Answer:
The time interval between the events in S its zero, and in S' its 4 min.
Explanation:
The space-time interval its defined as:
[tex]( \Delta S )^ 2 \ = (c \ \Delta t)^2 \ - \ ( \Delta \vec{x})^2[/tex]
(of course, using the (+ - - -) metric signature).
In the inertial frame S, the events are separated by 3 light - minutes in space, and occur at the same instant in time, so:
[tex]\Delta t \ = \ 0[/tex]
[tex] | \ \Delta \vec{x} \ | \ = \ 3 \ min \ * \ c[/tex]
[tex]( \Delta S )^ 2 \ = (c \ 0)^2 \ - \ ( 3 \ min \ * \ c)^2[/tex]
[tex]( \Delta S )^ 2 \ = \ - \ ( 3 \ min \ * \ c)^2[/tex]
In the inertial frame S', the events are separated by 5 light - minutes in space, and we don't know how far are they separated in time, so:
[tex] | \ \Delta \vec{x}' \ | \ = \ 5 \ min \ * \ c[/tex].
[tex]( \Delta S' )^ 2 \ = \ (c \Delta t')^2 \ - \ ( 5 \ min \ * \ c)^2[/tex]
Now, between inertial frames the spacetime interval for the same events must be constant. So:
[tex]\Delta S = \Delta S'[/tex]
[tex]\ - \ ( 3 \ min \ * \ c)^2 = \ (c \ \Delta t')^2 \ - \ ( 5 \ min \ * \ c)^2[/tex]
[tex]\ \ ( 5 \ min \ * \ c)^2 - \ ( 3 \ min \ * \ c)^2 = \ (c \ \Delta t')^2[/tex]
[tex]25 \ min^2 \ c ^ 2 - 9 \ min^2 \ c ^2 \ = \ (c \ \Delta t')^2[/tex]
[tex]16 \ min^2 \ c ^2 \ = \ (c \ \Delta t')^2[/tex]
[tex]\sqrt{16 \ min^2 \ c ^2 \ } = c \ \Delta t'[/tex]
[tex]4 \ min \ c \ = c \ \Delta t'[/tex]
[tex] \Delta t' \ = 4 \ min [/tex]
So, in the S' inertial frame, the time interval its 4 min.
