# Special relativity: Simple Lorentz transformation question

## Homework Statement

Observer O sees a fire-engine leave its station 6363 m due
north from Cape Canaveral, where the super-shuttle Lorentz had been
launched 10^-5 s earlier. A space-cruiser flying north-east sees these two events
also 10^-5 s apart, but with the shuttle launch occurring after the fire-engine
leaves the station.

(a) Show that the speed of the space-cruiser relative to the Earth is 12c/13.
[5 marks]

(b) How far apart does the navigator on the space-cruiser measure the two
events to be? [3 marks]

## Homework Equations

$$\Delta t ' = \gamma(\Delta t - v\Delta x/c^2)$$

## The Attempt at a Solution

I know this must be quite simple, i'm just not quite sure where to start. I've fiddled with the Lorentz transformation equations to no avail - i think its the time interval concept which confuses me here (what actually is the interval? 10^-5 s?)

I tried putting in the value of the speed given (*cos45 as it's heading north-east) to try obtain the times given or something related to them but also this was pretty useless.

I imagine part b is quite trivial once part a is complete

Ben

Last edited:

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vela
Staff Emeritus
Homework Helper
The equation for the Lorentz transformation you have there is for when the second reference frame moves with speed v in the +x direction and the spatial axes are parallel to each other. You need to orient your coordinate systems so that these conditions are met. Once you do that, figure out the coordinates of the two events in one reference frame and then use the transformation to find the interval in the second frame.

i understand the need to have the parallel spatial axes but i dont quite understand how to figure out the coordinates?

is the equation i stated the correct one to use in this example?

vela
Staff Emeritus