a question on special relativity

[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

please help, thank you

[Doc Al]

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

[stunner5000pt]

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

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

[jcsd]

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

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

[stunner5000pt]

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

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

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

[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.

[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