This is taken from my recent answer for
https://physics.stackexchange.com/questions/449583/invariance-of-the-relativistic-interval
Why is the invariant of the form S^2=\Delta t^2-\Delta x^2?
- A good motivation is a radar measurement of an event P=(t_P,x_P) not on your worldline.
Suppose you are an inertial observer.
To measure event P ,
imagine sending a light signal to P
and waiting for its echo, and
noting the times on your wristwatch
when you sent it t_{send}
and when you receive it t_{rec}.
From those two times, you would assign
event P the following coordinates:
- time coordinate t_P=\frac{1}{2}(t_{rec}+t_{send}) [the midway time during the round trip]
- spatial coordinate x_P=\frac{1}{2}(t_{rec}-t_{send}) [half of the roundtrip time (multiplied by c)]
Note that t_{rec}=(t_P+x_P) and t_{send}=(t_P-x_P).
Consider another inertial observer who met you when your wristwatch read zero and they set their wristwatch to zero.
They would make analogous measurements of event P .
Thus, note that
t'_{rec}=(t'_P+x'_P) and t'_{send}=(t'_P-x'_P).
Taking an image from Bondi's "Relativity and Common Sense"
It turns out for events joined by a future-directed light-signal
that t'_{send}=K t_{send} (where K is a proportionality constant
[which depends on the relative velocities of the observers]) and that
t_{rec}=K t'_{rec} (the same proportionality constant).
(Each is a Doppler effect.
This pair of Doppler effects is actually the Lorentz Boost transformation... in radar coordinates, which are related to the eigenbasis of the Lorentz boost transformation.)
So, it turns out that while t_P\neq t'_P and x_P\neq x'_P ,
it turns out
that $$ t_{rec}t_{send}=t'_{rec}t'_{send}.$$
(This is the product of times formula [Robb, Geroch],
which is the invariance of the square-interval in Radar coordinates.)
[This encodes the hyperbola, the “circle” of Minkowski Spacetime geometry. This also suggests the “area of a causal diamond”-interpretation of the square-interval that I use in my Relativity on Rotated Graph Paper approach. See my PF Insight for details.]
Expressing this back in terms of the t_Ps and x_Ps,
this says that $$({t_P}^2-{x_P}^2)=({t'_P}^2-{x'_P}^2).$$
(This is the invariance of the square-interval in standard form.)