Question

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$\mathit{t}{\mathbf{\text{'}}}_{\mathbf{3}}$ .

Short answer

Answer

(a) is equivalent to the following $\frac{v}{c}>\frac{u_0}{c}\frac{2}{1+u_0^2/c^2}$.

(b) If $u_0<c$, then should be positive for whatever value v is allowed to have.

(c) The right side of the expression does not exceed , 1is proved.

Step-by-step solution

Significance of the special relativity

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

(a) Determination of the statement t'3<0  is equivalent to the following vc>u0c21+u02/c2

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

(b) Determination of the argument

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.

(c) Determination of the right side of the expression

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.