# A question on special relativity

1. Oct 4, 2004

### stunner5000pt

two events occur at the same place in a certain inertial frame and are separated by a time interval of 5 microseconds. What is the spatical separation ebtween these two events in an inertial frame in which the events are separated by a time interval of 13 microseconds?

I must admit i am a a bit stumped on this one

on one hand i think it has something to do with simultaneity. But then the frame which observes the longer time must be in motion.

one the other hand i think that perhaps that the distance between the first and the second observer is so large that the doppler effect of light comes in to play. But i am not sure

2. Oct 4, 2004

### Staff: Mentor

Make use of the fact that the spacetime interval between any two events is an invariant.

3. Oct 4, 2004

### stunner5000pt

what is an invariant? perhaps i know it as something else please describe

4. Oct 4, 2004

### jcsd

You know that I in the equation below does not change:

$$I = dx^2 + dy^2 + dz^2 - dct^2$$

5. Oct 5, 2004

### stunner5000pt

ok that blows my mind even more...

someone suggested using a ratio between the two distances and times to get this answer but i was not usre about what he was saying

6. Oct 5, 2004

### jcsd

The Lornetz invariance of the interval is one of the most basic concepts in special relativity

you know that for any Minkowskian coordiantes:

$$dx^2 + dy^2 + dz^2 -dct^2 = dx'^2 + dy'^2 + dz'^2 - dct'^2$$

even if you haven't been given this it can be proved from the Lorentz transformation.

so in the first inertial frame you know that dct = 5c and dx = dy = dz = 0, in the second frame dct' = 13c, so use the above equation to find: $dx'^2 + dy'^2 + dz'^2$ which by the Pythagorean theorum is the square of the distance between the two events in the primed frame.

Last edited: Oct 5, 2004
7. Oct 5, 2004

### stunner5000pt

i understand this now, that light travls in a spherical front and the distnace travled by the light is given by ct (obviously)

so the distnacei s 12c which is 12 micorseconds times c so distance is 3.6 e +3

thank yo very much

Last edited: Oct 5, 2004