91Ó°ÊÓ

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

(a)

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

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

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

(b)

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

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

(c)

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

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

(d)

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

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

Solve for$\mathrm{β}$ :

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

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

(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 $0<\mathrm{β}<0.750$.

Thus, for $0<\mathrm{β}<0.750$, 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$0.750<\mathrm{β}<1$.

Thus, for $0.750<\mathrm{β}<1$, event $B$occurs before event $A$according to Bullwinkle.

(g)

Here, the value of $\frac{∆x}{∆t}$is $\frac{400m}{1\mathrm{μ}s}=4.00×{10}^{8}m/s$. 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.