91Ó°ÊÓ

We have given that, a 30-m-long rocket train car is traveling from Los Angeles to New York at 0.50c when a light at the center of the car flashes. When the light reaches the front of the car, it immediately rings a bell. Light reaching the back of the car immediately sounds a siren.

Let the earth be the S frame of reference and the train be S' frame of reference. S' move relative to S with velocity v = 0.5c.

As the origins S and S' coincide, the light flashes at t=t'=0.

For passenger on the train, light travel 15m in both direction at speed of light. The train moving relative to the earth doesn't affect the speed of light.

Thus the light flash arrives simultaneously at the both ends of the train, causing the bell and siren to be simultaneous. Since light flashed at t' = 0, the time of these two simultaneous events is :

$t{\text{'}}_B=t{\text{'}}_S=\frac{(15m)}{(300m/\mathrm{μ}s)}=0.050\mathrm{μ}s$

Let the earth be the S frame of reference and the train be S' frame of reference. S' move relative to S with velocity v = 0.5c.

As the origins S and S' coincide, the light flashes at t=t'=0.

The spacetime coordinates of the event bell rings are :

$({\mathrm{χ}}_B\text{'},t{\text{'}}_B)=(15m,0.050\mathrm{μ}s)$.

The coordinates of the event siren sounds are:

$(\mathrm{Î\S }{\text{'}}_{s,}t{\text{'}}_s)=(-15m,0.050\mathrm{μ}s)$

We will use the Lorentz time transformation to find the time of events in earth's frame of reference S.

We will first calculate the Lorentz factor.

The Lorentz factor will be :

$\mathrm{Î\yen }=\frac{1}{\sqrt{1-{\displaystyle \raisebox{1ex}{$v^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}}$

$\mathrm{Î\yen }=\frac{1}{\sqrt{1-{(0.50)}^2}}$

Time of Bell event :

$t_B=\mathrm{Î\yen }(t{\text{'}}_B+\raisebox{1ex}{$vx{\text{'}}_B$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.)$

$t_B=1.1547(0.050\mathrm{μ}s+\frac{(0.50)(15m)}{(300m/\mathrm{μ}s)}$

$t_B=0.087\mathrm{μ}s$

Time of siren event :

$t_s=\mathrm{Î\yen }(t{\text{'}}_S+\frac{vx{\text{'}}_s}{c^2})$

$t_s=1.1547(0.050\mathrm{μ}s-\frac{(0.50)\left(15m\right)}{(300m/\mathrm{μ}s)}$

$t_s=0.029\mathrm{μ}s$

Thus the siren sounds before the bell rings. The time interval between two event is :

$\mathrm{Δ}t=0.087\mathrm{μ}s-0.029\mathrm{μ}s$

$\mathrm{Δ}t=0.058\mathrm{μ}s$