91影视

Event $A$ happens at $x_A=5,y_A=5,z_A=0$ and time $c{t}_A=15$.

Event $B$ happens at $x_B=10,y_B=8,z_B=0$ and time $c{t}_B=5$.

Event $A$ happens at $x_A=2,y_A=0,z_A=0$ and time $c{t}_A=1$.

Event $B$ happens at $x_B=5,y_B=0,z_B=0$ and time $c{t}_B=3$.

The expression to calculate the change in the position along the axis is given as follows.

$\mathbf{螖虫}\mathbf{=}{\mathbf{x}}_{\mathbf{B}}\mathbf{-}{\mathbf{x}}_{\mathbf{A}}$ 鈥︹�� (1)

The expression to calculate the change in the position along the axis is given as follows.

$\mathbf{螖测}\mathbf{=}{\mathbf{y}}_{\mathbf{B}}\mathbf{-}{\mathbf{y}}_{\mathbf{A}}$ 鈥︹�� (2)

The expression to calculate the change in the position along the axis is given as follows.

$\mathbf{螖锄}\mathbf{=}{\mathbf{z}}_{\mathbf{B}}\mathbf{-}{\mathbf{z}}_{\mathbf{A}}$ 鈥︹�� (3)

The expression to calculate the time interval is given as follows.

$\mathbf{肠螖迟}\mathbf{=}{\mathbf{ct}}_{\mathbf{B}}\mathbf{-}{\mathbf{ct}}_{\mathbf{A}}$ 鈥︹�� (4)

The expression to calculate the invariant time interval is given as follows.

$\mathbf{I}\mathbf{=}\mathbf{-}{\mathbf{c}}^{\mathbf{2}}{\mathbf{\left(}\mathbf{螖迟}\mathbf{\right)}}^{\mathbf{2}}\mathbf{+}{\mathbf{\left(}\mathbf{螖虫}\mathbf{\right)}}^{\mathbf{2}}\mathbf{+}{\mathbf{\left(}\mathbf{螖测}\mathbf{\right)}}^{\mathbf{2}}\mathbf{+}{\mathbf{\left(}\mathbf{螖锄}\mathbf{\right)}}^{\mathbf{2}}$

(a)

Calculate the change in the position along the $X-$axis

Substitute $10$ for $X_B$, $5$ for $X_A$into equation (1).

$\begin{array}{l}螖x=10鈭�5\\ 螖x=5\end{array}$

Calculate the change in the position along the $y-$axis

Substitute $8$for $y_B$, $3$for $y_A$into equation (2).

$\begin{array}{l}螖y=8鈭�3\\ 螖y=5\end{array}$

Calculate the change in the position along the axis

Substitute $0$for $z_B$, $0$ for $z_A$ into equation (3).

$\begin{array}{l}螖z=0鈭�0\\ 螖z=0\end{array}$

Calculate the time interval.

Substitute $5$for $c{t}_B$, $15$for $c{t}_A$ into equation (4).

$\begin{array}{l}c螖t=5鈭�15\\ c螖t=鈭�10\end{array}$

(i) Calculate the invariant time interval.

Substitute $-10$ for $c^2{\left(螖t\right)}^2$ , $5$for $螖x$, $5$for $螖y$ and $0$for $螖z$ into equation (5)

$\begin{array}{l}I=鈭�{(鈭�10)}^2+5^2+5^2+0^2\\ I=鈭�100+25+25\\ I=鈭�50\end{array}$

Hence the invariant time interval is $-50$.

(ii) No, the invariant time interval is negative, therefore, there is no inertial system in which two events occurs simultaneously.

(iii) Yes, the invariant time is negative, therefore, there is inertial system for which two events occurs at same points.

Consider the diagram of the system $S$.

The velocity of the $S$ frame is given by,

$\begin{array}{l}v=\frac{鈭�5\widehat{x}鈭�5\widehat{y}}{\frac{10}{c}}\\ v=鈭�\frac{c}{2}\widehat{x}鈭�\frac{c}{2}\widehat{y}\end{array}$

The magnitude of the velocity is given by,

$\begin{array}{c}v=\sqrt{{\left(鈭�\frac{c}{2}\right)}^2+{\left(鈭�\frac{c}{2}\right)}^2}\\ v=\sqrt{2{\left(\frac{c}{2}\right)}^2}\\ v=\frac{c}{\sqrt{2}}\end{array}$

Hence, there is inertial system for which two events occurs at same points and the velocity relative to frame $S$ is $\frac{c}{\sqrt{2}}$.

(b)

Calculate the change in the position along the $X-$axis.

Substitute $5$for $X_B$, $2$ for $X_A$ into equation (1).

$\begin{array}{l}螖x=5鈭�2\\ 螖x=5\end{array}$

Calculate the change in the position along the $y-$axis.

Substitute $0$for $y_B$, $0$ for $Y_A$ into equation (2).

$\begin{array}{l}螖y=0鈭�0\\ 螖y=0\end{array}$

Calculate the change in the position along the $Z-$axis.

Substitute $0$for $Z_B$, $0$for $Z_A$ into equation (3).

$\begin{array}{l}螖z=0鈭�0\\ 螖z=0\end{array}$

Calculate the time interval.

Substitute $3$for $c{t}_B$, $1$for $c{t}_A$into equation (4).

$\begin{array}{l}c螖t=3鈭�1\\ c螖t=2\end{array}$

(i) Calculate the invariant time interval.

Substitute $2$ for $c^2{\left(螖t\right)}^2$, $3$for $螖x$,$0$ for $螖Y$ and $0$for $螖z$ into equation (5)

$\begin{array}{l}I=鈭�{\left(2\right)}^2+3^2+0^2+0^2\\ I=鈭�4+9\\ I=5\end{array}$

Hence the invariant time interval is $5$.

(ii) yes, the invariant time interval is positive, therefore, there is inertial system in which two events occurs simultaneously.

The relative velocity is given by,

$螖\left(\stackrel{炉}{ct}\right)=纬[螖\left(ct\right)鈭�尾螖x]$

Substitute $0$for $\stackrel{炉}{ct}$ into above equation.

$\begin{array}{l}0=纬[螖\left(ct\right)鈭�尾螖x]\\ 0=螖\left(ct\right)鈭�尾螖x\\ 尾=\frac{螖\left(ct\right)}{螖x}\end{array}$

Substitute $\frac{v}{c}$for $尾$into above equation

$\frac{v}{c}=\frac{螖\left(ct\right)}{螖x}$

Substitute $2$ for $螖\left(ct\right)$ and $3$ for $鈭�x$ into above equation.

$\begin{array}{l}\frac{v}{c}=\frac{2}{3}\\ v=\frac{2}{3}c\end{array}$

Hence there is inertial system in which two events occurs simultaneously and relative velocity is $\frac{2}{3}c$.

(iii) No, the invariant time is positive, therefore, there is no inertial system for which two events occurs at same points.