Question

Two events in reference frame S occur 10 ms apart at the same point in space. The distance between the two events is 2400 m in reference frame S′.

a. What is the time interval between the events in reference frame S′?

b. What is the velocity of S′ relative to S?

Short answer

The time interval between the events in reference frame S′ is $12.8\mu s$.

The velocity of S′ relative to S is 0.625 c.

Step-by-step solution

Given information

We have given that, two events in reference frame S occur 10 ms apart at the same point in space. The distance between the two events is 2400 m in reference frame S′.

Part(a) Step 1.

The spacetime interval between two events is invariant in all frames.

Step 2.

Equating the two spacetime intervals :

$c^2{\left(\Delta t\right)}^2-{\left(\Delta x\right)}^2=c^2{(\Delta t\text{'})}^2-{(\Delta x\text{'})}^2$

${\left(\frac{300m}{\mu s}\right)}^2{(10\mu s)}^2-{(0m)}^2={\left(\frac{300m}{\mu s}\right)}^2{(\Delta t\text{'})}^2-{(2400m)}^2$

$\Delta t\text{'}=12.8\mu s$

Part(b) Step 1

The event occurs at the same point in space, so $\Delta t\text{'}=\Delta \zeta$.

Step 2.

Hence :

$\Delta t\text{'}=\frac{\Delta \zeta }{\sqrt{1-{\displaystyle \raisebox{1ex}{$v^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}}$

$12.8\mu s=\frac{10\mu s}{\sqrt{1-{\displaystyle \raisebox{1ex}{$v^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}}$

$v=0.625c$