# A question on special relativity

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

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

Doc Al said:
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
Gold Member
You know that I in the equation below does not change:

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

jcsd said:
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
Gold Member
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.

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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

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