Relativistic reversal of events. Figures 37-25a and b show the (usual) situation in which a primed reference frame passes an unprimed reference frame, in the common positive direction of the $x$and $x$axes, at a constant relative velocity of magnitude . We are at rest in the unprimed frame; Bullwinkle, an astute student of relativity in spite of his cartoon upbringing, is at rest in the primed frame. The figures also indicate events $A$and $B$that occur at the following spacetime coordinates as measured in our unprimed frame and in Bullwinkle’s primed frame:

Event Unprimed Primed

$$\begin{gathered} A \\ B \end{gathered}$$ $$\begin{gathered} \left(x_A,t_A\right) \\ \left(x_B,t_B\right) \end{gathered}$$ $$\begin{gathered} \left({x^{\text{'}}}_A,{t^{\text{'}}}_A\right) \\ \left({x^{\text{'}}}_B,{t^{\text{'}}}_B\right) \end{gathered}$$

In our frame, event $A$occurs before event $B$, with temporal separation$∆t=t_A-t_B=1.00\mu s$and spatial separation$∆x=x_B-x_A=400m$. Let be the temporal separation of the events according to Bullwinkle. (a) Find an expression for$∆t^{\text{'}}$in terms of the speed parameter $\beta \left(=\frac{v}{c}\right)$and the given data. Graph$∆t^{\text{'}}$versus $\beta$for the following two ranges of$\beta$:

(b) $0$to$0.01$( $v$is low, from 0 to$0.01c$)

(c) 0.1 to 1 ( $v$is high, from to the limit$c$)

(d) At what value of $\beta$is $∆t^{\text{'}}=0$? For what range of $\beta$is the sequence of events $A$and $B$according to Bullwinkle (e) the same as ours and (f) the reverse of ours? (g) Can event $A$cause event $B$, or vice versa? Explain

The time dilation equation states that $∆t^{\text{'}}=\gamma \left(∆t-\frac{\beta ∆x}{c}\right)$.

(a)

Here, given that $∆t=1\mu s=1.00\times {10}^{-6}s$and $∆x=400m$. So, the value of$∆t^{\text{'}}$ becomes:

role="math" localid="1663224758577" $$\begin{gathered} ∆t^{\text{'}}=\gamma \left(∆t-\left(\frac{\beta ∆x}{c}\right)\right) \\ =\gamma \left(1.00\times {10}^{-6}s-\frac{\beta \left(400m\right)}{3.00\times {10}^{8}m/s}\right) \end{gathered}$$

Thus, the expression for$∆t^{\text{'}}$ is$∆t^{\text{'}}=\gamma \left(1.00\times {10}^{-6}s-\frac{\beta \left(400m\right)}{3.00\times {10}^{8}m/s}\right)$ .

(b)

In the table given below, there are some values of $∆t^{\text{'}}$for some $\beta$:

A plot of $∆t^{\text{'}}$as function of $\beta$for localid="1663228266560" $0<\beta <0.01$is shown below:

(c)

In the table given below, there are some values of $∆t^{\text{'}}$for some $\beta$:

A plot of $∆t^{\text{'}}$as function of $\beta$for $0.1<\beta <1$is shown below:

(d)

Put $∆t^{\text{'}}=0$and obtain the value of $\beta$as follows:

$$\begin{gathered} ∆t^{\text{'}}=0 \\ \gamma \left(∆t-\frac{\beta ∆x}{c}\right)=0 \\ \gamma \left(1.00\times {10}^{-6}-\frac{\beta \left(400\right)}{3.00\times {10}^8}\right)=0 \\ 1.00\times {10}^{-6}-\frac{\beta \left(400\right)}{3.00\times {10}^8}=0 \end{gathered}$$

Solve for$\beta$ :

$$\begin{gathered} \beta =\frac{1.00\times {10}^{-6}\left(3.00\times {10}^8\right)}{400} \\ \beta =0.750 \end{gathered}$$

Thus, the value of$∆t^{\text{'}}=0$ at$\beta =0.750$ .

(e)

In the graph shown in part (c), with the increase in $v$, the value of $∆t^{\text{'}}$becomes positive for the lower values in accordance with Bullwinkle. After that it approaches to zero and then becomes more negative progressively.

For the lower speeds with$∆t^{\text{'}}>0$implies that${t^{\text{'}}}_A<{t^{\text{'}}}_B$. This only happen for $0<\beta <0.750$.

Thus, for $0<\beta <0.750$, event $A$occurs before event $B$according to Bullwinkle.

(f)

For higher speeds, we can say that$∆t^{\text{'}}<0$implies that ${t^{\text{'}}}_A>{t^{\text{'}}}_B$. This can only happen if$0.750<\beta <1$.

Thus, for $0.750<\beta <1$, event $B$occurs before event $A$according to Bullwinkle.

(g)

Here, the value of $\frac{∆x}{∆t}$is $\frac{400m}{1\mu s}=4.00\times {10}^{8}m/s$. So, it is greater than $c$.

Any signal cannot go from event $A$to event $B$without exceeding $c$.

Thus, no, the event$A$ cannot cause event $B$or vice versa.