Lorentz Transformations and Reference Frames Problem

[zero13428]

Homework Statement

In the old West, a marshal riding on a train traveling 35.0 m/s sees a duel between two men standing on the Earth 55.0 m apart parallel to the train. The marshal's instruments indicate that in his reference frame the two men fire simultaneously. (a) Which of the two men, the first one the train passes (A) or the second one (B) should be arrested for firing the first shot? That is, in the gunfighter's frame of refernece, who fired first? (b) How much earlier did he fire? (c) Who was struck first?

Homework Equations

Lorentz transformations

The Attempt at a Solution

I'm not sure how to apply lorentz equations to this problem. I assume I need to find the distance between the duel and the train but beyond that I'm not sure how to begin solving this.

 

[member 553083]

A:You will need to apply the Lorentz transformation, but it's not too complicated.Let $x$ and $t$ be the coordinates of the two men in the reference frame of the marshal. Let $x'$ and $t'$ be the coordinates of the two men in the reference frame of the gunfighters.The transformation between the two frames is given by$$x'=\gamma\left(x-vt\right)$$$$t'=\gamma\left(t-\frac{vx}{c^2}\right)$$where $\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$Now, for the first man (A), we have $x=-55\ m$ and $t=0$. Therefore, his coordinates in the frame of the gunfighters are$$x'=-55\ m \ \gamma$$$$t'=0\ s \ \gamma$$For the second man (B), we have $x=55\ m$ and $t=0$. Therefore, his coordinates in the frame of the gunfighters are$$x'=55\ m \ \gamma$$$$t'=0\ s \ \gamma$$Therefore, man A fired first in the frame of the gunfighters, since his $x'$ is less than the $x'$ of man B.The time difference between the two shots is $\Delta t'=0\ s \ \gamma$, so he fired $\Delta t'$ earlier than man B.Finally, since man A fired first, he was also struck first.