3 ms·
It does if you put it that way, but another way of putting is that spacetime is hyperbolic (...well, lorentzian), and all (lightspeed) interactions are zero-ran
by Filligree 5y ago
It does if you put it that way, but another way of putting is that spacetime is hyperbolic (...well, lorentzian), and all (lightspeed) interactions are zero-ranged in 4D.
As in, photons that leave the surface of the sun always strike those specific points in space-time which are at a zero spacetime interval from said surface. If you take the described geometry seriously, then "spacetime interval" is just the square of the physical distance between the events.
(And any FTL path has a negative spacetime interval. If that's still the square of the distance, then I think we can confidently state that FTL is imaginary.)