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morrobay
#9
Feb20-12, 09:06 PM
P: 376
Quote Quote by elfmotat View Post
Yes, [itex](\Delta s)^2=(c\Delta \tau )^2=(c\Delta t)^2-(\Delta x)^2[/itex]. This comes from the fact that, for the clock, Δx=0.
So if c is taken to be 1 , then is the spacetime interval always numerically equal
to the proper time ?