MetricBrian said:
Thanks For Repyling, But really all I wanted to establish is this: is it correct (generally speaking) to say that the modified pythagorean theorem plays in an important role in calculating the distance of a curved spacetime.
I may be getting into more detail than what you want, sorry - but I'm not quite sure what you're asking.
In a flat space-time, the Lorentz interval ds is given by what you call the modified Pythagorean theorem:
ds^2 = -dt^2 + dx^2
where I've suppressed dy and dz for simplicity.
ds here is a very fundamental quantity - it's the same for all observers, it is an invariant. In fact, it can be regarded as the fundamental entity, whose description describes space-time. If that's all that you're asking, then you're on the right track. But that may not be all that you were asking.
We can get distances and times out of the lorentz interval as follows.
If ds^2 is positive, you have a spacelike interval, and sqrt(ds^2) is a distance interval. If ds^2 is negative, you have a timelike interval, and sqrt(-ds^2) is a time interval.
Now, how does this change in a general space-time?
In a general space-time, we would write instead a more general expression involving the metric coefficients g_{ij} to find the Lorentz interval, i.e. the modified Pythagorean theorem gets further modified.
<br />
ds^2 = g_{00}d t^2 + g_{01} (dt dx + dx dt) + g_{11} dx^2<br />
Here the metric coefficients g_{ij} are in general a function of (t,x)
If g_{00} = -1, g_{01} = 0, g_{11}=1 everywhere, then you have a flat space-time.