Time Intervals in Special Relativity

Nothing heckles our common sense notion of ‘time’ more than Special Relativity.

Consider a muon particle decaying as it enters the earth’s atmosphere. As per the muon, it should only live a few microseconds. Not enough to make it to the earth, it would seem.
Yet, not only does it make it to earth, it makes it without running out of time. This is because it lives LONGER when viewed from earth.
That should mess with your common sense notion of elapsed time –  which is it – does the muon live for … or does it live longer?

The answer is BOTH. There is no fixed lifetime for the muon – and to extrapolate things even more, there is no fixed time interval for ANYTHING. The actual time interval (for any phenomenon) depends on a) Who you ask (i.e. what reference frame is asking that question)?

So if time (intervals) AREN’T the same for everyone, what is? Is there something that replaces time’s special nature?

The SpaceTime Interval

If you take the time interval (squared) and subtract the space interval (squared) from it, you end up with this new INVARIANT quantity. This is called the SPACETIME Interval (as opposed to just TIME interval).

The SpaceTime Interval turns out to be IDENTICAL – regardless of who you ask (regardless of the reference frame of the observer).

More accurately, ({\rm SpaceTime\ Interval})^2 = (c \Delta t)^2 - (\Delta x)^2

Example

Say, the moon is 1 light second away. It takes 1 second for light from earth to reach the moon.

The two events (for which we will calculate our INTERVAL) are:

  1. Light LEAVES earth.
  2. Light ARRIVES on the moon.

If we take our example above, time (squared) is exactly 1 sec (squared). Hence (ct)^2 is just c^2

Space Interval (3 * 10 ^ 8 m)

Hence ((c \Delta t)^2 - (\Delta x)^2 = ( 9 (10)^{16} ) ) - (9 (10)^{16} ) = 0 So – for the earth observer, the spacetime interval is 0. Now, let us calculate this for a MOVING observer. Let us send an astronaut at HALF the speed of light – towards the moon. Intuitively, in one second, this astronaut would be HALF WAY to the moon – so when he sees the LIGHT pass him, it SHOULD seem slower than its regular speed. That is what we would expect if this were a speeding bullet – and the astronaut was on a moving train. The bullet would seem SLOWER to the moving train observer (since the train’s speed would be subtracted from the bullet speed).

As it turns out, the speed that the moving observer measures is still c (and not less than c). To understand this, remember that the spacetime interval has to stay unchanged.

Hence,  (c$'$ \Delta t$'$)^2 - (\Delta x$'$)^2) = 0.

Or  c$'$ = (\Delta x$'$ ) / (\Delta t$'$) = c !

Conclusion

 c$'$ = c !

The speed (of light) measured by the moving observer (even though he is moving REALLY fast) – is still exactly that measured by the stationary observer. The speed does not change – neither does the spacetime interval. So – what gives? Something’s gotta give somewhere – for all these things to remain unchanged !

Turns out, what changes are individual length and distance measurements. 

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