An event has spacetime coordinates $(x,t)=(1200m,2.0\mu s)$in reference frame $S$. What are the event’s spacetime coordinates (a) in reference frame $S\text{'}$that moves in the positive $x$-direction at $0.80c$and (b) in reference frame $S\text{'}\text{'}$that moves in the negative $x$-direction at$0.80c$?

We have given that an event has spacetime coordinates $(x,t)=(1200m,2.0\mu s)$in reference frame $S$.

We have to find the event’s spacetime coordinates in reference frame $S\text{'}$that moves in the positive $x$-direction at $0.80c$.

The beginnings of the $S,S\text{'},S\text{'}\text{'}$ reference frame coincide at $t=t\text{'}=t\text{'}\text{'}=0s$.

role="math" localid="1649705093133" $$\begin{gathered} y={\left(\sqrt{1-{\left(\frac{v}{c}\right)}^2}\right)}^{-1} \\ y=(1-{(0.8)}^2{)}^{-1/2} \\ y=1.667 \end{gathered}$$

Using Lorentz transformations,

$$\begin{gathered} x\text{'}=y(x-vt) \\ =1.667\left[1200m-(0.8)(3\times {10}^{8}m/s)(2\times {10}^{-6}s)\right] \\ =1200m \end{gathered}$$

role="math" localid="1649705215551" $$\begin{gathered} t\text{'}=y\left(t-\frac{v_x}{c^2}\right) \\ =1.667\left[2\times {10}^{-6}s-\frac{(0.8)(3\times {10}^{8}m/s)\left(1200m\right)}{{(3\times {10}^{8}m/s)}^2}\right] \\ =-2.0\mu s \end{gathered}$$

We have given that an event has spacetime coordinates $(x,t)=(1200m,2.0\mu s)$in reference frame $S$.

We have to find the event’s spacetime coordinates in reference frame $S\text{'}\text{'}$that moves in the negative $x$-direction at$0.80c$.

The beginnings of the $S,S\text{'},S\text{'}\text{'}$reference frame coincide at $t=t\text{'}=t\text{'}\text{'}=0s$.

role="math" localid="1649705552341" $$\begin{gathered} y={\left(\sqrt{1-{\left(\frac{v}{c}\right)}^2}\right)}^{-1} \\ y=(1-{(0.8)}^2{)}^{-1/2} \\ y=1.667 \end{gathered}$$

The above equations are fielded with $v=-0.8c$.

$$\begin{gathered} x\text{'}\text{'}=2800m \\ t\text{'}\text{'}=8.67\mu s \end{gathered}$$