Problem — Western duel: explicit calculation

Problem

The Ugly and the Bad are $L_{(A)}=4000$ ls (light-seconds) apart in the ground’s SR. A vulture flies at $v=0{,}8\,c$ from the Ugly towards the Bad. The two shots hit simultaneously in the ground’s SR. Applying the formula for the relativity of simultaneity $\Delta t_{12(B)} = -\gamma\,\frac{\vec\beta\cdot\vec L_{12(A)}}{c}$ calculate the time difference between «Ugly hit» ($E_1$) and «Bad hit» ($E_2$) in the vulture’s SR. Who is hit first?

Solution

Data and choice of events. $E_1$ = «Ugly hit», $E_2$ = «Bad hit», simultaneous in SRA (ground). The vector $\vec L_{12(A)}$ goes from $E_1$ to $E_2$, i.e. from the Ugly towards the Bad: the same direction as the vulture’s velocity $\vec\beta$. So $\vec\beta\cdot\vec L_{12(A)} = +\beta L_{(A)}$, with $\beta=0{,}8$.

Lorentz factor. $\gamma = \frac{1}{\sqrt{1-\beta^2}} = \frac{1}{\sqrt{1-0{,}64}} = \frac{1}{\sqrt{0{,}36}} = \frac{1}{0{,}6} = \frac{5}{3} \approx 1{,}667$

Applying the formula. The distance is expressed in light-seconds, so $L_{(A)}/c = 4000$ s: $\Delta t_{12(B)} = -\gamma\,\beta\,\frac{L_{(A)}}{c} = -\frac{5}{3}\cdot 0{,}8\cdot 4000\ \text{s} = -\frac{5}{3}\cdot 3200\ \text{s} \approx -5333\ \text{s}$

$\ev{\Delta t_{12(B)} \approx -5333\ \text{s}}$

Interpretation. The negative sign means $t_{2(B)} < t_{1(B)}$: in the vulture’s SR event $E_2$ («Bad hit») happens before $E_1$. The event downstream of the motion (in the direction of the velocity) happens first. In the vulture’s frame, therefore, the Bad is hit first, about $5333$ s ahead of the Ugly — even though the two shots are simultaneous for someone standing on the ground.

Links

Topics: Special relativity Concepts: Relativity of simultaneity Skills: Changing reference frame