac{u_0}{c} \\ 1⩾\frac{u_0}{c}\frac{2}{1+\frac{u_0^2}{c^2}} \end{gathered}$$

Hence, this expression is the same as the part (a) which shows that $t{\text{'}}_3<0$can be identified if the value of $u_0$does not exceed the light’s speed for having$v<c$.

Thus, the right side of the expression does not exceed 1, is proved.