Question

A rocket travels in the $x$-direction at speed $0.60c$ with respect to the earth. An experimenter on the rocket observes a collision between two comets and determines that the spacetime coordinates of the collision are $(x^{\text{'}},t^{\text{'}})=(3.0\times {10}^{10}m,200s)$. What are the spacetime coordinates of the collision in earth’s reference frame?

Short answer

$$\begin{gathered} \gamma =1.25 \\ x=8.25\times {10}^{10}m \\ t=325s \end{gathered}$$

Step-by-step solution

Given Information

We have given that a rocket travels in the $x$-direction at speed $0.60c$with respect to the earth.

The collision's spacetime coordinates are$(x\text{'},t\text{'})=(3.0\times {10}^{10}m,200s)$.

We have to find the spacetime coordinates of the collision in earth’s reference frame.

Simplify

We have,

$$\begin{gathered} \gamma ={\left[1-{\left(\frac{v}{c}\right)}^2\right]}^{-1/2} \\ \gamma ={\left[1-{(0.6)}^2\right]}^{-1/2} \\ \gamma =1.25 \end{gathered}$$

Lorentz transformation in Earth's frame yields:

$$\begin{gathered} x=\gamma (x\text{'}+vt\text{'})=1.2\left[3\times {10}^{10}m+(0.6)(3\times {10}^{8}m/s)\left(200\right)\right]=8.25\times {10}^{10}m \\ t=\gamma \left(t\text{'}+\frac{v_{x\text{'}}}{c^2}\right)=1.25\left[\frac{200+(0.6)(3\times {10}^{8}m/s)(3\times {10}^{10}m)}{{(3\times {10}^{8}m/s)}^2}\right]=325s \end{gathered}$$

Therefore,

$$\begin{gathered} \gamma =1.25 \\ x=8.25\times {10}^{10}m \\ t=325s \end{gathered}$$