Simultaneity of lasers homework

[zumulko]

Homework Statement

A spaceship (SH) moves with speed v=0.6c relative to a space station (SS) (sic!). Two lasers, A and B, on the SS are 5.00m apart as measured by the SS observers. The gamma factor for a speed v=0.6c is 1.25.

The lasers are fired simultaneously acc. to the SS observers. Light from each laser makes a mark on the SH. The SH observers measure the distance between the two marks to be 6.25m.
We are to calculate the difference in time between the firings of the two lasers acc. to the SH observers.

Homework Equations

I'm taking the equation 'for' simultaneity from http://en.wikipedia.org/wiki/Special_relativity

The Attempt at a Solution

\Delta t' = \gamma \left( \Delta t - \frac{v \Delta x}{c^2} \right),
where in our case:
\Delta t = 0s (events are simultaneous in S frame), \gamma = 0.6c, \Delta x = 6.25m.
It results in 15.6ns.

Please tell whether I'm correct with my attempt.

 

[americanforest]

I would use a relativistic invariant. r^{\mu}r_{\mu}=(c \Delta t)^{2}-{\Delta r}^{2}. This quantity is the inner product of two four vectors so it is invariant under Lorentz transformation. In this case, it would be equal both in the SH frame and the SS frame. Just set it equal for the two frames, plug in what you know, and solve for the unknown.

 

[vela]

Homework Statement

A spaceship (SH) moves with speed v=0.6c relative to a space station (SS) (sic!). Two lasers, A and B, on the SS are 5.00m apart as measured by the SS observers. The gamma factor for a speed v=0.6c is 1.25.

The lasers are fired simultaneously acc. to the SS observers. Light from each laser makes a mark on the SH. The SH observers measure the distance between the two marks to be 6.25m.
We are to calculate the difference in time between the firings of the two lasers acc. to the SH observers.

Homework Equations

I'm taking the equation 'for' simultaneity from http://en.wikipedia.org/wiki/Special_relativity

The Attempt at a Solution

\Delta t' = \gamma \left( \Delta t - \frac{v \Delta x}{c^2} \right),
where in our case:
\Delta t = 0s (events are simultaneous in S frame), \gamma = 0.6c, \Delta x = 6.25m.
It results in 15.6ns.

Please tell whether I'm correct with my attempt.

The unprimed quantities are measurements in the SS frame, so \Delta x\ne 6.25~\textrm{m}.

 

[zumulko]

The unprimed quantities are measurements in the SS frame, so \Delta x\ne 6.25~\textrm{m}.

To americanforest: I forgot to add that it should be solved with high school physics tools, therefore we mustn't use the invariant.

To vela: I agree. We should thus divide the 6.25m by gamma factor, right?

 

[vela]

No, not quite. Δt is the temporal difference between two events, right? Δx is the spatial distance between those same two events. What are the two events and how far apart in space are they?

 

[zumulko]

No, not quite. Δt is the temporal difference between two events, right? Δx is the spatial distance between those same two events. What are the two events and how far apart in space are they?

The two events are the firings. Since they are simultaneous in SS frame, dt=0. We also know that in SS frame their spatial separation is dx=5m.

 

[vela]

Yup, you got it.